MCQ 11 Mark
If $f ( x )=\lfloor x\rfloor$ for $x \in(-1,2)$, then $f$ is discontinuous at
- A
$x =-1,0,1,2$
- B
$x=-1,0,1$
- ✓
$x=0,1$
- D
$x=2$
AnswerCorrect option: C. $x=0,1$
(C) $x=0,1$
Hint:
$f(x)=\lfloor x\rfloor, x \in(-1,2)$
This function is discontinuous at all integer values of $x$ between -1 and 2 . $\therefore f ( x )$ is discontinuous at $x =0$ and $x =1$.
View full question & answer→MCQ 21 Mark
If $f(x)=\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}$, for $x \neq 0$ and $f(0)=k$, is continuous at $x=0$, then $k$ is
- A
$e ^7$
- B
$e^3$
- ✓
$e^{12}$
- D
$e^{\frac{3}{4}}$
AnswerCorrect option: C. $e^{12}$
(C) $e^{12}$
Hint:
$f (x)$ is continuous at $x=0$
$f(0)=\lim _{x \rightarrow 0} f(x)$
$=\lim _{x \rightarrow 0}\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}$
$=\lim _{x \rightarrow 0}\left[\frac{4\left(1+\frac{5 x}{4}\right)}{4\left(1-\frac{7 x}{4}\right)}\right]^{\frac{4}{x}}$
$=\frac{\lim _{x \rightarrow 0}\left[\left(1+\frac{5 x}{4}\right)^{\frac{4}{5 x}}\right]^5}{\lim _{x \rightarrow 0}\left[\left(1-\frac{7 x}{4}\right)^{\frac{-4}{7 x}}\right]^{-7}}$
$=\frac{ e ^5}{ e ^{-7}} \cdots\left[\because x \rightarrow 0, \frac{5 x}{4} \rightarrow 0, \frac{-7 x}{4} \rightarrow 0, \text { and } \lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}= e \right]$
$= e ^{12}$
View full question & answer→MCQ 31 Mark
If $f(x)=\frac{12^x-4^{x^2-3^x+1}}{1-\cos 2 x}$, for $x \neq 0$ is continuous at $x=0$ then the value of $f(0)$ is
AnswerCorrect option: B. $\log 2 \cdot \log 3$
(B) $\log 2 \cdot \log 3$
Hint:
$f (x)$ is continuous at $x=0$.
(given)
$f(0)=\lim _{x \rightarrow 0} f(x)$
$=\lim _{x \rightarrow 0} \frac{12^x-4^x-3^x+1}{1-\cos 2 x}$
$=\frac{1}{2} \lim _{x \rightarrow 0} \frac{4^x\left(3^x-1\right)\left(3^x-1\right)}{\sin ^2 x}$
$=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\left(3^x-1\right)\left(4^x-1\right)}{\sin ^2 x} $
$=\frac{1}{2} \frac{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right)}{\left(\lim \frac{\sin x}{x}\right)^2}$
$=\frac{1}{2} \times \frac{(\log 3) \times(\log 4)}{(1)^2} $
$=\frac{1}{2} \times \log 3 \times \log (2)^2$
$=\log 3 \cdot \log 2$
View full question & answer→MCQ 41 Mark
$f(x)=\frac{32^x-8^x-4^x+1}{4^x-2^{x+1}+1}, \text { for } x \neq 0$
$=k, \text { for } x=0$
is continuous at $x=0$, then value of ' $k$ ' is
- ✓
$6$
- B
$4$
- C
$(\log 2)(\log 4)$
- D
$3 \log 4$
Answer$(A) 6$
Hint:
$f (x)$ is continuous at $x=0$.
...(given)
$f(0)=\lim _{x \rightarrow 0} f(x)$
$k =\lim _{x \rightarrow 0} \frac{32^x-8^x-4^x+1}{4^x-2^{x+1}+1}$
$=\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)\left(8^x-1\right)}{\left(2^x-1\right)^2}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right)\left(\frac{8^x-1}{x}\right)}{\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)^2}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{8^x-1}{x}\right)}{\left(\lim _{x \rightarrow 0} \frac{2^x-1}{x}\right)^2}$
$=\frac{\log 4 \times \log 8}{(\log 2)^2}$
$=\frac{2 \log 2 \times 3 \log 2}{(\log 2)^2}=6$
View full question & answer→MCQ 51 Mark
$f(x)=\frac{\left(16^x-1\right)\left(9^x-1\right)}{\left(27^x-1\right)\left(32^x-1\right)}$, for $x \neq 0$ $=k_{\text {, }}$ for $x=0$
is continuous at $x=0$, then ' $k$ ' =
- A
$\frac{8}{3}$
- ✓
$\frac{8}{15}$
- C
$-\frac{8}{15}$
- D
$\frac{20}{3}$
AnswerCorrect option: B. $\frac{8}{15}$
(B) $\frac{8}{15}$
Hint:
$f (x) \text { is continuous at } x=0$
$f(0)=\lim _{x \rightarrow-0} f(x) $
$k =\lim _{x \rightarrow-0} \frac{\left(16^x-1\right)\left(9^x-1\right)}{\left(27^x-1\right)\left(32^x-1\right)} $
$=\frac{\lim _{x \rightarrow-0}\left(\frac{16^x-1}{x}\right) \times \lim _{x \rightarrow-0}\left(\frac{9^x-1}{x}\right)}{\lim _{x \rightarrow-0}\left(\frac{27^x-1}{x}\right) \times \lim _{x \rightarrow-0}\left(\frac{32^x-1}{x}\right)} $
$=\frac{\log 16 \times \log 9}{\log 27 \times \log 32} \quad \cdots\left[\because \lim _{x \rightarrow 0} \frac{ a ^x-1}{x}=\log a \right] $
$=\frac{4 \log 2 \times 2 \log 3}{3 \log 3 \times 5 \log 2}$
$=\frac{8}{15} $
View full question & answer→MCQ 61 Mark
If $f(x)=a x^2+b x+1$, for $|x-1| \geq 3$ and $=4 x+5$, for $-2is continuous everywhere then,
- ✓
$a =\frac{1}{2}, b =3$
- B
$a=-\frac{1}{2}, b=-3$
- C
$a=-\frac{1}{2}, b=3$
- D
$a=\frac{1}{2}, b=-3$
AnswerCorrect option: A. $a =\frac{1}{2}, b =3$
(A) $a=\frac{1}{2}, b=3$
Hint:
$f(x)=a x^2+b x+1,|x-1| \geq 3$
$=4 x+5 ;-2$
The first interval is
$|x-1| \geq 3 $
$\therefore x-1 \geq 3 \text { or } x-1 \leq-3 $
$\therefore x \geq 4 \text { or } x \leq-2$
$\therefore f ( x )$ is same for $x \leq-2$ as well as $x \geq 4$.
$\therefore f ( x )$ is defined as:
$f(x)=a x^2+b x+1 ; x \leq-2 $
$=4 x+5 ;-2=a x^2+b x+1 ; x \geq 4$
$f(x)$ is continuous everywhere.
$\therefore f ( x )$ is continuous at $x =-2$ and $x =4$.
As $f(x)$ is continuous at $x=-2$,
$\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x) $
$\therefore \lim _{x \rightarrow-2}\left(a x^2+b x+1\right)=\lim _{x \rightarrow-2}(4 x+5) $
$\therefore a(-2)^2+ b (-2)+1=4(-2)+5 $
$\therefore 4 a-2 b+1=-3$
$\therefore 4 a-2 b=-4$
$\therefore 2 a-b=-2 \ldots . .( i )$
$\because f ( x )$ is continuous at $x =4$,
$\lim _{x \rightarrow 4^{-}} f (x)=\lim\limits_{x \rightarrow 4^{+}} f (x) $
$\therefore \lim _{x \rightarrow 4}(4 x+5)=\lim _{x \rightarrow 4}\left(a x^2+b x+1\right) $
$4(4)+5=a(4)^2+ b (4)+1 $
$16 a +4 b +1=21$
$16 a +4 b =20$
$4 a + b =5 \ldots . .( ii )$
Adding $(i)$ and $(ii),$ we get
$6 a=3$
$\therefore a=\frac{1}{2}$
Substituting $a=\frac{1}{2}$ in (ii), we get
$4\left(\frac{1}{2}\right)+b=5 $
$\therefore 2+b=5 $
$\therefore b=3$
$\therefore a=\frac{1}{2}, b=3$
View full question & answer→MCQ 71 Mark
$f(x)=\frac{x^2-7 x+10}{x^2+2 x-8}$, for $x \in[-6,-3]$
- A
$f$ is discontinuous at $x=2$
- ✓
$f$ is discontinuous at $x=-4$
- C
$f$ is discontinuous at $x=0$
- D
$f$ is discontinuous at $x=2$ and $x=-4$
AnswerCorrect option: B. $f$ is discontinuous at $x=-4$
(B) $f$ is discontinuous at $x =-4$
Hint:
$f(x)=\frac{x^2-7 x+10}{x^2+2 x-8}, \text { for } x \in[-6,-3] $
$=\frac{x^2-7 x+10}{(x+4)(x-2)}$
Here $f(x)$ is a rational function and is continuous everywhere except at the points Where denominator becomes zero.
Here, denominator becomes zero when $x=-4$ or $x=2$
But $x=2$ does not lie in the given interval.
$\therefore x =-4$ is the point of discontinuity.
View full question & answer→MCQ 81 Mark
If $f(x)=\frac{(\sin 2 x) \tan 5 x}{\left(e^{2 x}-1\right)^2}$, for $x \neq 0$ is continuous at $x=0$, then $f(0)$ is
- A
$\frac{10}{e^2}$
- B
$\frac{10}{e^4}$
- C
$\frac{5}{4}$
- ✓
$\frac{5}{2}$
AnswerCorrect option: D. $\frac{5}{2}$
(D) $\frac{5}{2}$
Hint:
$f (x)$ is continuous at $x=0$
$f (0)$
$=\lim _{x \rightarrow 0} \frac{(\sin 2 x)(\tan 5 x)}{\left( e ^{2 x}-1\right)^2} $
$=\frac{\lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} \times \lim _{x \rightarrow 0} \frac{\tan 5 x}{5 x} \times 2 \times 5}{\left(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{2 x}\right)^2 \times(2)^2} $
$=\frac{1 \times 1 \times 2 \times 5}{(1)^2 \times 4} \cdots$
$[\because x \rightarrow 0,2 x \rightarrow 0,5 x \rightarrow 0 , \text { and } \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1, \lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1]$
$=\frac{5}{2}$
View full question & answer→MCQ 91 Mark
If $f(x)=\frac{1-\sqrt{2} \sin x}{\pi-4 x}$, for $x \neq \frac{\pi}{4}$ is continuous at $x=\frac{\pi}{4}$, then $f\left(\frac{\pi}{4}\right)=$
- A
$\frac{1}{\sqrt{2}}$
- B
$-\frac{1}{\sqrt{2}}$
- C
$-\frac{1}{4}$
- ✓
$\frac{1}{4}$
AnswerCorrect option: D. $\frac{1}{4}$
(D) $\frac{1}{4}$
Hint:
$f(x)$ is continuous at $x=\frac{\pi}{4}$
$\therefore \quad f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x) $
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x} $
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2}\left(\sin x-\frac{1}{\sqrt{2}}\right)}{4\left(x-\frac{\pi}{4}\right)} $
$=\frac{\sqrt{2}}{4} \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\sin \frac{\pi}{4}}{x-\frac{\pi}{4}} $
$=\frac{\sqrt{2}}{4} \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \cos \left(\frac{\left.x+\frac{\pi}{4}\right)}{2}\right) \cdot \sin \left(\frac{\left.x-\frac{\pi}{4}\right)}{2}\right)}{x-\frac{\pi}{4}}$
$=\frac{\sqrt{2}}{4} \cdot \lim _{x \rightarrow \frac{\pi}{4}} \cos \left(\frac{x}{2}+\frac{\pi}{8}\right) \cdot \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin \left(\frac{x-\frac{\pi}{4}}{2}\right)}{\frac{x-\frac{\pi}{4}}{2}}$
$=\frac{\sqrt{2}}{4} \cdot \cos \left(\frac{x}{8}+\frac{\pi}{8}\right) \times 1\ldots$
$\left[\because x \rightarrow \frac{\pi}{4}, x-\frac{\pi}{4} \rightarrow 0, \frac{x-\frac{\pi}{4}}{2} \rightarrow 0 ,\text { and } \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=\frac{\sqrt{2}}{4} \times \cos \frac{\pi}{4}$
$=\frac{1}{4} \quad$
View full question & answer→MCQ 101 Mark
$\begin{aligned}
& f(x)=\frac{2^{\mathrm{cotx}}-1}{\pi-2 x}, \text { for } x \neq \frac{\pi}{2} \\
& =\log \sqrt{2}, \text { for } x=\frac{\pi}{2}
\end{aligned}$
- ✓
$f$ is continuous at $x=\frac{\pi}{2}$
- B
$f$ has a jump discontinuity at $x=\frac{\pi}{2}$
- C
f has a removable discontinuity
- D
$\lim _{x \rightarrow \frac{\pi}{2}} f(x)=2 \log 3$
AnswerCorrect option: A. $f$ is continuous at $x=\frac{\pi}{2}$
(A) $f$ is continuous at $x=\frac{\pi}{2}$
Hint:
$
\begin{aligned}
& f\left(\frac{\pi}{2}\right)=\log \sqrt{2} \\
& \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2^{\cot x}-1}{\pi-2 x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2^{tan(\frac{\pi}{2}-xt)}-1}{2(\frac{\pi}{2}-x)}
\end{aligned}
$
Put $\frac{\pi}{2}-x=h$
As $x \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0$
$
\begin{aligned}
& \therefore \quad \lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} \frac{2^{\text{tan h}}-1}{2 \mathrm{~h}} \\
& =\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{2^{\text{tan h}}-1}{\tan h} \times \frac{\tan h}{h}\right) \\
& \ldots(\because \mathrm{h} \rightarrow 0, \tan h \rightarrow 0, \tan h \neq 0) \\
& =\frac{1}{2} \lim _{h \rightarrow 0} \frac{2^{\tanh }-1}{\tanh } \times \lim _{h \rightarrow 0} \frac{\tan h}{h} \\
& =\frac{1}{2} \cdot \log 2 \cdot(1) \\
& =\log \sqrt{2}=f\left(\frac{\pi}{2}\right) \\
&
\end{aligned}
$
$\therefore \quad \mathrm{f}(x)$ is continuous at $x=\frac{\pi}{2}$
View full question & answer→MCQ 112 Marks
The number of discontinuities of the greatest integer function $f(x)=[x], x \in\left(-\frac{7}{2}, 100\right)$ is
Answer(d) : Given $f(x)=[x], x \in\left(\frac{-7}{2}, 100\right)$ i.e.. $x \in(-3 \cdot 5,100)$
As we know that greatest integer function is discontinuous on integer value.
In interval $(-3.5,100)$, the integer values are $-3,-2,-1,0$, 99 i.e., 103 in number
$\therefore \quad f(x)$ is discontinuous at 103 points.
View full question & answer→MCQ 122 Marks
The function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x)=\left\{\begin{array}{cc}\frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right) & , x \neq 0 \\ k & , \quad x=0\end{array}\right.$ is continuous at $x=0$, then $k$ is
AnswerSo, $\lim _{x \rightarrow 0} f(x)=f(0)=k$
$\Rightarrow k=\lim _{x \rightarrow 0} \frac{\log \frac{(1+3 x)}{(1-2 x)}}{x}$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{(1+3 x)}{(1-2 x)}-1\right)}{\left(\frac{3 x+1}{1-2 x}-1\right) x} \times\left[\left(\frac{3 x+1}{1-2 x}\right)-1\right]$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{5 x}{1-2 x}\right)}{\frac{5 x}{1-2 x}} \times \frac{5}{1-2 x}$
$[$As we know that $\lim _{f(x) \rightarrow 0} \frac{\log (1+f(x))}{f(x)}=1]$
$\Rightarrow \lim _{x \rightarrow 0} \frac{5}{1-2 x}=5$
$\Rightarrow k=5$
View full question & answer→MCQ 132 Marks
The function $f(x)=[x] \cdot \cos \left(\frac{2 x-1}{2}\right) \pi$, where $[.]$ denotes the greatest integer function, is discontinuous at
AnswerCorrect option: B. no $x$.
(b) : As, $f(x)=[x] \cos \left(\frac{2 x-1}{2}\right) \pi$
For a function to be continuous LHL $=$ RHL $=f(x)$
So, let us check continuity for $x=a$ where $a$ is an integer
L.HL at $x=a$ is given by
$
\begin{aligned}
& \lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{-}}[x] \cos \left(\frac{2 x-1}{2}\right) \pi \\
& =0 \\
&
\end{aligned}
$
RHL at $x=a$ is given by
$
\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}}[x] \cos \left(\frac{2 x-1}{2}\right) \pi=0
$
Also $f(a)=[a] \cos \left(\frac{2 a-1}{2}\right) \pi=0$
$
\Rightarrow LHL = RHL =f(x), \text { for } x=a
$
So, $f(x)$ will be continuous for all $x$.
View full question & answer→MCQ 142 Marks
If $f(x)=\left\{\begin{array}{cc}\frac{x-4}{|x-4|}+a, & \text { for } x<4 \\ a+b, & \text { for } x=4 \\ \frac{x-4}{|x-4|}+b, & \text { for } x>4\end{array}\right.$
is continuous at $x=4$, then
- A
$a=1, b=1$
- ✓
$a=1, b=-1$
- C
$a=0, b=0$
- D
$a=-1, b=1$
AnswerCorrect option: B. $a=1, b=-1$
(b) : Since $f(x)$ is continuous at $x=4$ So, $\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=f(x)$ at $x=4-1+a=a+b=1+b$ Now, $a-1=a+b \Rightarrow b=-1$ and $a+b=1+b \Rightarrow a=1$ Hence, $a=1, b=-1$.
View full question & answer→MCQ 152 Marks
If $f(x)\left\{\begin{array}{rr}\log \left(\sec ^2 x\right)^{\cot 2 x}, & \text { for } x \neq 0 \\ K+1, & \text { for } x=0\end{array}\right.$ is continuous at $x=0$, then value of $K$ is
Answer(a) : $f(x)=\left\{\begin{array}{cc}\log \left(\sec ^2 x\right)^{\cot ^2 x}, & x \neq 0 \\ K+1 & , x=0\end{array}\right.$ is continuous at $x=0$.
$
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \lim _{x \rightarrow 0} \log \left(\sec ^2 x\right)^{\cot ^2 x}=K+1 \\
& \Rightarrow \lim _{x \rightarrow 0} \cot ^2 x \log \left(1+\tan ^2 x\right)=K+1 \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^2 x\right)}{\tan ^2 x}=K+1 \\
& \Rightarrow K+1=1 \Rightarrow K=0 \quad \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]
\end{aligned}
$
View full question & answer→MCQ 162 Marks
If $f(x)=\left\{\begin{array}{ll}x, & \text { if } x \leq 1 \\ 7, & \text { if } x>1\end{array}\right.$ then at $x=1$
AnswerCorrect option: D. $f$ is discontinuous
(d) $: f(1)=1$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} x=1, \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 7=7$
Since, $f(1) \neq \lim _{x \rightarrow 1^{+}} f(x)$
$\therefore \quad f$ is discontinuous at $x=1$
View full question & answer→MCQ 172 Marks
Determine the value of $k$ for which the function $f(x)$ is continuous at $x=4$.$f(x)= \begin{cases}\frac{x^2-16}{x-4}, & x \neq 4 \\ k, & x=4\end{cases}$
Answer(d) : Since $f(x)$ is continuous at $x=4$. Therefore, $\lim _{x \rightarrow 4} f(x)=f(4)$
$\Rightarrow \lim _{x \rightarrow 4} f(x)=k \quad[\because f(4)=k]$
$\Rightarrow \lim _{x \rightarrow 4} \frac{x^2-16}{x-4}=k \Rightarrow \lim _{x \rightarrow 4} \frac{(x-4)(x+4)}{x-4}=k$
$\Rightarrow \lim _{x \rightarrow 4}(x+4)=k \Rightarrow k=8$
View full question & answer→MCQ 182 Marks
Which of the following function is not continuous at $x=0$ ?
- A
$f(x)= \begin{cases}(1+2 x)^{1 / x}, & x \neq 0 \\ e^2, & x=0\end{cases}$
- B
$f(x)= \begin{cases}\sin x-\cos x, & x \neq 0 \\ -1, & x=0\end{cases}$
- ✓
$f(x)= \begin{cases}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ -1, & x=0\end{cases}$
- D
$f(x)= \begin{cases}\frac{e^{5 x}-e^{2 x}}{\sin 3 x}, & x \neq 0 \\ 1, & x=0\end{cases}$
AnswerCorrect option: C. $f(x)= \begin{cases}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ -1, & x=0\end{cases}$
(c) $f(x)= \begin{cases}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ -1, & x=0\end{cases}$
View full question & answer→MCQ 192 Marks
If the function $f(x)=\frac{\log (1+a x)-\log (1-b x)}{x}$, $x \neq 0$ is continuous at $x=0$, then $f(0)=$
- A
$\log a-\log b$
- ✓
$a+b$
- C
$\log a+\log b$
- D
$a-b$
Answer(b) : If $f(x)$ is continuous at $x=0$, then $\lim _{x \rightarrow 0} f(x)=f(0)$
$
\therefore f(0)=\lim _{x \rightarrow 0} \frac{\log (1+a x)-\log (1-b x)}{x}
$
$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\log (1+a x)}{x}-\lim _{x \rightarrow 0} \frac{\log (1-b x)}{x} \\
& =\lim _{x \rightarrow 0} \frac{a}{1+a x}+\lim _{x \rightarrow 0} \frac{b}{1+b x} \quad \text { [By L.H. Rule] } \\
& =a+b
\end{aligned}
$
View full question & answer→MCQ 202 Marks
If function $f(x)=\left\{\begin{array}{ll}x-\frac{|x|}{x}, & x<0 \\ x+\frac{|x|}{x}, & x>0 \\ 1, & x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{-}} f(x)$ does not exist
- B
$\lim _{x \rightarrow 0^{+}} f(x)$ does not exist
- ✓
$f(x)$ is continuous at $x=0$
- D
$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$
AnswerCorrect option: C. $f(x)$ is continuous at $x=0$
(c) : We have, $f(0)=1$
$
\begin{aligned}
& \begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left(x-\frac{|x|}{x}\right)=\lim _{x \rightarrow 0}\left(x+\frac{x}{x}\right) \\
&=\lim _{x \rightarrow 0}(x+1)=1 \\
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}\left(x+\frac{|x|}{x}\right)=\lim _{x \rightarrow 0}\left(x+\frac{x}{x}\right) \\
&=\lim _{x \rightarrow 0}(x+1)=1
\end{aligned}
\end{aligned}
$
Thus, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\therefore f(x)$ is continuous at $x=0$.
View full question & answer→MCQ 212 Marks
If the function $f(x)=\left\{\begin{array}{l}\frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}, x \neq 0 \\ 16, x=0\end{array}\right.$ is continuous at $x=0$, then $k=$
- A
$\pm \frac{1}{8}$
- B
$\pm 4$
- C
$\pm 2$
- ✓
$\pm 8$
AnswerCorrect option: D. $\pm 8$
Since, $f(x)$ is continuous at $x=0$.
$\therefore \lim _{x \rightarrow 0} f(x)=f(0) $
$\Rightarrow \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}=16$
$ \Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right)}{k x} \times \frac{\tan k x}{k x} \times k^2=16$
$ \Rightarrow \frac{k^2}{4} \lim _{x \rightarrow 0} \frac{e^{k x}-1}{k x} \times \lim _{x \rightarrow 0} \frac{\tan k x}{k x}=16 $
$\Rightarrow \frac{k^2}{4} \times 1 \times 1=16$
$ \Rightarrow k^2=64 $
$\Rightarrow k= \pm 8$
View full question & answer→MCQ 222 Marks
If $f(x)=\frac{e^{x^2}-\cos x}{x^2}$, for $x \neq 0$ is continuous at $x=0$, then value of $f(0)$ is
- A
$\frac{2}{3}$
- B
$\frac{5}{2}$
- C
- ✓
$\frac{3}{2}$
AnswerCorrect option: D. $\frac{3}{2}$
(d) : Given, $f(x)=\frac{e^{x^2}-\cos x}{x^2}$ is continuous at $x=0$
$\Rightarrow$ L.H.L. $=$ R.H.L $=f(0)$ at $x=0$
L.H.L. $($ at $x=0)=\lim _{h \rightarrow 0} \frac{e^{(0-h)^2}-\cos (0-h)}{(0-h)^2}$
$=\lim _{h \rightarrow 0} \frac{e^{h^2}-\cos h-1+1}{h^2}=\lim _{h \rightarrow 0}\left(\frac{e^{h^2}-1}{h^2}+\frac{1-\cos h}{h^2}\right)$
$=\lim _{h \rightarrow 0} \frac{e^{h^2}-1}{h^2}+\lim _{h \rightarrow 0} \frac{2 \sin ^2\left(\frac{h}{2}\right)}{h^2}$
$=1+2 \lim _{h \rightarrow 0} \frac{\sin ^2(h / 2)}{(h / 2)^2} \times \frac{1}{4}=1+\frac{2 \times 1}{4}=\frac{3}{2}$
Hence, $f(0)=\frac{3}{2}$
View full question & answer→MCQ 232 Marks
If $f(x)=x^2+\alpha$ for $x \geq 0$ $=2 \sqrt{x^2+1}+\beta$ for $x<0$ is continous at $x=0$ and $f\left(\frac{1}{2}\right)=2$, then $\alpha^2+\beta^2$ is
- A
- B
$\frac{8}{25}$
- ✓
$\frac{25}{8}$
- D
$\frac{1}{3}$
AnswerCorrect option: C. $\frac{25}{8}$
(c) : We have $f(x)=\left\{\begin{array}{c}x^2+\alpha \text { for } x \geq 0 \\ 2 \sqrt{x^2+1}+\beta \text { for } x<0\end{array}\right.$
$\therefore \quad$ L.H.L. $($ at $x=0)=\lim _{h \rightarrow 0} 2 \sqrt{(0-h)^2+1}+\beta$
$=\lim _{h \rightarrow 0} 2 \sqrt{h^2+1}+\beta=2+\beta$
and R.H.L. $($ at $x=0)=\lim _{h \rightarrow 0}(0+h)^2+\alpha=\lim _{h \rightarrow 0} h^2+\alpha=\alpha$
Now, $f(0)=(0)^2+\alpha=\alpha$
Since $f(x)$ is continuous at $x=0$.
$\Rightarrow$ L.H.L. at $x=0=$ R.H.L at $x=0=f(0)$
$\Rightarrow 2+\beta=\alpha$
Also $f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+\alpha \Rightarrow \frac{1}{4}+\alpha=2 \Rightarrow \alpha=\frac{7}{4}$.
Put $\alpha=\frac{7}{4}$ in (i), we get $2+\beta=\frac{7}{4} \Rightarrow \beta=\frac{-1}{4}$
Now, $\alpha^2+\beta^2=\left(\frac{7}{4}\right)^2+\left(\frac{-1}{4}\right)^2=\frac{49}{16}+\frac{1}{16}=\frac{25}{8}$
View full question & answer→MCQ 242 Marks
If $f(x)=\left\{\begin{array}{cc}\log \left(\sec ^2 x\right)^{\cot ^2 x} & \text { for } x \neq 0 \\ K & \text { for } x=0\end{array}\right.$ is continuous at $x=0$ then $K$ is
Answer(b) : Since $f(x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} \log \left(\sec ^2 x\right)^{\cot ^2 x}$
$\Rightarrow K=\lim _{x \rightarrow 0} \cot ^2 x \cdot \log \left(1+\tan ^2 x\right) \Rightarrow K=\lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^2 x\right)}{\tan ^2 x}$
$\Rightarrow K=1 \quad\left[\because \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
View full question & answer→MCQ 252 Marks
If the function $f(x)=\left\{\left[\begin{array}{cc}{\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{1 / x}} & \text { for } x \neq 0 \\ K & \text { for } x=0\end{array}\right.\right.$ is continuous at $x=0$, then $K=$ ?
- A
$e$
- B
$e^{-1}$
- ✓
$e^2$
- D
$e^{-2}$
Answer(c) : We are given that $f(x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow K=\lim _{x \rightarrow 0}\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\left[(1+\tan x)^{\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}{\left.(1-\tan x)^{-\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}$ $=\frac{e}{e^{-1}}=e^2$
View full question & answer→MCQ 262 Marks
For what value of $k$, the function defined by $f(x)=\left\{\begin{array}{cc}\frac{\log (1+2 x) \sin x^{\circ}}{x^2} & \text { for } x \neq 0 \\ k & \text { for } x \neq 0\end{array}\right.$ is continuous at $x=0$ ?
- A
- B
$\frac{1}{2}$
- ✓
$\frac{\pi}{90}$
- D
$\frac{90}{\pi}$
AnswerCorrect option: C. $\frac{\pi}{90}$
(c) : $f(x)= \begin{cases}\frac{\log (1+2 x) \sin x^0}{x^2} & , x \neq 0 \\ k & , x=0\end{cases}$
Since, $f(x)$ is continuous at $x=0$
$
\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \\
& \therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left[2 \cdot \frac{\log (1+2 x)}{2 x} \frac{\sin \left(\frac{\pi x}{180}\right)}{x}\right] \\
& =2 \times \lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x} \frac{\sin \left(\frac{\pi x}{180}\right)}{\left(\frac{\pi x}{180}\right)} \times \frac{\pi}{180} \\
& \pi
\end{aligned}
$
$
=2 \times \frac{\pi}{180}=\frac{\pi}{90}\left[\because \lim _{x \rightarrow 0} \frac{\log (1+2 x)}{2 x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]
$
Also $f(0)=k$
$\therefore k=\frac{\pi}{90}$
View full question & answer→MCQ 272 Marks
If the function $f(x)$ defined by$f(x)= \begin{cases}x \sin \frac{1}{x} & \text { for } x \neq 0 \\ k & \text { for } x \neq 0\end{cases}$ is continuous at $x=0$, then $k=$
AnswerGiven $f(x)=\left\{\begin{array}{ll}x \sin \frac{1}{x} & , x \neq 0 \\ k & , x=0\end{array}\right.$
Also, $f(x)$ is continuous at $x=0$
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left[x \sin \frac{1}{x}\right]...(i)$
We know, $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$
$\Rightarrow-x \leq x \sin \left(\frac{1}{x}\right) \leq x$
$\Rightarrow \lim _{x \rightarrow 0}(-x) \leq \lim _{x \rightarrow 0}\left(x \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0}(x)$
$\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left[x \sin \frac{1}{x}\right] \leq 0 $
$\Rightarrow \lim _{x \rightarrow 0} x \sin \frac{1}{x}=0$
So, from $ (i) \lim _{x \rightarrow 0^{-}} f (x)=0 \quad$ and $f(0)=k ...(ii)$
$\therefore$ From $(i)$ and $(ii),$ we get $k=0$
View full question & answer→MCQ 282 Marks
For every integer $n$, let $a_n$ and $b_n$ be real number
Let function $f : R \rightarrow R$ be given by
$f(x)=\left\{\begin{array}{lll}a_n+\sin \pi x & \text { for } & x \in[2 n, 2 n+1] \\b_n+\cos \pi x & \text { for } & x \in(2 n-1,2 n)\end{array}\right.$
for all integers $n$. If $f$ is continuous, then which of the following does not hold for all $n$ ?
- A
$a_n-b_{n+1}=-1$
- ✓
$a_{n-1}-b_{n-1}=0$
- C
$a_n-b_n=1$
- D
$a_{n-1}-b_n=-1$
AnswerCorrect option: B. $a_{n-1}-b_{n-1}=0$
(B)
Since $f$ is continuous at every point in $R$.
$\therefore \quad f$ is continuous at $x=2 n$.
$\therefore \quad \lim _{x \rightarrow(2 n )^{-}} f (x)=\lim _{x \rightarrow(2 n )^{+}} f (x)= f (2 n )$
$\Rightarrow \lim _{x \rightarrow(2 \pi)^{-}}\left( b _{ n }+\cos \pi x\right)=\lim _{x \rightarrow(2 \pi)^{+}}\left( a _{ n }+\sin \pi x\right)$
$\Rightarrow b_n+\cos 2 n \pi=a_n+\sin 2 n \pi$
$\Rightarrow b _{ n }+ l = a _{ n } \Rightarrow a _{ n }- b _{ n }=1$
So, option (C) is correct.
Also, f is continuous at $x=2 n +1$.
$\lim _{x \rightarrow(2 n +1)^{-}} f (x)=\lim _{x \rightarrow(2 n +1)^{+}} f (x)= f (2 n +1)$
$\Rightarrow \lim _{x \rightarrow(2 n+1)^{-}}\left(a_n+\sin \pi x\right)=\lim _{x \rightarrow(2 n+1)^{+}}\left(b_{n+1}+\cos \pi x\right)$
$\Rightarrow a_n+\sin (2 n+1) \pi=b_{n+1}+\cos (2 n+1) \pi$
$\ldots .\left[\because f (x)= b _{ n +1}+\cos \pi x, x \in(2 n +1,2 n +2)\right]$
$\Rightarrow a _{ n }= b _{ n +1}-1$
$\Rightarrow a _{ n }- b _{ n +1}=-1$
Replacing $n$ by $n-1$, we get
$a _{ n -1}- b _{ n }=-1$
So, options (A) and (D) are correct.
Hence, option (B) does not hold.
View full question & answer→MCQ 292 Marks
The value of $p$ and $q$ for which the function
$f(x)=\left\{\begin{array}{ll}\frac{\sin (p+1) x+\sin x}{x}, & x<0 \\q, & x=0 \\\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}}, & x>0\end{array}\right.$
is continuous for all x in R, are
- A
$p=\frac{1}{2}, q=-\frac{3}{2}$
- B
$p=\frac{5}{2}, q=\frac{1}{2}$
- ✓
$p=-\frac{3}{2}, q=\frac{1}{2}$
- D
$p=\frac{1}{2}, q=\frac{1}{2}$
AnswerCorrect option: C. $p=-\frac{3}{2}, q=\frac{1}{2}$
(C)
Since $f (x)$ is continuous for all $x$ in R .
$\therefore f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0^{-}} f(x)$
$\Rightarrow q =\lim _{x \rightarrow 0} \frac{\sin ( p +1) x+\sin x}{x}$
$\Rightarrow q =\lim _{x \rightarrow 0}\left[( p +1) \times \frac{\sin ( p +1) x}{( p +1) x}+\frac{\sin x}{x}\right]$
$\Rightarrow q =( p +1)+1$
$\Rightarrow q=p+2$
The values of $p$ and $q$ in option (C) satisfies this condition.
View full question & answer→MCQ 302 Marks
If the function
$f(x)=\left\{\begin{array}{c}1+\sin \frac{\pi x}{2}, \text { for }-\infty < x \leq 1 \\a x+b, \text { for } 1 < x < 3 \\6 \tan \frac{\pi x}{12}, \text { for } 3 \leq x < 6\end{array}\right.$
is continuous in the interval $(-\infty, 6)$, then the values of a and b are respectively
Answer(C)
Since $f(x)$ is continuous in $(-\infty, 6)$.
$\therefore $ it is continuous at $x=1$ and $x=3$.
$\therefore \lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}\left(1+\sin \frac{\pi x}{2}\right)=\lim _{x \rightarrow 1^{+}}(a x+b)$
$\Rightarrow 1+\sin \frac{\pi}{2}=a+b$
$\Rightarrow a+b=2$ ...(i)
Also, $\lim _{x \rightarrow 3^{-}} f (x)=\lim _{x \rightarrow 3^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 3^{-}}(a x+b)=\lim _{x \rightarrow 3^{+}}\left(6 \tan \frac{\pi x}{12}\right)$
$\Rightarrow 3 a+b=6 \tan \frac{3 \pi}{12}$
$\Rightarrow 3 a+b=6$ ...(ii)
From (i) and (ii), we get $a=2, b=0$
View full question & answer→MCQ 312 Marks
If $f(x)$ is continuous in $[0, \pi]$, where
$f(x)=\left\{\begin{array}{ll}x+a \sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\2 x \cot x+b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2}, \text { then } \\a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi\end{array}\right.$
- A
$a=\frac{\pi}{6}, b=\frac{\pi}{12}$
- B
$a=-\frac{\pi}{6}, b=\frac{\pi}{12}$
- ✓
$a=\frac{\pi}{6}, b=-\frac{\pi}{12}$
- D
$a=-\frac{\pi}{6}, b=-\frac{\pi}{12}$
AnswerCorrect option: C. $a=\frac{\pi}{6}, b=-\frac{\pi}{12}$
(C)
Sincc $f(x)$ is continuous in $[0, \pi]$.
$\therefore $ it is continuous at $x=\frac{\pi}{4}$ and $x=\frac{\pi}{2}$.
$\therefore \lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}} f (x)=\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{-}}(x+ a \sqrt{2} \sin x)=\lim _{x \rightarrow\left(\frac{\pi}{4}\right)^{+}}(2 x \cot x+ b )$
$\Rightarrow \frac{\pi}{4}+ a \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)=2\left(\frac{\pi}{4}\right)(1)+ b$
$\Rightarrow \frac{\pi}{4}+ a =\frac{\pi}{2}+ b$
$\Rightarrow a-b=\frac{\pi}{4}$ $\quad\ldots(i)$
Also, $\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} f (x)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(2 x \cot x+ b )=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}}( a \cos 2 x- b \sin x)$
$\Rightarrow 2\left(\frac{\pi}{2}\right)(0)+ b = a (-1)- b (1)$
$\Rightarrow b =- a - b$
$\Rightarrow a+2 b=0$ $\quad\ldots(ii)$
From (i) and (ii), we get
$a=\frac{\pi}{6}$ and $b=\frac{-\pi}{12}$
View full question & answer→MCQ 322 Marks
Let $f (x)=\left\{\begin{array}{cc}\frac{x^4-5 x^2+4}{|(x-1)(x-2)|} ; & x \neq 1,2 \\ 6 ; & x=1 \\ 12 ; & x=2\end{array}\right.$
then $f (x)$ is continuous on the set
- A
- B
$R -\{1\}$
- C
$R -\{2\}$
- ✓
$R -\{1, 2\}$
AnswerCorrect option: D. $R -\{1, 2\}$
(D)
$f (x)=\frac{(x-1)(x+1)(x-2)(x+2)}{|x-1||x-2|}$
Since $\lim _{x \rightarrow 1} \frac{x-1}{|x-1|}$ does not exist.
Also, $\lim _{x \rightarrow 2} \frac{x-2}{|x-2|}$ does not exist
$\therefore f (x)$ is discontinuous at $x=1,2$.
For any $x \neq 1,2, f (x)$ is the quotient of two polynomials and a polynomial is everywhere continuous. Therefore, $f (x)$ is continuous for all $x \neq 1,2$.
$\therefore f (x)$ is continuous on $R -\{1,2\}$.
View full question & answer→MCQ 332 Marks
A function $f (x)$ is defined by $f (x)=\frac{ e ^x+ e ^{-x}-2}{x \sin x}$ for $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$. The value of $f(0)$ so that $f$ will be continuous in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is
Answer(B)
$f (x)$ is continuous in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ except at $x=0$.
For $f (x)$ to be continuous in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,
$f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{ e ^x+ e ^{-x}-2}{x \sin x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{ e ^x- e ^{-x}}{x \cos x+\sin x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{ e ^x+ e ^{-x}}{-x \sin x+\cos x+\cos x}$
$=\frac{ e ^0+ e ^0}{0+2 \cos 0}=\frac{1+1}{2(1)}=1$
View full question & answer→MCQ 342 Marks
The value of $f(0)$ so that the function
$f(x)=\frac{\sqrt{1+x}-(1+x)^{\frac{1}{3}}}{x}$
becomes continuous is equal to
- ✓
$\frac{1}{6}$
- B
$\frac{1}{4}$
- C
- D
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{1}{6}$
(A)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{2}}-(1+x)^{\frac{1}{3}}}{x}$
$=\lim _{x \rightarrow 0} \frac{\left(1+\frac{1}{2} x-\frac{1}{8} x^2+\ldots .\right)-\left(1+\frac{1}{3} x-\frac{1}{9} x^2+\ldots .\right)}{x}$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{1}{2}-\frac{1}{3}\right) x+\left(\frac{1}{9}-\frac{1}{8}\right) x^2+\ldots .}{x}$
$=\lim _{x \rightarrow 0}\left[\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{9}-\frac{1}{8}\right) x+\ldots\right]$
$\therefore f(0)=\frac{1}{6}$
View full question & answer→MCQ 352 Marks
The values of $a$ and $b$ such that the function defined by
$f(x)=\left\{\begin{array}{cl}7 & , \text { if } x \leq 2 \\ax+b & , \text { if } 2 < x < 9 \text { is continuous on its } \\21 & , \text { if } x \geq 9\end{array}\right.$
domain are
Answer(B)
Since $f (x)$ is continuous on its domain.
$\therefore $ it is continuous at $x=2$ and $x=9$.
$\therefore \lim _{x \rightarrow 2^{+}} f (x)=\lim _{x \rightarrow 2^{-}} f (x)$
$\Rightarrow \lim _{x \rightarrow 2^{+}}( ax + b )=7$
$\Rightarrow 2 a+b=7$ $\quad\ldots(i)$
Also, $\lim _{x \rightarrow 9^{-}} f (x)=\lim _{x \rightarrow 9^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 9^{-}}( ax + b )=21$
$\Rightarrow 9 a+b=21$ $\quad\ldots(ii)$
Solving (i) and (ii), we get $a=2, b=3$
View full question & answer→MCQ 362 Marks
If $f(x)$ is continuous in $[-2,2]$, where
$f(x)=\left\{\begin{array}{ll}\frac{\sin a x}{x}-2, & \text { for }-2 \leq x < 0 \\2 x+1, & \text { for } 0 \leq x \leq 1 \\2 b \sqrt{x^2+3}-1, & \text { for } 1 < x \leq 2\end{array}\right.$
then the value of $(a+b)$ is
Answer(B)
Since $f (x)$ is continuous in $[-2,2]$.
∴ $\quad$ it is continuous at $x=0$ and $x=1$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(\frac{\sin a x}{x}-2\right)=\lim _{x \rightarrow 0^{+}}(2 x+1)$
$\Rightarrow a-2=0+1 \Rightarrow a=3$
Also, $\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}(2 x+1)=\lim _{x \rightarrow 1^{+}}\left(2 b \sqrt{x^2+3}-1\right)$
$\Rightarrow 2(1)+1=2 b \sqrt{1+3}$
$\Rightarrow 3=4 b-1$
$\Rightarrow b =1$
$\therefore \quad a+b=3+1=4$
View full question & answer→MCQ 372 Marks
If the function $f (x)$ is continuous in $[0,8]$, where
$f(x)$ $=\left\{\begin{array}{ll}x^2+a x+6, & 0 \leq x< 2 \\3 x+2, & 2 \leq x \leq 4 \\2 a x+5 b, & 4< x \leq 8\end{array}\right. \text {, then }$
- A
$a=1, b=\frac{22}{5}$
- ✓
$a=-1, b=\frac{22}{5}$
- C
$a=1, b=\frac{-22}{5}$
- D
$a=-1, b=\frac{-22}{5}$
AnswerCorrect option: B. $a=-1, b=\frac{22}{5}$
(B)
Since $f(x)$ is continuous in $[0,8]$.
$\therefore \quad$ it is continuous at $x=2$ and $x=4$.
$\therefore \quad \lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 2^{-}}\left(x^2+a x+6\right)=\lim _{x \rightarrow 2^{+}}(3 x+2)$
$\Rightarrow(2)^2+2 a+6=3(2)+2$
$\Rightarrow 10+2 a =8$
$\Rightarrow a=-1$ ...(i)
Also, $\lim _{x \rightarrow 4^{-}} f (x)=\lim _{x \rightarrow 4^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 4^{-}}(3 x+2)=\lim _{x \rightarrow 4^{+}}(2 a x+5 b)$
$\Rightarrow 3(4)+2=2 a (4)+5 b$
$\Rightarrow 14=8 a+5 b$
$\Rightarrow b=\frac{22}{5} \quad\ldots[From (i)]$
View full question & answer→MCQ 382 Marks
If $f (x)=\left\{\begin{array}{ll}\sin x & \text { if } x \leq 0 \\ x ^ 2+ a ^2 & \text { if } 0 < x<1 \\ b x+2 & \text { if } 1 \leq x \leq 2 \\ 0 & \text { if } x>2\end{array}\right.$
is continuous on R , then $a + b + ab =$
Answer(D)
$f (x)=\sin x$
$\therefore \quad f(0)=\sin 0=0$
$\lim _{x \rightarrow 0^{+}} x^2+a^2=0^2+a^2=a^2$
Since the function is continuous at $x=0$,
$\lim _{x \rightarrow 0^{+}} f (x)= f (0)$
$\Rightarrow 0= a ^2$
$\Rightarrow a =0$
$\lim _{x \rightarrow 1^{-}} x^2+a^2=1^2+a^2=1$
$f (x)= b x+2$
$\therefore f (1)= b +2$
Now, $\lim _{x \rightarrow 1^{-}} f (x)= f (1)$
$\Rightarrow 1=b+2$
$\Rightarrow b =-1$
$a+b+a b=0-1+0(-1)=-1$
View full question & answer→MCQ 392 Marks
The function $f (x)=\sin |x|$ is
- ✓
- B
Continuous only at certain points
- C
Differentiable at all points
- D
Answer(A)
Let $g (x)=|x|$ and $h (x)=\sin x$.
Then, $f (x)=(\operatorname{hog})(x)$ for all $x \in R$.
As both $g$ and $h$ are continuous functions on $R$.
$\therefore \quad f (x)$ is also continuous for all $x \in R$.
View full question & answer→MCQ 402 Marks
The function $f(x)=\frac{x^2-4}{\sin x-2}$ is
- ✓
continuous for all real values of x
- B
- C
discontinuous when sin x = 2
- D
AnswerCorrect option: A. continuous for all real values of x
(A)
For all $x \in R ,-1 \leq \sin x \leq 1$
$\therefore f (x)$ is continuous for all real values of $x$.
View full question & answer→MCQ 412 Marks
The function $f (x)=[x]$, where $[x]$ the greatest function is continuous at
Answer(C)
Since $f (x)=[x]$ is continuous at every non integer points.
∴ option (C) is the correct answer.
View full question & answer→MCQ 422 Marks
The function $f (x)=\left\{\begin{array}{ll}x+2, & 1 \leq x<2 \\ 4, & x=2 \\ 3 x-2, & x>2\end{array}\right.$ is continuous at
- A
$x=2$ only
- B
$x \leq 2$
- ✓
$x \geq 2$
- D
AnswerCorrect option: C. $x \geq 2$
(C)
The given function is defined only in the interval $[1, \infty)$. For $x>2, y=3 x-2$ which is a straight line, hence continuous. Also, the given function is continuous at $x=2$.
∴ option (C) is the correct answer.
View full question & answer→MCQ 432 Marks
Let $f (x)=\left\{\begin{aligned} \frac{x^3+x^2-16 x+20}{(x-2)^2} & ; \text { if } x \neq 2 \\ k & ; \text { if } x=2\end{aligned}\right.$. If $f (x)$ is continuous for all $x$, then $k =$
Answer(A)
Since $f (x)$ is continuous for all $x$.
$\therefore f (x)$ is continuous at $x=2$.
$\therefore f (2)=\lim _{x \rightarrow 2} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 2} \frac{x^3+x^2-16 x+20}{(x-2)^2}$
$=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^2+3 x-10\right)}{(x-2)^2}$
$=\lim _{x \rightarrow 2} \frac{(x-2)^2(x+5)}{(x-2)^2}$
$=7$
View full question & answer→MCQ 442 Marks
If $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & -1 \leq x<0 \\ \frac{2 x+1}{x-2}, & 0 \leq x \leq 1\end{array}\right.$ is continuous in $[-1,1]$, then $p$ is equal to
- A
- ✓
$\frac{-1}{2}$
- C
$\frac{1}{2}$
- D
AnswerCorrect option: B. $\frac{-1}{2}$
(B)
Since $f (x)$ is continuous in $[-1,1]$.
∴ it is continuous at $x=0$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+ p x}-\sqrt{1- p x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x-2}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{(1+ p x)-(1- p x)}{x(\sqrt{1+ p x}+\sqrt{1- p x})}=\frac{-1}{2}$
$\Rightarrow p =\frac{-1}{2}$
View full question & answer→MCQ 452 Marks
Let $f (x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right]$. If $f (x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$, then $f \left(\frac{\pi}{4}\right)$ is
- A
- B
$\frac{1}{2}$
- ✓
$-\frac{1}{2}$
- D
AnswerCorrect option: C. $-\frac{1}{2}$
(C)
Since $f(x)$ is continuous in $\left[0, \frac{\pi}{2}\right]$
$\therefore \quad$ it is continuous at $x=\frac{\pi}{4}$
$\therefore f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}$
Applying L'Hospital rule on R.H.S., we get
$f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^2 x}{4}$
$\Rightarrow f \left(\frac{\pi}{4}\right)=\frac{-2}{4}=\frac{-1}{2}$
View full question & answer→MCQ 462 Marks
$\begin{aligned}\text {If } f (x) & =\frac{\left(2^x-1\right)^2}{\tan x \cdot \log (1+x)}, \text { for } x \neq 0 \\ & =\log 4 \quad, \text { for } x=0, \text { then }\end{aligned}$
- A
$f (x)$ is continuous at $x=0$
- ✓
$f (x)$ has removable discontinuity at $x=0$
- C
$f (x)$ has irremovable discontinuity at $x=0$
- D
AnswerCorrect option: B. $f (x)$ has removable discontinuity at $x=0$
(B)
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)^2}{\tan x \cdot \log (1+x)}$
$=\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)^2}{x^2} \times \frac{1}{\frac{\tan x}{x}} \times \frac{1}{\frac{\log (1+x)}{x}}$
$=(\log 2)^2 \times 1=(\log 2)^2$
and $f(0)=\log 4$
$\therefore f (x)$ is discontinuous at $x=0$.
Here, $\lim _{x \rightarrow 0} f(x)$ exists but not equal to $f(0)$.
∴ the discontinuity at $x=0$ is removable.
View full question & answer→MCQ 472 Marks
If the function is defined as
$\begin{aligned}f(x) & =\frac{5^{\cos x}-1}{\frac{\pi}{2}-x}, \text { when } x \neq \frac{\pi}{2} \\& =2 \log 5, \text { when } x=\frac{\pi}{2}, \text { then }\end{aligned}$
- A
$f (x)$ is continuous at $x=\frac{\pi}{2}$
- ✓
$f (x)$ has removable discontinuity at $x=\frac{\pi}{2}$
- C
$f (x)$ has irremovable discontinuity at $x=\frac{\pi}{2}$
- D
AnswerCorrect option: B. $f (x)$ has removable discontinuity at $x=\frac{\pi}{2}$
(B)
Applying L'Hospital rule, we get
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{5^{\cos x}-1}{\frac{\pi}{2}-x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{5^{\cos x} \cdot \log 5(-\sin x)}{-1}$
$=5^{\cos \frac{\pi}{2}} \cdot \log 5 \sin \frac{\pi}{2}=\log 5$
and $f\left(\frac{\pi}{2}\right)=2 \log 5$
$\therefore f (x)$ is discontinuous at $x=\frac{\pi}{2}$.
Here, $\lim _{x \rightarrow \frac{\pi}{2}} f (x)$ exists but not equal to $f \left(\frac{\pi}{2}\right)$.
$\therefore $ the discontinuity at $x=\frac{\pi}{2}$ is removable.
View full question & answer→MCQ 482 Marks
If $f(x)=\left\{\begin{array}{ll}\frac{5^x- e ^{ z }}{\sin 2 x} \quad ; \quad x \neq 0 \\ \frac{1}{2}(\log 5+1) \quad ; \quad x=0\end{array}\right.$ then
- A
$f (x)$ is continuous at $x=0$
- ✓
$f (x)$ is discontinuous at $x=0$
- C
$\lim _{x \rightarrow 0} f (x)$ does not exist
- D
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=0$
(B)
$\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{5^x- e ^x}{\sin 2 x}=\lim _{x \rightarrow 0} \frac{5^x-1+1- e ^x}{\sin 2 x}$
$=\lim _{x \rightarrow 0} \frac{\frac{5^x-1}{x}-\frac{ e ^x-1}{x}}{\frac{\sin 2 x}{2 x} \times 2}$
$=\frac{\log 5-\log e }{2}=\frac{1}{2}(\log 5-1)$
$\therefore \lim _{x \rightarrow 0} f (x) \neq f (0)$
$\therefore f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 492 Marks
$\begin{aligned}\text {If } f(x) & =\frac{x \cos x-3 \tan x}{x^2+2 \sin x}, x \neq 0 \\ & =1, \quad x=0, \text { then }\end{aligned}$
- ✓
$f ( x )$ is discontinuous at $x=0$
- B
$f(x)$ is continuous at $x=0$
- C
$\lim _{x \rightarrow 0} f(x)$ does not exist
- D
AnswerCorrect option: A. $f ( x )$ is discontinuous at $x=0$
(A)
$\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{x \cos x-3 \tan x}{x^2+2 \sin x}$
$=\lim _{x \rightarrow 0} \frac{\frac{x \cos x-3 \tan x}{x}}{\frac{x^2+2 \sin x}{x}}$
$=\lim _{x \rightarrow 0} \frac{\cos x-\frac{3 \tan x}{x}}{x+\frac{2 \sin x}{x}}$
$=\frac{1-3}{0+2}=-1$
$\therefore \quad \lim _{x \rightarrow 0} f (x) \neq f (0)$
$\therefore \quad f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 502 Marks
$\begin{aligned}\text {If } f(x) & =\frac{\sin 2 x}{\sqrt{1-\cos 2 x}}, 0 < x \leq \frac{\pi}{2} \\ & =\frac{\cos x}{\pi-2 x}, \frac{\pi}{2}< x< \pi, \text { then }\end{aligned}$
- A
$\lim _{x\rightarrow\frac{\pi^{-}}{2}} f(x)=1$
- B
$\lim _{x\rightarrow{\frac{\pi^+}{2}}} f(x)=1$
- C
$f ( x )$ is continuous at $x=\frac{\pi}{2}$
- ✓
$f ( x )$ is discontinuous at $x=\frac{\pi}{2}$
AnswerCorrect option: D. $f ( x )$ is discontinuous at $x=\frac{\pi}{2}$
(D)
$\lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin 2 x}{\sqrt{1-\cos 2 x}}=\frac{\sin \pi}{\sqrt{1-\cos \pi}}=0$
$\lim _{x \rightarrow \frac{\pi}{2}^{+}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\pi-2 x}$
$=\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}$
$=\lim _{h \rightarrow 0} \frac{-\sin h}{-2 h}=\frac{1}{2}(1)=\frac{1}{2}$
$\therefore \lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x) \neq \lim _{x \rightarrow \frac{\pi^{+}}{2}} f (x)$
$\therefore f (x)$ is discontinuous at $x=\frac{\pi}{2}$.
View full question & answer→MCQ 512 Marks
The points of discontinuity of $\tan x$ are
AnswerCorrect option: C. $(2 n+1) \frac{\pi}{2}, n \in I$
(C)
Let $f (x)=\tan x$
The point of discontinuity of $f (x)$ are those points where $\tan x$ is infinite.
i.e., $\tan x=\infty$
$\Rightarrow x=(2 n +1) \frac{\pi}{2}, n \in I$
View full question & answer→MCQ 522 Marks
The number of discontinuities of the greatest integer function $f (x)=[x], x \in\left(-\frac{7}{2}, 100\right)$ is equal to
Answer(D)
Given, $f (x)=[x], x \in(-3.5,100)$
As we know greatest integer function is discontinuous on integer values.
In given interval, the integer values are $(-3,-2,-1,0, \ldots, 99)$.
$\therefore \quad$ the total number of integers are 103 .
View full question & answer→MCQ 532 Marks
For the function $f(x)=\left\{\begin{array}{r}\frac{\sin ^2 a x}{x^2}, \text { when } x \neq 0 \\ 1, \text { when } x=0\end{array}\right.$ which one is a true statement?
- A
$f (x)$ is continuous at $x=0$, when $a \neq \pm 1$
- ✓
$f (x)$ is discontinuous at $x=0$, when $a \neq \pm 1$
- C
$\lim _{x \rightarrow 0} f (x)= a$
- D
$\lim _{x \rightarrow 0} f(x)=a^3$
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=0$, when $a \neq \pm 1$
(B)
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin ^2 a x}{(a x)^2} a^2=a^2$ and $f(0)=1$.
$\therefore f (x)$ is discontinuous at $x=0$, when $a \neq \pm 1$
View full question & answer→MCQ 542 Marks
If $f (y)=\frac{\left( e ^{2 y}-1\right) \cdot \sin y}{y^2} $, for $y \neq 0$
$=4$ , for $y=0$, then
- ✓
$f (y)$ is discontinuous at $y=0$
- B
$f (y)$ is continuous at $y=0$
- C
$\lim _{y \rightarrow 0} f (y)$ does not exist
- D
AnswerCorrect option: A. $f (y)$ is discontinuous at $y=0$
(A)
$\lim _{y \rightarrow 0} f (y)=\lim _{y \rightarrow 0} \frac{\left( e ^{2 y}-1\right)}{2 y} \times 2 \times \frac{\sin y}{y}$
$=\log e \times 2 \times 1=2$
and $f(0)=4$
$\therefore f (y)$ is discontinuous at $y=0$.
View full question & answer→MCQ 552 Marks
If $f: R \rightarrow R$ is defined by
$f(x)=\left\{\begin{array}{ll}x-1, & \text { for } x \leq 1 \\2-x^2, & \text { for } 1 < x \leq 3 \\x-10, & \text { for } 3 < x< 5 \\2 x, & \text { for } x \geq 5\end{array}\right.$
then the set of points of discontinuity of $f$ is
- A
$R -\{1,3,5\}$
- B
$\{1,3,5\}$
- C
$R-\{1,5\}$
- ✓
$\{1,5\}$
AnswerCorrect option: D. $\{1,5\}$
(D)
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 2-x^2=1$
$f(1)=1-1=0$
$\lim _{x \rightarrow 1} f(x) \neq f(1)$
$\therefore \quad$ The function is discontinuous at $x=1$
$\lim _{x \rightarrow 5} f (x)=\lim _{x \rightarrow 5} x-10=-5$
$f(5)=2(5)=10$
$\lim _{x \rightarrow 5} f(x)+f(5)$
$\therefore \quad$ The function is discontinuous at $x=5$
$\lim _{x \rightarrow 3^{+}} f (x)=x-10=-7$
$f(3)=2-3^2=-7$
$\lim _{x \rightarrow 3} f(x)=f(3)$
$\therefore \quad$ The function is continuous at $x=3$
View full question & answer→MCQ 562 Marks
If $f(x)=\left\{\begin{array}{ll}4-3 x ; & 0 < x \leq 2 \\ 2 x-6 ; & 2 < x \leq 3 \\ x+5 ; & 3 < x \leq 6\end{array}\right.$, then $f(x)$ is
- ✓
continuous at $x=2$ and discontinuous at $x=3$
- B
continuous at $x=3$ and discontinuous at $x=2$
- C
continuous at $x=2$ and $x=3$
- D
discontinuous at $x=2$ and $x=3$
AnswerCorrect option: A. continuous at $x=2$ and discontinuous at $x=3$
(A)
$\lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2^{-}}(4-3 x)=4-6=-2$
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}(2 x-6)=4-6=-2$
$f(2)=4-3(2)=-2$
$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}(2 x-6)=6-6=0$
$\lim _{x \rightarrow 3^{+}} f (x)=\lim _{x \rightarrow 3^{+}}(x+5)=3+5=8$
$\therefore \quad \lim _{x \rightarrow 3^{-}} f (x) \neq \lim _{x \rightarrow 3^{+}} f (x)$
$\therefore f (x)$ is continuous at $x=2$ and discontinuous at $x=3$.
View full question & answer→MCQ 572 Marks
The value(s) of r for which the function
$f(x)=\left\{\begin{array}{cc}1-x , & x<1 \$1-x)(2-x) , & 1 \leq x \leq 2 \\3-x , & x>2\end{array}\right.$
fails to be continuous is (are)
Answer(B)
$\begin{array}{l}\lim _{x \rightarrow 1^{-}} f(x)=0, \lim _{x \rightarrow 1^{+}} f(x)=0 \text { and } f(1)=0\end{array}$
$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
$\therefore f (x)$ is continuous at $x=1$.
$\lim _{x \rightarrow 2^{-}} f (x)=0$ and $\lim _{x \rightarrow 2^{+}} f (x)=1$
$\therefore \lim _{x \rightarrow 2^{-}} f (x) \neq \lim _{x \rightarrow 2^{+}} f (x) =1$
$\therefore f (x)$ is not continuous at $x=2$.
View full question & answer→MCQ 582 Marks
The number of discontinuities in R for the function $f(x)=\frac{x-1}{x^3+6 x^2+11 x+6}$ is
Answer(A)
$f (x)=\frac{x-1}{x^3+6 x^2+11 x+6}$
$=\frac{x-1}{(x+1)(x+2)(x+3)}$
∴ the points of discontinuity are $x=-1, x=-2$ and $x=-3$.
$\therefore \quad$ The number of discontinuities is 3 .
View full question & answer→MCQ 592 Marks
The function $f (x)=\frac{2 x^2+7}{x^3+3 x^2-x-3}$ is discontinuous for
AnswerCorrect option: C. $x=1, x=-1, x=-3$ only
(C)
$f (x)=\frac{2 x^2+7}{x^2(x+3)-1(x+3)}=\frac{2 x^2+7}{\left(x^2-1\right)(x+3)}$
$=\frac{2 x^2+7}{(x-1)(x+1)(x+3)}$
∴ the points of discontinuity are
$x=1, x=-1$ and $x=-3$ only.
View full question & answer→MCQ 602 Marks
The function $f(x)=\frac{4-x^2}{4 x-x^3}$ is
- A
discontinuous at only one point
- B
discontinuous exactly at two points
- ✓
discontinuous exactly at three points
- D
AnswerCorrect option: C. discontinuous exactly at three points
(C)
$f (x)=\frac{4-x^2}{x\left(4-x^2\right)}=\frac{4-x^2}{x(2+x)(2-x)}$
Since $f (x)$ does not exist at $x=0,2,-2$.
∴ there are three points of discontinuity.
View full question & answer→MCQ 612 Marks
The number of points at which the function $f ( x )=\frac{1}{\log |x|}$ is discontinuous are
Answer(C)
Since $f (x)$ is not defined at $x=0,1,-1$ and at all other points $f (x)$ is continuous.
∴ the given function is discontinuous at 3 points.
View full question & answer→MCQ 622 Marks
For the function $f (x)=\left\{\begin{array}{ll}\frac{ e ^{1 / x}-1}{ e ^{1 / x}+1}, & , x \neq 0 \\ 1 & , x=0\end{array}\right.$, which of the following is correct?
- A
$\lim _{x \rightarrow 0^{+}} f(x)=-1$
- B
$f (x)$ is continuous at $x=0$
- C
$\lim _{x \rightarrow 0^{-}} f(x)=1$
- ✓
$f (x)$ is not continuous at $x=0$
AnswerCorrect option: D. $f (x)$ is not continuous at $x=0$
(D)
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \frac{e^{\frac{-1}{h}}-1}{e^{\frac{-1}{h}}+1}=\lim _{h \rightarrow 0} \frac{e^{\frac{1}{{ }^{\frac{1}{h}}}-1}}{\frac{1}{e^{\frac{1}{h}}}+1}=\frac{0-1}{0+1}=-1$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} \frac{e^{\frac{1}{h}}-1}{e^{\frac{1}{h}}+1}=\lim _{h \rightarrow 0} \frac{1-\frac{1}{e^{\frac{1}{h}}}}{1+\frac{1}{e^{\frac{1}{h}}}}=\frac{1-0}{1+0}=1$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x) \neq \lim _{x \rightarrow 0^{+}} f (x)$
$\therefore f (x)$ is not continuous at $x=0$.
View full question & answer→MCQ 632 Marks
The function $f(x)=|x|+\frac{|x|}{x}$ is
- A
- B
Discontinuous at the origin because $|x|$ is discontinuous there
- ✓
Discontinuous at the origin because $\frac{|x|}{x}$ is discontinuous there
- D
Discontinuous at the origin because both $|x|$ and $\frac{|x|}{x}$ are discontinuous there
AnswerCorrect option: C. Discontinuous at the origin because $\frac{|x|}{x}$ is discontinuous there
(C)
$|x|$ is continuous at $x=0$ and $\frac{|x|}{x}$ is discontinuous at $x=0$.
$\therefore f (x)=|x|+\frac{|x|}{x}$ is discontinuous at $x=0$.
View full question & answer→MCQ 642 Marks
The function $f(x)=\frac{|x|}{x^2+2 x}, x \neq 0$ and $f(0)=0$ is not continuous at $x=0$ because
- A
$\lim _{x \rightarrow 0} f(x) \neq f(0)$
- B
$\lim _{x \rightarrow 0^{+}} f(x)$ does not exist
- C
$\lim _{x \rightarrow 0^{-}} f(x)$ does not exist
- ✓
$\lim _{x \rightarrow 0} f (x)$ does not exist
AnswerCorrect option: D. $\lim _{x \rightarrow 0} f (x)$ does not exist
(D)
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}} \frac{|x|}{x^2+2 x}$
$=\lim _{x \rightarrow 0} \frac{-x}{x^2+2 x}=-\frac{1}{2}$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{|x|}{x^2+2 x}$
$=\lim _{x \rightarrow 0} \frac{x}{x^2+2 x}=\frac{1}{2}$
$\therefore \quad \lim _{x \rightarrow 0} f (x)$ does not exist.
View full question & answer→MCQ 652 Marks
If $f (x)=\left\{\begin{array}{r}\frac{x-|x|}{x} ; \text { when } x \neq 0 \\ 2 ; \text { when } x=0\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=0$
- ✓
$f (x)$ is discontinuous at $x=0$
- C
$\lim _{x \rightarrow 0} f(x)=2$
- D
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=0$
(B)
$\lim _{x \rightarrow 0^{-}} f (x)=1+1=2$
$\lim _{x \rightarrow 0^{+}} f(x)=0$
$\therefore \quad f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 662 Marks
The function $\frac{\sin x}{|x|}$
AnswerCorrect option: B. is discontinuous at $x=0$
(B)
$\lim _{x \rightarrow 0^{+}} \frac{\sin x}{|x|}=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
and $\lim _{x \rightarrow 0^{-}} \frac{\sin x}{|x|}=\lim _{x \rightarrow 0} \frac{\sin x}{-x}=-1$
∴ the given function is discontinuous at $x=0$.
View full question & answer→MCQ 672 Marks
The function $f (x)=\frac{|3 x-4|}{3 x-4}$ is discontinuous at
- A
$x=4$
- B
$x=\frac{3}{4}$
- ✓
$x=\frac{4}{3}$
- D
$x=\frac{2}{3}$
AnswerCorrect option: C. $x=\frac{4}{3}$
(C)
As $\frac{|x|}{x}$ is discontinuous at $x=0$.
$\therefore \quad \frac{|3 x-4|}{3 x-4}$ is discontinuous at $3 x-4=0$.
$\therefore \quad x=\frac{4}{3}$
View full question & answer→MCQ 682 Marks
$\begin{aligned}\text {If} f (x) & =\frac{|x|}{x}, \text { for } x \neq 0 \\ & =1, \text { for } x=0,\end{aligned}$
then $f (x)$ is
AnswerCorrect option: B. discontinuous at $x=0$
(B)
When $x<0,|x|=-x$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0} \frac{-x}{x}=\lim _{x \rightarrow 0}(-1)=-1$
When $x>0,|x|=x$
$\therefore \quad \lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0} \frac{x}{x}=\lim _{x \rightarrow 0}(1)=1$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x) \neq \lim _{x \rightarrow 0^{+}} f (x)$
$\therefore f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 692 Marks
If $f(x)=\left\{\begin{array}{r}\frac{x^4-16}{x-2}, \text { when } x \neq 2 \\ 16, \text { when } x=2\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=2$
- ✓
$f (x)$ is discontinuous at $x=2$
- C
$\lim _{x \rightarrow 2} f(x)=16$
- D
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=2$
(B)
$\lim _{x \rightarrow 2} f (x)=\lim _{x \rightarrow 2} \frac{x^4-16}{x-2}$
$=\lim _{x \rightarrow 2} \frac{(x-2)(x+2)\left(x^2+4\right)}{x-2}$
$=\lim _{x \rightarrow 2}(x+2)\left(x^2+4\right)=32$ and $f(2)=16$
$\therefore \quad \lim _{x \rightarrow 2} f (x) \neq f (2)$
$\therefore f (x)$ is discontinuous at $x=2$.
View full question & answer→MCQ 702 Marks
If $f(x)=\left\{\begin{array}{ll}e^{\frac{1}{x}}, & \text { when } x \neq 0 \\ 1, & \text { when } x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{+}} f(x)=e$
- B
$\lim _{x \rightarrow 0^{+}} f(x)=0$
- ✓
$f (x)$ is discontinuous at $x=0$
- D
$f (x)$ is continuous at $x=0$
AnswerCorrect option: C. $f (x)$ is discontinuous at $x=0$
(C)
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} e^{-1 / h}=0$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} e^{1 / h}=\infty$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x) \neq \lim _{x \rightarrow 0^{+}} f (x)$
$\therefore f (x)$ is discontinuous at $x=0$.
View full question & answer→MCQ 712 Marks
$\text { If } f(x)=\left\{\begin{array}{ll}\frac{\sin (a+1) x+\sin x}{x}, & x<0 \\c, & x=0 \\\frac{\sqrt{x+b x^2}-\sqrt{x}}{b \sqrt{x}}, & x>0\end{array}\right.$
is continuous at $x=0$, then
- A
$a=-2, b=0, c \neq 0$
- ✓
$a=-2, b \neq 0, c=0$
- C
$a=2, b=0, c \neq 0$
- D
$a=2, b \neq 0, c=0$
AnswerCorrect option: B. $a=-2, b \neq 0, c=0$
(B)
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}} \frac{\sin ( a +1) x+\sin x}{x}$
$=\lim _{x \rightarrow 0^{-}}\left[\frac{\sin (a+1) x}{(a+1) x} \times(a+1)+\frac{\sin x}{x}\right]$
$=a+1+1$
$=a+2$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x+b x^2}-\sqrt{x}}{b \sqrt{x}}$
$=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}\left(\begin{array}{ll}\sqrt{1+b x}-1\end{array}\right)}{b \sqrt{x}}$
$=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1+ b x}-1}{b}=\frac{0}{b}=0$, if $b \neq 0$
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow a +2=0= c$
$\Rightarrow a =-2, c =0$
$\therefore \quad a=-2, b \neq 0$ and $c=0$
View full question & answer→MCQ 722 Marks
If $f (x)=|x|+|x-1|$, then
- A
$f (x)$ is continuous at $x=0$ only
- B
$f (x)$ is continuous at $x=1$ only
- ✓
$f (x)$ is continuous at both $x=0$ and $x=1$
- D
$f (x)$ is discontinuous
AnswerCorrect option: C. $f (x)$ is continuous at both $x=0$ and $x=1$
(C)
Given, $f (x)=|x|+|x-1|$
$\therefore f (x)=\left\{\begin{array}{ll}-x-(x-1), & \text { if } x<0 \\ x-(x-1), & \text { if } 0 \leq x<1 \\ x+(x-1), & \text { if } x \geq 1\end{array}\right.$
$\therefore f (x)=\left\{\begin{array}{ll}-2 x+1, & \text { if } x<0 \\ 1, & \text { if } 0 \leq x<1 \\ 2 x-1, & \text { if } x \geq 1\end{array}\right.$
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0}(-2 x+1)=1$
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0} 1=1$
$f(0)=1$
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)= f (0)$
$\therefore f (x)$ is continuous at $x=0$.
$\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1} 1=1$
$\lim _{x \rightarrow 1^{+}} f (x)=\lim _{x \rightarrow 1}(2 x-1)=1$
$f(1)=2(1)-1=1$
$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
$\therefore f (x)$ is continuous at $x=1$.
View full question & answer→MCQ 732 Marks
If $f(x)=\left\{\begin{array}{ll}\frac{x-2}{|x-2|}+a, & x<2 \\ a+b, & x=2 \\ \frac{x-2}{|x-2|}+b, & x>2\end{array}\right.$ is continuous at $x=2$, then $a + b =$
Answer(C)
$\lim _{x \rightarrow 2^{-}} f (x)=\lim _{ h \rightarrow 0} f (2- h )$
$=\lim _{h \rightarrow 0} \frac{2-h-2}{|2-h-2|}+a$
$=\lim _{h \rightarrow 0}\left(-\frac{h}{h}+a\right)=a-1$
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)$
$=\lim _{h \rightarrow 0} \frac{2+h-2}{|2+h-2|}+b=b+1$ and $f(2)=a+b$
Since $f (x)$ is continuous at $x=2$,
$\therefore \quad \lim _{x \rightarrow 2^{-}} f (x)= f (2)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow a -1= a + b = b +1$
$\Rightarrow b =-1$ and $a =1$
View full question & answer→MCQ 742 Marks
If $f ( x )$ is continuous at $x=0$, where$f(x)=\left\{\begin{array}{ll}x^2+a, & x \geq 0 \\2 \sqrt{x^2+1}+b, & x<0\end{array}\right.$and $f\left(\frac{1}{2}\right)=2$, then the values of $a$ and $b$ are respectively
- A
$\frac{7}{4}, \frac{1}{4}$
- ✓
$\frac{7}{4},-\frac{1}{4}$
- C
$\frac{-1}{4}, \frac{7}{4}$
- D
$-\frac{7}{4},-\frac{1}{4}$
AnswerCorrect option: B. $\frac{7}{4},-\frac{1}{4}$
(B)
$f \left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2+ a$
$\Rightarrow 2=\frac{1}{4}+a \Rightarrow a=\frac{7}{4}\quad\ldots(i)$
Since $f (x)$ is continuous at $x=0$
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(2 \sqrt{x^2+1}+ b \right)=\lim _{x \rightarrow 0^{+}}\left(x^2+ a \right)$
$\Rightarrow 2 \sqrt{0+1}+b=0+a$
$\Rightarrow 2+b=\frac{7}{4} \quad \ldots[From (i)]$
$\Rightarrow b =-\frac{1}{4}$
View full question & answer→MCQ 752 Marks
$f(x)=\left\{\begin{array}{ll}\frac{x^2-9}{x-3}+a , &x>3 \\ 5 , &x=3 \\ 2 x^2+3 x+b, &x<3\end{array}\right.$
is continuous at $x=3$, then
- A
$a=1, b=-22$
- B
$a=1, b=22$
- C
$a=-1, b=22$
- ✓
$a=-1, b=-22$
AnswerCorrect option: D. $a=-1, b=-22$
(D)
Since $f (x)$ is continuous at $x=3$.
$\therefore \quad \lim _{x \rightarrow 3^{-}} f (x)=\lim _{x \rightarrow 3^{+}} f (x)= f (3)$
$\lim _{x \rightarrow 3^{-}} f (x)= f (3)$
$\Rightarrow \lim _{x \rightarrow 3^{-}}\left(2 x^2+3 x+ b \right)=5$
$\Rightarrow 2(3)^2+3(3)+b=5$
$\Rightarrow b =-22$
Also, $\lim _{x \rightarrow 3^{+}} f (x)= f (3)$
$\Rightarrow \lim _{x \rightarrow 3^{+}}\left(\frac{x^2-9}{x-3}+ a \right)=5$
$\Rightarrow(3+3+a)=5$
$\Rightarrow a =-1$
View full question & answer→MCQ 762 Marks
The value of $p$ for which the function
$f(x)=\left\{\begin{array}{ll}\frac{\left(4^{\prime}-1\right)^3}{\sin \left(\frac{x}{p}\right) \log \left(1+\frac{x^2}{3}\right)} & , x \neq 0 \$12)(\log 4)^3 & , x=0\end{array}\right. $
may be continuous at $x=0$, is
Answer(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 12(\log 4)^3=\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)^3}{\sin \frac{x}{p} \log \left(1+\frac{x^2}{3}\right)}$
$\Rightarrow 12(\log 4)^3$ $=\lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right)^3 \times \frac{\left(\frac{x}{ p }\right)}{\left(\sin \frac{x}{ p }\right)} \times \frac{ p }{\frac{\log \left(1+\frac{1}{3} x^2\right)}{\frac{x^2}{3} \times 3}}$
$\Rightarrow 12(\log 4)^3=(\log 4)^3(1)\left(\frac{3 p }{1}\right)$
$\Rightarrow p =4$
View full question & answer→MCQ 772 Marks
The value of $f(0)$ so that the function $f (x)=\frac{\log \left(\sec ^2 x\right)}{x \sin x}, x \neq 0$, is continuous at $x=0$ is
Answer(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{\log \left(\sec ^2 x\right)}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^2 x\right)}{\tan ^2 x} \times \frac{\tan ^2 x}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^2 x\right)}{\tan ^2 x} \times \frac{\frac{\tan ^2 x}{x^2}}{\frac{\sin x}{x}}$
$=1 \times \frac{1^2}{1}=1$
View full question & answer→MCQ 782 Marks
If $f (x)=\frac{\log _{ e }\left(1+x^2 \tan x\right)}{\sin x^3}, x \neq 0$ is continuous at $x=0$, then $f (0)$ must be defined as
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{\log _e\left(1+x^2 \tan x\right)}{\sin x^3}$
$=\lim _{x \rightarrow 0}\left(\frac{\log _{ e }\left(1+x^2 \tan x\right)}{x^2 \tan x} \cdot \frac{x^2 \tan x}{\sin x^3}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{\log _{ e }\left(1+x^2 \tan x\right)}{x^2 \tan x} \cdot \frac{x^3}{\sin x^3} \cdot \frac{\tan x}{x}\right)$
$\therefore \quad f(0)=1$
View full question & answer→MCQ 792 Marks
$\begin{aligned}f(x) & =\frac{\left(3^{\sin x}-1\right)^2}{x \log (1+x)}, x \neq 0 \\& =k \quad, x=0\end{aligned}$
If f is continuous at $x=0$, then $k =$
- A
$\frac{1}{2} \log 3$
- B
$\log 3$
- C
$2 \log 3$
- ✓
$(\log 3)^2$
AnswerCorrect option: D. $(\log 3)^2$
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\left(3^{\sin x}-1\right)^2}{x \log (1+x)}$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\left(\frac{3^{\sin x}-1}{\sin x}\right)^2 \cdot\left(\frac{\sin x}{x}\right)^2}{\frac{\log (1+x)}{x}}$
$\Rightarrow k =\frac{(\log 3)^2 \times(1)^2}{1}=(\log 3)^2$
View full question & answer→MCQ 802 Marks
$\text { If } f(x)=\left\{\begin{array}{cl}\frac{\log (1+2 a x)-\log (1-b x)}{x}, & x \neq 0 \\k, & x=0 \end{array}\right. \text { is }$
continuous at $x=0$, then k is equal to
- ✓
$2 a+b$
- B
$2 a-b$
- C
$b -2 a$
- D
$a+b$
AnswerCorrect option: A. $2 a+b$
(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\log (1+2 a x)-\log (1- b x)}{x}$
$\Rightarrow k =\lim _{x \rightarrow 0}\left[\frac{\log (1+2 a x)}{2 a x} \times 2 a +\frac{\log (1- b x)}{- b x} \times b \right]$
$\Rightarrow k =2 a + b$
View full question & answer→MCQ 812 Marks
$\begin{array}{rlrl}\text {If } f(x) & =\frac{\log x-\log 7}{x-7}, & x \neq 7 \\& =k ,& x=7\end{array}$
is continuous at $x=7$, then the value of $k$ is
- A
$\frac{1}{3}$
- B
$\frac{1}{5}$
- ✓
$\frac{1}{7}$
- D
$\frac{1}{9}$
AnswerCorrect option: C. $\frac{1}{7}$
(C)
Since $f (x)$ is continuous at $x=7$.
$\therefore \quad f(7)=\lim _{x \rightarrow 7} f(x)$
$\Rightarrow k =\lim _{x \rightarrow 7} \frac{\log x-\log 7}{x-7}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow 7} \frac{\frac{1}{x}}{1}=\frac{1}{7}$
View full question & answer→MCQ 822 Marks
$\begin{array}{rlrl}f(x) & =\frac{\log (1+k x)}{\sin x}, & x \neq 0 \\& =5 , & x=0\end{array}$
If f is continuous at $x=0$, then $k =$
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 5=\lim _{x \rightarrow 0} \frac{\log (1+ k x)}{\sin x}$
$\Rightarrow 5=\lim _{x \rightarrow 0} \frac{\frac{\log (1+ k x)}{ k x} \times k }{\frac{\sin x}{2}}$
$\Rightarrow 5=\frac{1 \times k }{1} \Rightarrow k =5$
View full question & answer→MCQ 832 Marks
Function $f(x)=\left\{\begin{array}{ll}\left(\log _2 2 x\right)^{\log _x 8} ; & x \neq 1 \\ (k-1)^3 ; & x=1\end{array}\right.$ is continuous at $x=1$, then $k =$ __________
- ✓
$e+1$
- B
$e^{1 / 3}$
- C
$e ^3$
- D
$e-1$
Answer(A)
$\lim _{x \rightarrow 1}\left(\log _2 2 x\right)^{\log _x 8}$
$=\lim _{x \rightarrow 1}\left[\log _2 2+\log _2 x\right]^{\log _x 2^3}$
$=\lim _{x \rightarrow 1}\left[1+\log _2 x\right]^{3 \log _x 2}=\lim _{x \rightarrow 1}\left[1+\log _2 x\right]^{\frac{3}{\log _2 x}}$
$= e ^{\lim _{x \rightarrow 1} \log _2 x \times \frac{3}{\log _2 x}}= e ^3$
Since the function is continuous at $x=1$,
$\therefore \quad \lim _{x \rightarrow 1} f(x)=f(1)$
$\Rightarrow e ^3=( k -1)^3$
$\Rightarrow e = k -1$
$\Rightarrow k = e + l$
View full question & answer→MCQ 842 Marks
If $f (x)$ is continuous at $x=\frac{\pi}{2}$, where $f(x)=(\sin x)^{\frac{1}{\pi-2x}}$, for $x \neq \frac{\pi}{2}$, then $f\left(\frac{\pi}{2}\right)=$
Answer(C)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$,
$\therefore f \left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f (x)$
$=\lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\frac{1}{\pi-2 x}}$
$=\lim _{x \rightarrow \frac{\pi}{2}}[1+(\sin x-1)]^{\frac{1}{\pi-2 x}}$
$= e ^{\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\sin x-1}{\pi-2 x}\right)}$
$= e ^{\left(-\frac{1}{2} \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\cos \left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)}\right)}$
$= e ^0$
$=1$
View full question & answer→MCQ 852 Marks
$\begin{aligned}\text { If }f(x) & =\left(\sec ^2 x\right)^{\cot ^2 x}, & x \neq 0 \\& =k , &x=0 \end{aligned}$
is continuous at $x=0$, then k is equal to
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim_{x \rightarrow 0} \left(\sec ^2 x\right)^{\cot ^2 x}$
$\Rightarrow k =\lim _{x \rightarrow 0}\left(1+\tan ^2 x\right)^{\frac{1}{\tan ^2 x}}$
$\Rightarrow k = e$
View full question & answer→MCQ 862 Marks
If $f (x)=\left\{\begin{array}{ll}\frac{\sin 3 x}{ e ^{2 x}-1} & ; x \neq 0 \\ k -2 & ; x=0\end{array}\right.$ is continuous at $x=0$, then $k =$
- A
$\frac{9}{5}$
- B
$\frac{2}{3}$
- ✓
$\frac{7}{2}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $\frac{7}{2}$
(C)
f is continuous at $x=0$
$\Rightarrow \lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x}{ e ^{2 x}-1}= k -2$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{x}}{\frac{ e ^{2 x}-1}{x}}= k -2$
$\Rightarrow \frac{3 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}}{2 \lim _{x \rightarrow 0} \frac{ e ^{2 x}-1}{2 x}}= k -2$
$\Rightarrow \frac{3}{2}= k -2$
$\Rightarrow k=\frac{7}{2}$
View full question & answer→MCQ 872 Marks
$\begin{aligned}\text { If } f(x) & =\left(\frac{4 x+1}{1-4 x}\right)^{\frac{1}{x}}, x \neq 0 \\& =k, \quad x=0\end{aligned}$
is continuous at $x=0$, then $k =$
- A
$e ^2$
- B
$e ^4$
- C
$e^6$
- ✓
$e^8$
Answer(D)
Since $f (x)$ is continuous at $x=0$.)
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0}\left(\frac{4 x+1}{1-4 x}\right)^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0} \frac{\left[(1+4 x)^{\frac{1}{4 x}}\right]^4}{\left[(1-4 x)^{-\frac{1}{4 x}}\right]^{-4}}$
$=\frac{e^4}{e^{-4}}=e^8$
View full question & answer→MCQ 882 Marks
In order that the function $f (x)=(x+1)^{\cot x}$ is continuous at $x=0, f (0)$ must be defined as
- A
$f(0)=\frac{1}{e}$
- B
$f(0)=0$
- ✓
$f(0)=e$
- D
$f(0)=e^2$
AnswerCorrect option: C. $f(0)=e$
(C)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0}(x+1)^{\cot x}$
$=\lim _{x \rightarrow 0}\left\{(1+x)^{\frac{1}{x}}\right\}^{x \cot x}$
$=\lim _{x \rightarrow 0}\left\{(1+x)^{\frac{1}{x}}\right\}^{\lim _{x \rightarrow 0}\left(\frac{x}{\tan x}\right)}$
$= e ^1= e$
View full question & answer→MCQ 892 Marks
The function $f : R -\{0\} \rightarrow R$ given by $f (x)=\frac{1}{x}-\frac{2}{ e ^{2 x}-1}$ can be made continuous at $x=0$ by defining $f(0)$ as
Answer(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{2}{ e ^{2 x}-1}\right)$
$=\lim _{x \rightarrow 0} \frac{ e ^{2 x}-1-2 x}{x\left( e ^{2 x}-1\right)}$
Applying L'Hospital rule on R.H.S., we get
$f(0)=\lim _{x \rightarrow 0} \frac{2 e^{2 x}-2}{x\left(2 e^{2 x}\right)+1\left(e^{2 x}-1\right)}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{4 e ^{2 x}}{2 x\left(2 e ^{2 x}\right)+ e ^{2 x}(2)+2 e ^{2 x}}$
$\Rightarrow f (0)=\frac{4}{2+2}=1$
View full question & answer→MCQ 902 Marks
If $f (x)=\frac{ e ^x- e ^{ e }}{2(x-\sin x)}, x \neq 0$ is continuous at $x=0$, then $f (0)=$
AnswerCorrect option: C. $\frac{1}{2}$
(C)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{ e ^x e ^{\sin x}}{2(x-\sin x)}$
$=\frac{1}{2} \lim _{x \rightarrow 0} e ^{\sin x}\left(\frac{ e ^{x-\sin x}-1}{x-\sin x}\right)$
$=\frac{1}{2} \times e ^0 \times 1 \quad \ldots\left[\because \lim _{x \rightarrow 0} \frac{ e ^x-1}{x}=1\right]$
$=\frac{1}{2}$
View full question & answer→MCQ 912 Marks
$\begin{aligned}\text { If } f(x) & =\frac{e^{5 x}-e^{2 x}}{\sin 3 x}, x \neq 0 \\& =\frac{k}{2}, \quad x=0 \end{aligned}$
is continuous at $x=0$, the value of k is
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow \frac{ k }{2}=\lim _{x \rightarrow 0} \frac{ e ^{5 x}- e ^{2 x}}{\sin 3 x}$
Applying L'Hospital rule on R.H.S., we get
$\frac{ k }{2}=\lim _{x \rightarrow 0} \frac{5 e ^{5 x}-2 e ^{2 x}}{3 \cos 3 x}$
$\Rightarrow \frac{ k }{2}=\frac{5 e ^0-2 e ^0}{3 \cos 0}=\frac{5-2}{3}=1$
$\Rightarrow k =2$
View full question & answer→MCQ 922 Marks
If f is continuous at $x=0$, where $f (x)=\frac{\left( e ^{3 x}-1\right) \sin x^{\circ}}{x^2}, x \neq 0$, then $f (0)$ is
- A
$\frac{\pi}{30}$
- B
$\frac{\pi}{45}$
- ✓
$\frac{\pi}{60}$
- D
$\frac{\pi}{90}$
AnswerCorrect option: C. $\frac{\pi}{60}$
(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{\left( e ^{3 x}-1\right) \sin x^{\circ}}{x^2}$
$=\lim _{x \rightarrow 0} \frac{ e ^{3 x}-1}{3 x} \times 3 \times \frac{\sin \frac{\pi x}{180}}{\frac{\pi x}{180}} \times \frac{\pi}{180}$
$=1 \times 3 \times 1 \times \frac{\pi}{180}=\frac{\pi}{60}$
View full question & answer→MCQ 932 Marks
If $f (x)$ is continuous at $x=0$, where $f(x)=\frac{8^x-2^x}{k^x-1}$, for $x \neq 0 =2$, for $x=0$, then k is equal to
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 2=\lim _{x \rightarrow 0} \frac{8^x-2^x}{k^x-1}$
$\Rightarrow 2=\lim _{x \rightarrow 0} \frac{2^x\left(\frac{4^x-1}{x}\right)}{\frac{k^x-1}{x}}$
$\Rightarrow 2=\frac{2^0 \log 4}{\log k }$
$\Rightarrow 2 \log k =\log 4$
$\Rightarrow 2 \log k =2 \log 2$
$\Rightarrow k =2$
View full question & answer→MCQ 942 Marks
If $f (x)=\frac{3^x+3^{-x}-2}{x^2}, x \neq 0$, is continuous at $x=0$, then $f(0)$ equals
AnswerCorrect option: B. $(\log 3)^2$
(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{3^x+3^{-x}-2}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(3^x-1\right)^2}{x^2}}{3^x}=\frac{(\log 3)^2}{1}=(\log 3)^2$
View full question & answer→MCQ 952 Marks
The value of f at $x=0$ so that the function $f (x)=\frac{2^x-2^{-x}}{x}, x \neq 0$ is continuous at $x=0$, is
- A
$\log 2$
- B
- C
$e^4$
- ✓
$\log 4$
AnswerCorrect option: D. $\log 4$
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{2^x-2^{-x}}{x}\right)$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0}\left[\frac{\left(2^x+2^{-x}\right) \log _e 2}{1}\right]$
$=\left(2^0+2^0\right) \log _{ e } 2$
$\therefore \quad f(0)=2 \log _e 2=\log _e 4$
View full question & answer→MCQ 962 Marks
The function defined by
$f (x)=\left\{\begin{array}{cc}\left(x^2+ e ^{\frac{1}{2-x}}\right)^{-1} & , x \neq 2 \\ k & , x=2\end{array}\right.$, is continuous from right at the point $x=2$, then k is equal to
- A
$0$
- ✓
$\frac{1}{4}$
- C
$-\frac{1}{4}$
- D
AnswerCorrect option: B. $\frac{1}{4}$
(B)
If $f (x)$ is continuous from right at $x=2$, then
$f (2)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow k =\lim _{ h \rightarrow 0} f (2+ h )$
$\Rightarrow k =\lim _{ h \rightarrow 0}\left[(2+ h )^2+ e ^{\frac{1}{2-(2+ h )}}\right]^{-1}$
$\Rightarrow k =\lim _{ h \rightarrow 0}\left[4+ h ^2+4 h+ e ^{\frac{-1}{h}}\right]^{-1}$
$\Rightarrow k =\left(4+0+0+ e ^{-\infty}\right)^{-1}$
$\Rightarrow k =\frac{1}{4}$
View full question & answer→MCQ 972 Marks
If the function $f (x)=\frac{\cos ^2 x-\sin ^2 x-1}{\sqrt{x^2+1}-1}, x \neq 0$, is continuous at $x=0$, then $f (0)$ is equal to
Answer(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{\cos ^2 x-\sin ^2 x-1}{\sqrt{x^2+1}-1}$
$=\lim _{x \rightarrow 0} \frac{\left(\cos ^2 x-1\right)-\sin ^2 x}{\sqrt{x^2+1}-1} \times \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$
$=\lim _{x \rightarrow 0} \frac{\left(-\sin ^2 x-\sin ^2 x\right)\left(\sqrt{x^2+1}+1\right)}{x^2+1-1}$
$=\lim _{x \rightarrow 0} \frac{-2 \sin ^2 x}{x^2} \times\left(\sqrt{x^2+1}+1\right)$
$=-2(1)^2\left(\sqrt{0^2+1}+1\right)=-4$
View full question & answer→MCQ 982 Marks
If $f (x)=\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}, x \neq 0$, is continuous at $x=0$, then $f (0)$ is
Answer(A)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}$
Applying L'Hospital rule on R.H.S., we get
$f(0)=\lim _{x \rightarrow 0} \frac{\frac{\cos x}{2 \sqrt{1+\sin x}}+\frac{\cos x}{2 \sqrt{1-\sin x}}}{1}$
$=\frac{1}{2}(1+1)=1$
View full question & answer→MCQ 992 Marks
$f (x)=\frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos ^2 x}, x \neq \frac{\pi}{2}$. The value of $f \left(\frac{\pi}{2}\right)$ so that f is continuous at $x=\frac{\pi}{2}$ is
- A
$\frac{1}{\sqrt{2}}$
- B
$\frac{1}{2 \sqrt{2}}$
- C
$\frac{1}{3 \sqrt{2}}$
- ✓
$\frac{1}{4 \sqrt{2}}$
AnswerCorrect option: D. $\frac{1}{4 \sqrt{2}}$
(D)
For $f (x)$ to be continuous at $x=\frac{\pi}{2}$,
$f \left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f (x)$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos ^2 x}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2-(1+\sin x)}{\left(1-\sin ^2 x\right)(\sqrt{2}+\sqrt{1+\sin x})}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{(1-\sin x)(1+\sin x)(\sqrt{2}+\sqrt{1+\sin x})}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1}{(1+\sin x)(\sqrt{2}+\sqrt{1+\sin x})}$
$=\frac{1}{(1+1)(\sqrt{2}+\sqrt{1+1})}=\frac{1}{4 \sqrt{2}}$
View full question & answer→MCQ 1002 Marks
If the function $f(x)=\left\{\begin{aligned}(\cos x)^{\frac{1}{x}} & ; x \neq 0 \\ k & ; x=0\end{aligned}\right.$ is continuous at $x=0$, then the value of k is
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow k =\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}$
$\Rightarrow \log k =\lim _{x \rightarrow 0} \frac{1}{x} \log (\cos x)$
Applying L'Hospital rule on R.H.S., we get
$\log k =\lim _{x \rightarrow 0} \frac{-\frac{\sin x}{\cos x}}{1}$
$\Rightarrow \log k =0$
$\Rightarrow k = e ^0=1$
View full question & answer→MCQ 1012 Marks
The value of $f(0)$ so that the function $f(x)=\frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}, x \neq 0$, is continuous at $x=0$, is
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{\frac{1}{3}}-3}{9-3(243+5 x)^{\frac{1}{5}}}$
Applying L'Hospital rule on R.H.S., we get
$f(0)=\lim _{x \rightarrow 0} \frac{\frac{1}{3}(27-2 x)^{\frac{-2}{3}}(-2)}{-\frac{3}{5}(243+5 x)^{\frac{-4}{5}}(5)}=2$
View full question & answer→MCQ 1022 Marks
If $f (x)=\frac{2-\sqrt{x+4}}{\sin 2 x} ;(x \neq 0)$, is continuous function at $x=0$, then $f(0)$ equals
- A
$\frac{1}{4}$
- B
$-\frac{1}{4}$
- C
$\frac{1}{8}$
- ✓
$-\frac{1}{8}$
AnswerCorrect option: D. $-\frac{1}{8}$
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{\left(-\frac{1}{2 \sqrt{x+1}}\right)}{2 \cos 2 x}=-\frac{1}{8}$
View full question & answer→MCQ 1032 Marks
If $f (x)=\frac{( a +x)^2 \sin ( a +x)- a ^2 \sin a }{x}, x \neq 0$, then the value of $f (0)$ so that f is continuous at $x=0$ is
AnswerCorrect option: B. $a^2 \cos a+2 a \sin a$
(B)
For $f (x)$ to be continuous at $x=0$,
$f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{( a +x)^2 \sin ( a +x)- a ^2 \sin a }{x}$
Applying L'Hospital rule on R.H.S., we get
$f(0)=\lim _{x \rightarrow 0} \frac{2(a+x) \sin (a+x)+(a+x)^2 \cos (a+x)}{1}$
$\Rightarrow f(0)=2 a \sin a+a^2 \cos a$
View full question & answer→MCQ 1042 Marks
If $f (x)=\frac{1-\sin x}{(\pi-2 x)^2}$, when $x \neq \frac{\pi}{2}$ and $f \left(\frac{\pi}{2}\right)=\lambda$, then $f (x)$ will be continuous function at $x=\frac{\pi}{2}$, when $\lambda=$
- ✓
$\frac{1}{8}$
- B
$\frac{1}{4}$
- C
$\frac{1}{2}$
- D
AnswerCorrect option: A. $\frac{1}{8}$
(A)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$.
$\therefore \quad f\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)$
$\Rightarrow \lambda=\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{(\pi-2 x)^2}$
Applying L'Hospital rule on R.H.S., we get
$\lambda=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-4(\pi-2 x)}$
Applying L'Hospital rule on R.H.S., we get
$\lambda=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x}{-4(-2)}$
$\Rightarrow \lambda=\frac{1}{8}$
View full question & answer→MCQ 1052 Marks
If $f(x)=\left\{\begin{array}{ll}\frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}, & x \neq \frac{\pi}{6} \\ a, & x=\frac{\pi}{6}\end{array}\right.$ is continuous at $x=\frac{\pi}{6}$, then $a=$
- A
$\sqrt{3}$
- ✓
$\frac{1}{\sqrt{3}}$
- C
$-\sqrt{3}$
- D
$-\frac{1}{\sqrt{3}}$
AnswerCorrect option: B. $\frac{1}{\sqrt{3}}$
(B)
Since $f(x)$ is continuous at $x=\frac{\pi}{6}$,
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{6}} f(x)=f\left(\frac{\pi}{6}\right)$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}= a$
Applying L'Hospital rule to L.H.S, we get
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \cos x+\sqrt{3} \sin x}{6}=a$
$\Rightarrow \frac{3\left(\frac{\sqrt{3}}{2}\right)+\sqrt{3}\left(\frac{1}{2}\right)}{6}= a$
$\Rightarrow \frac{4 \sqrt{3}}{12}=a \Rightarrow a=\frac{1}{\sqrt{3}}$
View full question & answer→MCQ 1062 Marks
If $f(x)=\frac{1-\tan x}{1-\sqrt{2} \sin x}, x \neq \frac{\pi}{4}$, is continuous $x=\frac{\pi}{4}$ and $f\left(\frac{\pi}{4}\right)=k$, then $k=$
- ✓
- B
$2 \sqrt{2}$
- C
- D
$4 \sqrt{2}$
Answer(A)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$.
$\therefore \quad f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{1-\sqrt{2} \sin x}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^2 x}{-\sqrt{2} \cos x}=\frac{-2}{-1}=2$
View full question & answer→MCQ 1072 Marks
Let $f(x)=\left\{\begin{array}{cl}\frac{\tan x-\cot x}{x-\frac{\pi}{4}}, & x \neq \frac{\pi}{4} \\ \text { a, } & x=\frac{\pi}{4}\end{array}\right.$
The value of a so that $f(x)$ is continuous , $x=\frac{\pi}{4}$, is
Answer(B)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$.
$\therefore f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x)$
$\Rightarrow a =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan x-\cot x}{x-\frac{\pi}{4}}$
Applying L'Hospital rule on R.H.S., we get
$a =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sec ^2 x+\operatorname{cosec}^2 x}{1}$
$\Rightarrow a =(\sqrt{2})^2+(\sqrt{2})^2=4$
View full question & answer→MCQ 1082 Marks
Let $f: R \rightarrow R$ be defined by
$f(x)=\left\{\begin{array}{cc}\frac{\cos 3 x-\cos x}{x^2}, & x \neq 0 \\\lambda, & x=0\end{array} .\right.$
If f is continuous at $x=0$, then $\lambda$ is equal to
Answer(B)
Since $f (x)$ is continuous $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow \lambda=\lim _{x \rightarrow 0} \frac{\cos 3 x-\cos x}{x^2}$
Applying L'Hospital rule on R.H.S., we get
$\lambda=\lim _{x \rightarrow 0} \frac{-3 \sin 3 x+\sin x}{2 x}$
Applying L'Hospital rule on R.H.S., we get
$\lambda=\lim _{x \rightarrow 0} \frac{-9 \cos 3 x+\cos x}{2} \Rightarrow \lambda=\frac{-9+1}{2}=-4$
View full question & answer→MCQ 1092 Marks
If a function of defined by
$f(x)=\left\{\begin{array}{cc}\frac{1-\sqrt{2} \sin x}{\pi-4 x}, & \text { if } x \neq \frac{\pi}{4} \\k, & \text { if } x=\frac{\pi}{4} \end{array}\right.$
is continuous at $x=\frac{\pi}{4}$, then $k =$
- ✓
$\frac{1}{4}$
- B
- C
$-\frac{1}{4}$
- D
AnswerCorrect option: A. $\frac{1}{4}$
(A)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$,
$\therefore \quad f\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f(x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4} \quad \Rightarrow k =\frac{1}{4}$
View full question & answer→MCQ 1102 Marks
If the function $f(x)=\left\{\begin{array}{cc}\frac{k \cos x}{\pi-2 x}, & \text { when } x \neq \frac{\pi}{2} \\ 3, & \text { when } x=\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then $k =$
Answer(B)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$.
$\therefore \quad f \left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f (x)$
$\Rightarrow 3=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{ k \cos x}{\pi-2 x}\right)$
Applying L'Hospital rule on R.H.S., we get
$3=\lim _{x \rightarrow \frac{\pi}{2}} \frac{ k (-\sin x)}{-2}$
$\Rightarrow 3-\frac{ k }{2} \Rightarrow k -6$
View full question & answer→MCQ 1112 Marks
$\begin{array}{lc}\text {Let } f (x)=\left\{\begin{array}{rc}(1+|\sin x|)^{\frac{1}{|\sin x|}}, & -\frac{\pi}{6} < x<0 \\ b, & x=0 \\ e ^{\frac{\tan 2 x}{\tan 3 x}}, & 0 < x<\frac{\pi}{6}\end{array}\right. \end{array}$
Then the values of $a$ and $b$ if $f$ is continuous at $x=0$, are respectively
- A
$\frac{2}{3}, \frac{3}{2}$
- ✓
$\frac{2}{3}, e ^{\frac{2}{3}}$
- C
$\frac{3}{2}, e ^{\frac{3}{2}}$
- D
AnswerCorrect option: B. $\frac{2}{3}, e ^{\frac{2}{3}}$
(B)
For $f (x)$ to be continuous at $x=0$, we must have
$\lim _{x \rightarrow 0^{-}} f (x)= f (0)=\lim _{x \rightarrow 0^{+}} f (x)$
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0^{+}} e ^{\tan 2 x / \tan 3 x}$
$=\lim _{x \rightarrow 0^{+}} e ^{\left(\frac{\tan 2 x}{2 x} \times 2 x\right) /\left(\frac{\tan 3 x}{3 x} \times 3 x\right)}$
$= e ^{\frac{2}{3}}$
$f (0)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow b = e ^{\frac{2}{3}}$
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{ a /|\sin x|}$
$= e ^{\lim _{x \rightarrow 0}\left(|\sin x| \times \frac{ a }{|\sin x|}\right)}= e ^{ a }$
$f (0)=\lim _{x \rightarrow 0^{-}} f (x)$
$\Rightarrow b = e ^{ a } \Rightarrow e ^{\frac{2}{3}}= e ^{ a }$
$\Rightarrow a =\frac{2}{3}$
View full question & answer→MCQ 1122 Marks
If $f(x)=\left\{\begin{array}{cl}\frac{3 \sin \pi x}{5 x}, & x \neq 0 \\ 2 k, & x=0\end{array}\right.$ is continuous at $x=0$, then the value of k is equal to
- ✓
$\frac{3 \pi}{10}$
- B
$\frac{3 \pi}{5}$
- C
$\frac{\pi}{10}$
- D
$\frac{3 \pi}{2}$
AnswerCorrect option: A. $\frac{3 \pi}{10}$
(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow 2 k =\lim _{x \rightarrow 0} \frac{3 \sin \pi x}{5 x}$
$=\lim _{x \rightarrow 0} \frac{3 \sin \pi x}{5(\pi x)} \times \pi=\frac{3 \pi}{5}$
$\therefore \quad k =\frac{3 \pi}{10}$
View full question & answer→MCQ 1132 Marks
$\begin{array}{rlr}\text {If } f(x) & =\frac{\cos x-\sin x}{\cos 2 x}, x \neq \frac{\pi}{4} \\& =k, \quad x=\frac{\pi}{4}\end{array}$
is continuous at $x=\frac{\pi}{4}$, then the value of $k$ is
- A
$\sqrt{2}$
- ✓
$\frac{1}{\sqrt{2}}$
- C
$2 \sqrt{2}$
- D
$\frac{1}{2 \sqrt{2}}$
AnswerCorrect option: B. $\frac{1}{\sqrt{2}}$
(B)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$.
$\therefore \quad f \left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f (x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{\cos 2 x}$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{\cos ^2 x-\sin ^2 x}$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{(\cos x-\sin x)(\cos x+\sin x)}$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1}{\cos x+\sin x}=\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 1142 Marks
If $f(x)=\left\{\begin{array}{c}\frac{\sin [x]}{[x]+1}, \text { for } x>0 \\ \frac{\cos \frac{\pi}{2}[x]}{[x]}, \text { for } x<0 ; \\ k, \text { for } x=0\end{array}\right.$
where $[x]$ denotes the greatest integer less than or equal to $x$, then in order that f be continuous at $x=0$, the value of k is
AnswerCorrect option: A. Equal to $0$
(A)
For $f (x)$ to be continuous at $x=0$,
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0^{-}} f (x)$
$\Rightarrow k =\lim _{ h \rightarrow 0} f (0- h )=\lim _{ h \rightarrow 0} \frac{\cos \frac{\pi}{2}[0- h ]}{[0- h ]}$
$\Rightarrow k =\lim _{ h \rightarrow 0} \frac{\cos _2^\pi[ h ]}{[- h ]}$
$\Rightarrow k =\lim _{ h \rightarrow 0} \frac{\cos \left(-\frac{\pi}{2}\right)}{-1}$
$\Rightarrow k =0$
View full question & answer→MCQ 1152 Marks
If $f (x)=\left\{\begin{array}{c} ax ^2- b , \text { when } 0 \leq x<1 \\ 2, \text { when } x=1 \\ x+1, \text { when } l < x \leq 2\end{array}\right.$ is continuous at
$x=1$, then the most suitable values of $a , b$ are
- A
$a=2, b=0$
- B
$a=1, b=-1$
- C
$a=4, b=2$
- ✓
Answer(D)
Since $f (x)$ is continuous at $x=1$.
$\therefore \quad f (1)=\lim _{x \rightarrow 1^{-}} f (x)$
$\Rightarrow 2=\lim _{x \rightarrow 1}\left(a x^2-b\right)$
$\Rightarrow 2=a-b$
The values of $a$ and $b$ in options (A), (B) and (C) satisfies this relation.
$\therefore $ option (D) is the correct answer.
View full question & answer→MCQ 1162 Marks
The function $f (x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$ is not defined at $x=\pi$. The value of $f (\pi)$, so that $f (x)$ is continuous at $x=\pi$, is
- A
$-\frac{1}{2}$
- B
$\frac{1}{2}$
- ✓
- D
Answer(C) For $f (x)$ to be continuous at $x=\pi$,
$f(\pi)=\lim _{x \rightarrow \pi}
f(x)=\lim _{x \rightarrow \pi} \frac{1-\sin x+\cos x}{1+\sin x+\cos x}$
Applying L'Hospital rule on R.H.S., we get
$f(\pi)=\lim _{x \rightarrow \pi} \frac{-\cos x-\sin x}{\cos x-\sin x}$
$\Rightarrow f (\pi)=-1$
View full question & answer→MCQ 1172 Marks
The value of k which makes $f(x)=\left\{\begin{array}{c}\sin \frac{1}{x}, x \neq 0 \\ k , x=0\end{array}\right.$ continuous at $x=0$ is
Answer(D) If $x \rightarrow 0$, then the value of $\sin \frac{1}{x}$ passes through $[-1,1]$ infinitely many ways, therefore, limit of the function docs not exist at $x=0$. Hence, there is no value of k for which the function is continuous at $x=0$.
View full question & answer→MCQ 1182 Marks
$\begin{aligned}f(x) & =\frac{1-\cos 3 x}{x \tan x}, \text { for } x \neq 0 \\& =k, \text { for } x=0\end{aligned}$
If $f (x)$ is continuous at $x=0$, the value of k is
- A
$\frac{3}{2}$
- B
$\frac{5}{2}$
- C
$\frac{7}{2}$
- ✓
$\frac{9}{2}$
AnswerCorrect option: D. $\frac{9}{2}$
(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{1-\cos 3 x}{x \tan x}$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{1-\cos 3 x}{x^2} \times \frac{1}{\frac{\tan x}{x}}$
$\Rightarrow k =\frac{3^2}{2} \times l \quad \ldots .\left[\because \lim _{x \rightarrow 0}\left(\frac{1-\cos k x}{x^2}\right)=\frac{ k ^2}{2}\right]$
$\Rightarrow k =\frac{9}{2}$
View full question & answer→MCQ 1192 Marks
If $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & ; \text { when } x<0 \\ a & ; \text { when } x=0 \\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4} & ; \text { when } x>0\end{array}\right.$, is continuous at $x=0$, then the value of ' $a$ ' will be
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0^{-}} f (x)$
$\Rightarrow a =\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{x^2}$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^2 2 x}{x^2}$
$=2 \lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{(2 x)^2} \times 4=2 \times 4=8$
View full question & answer→MCQ 1202 Marks
Function $f (x)=\frac{1-\cos 4 x}{8 x^2}$, where $x \neq 0$ and $f (x)= k$, where $x=0$ is a continuous function at $x=0$, then the value of k will be
Answer(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{8 x^2}=\lim _{x \rightarrow 0} \frac{2 \sin ^2 2 x}{8 x^2}$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{4 x^2}=1$
View full question & answer→MCQ 1212 Marks
If $f (x)=\frac{\tan \left(x^2-x\right)}{x}, x \neq 0$, is continuous at $x=0$, then $f(0)$ is
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{\tan \left(x^2-x\right)}{x}$
$=\lim _{x \rightarrow 0} \frac{\tan [x(x-1)]}{x(x-1)} \times(x-1)$
$=1 \times(-1)=-1$
View full question & answer→MCQ 1222 Marks
$\begin{aligned}\text { If } f(x) & =\frac{x^4-64 x}{\sqrt{x^2+9}-5}, x \neq 4 \\& =k \quad, x=4\end{aligned}$
is continuous at $x=4$, then $k =$
Answer(D)
Since $f (x)$ is continuous at $x=4$.
$\therefore \quad f (4)=\lim _{x \rightarrow 4} f (x)$
$=\lim _{x \rightarrow 4} \frac{x^4-64 x}{\sqrt{x^2+9}-5}$
$=\lim _{x \rightarrow 4} \frac{x\left(x^3-64\right)\left(\sqrt{x^2+9}+5\right)}{\left(x^2+9\right)-25}$
$=\left(\lim _{x \rightarrow 4} \frac{x^3-4^3}{x^2-4^2}\right)\left[\lim _{x \rightarrow 4} x\left(\sqrt{x^2+9}+5\right)\right]$
$=\frac{3}{2}(4)[4(\sqrt{16+9}+5)]$
$=240$
View full question & answer→MCQ 1232 Marks
If $f(x)=\left\{\begin{array}{lll}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & \text { if } & -1 \leq x<0 \\ \frac{2 x+1}{x-1} & \text { if } & 0 \leq x \leq 1\end{array}\right.$ is continuous at $x=0$, then the value of k is
- A
$k=1$
- ✓
$k=-1$
- C
$k =0$
- D
$k=2$
AnswerCorrect option: B. $k=-1$
(B)
Since $f (x)$ is continuous at $x=0$,
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{1+ k x}-\sqrt{1- k x}}{x}=\lim _{x \rightarrow 0} \frac{2 x+1}{x+1}$
Applying L'Hospital rule on L.H.S, we get
$\lim _{x \rightarrow 0} \frac{\frac{ k }{2 \sqrt{1+ k x}}-\frac{(- k )}{2 \sqrt{1- k x}}}{1}=-1$
$\Rightarrow \frac{ k }{2}+\frac{ k }{2}=-1 \quad \Rightarrow k =-1$
View full question & answer→MCQ 1242 Marks
If $f (x)=\frac{\sqrt{x+3}-2}{x^3-1}, x \neq 1$, is continuous at $x=1$, then $f (1)$ is
- A
- B
$\frac{1}{8}$
- ✓
$\frac{1}{12}$
- D
AnswerCorrect option: C. $\frac{1}{12}$
(C)
Since $f (x)$ is continuous at $x=1$.
$\therefore f (1)=\lim _{x \rightarrow 1} f (x)=\lim _{x \rightarrow 1} \frac{\sqrt{x+3-2}}{x^3-1}$
$=\lim _{x \rightarrow 1} \frac{\sqrt{x+3}-2}{x^3-1^3} \times \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}$
$=\lim _{x \rightarrow 1} \frac{x-1}{(x-1)\left(x^2+x+1\right)(\sqrt{x+3}+2)}$
$=\frac{1}{3(4)}=\frac{1}{12}$
View full question & answer→MCQ 1252 Marks
If $f (x)=\frac{x- a }{\sqrt{x}-\sqrt{ a }}, x \neq a$, is continuous at $x= a$, then $f(a)$ is equal to
- A
$\sqrt{ a }$
- ✓
$2 \sqrt{ a }$
- C
- D
AnswerCorrect option: B. $2 \sqrt{ a }$
(B)
Since $f (x)$ is continuous at $x= a$.
$\therefore \quad f(a)=\lim _{x \rightarrow a} f(x)$
$=\lim _{x \rightarrow a } \frac{x- a }{\sqrt{x}-\sqrt{ a }} \times \frac{\sqrt{x}+\sqrt{ a }}{\sqrt{x}+\sqrt{ a }}$
$=\lim _{x \rightarrow a}(\sqrt{x}+\sqrt{a})=\sqrt{a}+\sqrt{a}=2 \sqrt{a}$
View full question & answer→MCQ 1262 Marks
If $f (x)=\left\{\begin{array}{cc}\frac{x^6-\frac{1}{64}}{x^3-\frac{1}{8}}, & x \neq \frac{1}{2} \\ k , & x=\frac{1}{2}\end{array}\right.$ is continuous at $x=\frac{1}{2}$, then the value of k is
- A
$\frac{1}{2}$
- B
$\frac{1}{3}$
- ✓
$\frac{1}{4}$
- D
$\frac{1}{5}$
AnswerCorrect option: C. $\frac{1}{4}$
(C)
Since $f (x)$ is continuous at $x=\frac{1}{2}$.
$\therefore \quad f \left(\frac{1}{2}\right)=\lim _{x \rightarrow \frac{1}{2}} f (x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{1}{2}} \frac{x^6-\frac{1}{64}}{x^3-\frac{1}{8}}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow \frac{1}{2}} \frac{6 x^5}{3 x^2}=\lim _{x \rightarrow \frac{1}{2}} 2 x^3=2\left(\frac{1}{2}\right)^3=\frac{1}{4}$
View full question & answer→MCQ 1272 Marks
Let $f (x)=\left\{\begin{array}{lc}5^{\frac{1}{x}} ; & x<0 \\ \lambda[x] ; & x \geq 0, \lambda \in R \end{array}\right.$, then at $x=0$
AnswerCorrect option: A. f is continuous whatever $\lambda$ may be
(A)
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{-}} 5^{\frac{1}{x}}=\lim _{ h \rightarrow 0} 5^{-\frac{1}{h}}=0$
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0^{+}} \lambda[x]=0$, for all $\lambda \in R$
$f(0)=\lambda(0)=0$
$\therefore f$ is continuous at $x=0$, whatever $\lambda$ may be.
View full question & answer→MCQ 1282 Marks
Which of the following functions is continuous at $x=0$ ?
- A
$f(x)=\left\{\begin{array}{cl}\frac{\sin 2 x}{x} ; & x \neq 0 \\ 1 ; & x=0\end{array}\right.$
- B
$f(x)=\left\{\begin{array}{cl}(1+x)^{\frac{1}{x}} ; & x \neq 0 \\ 1 ; & x=0\end{array}\right.$
- C
$f (x)=\left\{\begin{array}{cc} e ^{\frac{-1}{x}} ; & x \neq 0 \\ 1 ; & x=0\end{array}\right.$
- ✓
$f(x)=\left\{\begin{array}{cc}\frac{3 x+4 \tan x}{x} ; & x \neq 0 \\ 7 ; & x=0\end{array}\right.$
AnswerCorrect option: D. $f(x)=\left\{\begin{array}{cc}\frac{3 x+4 \tan x}{x} ; & x \neq 0 \\ 7 ; & x=0\end{array}\right.$
(D)
$\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=2 \neq f (0)$
$\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e \neq f(0)$
$\lim _{x \rightarrow 0} e ^{\frac{-1}{x}}=\lim _{x \rightarrow 0} \frac{1}{ e ^{\frac{1}{x}}}=\frac{1}{ e ^{\infty}}=\frac{1}{\infty}=0 \neq f (0)$
$\lim _{x \rightarrow 0}\left(\frac{3 x}{x}+\frac{4 \tan x}{x}\right)=3+4=7= f (0)$
$\therefore f (x)$ is continuous at $x=0$.
View full question & answer→MCQ 1292 Marks
If $f(x)=\left\{\begin{array}{c}\frac{x}{e^{\frac{1}{x}}+1}, \text { when } x \neq 0 \\ 0, \text { when } x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{+}} f(x)=1$
- B
$\lim _{x \rightarrow 0^{-}} f(x)=1$
- ✓
$f (x)$ is continuous at $x=0$
- D
$f (x)$ is not continuous at $x=0$
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)$
$=\lim _{h \rightarrow 0} \frac{-h}{e^{\frac{-1}{h}}+1}=\lim _{h \rightarrow 0} \frac{-h}{1+\frac{1}{e^{\frac{1}{h}}}}=0$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} \frac{h}{e^{\frac{1}{h}}+1}=0$
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)= f (0)$
$\therefore \quad f (x)$ is continuous at $x=0$.
View full question & answer→MCQ 1302 Marks
If $f (x)=\left\{\begin{array}{cc}x^* \sin \frac{1}{x} ; & x \neq 0 \\ 0 ; & x=0\end{array}\right.$ is continuous an $x=0$, then
Answer(B)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad \lim _{x \rightarrow 0} f (x)= f (0)$
$\therefore \quad \lim _{x \rightarrow 0} x^{ a } \sin \frac{1}{x}=0$, if $a >0$
View full question & answer→MCQ 1312 Marks
If $f(x)=\left\{\begin{array}{c}x^2 \sin \frac{1}{x} ; \text { when } x \neq 0 \\ 0 ; \text { when } x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{+}} f(x)=1$
- B
$\lim _{x \rightarrow 0^{-}} f(x)=-1$
- ✓
$f (x)$ is continuous at $x=0$
- D
$f (x)$ is discontinuous at $x=0$
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f(x)-\lim _{x \rightarrow 0} x^2 \sin \frac{1}{x}$, but $-1 \leq \sin \frac{1}{x} \leq 1$ and $x \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow 0^{+}} f (x)=0=\lim _{x \rightarrow 0^{-}} f (x)= f (0)$
$\therefore f (x)$ is continuous at $x=0$.
View full question & answer→MCQ 1322 Marks
If $f (x)=\left\{\begin{array}{r}\sin ^{-1}|x| \text {; when } x \neq 0 \\ 0 \text {; when } x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0^{+}} f(x) \neq 0$
- B
$\lim _{x \rightarrow 0^{-}} f(x) \neq 0$
- ✓
$f (x)$ is continuous at $x=0$
- D
$f (x)$ is not continuous at $x=0$
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f(x)=\sin ^{-1}(0)=0=f(0)$
$\therefore \quad f (x)$ is continuous at $x=0$.
View full question & answer→MCQ 1332 Marks
If $f(x)=\left\{\begin{array}{c}1 ; \text { when } 0< x \leq \frac{3 \pi}{4} \\ 2 \sin \frac{2}{9} x ; \text { when } \frac{3 \pi}{4}< x<\pi\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=0$
- B
$f (x)$ is continuous at $x=\pi$
- ✓
$f(x)$ is continuous at $x=\frac{3 \pi}{4}$
- D
$f (x)$ is discontinuous at $x=\frac{3 \pi}{4}$
AnswerCorrect option: C. $f(x)$ is continuous at $x=\frac{3 \pi}{4}$
(C)
Here, $f \left(\frac{3 \pi}{4}\right)=1$ and $\lim _{x \rightarrow \frac{3 \pi^{-}}{4}} f (x)=1$
$\lim _{x \rightarrow \frac{3 \pi^{+}}{4}} f (x)=\lim _{ h \rightarrow 0} 2 \sin \frac{2}{9}\left(\frac{3 \pi}{4}+ h \right)$
$=2 \sin \frac{\pi}{6}=1$
$\therefore \quad f (x)$ is continuous at $x-\frac{3 \pi}{4}$.
View full question & answer→MCQ 1342 Marks
If $f (x)=|x-2|$, then
- A
$\lim _{x \rightarrow z^{+}} f(x) \neq 0$
- B
$\lim _{x \rightarrow 2^{-}} f (x) \neq 0$
- C
$\lim _{x \rightarrow 2^{+}} f(x) \neq \lim _{x \rightarrow 2^{-}} f(x)$
- ✓
$f (x)$ is continuous at $x=2$
AnswerCorrect option: D. $f (x)$ is continuous at $x=2$
(D)
Here, f(2) = 0
$\lim _{x \rightarrow 2^{-}} f (x)=\lim _{ h \rightarrow 0} f (2- h )=\lim _{ h \rightarrow 0}|2- h -2|=0$
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}|2+h-2|=0$
$\therefore \quad f (x)$ is continuous at $x=2$.
View full question & answer→MCQ 1352 Marks
If $f(x)$ is continuous on $[-4,2]$, where $f(x)=\left\{\begin{array}{ll}6 b-3 a x, & \text { for }-4 \leq x<-2 \\ 4 x+1, & \text { for }-2 \leq x \leq 2\end{array}\right.$, then $a+b=$
- A
$\frac{1}{6}$
- B
$-\frac{1}{6}$
- C
$\frac{7}{6}$
- ✓
$-\frac{7}{6}$
AnswerCorrect option: D. $-\frac{7}{6}$
(D)
Since $f(x)$ is continuous on $[-4,2]$.
$\therefore \quad$ it is continuous at $x=-2$.
$\therefore \quad \lim _{x \rightarrow-2^{-}} f (x)=\lim _{x \rightarrow-2^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow-2^{-}}(6 b-3 a x)=\lim _{x \rightarrow-2^{+}}(4 x+1)$
$\Rightarrow 6 b-3 a (-2)=4(-2)+1$
$\Rightarrow 6 b+6 a=-7$
$\Rightarrow a + b =-\frac{7}{6}$
View full question & answer→MCQ 1362 Marks
If $f(x)$ is continuous in $[-2,2]$. where$f(x)=\left\{\begin{array}{ll}x+a, & x<0 \\x, & 0 \leq x<1, \text { then } \\b-x, & x \geq 1\end{array}\right.$
Answer(A)
Since $f(x)$ is continuous in $[-2,2]$.
$\therefore $ it is continuous at $x=0$ and $x=1$.
$\therefore \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}(x+a)=\lim _{x \rightarrow 0^{+}} x$
$\Rightarrow a=0$
Also, $\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1^{-}} x=\lim _{x \rightarrow 1^{+}}( b -x)$
$\Rightarrow l = b -1 \Rightarrow b=2$
View full question & answer→MCQ 1372 Marks
If the function $f(x)=\left\{\begin{array}{cc}5 x-4, & \text { if } 0< x \leq 1 \\ 4 x^2+3 b x, & \text { if } 1< x< 2\end{array}\right.$ is continuous at every point of its domain, they value of $b$ is
Answer(A)
Since $f (x)$ is continuous at every point of its domain.
∴ it is continuous at $x=1$.
$\therefore \quad \lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 1}(5 x-4)=\lim _{x \rightarrow 1}\left(4 x^2+3 b x\right)$
$\Rightarrow 1=4+3 b \Rightarrow b=-1$
View full question & answer→MCQ 1382 Marks
If f(x) = $\frac{x+1}{(x-2)(x-5)}$, then in $[0,1], f(x)$ is
- ✓
- B
- C
continuous except at x = 0
- D
discontinuous except at x = 0
Answer(A)
$f (x)$ being a rational function, is continuous in $[0,1]$ except at those points where the denominator $(x-2)(x-5)=0$
i.e. when $x=2$ or $x=5$
Since $2,5 \notin[0,1]$
$\therefore f (x)$ is continuous in $[0,1]$.
View full question & answer→MCQ 1392 Marks
If $f (x)=\sqrt{x-2}, 2< x< 4$, then
- ✓
f(x) is continuous in (2, 4)
- B
f(x) is discontinuous in (2, 4)
- C
f(x) is continuous in (2, 4) except at x = 3
- D
f(x) is discontinuous in (2, 4) except at x = 3
AnswerCorrect option: A. f(x) is continuous in (2, 4)
(A)
$\lim _{x \rightarrow 3} f (x)=\lim _{x \rightarrow 3} \sqrt{x-2}=1$
$f(3)=\sqrt{3-2}=1$
$\therefore \quad \lim _{x \rightarrow 3} f (x)= f (3)$
$\therefore f (x)$ is continuous at $x=3$.
Since $3 \in(2,4)$
$\therefore f (x)$ is continuous in $(2,4)$.
View full question & answer→MCQ 1402 Marks
Function $f (x)=\left\{\begin{array}{cc}x-1, & x<2 \\ 2 x-3, & x \geq 2\end{array}\right.$ is continuous
Answer(A)
For $x<2, f (x)=x-1$
Since f is a polynomial function, it is continuous for all $x<2$.
For $x>2, f (x)=2 x-3$
Since f is a polynomial function, it is continuous for all $x>2$.
$\lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2}(x-1)=1$
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2}(2 x-3)=1$
$f(2)=1$
$\therefore f (x)$ is continuous for all real values of $x$.
View full question & answer→MCQ 1412 Marks
If $f (x)=\left\{\begin{array}{l}x, x \geq 0 \\ x^2, x<0\end{array}\right.$, then $f (x)$ is
- ✓
- B
- C
continuous on R except at x = 0
- D
discontinuous on R except at x = 0
Answer(A)
For $x>0, f (x)=x$
Since f is a polynomial function, it is continuous for all $x>0$.
For $x<0, f (x)=x^2$
Since f is a polynomial function, it is continuous for all $x<0$.
$\lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0} x^2=0$
$\lim _{x \rightarrow 0^{+}} f (x)=\lim _{x \rightarrow 0^{+}} x=0$
$f(0)=0$
$\therefore f (x)$ is continuous at $x=0$.
$\therefore f (x)$ is continuous on R .
View full question & answer→MCQ 1422 Marks
If f and g are both continuous at $x= a$, then $f - g$ is
View full question & answer→MCQ 1432 Marks
The points at which the function $f (x)=\frac{x+1}{x^2+x-12}$ is discontinuous are
- A
$-3, 4$
- ✓
$3, -4$
- C
- D
$-3, -4$
AnswerCorrect option: B. $3, -4$
(B)
$f (x)=\frac{x+1}{(x-3)(x+4)}$
$\therefore f (x)$ is discontinuous at $x=3,-4$.
View full question & answer→MCQ 1442 Marks
$f (x)=\frac{x^2+x-2}{x^2-3 x+2}$ is discontinuous at $x=$
Answer(B)
$f (x)$ is discontinuous, when $x^2-3 x+2=0$
i.e., $(x-1)(x-2)=0 \Rightarrow x=1, x=2$
View full question & answer→MCQ 1452 Marks
If $f (x)=\left\{\begin{array}{l}x \sin x ; 0 < x \leq \frac{\pi}{2} \\ \frac{\pi}{2} \sin (\pi+x) ; \quad \frac{\pi}{2}< x< \pi\end{array}\right.$, then
- ✓
$f (x)$ is discontinuous at $x=\frac{\pi}{2}$
- B
$f (x)$ is continuous at $x=\frac{\pi}{2}$
- C
$f (x)$ is continuous at $x=0$
- D
AnswerCorrect option: A. $f (x)$ is discontinuous at $x=\frac{\pi}{2}$
(A)
$\lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x)=\lim _{x \rightarrow \frac{\pi}{2}} x \sin x=\frac{\pi}{2}$
$\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\pi}{2} \sin (\pi+x)=\frac{-\pi}{2}$
$\therefore f (x)$ is discontinuous at $x=\frac{\pi}{2}$.
View full question & answer→MCQ 1462 Marks
If $f (x)=\left\{\begin{array}{c}\frac{x^2-4 x+3}{x^2-1}, \text { for } x \neq 1 \\ 2, \text { for } x=1\end{array}\right.$, then
- A
$\lim _{x \rightarrow 1^{+}} f(x)=2$
- B
$\lim _{x \rightarrow 1^{-}} f(x)=3$
- ✓
$f (x)$ is discontinuous at $x=1$
- D
$f (x)$ is continuous at $x=1$
AnswerCorrect option: C. $f (x)$ is discontinuous at $x=1$
(C)
$\lim _{x \rightarrow 1} f (x)=\lim _{x \rightarrow 1} \frac{x^2-4 x+3}{x^2-1}$
$=\lim _{x \rightarrow 1} \frac{(x-3)}{(x+1)}=-1$
$f(1)=2$
$\therefore \quad \lim _{x \rightarrow 1} f (x) \neq f (1)$
$\therefore f (x)$ is discontinuous at $x=1$.
View full question & answer→MCQ 1472 Marks
$\begin{aligned}\text {If} \ f (y) & =y^2-y-1, & & \text { for } 0 \leq y<2 \\ & =4 y+1, & & \text { for } 2 \leq y \leq 4, \text { then }\end{aligned}$
- A
$f (y)$ is continuous at $y=2$
- ✓
$f (y)$ is discontinuous at $y=2$
- C
$\lim _{y \rightarrow 2^{-}} f (y)=9$
- D
$\lim _{y \rightarrow 2^{+}} f (y)=1$
AnswerCorrect option: B. $f (y)$ is discontinuous at $y=2$
(B)
$\lim _{y \rightarrow 2^{-}} f (y)=\lim _{y \rightarrow 2^{-}}\left(y^2-y-1\right)$
$=4-2-1=1$
$\lim _{y \rightarrow 2^{+}} f (y)=\lim _{y \rightarrow 2^{+}}(4 y+1)$
$=8+1=9$
$\therefore \quad \lim _{y \rightarrow 2^{-}} f (y) \neq \lim _{y \rightarrow 2^{+}} f (y)$
$\therefore f (y)$ is discontinuous at $y=2$.
View full question & answer→MCQ 1482 Marks
If $f(x)=\left\{\begin{array}{l}1+x^2, \text { when } 0 \leq x \leq 1 \\ 1-x, \text { when } x>1\end{array}\right.$, then
- A
$\lim _{x \rightarrow 1^{+}} f(x) \neq 0$
- B
$\lim _{x \rightarrow 1^{-}} f(x) \neq 2$
- ✓
$f (x)$ is discontinuous at $x=1$
- D
$f (x)$ is continuous at $x=1$
AnswerCorrect option: C. $f (x)$ is discontinuous at $x=1$
(C)
$\lim _{x \rightarrow 1^{+}} f (x)=\lim _{x \rightarrow 1}(1-x)=0$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(1+x^2\right)$
$=1+1^2$
$=2$
$\therefore \quad \lim _{x \rightarrow 1^{+}} f (x) \neq \lim _{x \rightarrow 1^{-}} f (x)$
$\therefore f (x)$ is discontinuous at $x=1$.
View full question & answer→MCQ 1492 Marks
If $f(x)=\left\{\begin{array}{l}\frac{5}{2}-x, \text { when } x<2 \\ 1, \text { when } x=2 \\ x-\frac{3}{2}, \text { when } x>2\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=2$
- ✓
$f (x)$ is discontinuous at $x=2$
- C
$\lim _{x \rightarrow 2} f (x)=1$
- D
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=2$
(B)
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{5}{2}-x\right)=\frac{1}{2}$
$\lim _{x \rightarrow 2^{+}} f (x)=\lim _{x \rightarrow 2}\left(x-\frac{3}{2}\right)=\frac{1}{2}$ and $f (2)=1$
$\therefore \quad \lim _{x \rightarrow 2^{-}} f (x)=\lim _{x \rightarrow 2^{+}} f (x) \neq f (2)$
$\therefore f (x)$ is discontinuous at $x=2$.
View full question & answer→MCQ 1502 Marks
If $f (x)=\left\{\begin{array}{c}x^2 ; \text { when } x \leq 1 \\ x+5 ; \text { when } x>1\end{array}\right.$, then
- A
$f (x)$ is continuous at $x=1$
- ✓
$f (x)$ is discontinuous at $x=1$
- C
$\lim _{x \rightarrow 1^{+}} f(x)=1$
- D
$\lim _{x \rightarrow 1^{-}} f(x)=6$
AnswerCorrect option: B. $f (x)$ is discontinuous at $x=1$
(B)
$\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1} x^2=1$
$\lim _{x \rightarrow 1^{+}} f (x)=\lim _{x \rightarrow 1}(x+5)=6$
$\therefore \quad f (x)$ is discontinuous at $x=1$.
View full question & answer→MCQ 1512 Marks
If $f (x)=\left\{\begin{array}{ll} ax +1 & , x \leq \frac{\pi}{2} \\ \sin x+ b & , x>\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then
- A
$a=1, b=0$
- B
$a=b \frac{\pi}{2}+1$
- ✓
$b =\frac{ a \pi}{2}$
- D
$a = b =\frac{\pi}{2}$
AnswerCorrect option: C. $b =\frac{ a \pi}{2}$
(C)
Since $f (x)$ is continuous at $x=\frac{\pi}{2}$.
$\therefore \quad \lim _{x \rightarrow \frac{\pi^{-}}{2}} f (x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f (x)$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}}(a x+1)=\lim _{x \rightarrow \frac{\pi}{2}}(\sin x+b)$
$\Rightarrow a \cdot \frac{\pi}{2}+1=1+ b \quad \Rightarrow b =\frac{ a \pi}{2}$
View full question & answer→MCQ 1522 Marks
$f (x)=\left\{\begin{array}{ll}\frac{x^2-4}{x-2}+ a , & \text { for } x<2 \\ 8, & \text { for } x=2 \\ x+ b +4, & \text { for } x>2\end{array}\right.$ is continuous at $x=2$, then the values of a and b are respectively
Answer(B)
Since $f (x)$ is continuous at $x=2$.
$\therefore \quad f (2)=\lim _{x \rightarrow 2^{-}} f (x)$
$\Rightarrow f (2)=\lim _{x \rightarrow 2}\left(\frac{x^2-4}{x-2}+ a \right) \Rightarrow 8=4+ a$
$\Rightarrow a =4$
Also, $f (2)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow f (2)=\lim (x+ b +4) \Rightarrow 8=6+ b$
$\Rightarrow b =2$
View full question & answer→MCQ 1532 Marks
If $f (x)=\left\{\begin{array}{r}x+\lambda, x<3 \\ 4, x=3 \\ 3 x-5, x>3\end{array}\right.$ is continuous at $x=3$, then
Answer(D)
Since $f (x)$ is continuous at $x=3$.
$\therefore f (3)=\lim _{x \rightarrow 3^{-}} f (x)$
$\Rightarrow 4=3+\lambda$
$\Rightarrow \lambda=1$
View full question & answer→MCQ 1542 Marks
If $f (x)=\left\{\begin{array}{rc}x^2+ k & ; \quad x \geq 0 \\ -x^2- k & ; \quad x<0\end{array}\right.$ is continuous at $x=0$, then k is equal to
Answer(A)
Since $f (x)$ is continuous at $x=1$.
$\therefore \quad \lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\Rightarrow \lim (2 x+1)=\lim \left(3-k x^2\right)$
$\Rightarrow 2+1=3-k(1)^2$
$\Rightarrow k =0$
View full question & answer→MCQ 1552 Marks
If $f (x)=\left\{\begin{array}{ll}2 x+1, & x \leq 1 \\ 3- k x^2, & x>1\end{array}\right.$ is continuous at $x=1$, then the value of k is
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad \lim _{x \rightarrow 0^{-}} f (x)=\lim _{x \rightarrow 0^{+}} f (x)$
$\Rightarrow \lim _{x \rightarrow 0^{-}}\left(-x^2- k \right)=\lim _{x \rightarrow 0^{+}}\left(x^2+ k \right)$
$\Rightarrow- k = k$
$\Rightarrow k =0$
View full question & answer→MCQ 1562 Marks
$f(x)=\left\{\begin{array}{cc}3 x-8, & \text { if } x \leq 5 \\ 2 k, & \text { if } x>5\end{array}\right.$ is continuous at $x=5$, find k .
- A
$\frac{4}{7}$
- B
$\frac{2}{7}$
- ✓
$\frac{7}{2}$
- D
$\frac{3}{7}$
AnswerCorrect option: C. $\frac{7}{2}$
(C)
$\lim _{x \rightarrow 5^{-}} f (x)=\lim _{x \rightarrow 5}(3 x-8)=7$
$\lim _{x \rightarrow 5^{+}} f (x)=\lim _{x \rightarrow 5} 2 k =2 k$
Since $f (x)$ is continuous at $x=5$.
$\therefore \quad \lim _{x \rightarrow 5^{-}} f (x)=\lim _{x \rightarrow 5^{+}} f (x)$
$\Rightarrow 7=2 k \Rightarrow k =\frac{7}{2}$
View full question & answer→MCQ 1572 Marks
If $f(x)=\left\{\begin{array}{ll}k x^2 & \text { if } x \leq 2 \\ 3 & \text { if } x>2\end{array}\right.$ is continuous at $x=2,$ then the value of $k$ is
- A
- B
- ✓
$\frac{3}{4}$
- D
$\frac{4}{3}$
AnswerCorrect option: C. $\frac{3}{4}$
(C)
$f(2)=k(2)^2=4 k$
$\lim _{x \rightarrow 2^{+}} f (x)=\lim _{x \rightarrow 2^{+}} 3=3$
Since the function is continous at $x=2$,
$\lim _{x \rightarrow 2^{+}} f (x)= f (2)$
$\Rightarrow 4 k =3$
$\Rightarrow k =\frac{3}{4}$
View full question & answer→MCQ 1582 Marks
If $f (x)=\left\{\begin{array}{cl}2 x+5 & ; x>1 \\ k & ; x=1, \\ 8 x-1 & ; x<1\end{array}\right.$ is continuous at $x=1$, then the value of k is
Answer(C)
Since $f (x)$ is continuous at $x=1$.
$\therefore \quad \lim _{x \rightarrow 1^{-}} f (x)= f (1)$
$\Rightarrow \lim _{x \rightarrow 1^{-}}(8 x-1)= k$
$\Rightarrow k =7$
View full question & answer→MCQ 1592 Marks
The function ' f ' is defined by $f (x)=2 x-1 \ if \ x>2, f (x)= k$ if $x=2$ and $x^2-1$ if $x<2$ continuous, then the value of $k$ is equal $t_0$
Answer(B)
Since $f (x)$ is continuous at $x=2$.
$\therefore \lim _{x \rightarrow 2^{-}} f (x)= f (2)$
$\Rightarrow \lim _{x \rightarrow 2^{-}}\left(x^2-1\right)= k$
$\Rightarrow k =3$
View full question & answer→MCQ 1602 Marks
If $f(x)=\left\{\begin{array}{cl}x: & 0 \leq x<\frac{1}{2} \\ 1-x: & \frac{1}{2} \leq x<1\end{array}\right.$, then
- ✓
$f ( x )$ is continuous at $x=\frac{1}{2}$
- B
$f ( x )$ is discontinuous at $x=\frac{1}{2}$
- C
$\lim _{x \rightarrow \frac{1}{2}^{-}} f(x)=1$
- D
$\lim _{x \rightarrow \frac{1}{2}^{+}} f(x)=1$
AnswerCorrect option: A. $f ( x )$ is continuous at $x=\frac{1}{2}$
(A)
$f \left(\frac{1}{2}\right)=1-\frac{1}{2}=\frac{1}{2}$
$\lim _{x \rightarrow \frac{1}{2}^{-}} f (x)=\lim _{x \rightarrow \frac{1}{2}^{-}}(x)=\frac{1}{2}$
$\lim _{x \rightarrow \frac{1}{2}^{+}} f(x)=\lim _{x \rightarrow \frac{1^{+}}{2}}(1-x)=1-\frac{1}{2}=\frac{1}{2}$
$\therefore \lim _{x \rightarrow \frac{1}{2}^{-}} f (x)=\lim _{x \rightarrow \frac{1}{2}^{+}} f (x)= f \left(\frac{1}{2}\right)$
$\therefore f (x)$ is continuous at $x=\frac{1}{2}$.
View full question & answer→MCQ 1612 Marks
$f(x)=\left\{\begin{array}{cl}2 x+1, & x<1 \\ 2, & x=1 \ is\\ x^2+1, & x>1\end{array}\right.$
Answer(C)
$\lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{-}}(2 x+1)=3 \neq f (1)$
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^2+1\right)=2=f(1)$
View full question & answer→MCQ 1622 Marks
The function $f (x)=\frac{\log (1+ ax )-\log (1- bx )}{x}$ is not defined at $x=0$. The value which should be assigned to f at $x=0$ so that it is continuous at $x=0$, is
- A
$a - b$
- ✓
$a+b$
- C
$\log a+\log b$
- D
$\log a-\log b$
Answer(B)
For $f (x)$ to be continuous at $x=0$,
$f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow f (0)=\lim _{x \rightarrow 0} \frac{\log (1+ ax )-\log (1- bx )}{x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \frac{\frac{ a }{1+ a x}+\frac{ b }{1- b x}}{1}$
$\Rightarrow f (0)= a + b$
View full question & answer→MCQ 1632 Marks
If $f(x)=\left\{\begin{array}{c}\frac{1-\cos x}{x}, x \neq 0 \\ k , x=0\end{array}\right.$ is continuous at $x=0$, then $k =$
- ✓
$0$
- B
$\frac{1}{2}$
- C
$\frac{1}{4}$
- D
$-\frac{1}{2}$
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{1-\cos x}{x}=\lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{\frac{x^2}{4}} \times \frac{x}{4}$
$\therefore \quad f(0)=2(1)(0)=0$
Alternate method:
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$
$\rightarrow f (0)=\lim _{x \rightarrow 0} \frac{1-\cos x}{x}$
Applying L'Hospital rule on R.H.S., we get
$f (0)=\lim _{x \rightarrow 0} \sin x=0$
View full question & answer→MCQ 1642 Marks
If f is continuous at $x=0$, where $f (x)=\frac{\left( e ^{3 x}-1\right) \sin x}{x^2}, x \neq 0$, then $f (0)=$
Answer(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{\left(e^{3 x}-1\right) \sin x}{x^2}$
$=\lim _{x \rightarrow 0} \frac{ e ^{3 x}-1}{3 x} \times 3 \times \frac{\sin x}{x}=1 \times 3 \times 1$
$\therefore f(0)=3$
View full question & answer→MCQ 1652 Marks
In order that the function $f (x)=(x+1)^{\frac{1}{x}}$ is continuous at $x=0, f (0)$ must be defined as
- A
$f(0)=0$
- ✓
$f(0)=e$
- C
$f(0)=\frac{1}{e}$
- D
$f(0)=1$
AnswerCorrect option: B. $f(0)=e$
(B)
For $f (x)$ to be continuous at $x=0$,
$f (0)=\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}= e$
View full question & answer→MCQ 1662 Marks
If $f (x)=\left\{\begin{array}{cl}\frac{\log _e x}{x-1}, & x \neq 1 \\ k , & x=1\end{array}\right.$ is continuous at $x=1$, then the value of k is
Answer(B)
Since $f (x)$ is continuous at $x=1$.
$\therefore f (1)=\lim _{x \rightarrow 1} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 1} \frac{\log x}{x-1}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow 1} \frac{\frac{1}{x}}{1}=1$
View full question & answer→MCQ 1672 Marks
Let $f(x)=\left\{\begin{array}{ll}\frac{\sin \pi x}{5 x} ; & x \neq 0 \\ k ; & x=0\end{array}\right.$.If $f(x)$ is continuous at $x=0$, then $k =$
- ✓
$\frac{\pi}{5}$
- B
$\frac{5}{\pi}$
- C
- D
$0$
AnswerCorrect option: A. $\frac{\pi}{5}$
(A)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 0} \frac{\sin \pi x}{5 x}$
$\Rightarrow k =\lim _{x \rightarrow 0}\left(\frac{\sin \pi x}{\pi x}\right) \cdot \frac{\pi}{5}$
$\Rightarrow k =(1) \cdot \frac{\pi}{5}$
$\Rightarrow k =\frac{\pi}{5}$
View full question & answer→MCQ 1682 Marks
If $f(x)=\left\{\begin{array}{cc}\frac{\sin 3 x}{x}, & x \neq 0 \\ \frac{k}{2}, & x=0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ is
Answer(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$\Rightarrow \frac{ k }{2}=\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \cdot 3$
$\Rightarrow \frac{k}{2}=3 \Rightarrow k=6$
View full question & answer→MCQ 1692 Marks
If $f (x)=\frac{2 x+\tan x}{x}, x \neq 0$, is continuous at $x=0$, then $f (0)$ equals
Answer(D)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0} \frac{2 x+\tan x}{x}$
$=\lim _{x \rightarrow 0}\left(2+\frac{\tan x}{x}\right)=2+1=3$
View full question & answer→MCQ 1702 Marks
If $f (x)=\sin x-\cos x, x \neq 0$, is continuous at $x=0$, then $f (0)$ is equal to
Answer(B)
Since $f (x)$ is continuous $x=0$.
$\therefore \quad f (0)=\lim _{x \rightarrow 0} f (x)$
$=\lim _{x \rightarrow 0}(\sin x-\cos x)$
$=\sin 0-\cos 0=-1$
View full question & answer→MCQ 1712 Marks
If the function$f(x)=\left\{\begin{array}{cc}\frac{x^2-(A+2) x+A}{x-2}, & \text { for } x \neq 2 \\2, & \text { for } x=2\end{array}\right. $is continuous at $x=2$, then
- ✓
$A=0$
- B
$A=1$
- C
$A=-1$
- D
$A=2$
Answer(A)
Since $f (x)$ is continuous at $x=2$.
$\therefore f (2)=\lim _{x \rightarrow 2} f (x)$
$\Rightarrow 2=\lim _{x \rightarrow 2} \frac{x^2-( A +2) x+ A }{x-2}$
$\Rightarrow 2=\lim _{x \rightarrow 2} \frac{x(x-2)- A (x-1)}{x-2}$,
which is true if $A =0$
View full question & answer→MCQ 1722 Marks
For the function $f (x)=\left\{\begin{array}{cc}\frac{x^3- a ^3}{x- a }, & x \neq a \\ b , & x= a \end{array}\right.$
If $f (x)$ is continuous at $x= a$, then b is equal to
- A
$a^2$
- B
$2 a ^2$
- ✓
$3 a ^2$
- D
$4 a^2$
AnswerCorrect option: C. $3 a ^2$
(C)
Since $f (x)$ is continuous at $x= a$.
$\therefore \quad f ( a )=\lim _{x \rightarrow a } f (x)$
$\Rightarrow b =\lim _{x \rightarrow a } \frac{x^3- a ^3}{x_{-} a }$
$\Rightarrow b =3 a ^{3-1}=3 a ^2$
View full question & answer→MCQ 1732 Marks
If $f(x)=\left\{\begin{array}{ll}\frac{x^2-9}{x-3} ; & \text { if } x \neq 3 \\ 2 x+k ; & \text { otherwise }\end{array}\right.$, is continuous at $x=3$, then $k =$
Answer(B)
Since $f (x)$ is continuous at $x=3$.
$\therefore f (3)=\lim _{x \rightarrow 3} f (x)$
$\Rightarrow 2(3)+ k =\lim _{x \rightarrow 3} \frac{x^2-9}{x-3}$
$\Rightarrow 6+ k =\lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{x-3}$
$\Rightarrow 6+ k =\lim _{x \rightarrow 3}(x+3)$
$\Rightarrow 6+ k =6 \Rightarrow k =0$
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If function $f(x)=\left\{\begin{array}{r}\frac{x^2-1}{x-1}, \text { when } x \neq 1 \\ k, \text { when } x=1\end{array}\right.$ is continuous at $x=1$, then the value of k will be
Answer(B)
Since $f (x)$ is continuous at $x=1$.
$\therefore \quad f (1)=\lim _{x \rightarrow 1} f (x)$
$\Rightarrow k =\lim _{x \rightarrow 1} \frac{x^2-1}{x-1}$
$\Rightarrow k =\lim _{x \rightarrow 1}(x+1)$
$\Rightarrow k =2$
View full question & answer→MCQ 1752 Marks
If $f(x)=\frac{x^2-10 x+25}{x^2-7 x+10}$ for $x \neq 5$ and $f$ is continuous at $x=5$, then $f (5)=$
Answer(A)
Since $f (x)$ is continuous at $x=5$.
$\therefore f (5)=\lim _{x \rightarrow 5} f (x)=\lim _{x \rightarrow 5} \frac{x^2-10 x+25}{x^2-7 x+10}$
$=\lim _{x \rightarrow 5} \frac{(x-5)^2}{(x-2)(x-5)}=\frac{5-5}{5-2}=0$
View full question & answer→MCQ 1762 Marks
If $f (x)=\left\{\begin{array}{cl}\left(1+\frac{4 x}{5}\right)^{\frac{1}{x}}, & x \neq 0 \\ e ^{\frac{4}{5}}, & x=0\end{array}\right.$, then
- A
$\lim _{x \rightarrow 0} f(x)=e^{\frac{2}{5}}$
- B
$\lim _{x \rightarrow 0} f(x)$ does not exist
- ✓
$f (x)$ is continuous at $x=0$
- D
$f (x)$ is discontinuous at $x=0$
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0}\left(1+\frac{4 x}{5}\right)^{\frac{1}{x}}$
$=\left[\lim _{x \rightarrow 0}\left(1+\frac{4 x}{5}\right)^{\frac{5}{4 x}}\right]^{\frac{4}{5}}= e ^{\frac{4}{5}}= f (0)$
View full question & answer→MCQ 1772 Marks
If $f (x)=\left\{\begin{array}{cc}\frac{\sin x}{x}+\cos x, & x \neq 0 \\ 2, & x=0\end{array}\right.$, then
- A
$f (x)$ is discontinuous at $x=0$
- B
$\lim _{x \rightarrow 0} f (x)=1$
- ✓
$f (x)$ is continuous at $x=0$
- D
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}+\cos x\right)$
$\therefore f (x)$ is continuous at $x=0$
View full question & answer→MCQ 1782 Marks
$\text {If } f(x)=\left\{\begin{array}{ll}\frac{1}{2} \sin x^2, & x \neq 0 \\0, & x=0\end{array}\right. \text {, then }$
- A
$\lim _{x \rightarrow 0} f(x)=\frac{1}{2}$
- B
$f (x)$ is discontinuous at $x=0$
- ✓
$f (x)$ is continuous at $x=0$
- D
AnswerCorrect option: C. $f (x)$ is continuous at $x=0$
(C)
$\lim _{x \rightarrow 0} f (x)=\lim _{x \rightarrow 0} \frac{1}{2} \sin x^2=0= f (0)$
$\therefore \quad f (x)$ is continuous at $x=0$.
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