MCQ 1011 Mark
The function $f(x) = x − [x],$ where $[⋅]$ denotes the greatest integer function is :
AnswerCorrect option: C. Continuous at non $-$ integer points only.
$f(x) = x - x$
Consider $n$ be an integer.
$\text{f(x)}=\text{x}-[\text{x}]=\begin{cases}\text{x}-(\text{n}-1)&\text{n}-1\leq\text{x}<\text{n}\\0&\text{x}=\text{n}\\\text{x}-\text{n}&\text{n}\leq\text{x}<\text{n}+1\end{cases}$
Now,
$\text{LHL}$ at $x = n$
$=\lim\limits_{\text{x}\rightarrow\text{n}^{-}}\text{f(x)}=\text{x}-\text{n}-1=\text{x}-\text{n}+1$
RHL at x = n
$=\lim\limits_{\text{x}\rightarrow\text{n}^{+}}\text{f(x)}=\text{x}-\text{n}=\text{x}-\text{n}$ As,
$\text{LHL}\neq\text{RHL}$ at $x = n$
i.e., given function is not continuous at $n.$
Now, $n$ is any integer.
Therefore, given function is not continuous at integers.
Therefore, given points are continuous at non $-$ integer points only.
View full question & answer→MCQ 1021 Mark
Choose the correct answers from the given four options : The function $\text{f(x)}=\cot\text{x}$ is discontinuous on the set
- ✓
$\big\{\text{x}=\text{n}\pi:\text{n}\in\text{Z}\big\}$
- B
$\big\{\text{x}=2\text{n}\pi:\text{n}\in\text{Z}\big\}$
- C
$\Big\{\text{x}=(2\text{n}+1)\frac{\pi}{2};\text{n}\in\text{Z}\Big\}$
- D
$\Big\{\text{x}=\frac{\text{n}\pi}{2};\text{n}\in\text{Z}\Big\}$
AnswerCorrect option: A. $\big\{\text{x}=\text{n}\pi:\text{n}\in\text{Z}\big\}$
Consider, $\text{f(x)}=\cos\text{x}=\frac{\cos\text{x}}{\sin\text{x}}$
We know that, $\big[\sin\text{x}=0\text{ at }\text{x}=\text{n}\pi,\text{n}\in\text{Z}\big]$
Hence, $\text{f(x)}=\cot\text{x}$ is discontinuous on the set $\big\{\text{x}=\text{n}\pi:\text{n}\in\text{Z}\big\}.$
View full question & answer→MCQ 1031 Mark
The function $\text{f(x)}=\frac{\text{x}^3+\text{x}^2-16\text{x}+20}{\text{x}-2}$ is not defind for $x = 2$. in order to make $f(x)$ continuous at $x = 2$, here $f(2)$ should be defined as :
AnswerHere,
$x^3+x^2-16 x+20$
$=x^3-2 x^2+3 x^2-6 x-10 x+20$
$=x^2(x-2)+3 x(x-2)-10(x-2)$
$=(x-2)\left(x^2+3 x-10\right)$
$=(x-2)(x-2)(x-5)$
$=(x-2)^2(x+5)$
So, the given function can be rewritten as
$\text{f(x)}=\frac{(\text{x}-2)^2(\text{x}+5)}{\text{x}-2}$
$\Rightarrow\text{f(x)}=(\text{x}-2)(\text{x}+5)$
If $f(x)$ is continuous at $x = 2,$ then
$\lim\limits_{\text{x}\rightarrow2}\text{f(x)}=\text{f}(2)$
$\Rightarrow\lim\limits_{\text{x}\rightarrow2}(\text{x}-2)(\text{x}+5)=\text{f}(2)$
$\Rightarrow\text{f}(2)=0$
Hence, in order to make $f(x)$ continuous at $x = 2, f(2)$ should be defined as $0$.
View full question & answer→MCQ 1041 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) : $ if $\text{y}=\tan5\text{x}^\circ,\text{then}\frac{\text{dy}}{\text{dx}}=\frac{5\pi}{180}\sec^2(5\text{x}^\circ)$
Reason $(R) : \pi^\text{c}=90^\circ$
- A
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
- B
Both $A$ and $R$ are true but $R$ is $\text{NOT}$ the correct explanation of $A.$
- ✓
$A$ is true but $R$ is false
- D
$A$ is false but $R$ is true
AnswerCorrect option: C. $A$ is true but $R$ is false
View full question & answer→MCQ 1051 Mark
If $\text{f(x)}=\begin{cases}\frac{\sin(\text{a}+1)}{\text{x}},&\text{x}<0\\\text{c},&\text{x}=0\\\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}},&\text{x}>0&\end{cases}$ is continuouse at $x = 0,$ then :
- A
$\text{a}=-\frac{3}{2},\text{b}=0,\text{c}=\frac{1}{2}$
- B
$\text{a}=-\frac{3}{2},\text{b}=1,\text{c}=-\frac{1}{2}$
- ✓
$\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$
- D
AnswerCorrect option: C. $\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}\times\frac{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c }$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x+bx}^2-\text{x}}{\text{bx}\sqrt{\text{x}}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{bx}^2}{\text{bx}\sqrt{\text{x}}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\sqrt{\text{x}}\times\frac{1}{\sqrt{\text{x+bx}^2}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}\Big(1+\text{bx}^\frac{3}{2}\Big)}+\sqrt{\text{x}}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}{\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\Bigg[\sqrt{\Big(1+\text{b}^\frac{3}{2}}\Big)+1\Bigg]}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{1}{\Bigg[\sqrt{\Big(1+\text{bx}^{\frac{3}{2}}\Big)}+1\Bigg]}=\text{c}$
$\text{c}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin(\text{a+1})\text{x}+\sin\text{x}}{\text{x}}=\text{c}$
$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\bigg[\frac{(\text{a+1)x+x}}{2}\bigg]\cos\bigg(\frac{\text{ax}}{2}\bigg)}{\text{x}}=\frac{1}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin\Big[\frac{(\text{a+2)x}}{2}\Big]}{\frac{(\text{a+2)x}}{2}}\times\frac{(\text{a}+2)}{2}\cos\Big(\frac{\text{ax}}{2}\Big)=\frac{1}{4}$
$1\times\frac{(\text{a+2)}}{2}\times1=\frac{1}{4}$
$\text{a}+2=\frac{1}{2}$
$\text{a}=\frac{-3}{2}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}}$ exist if $\text{b}\neq0$
$\text{b }\in\text{ R}-(0)$
View full question & answer→MCQ 1061 Mark
The function $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$ is continuous at $x = 0,$ then $k =$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{x}\ne0\\\frac{\text{k}}{2},&\text{x}=0\end{cases}$
If $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{\text{x}}=\text{f}(0)$
$\Rightarrow3\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}=\frac{\text{k}}{2}$
$\Rightarrow3\times1=\frac{\text{k}}{2}$
$\Rightarrow\frac{\text{k}}{2}=3$
$\Rightarrow\text{k}=6$
View full question & answer→MCQ 1071 Mark
Choose the correct answers from the given four options : If $\text{f(x)}=\text{x}^2\sin\frac{1}{\text{x}},$ where $\text{x}\neq0,$ then the value of the function $f$ at $x = 0,$ so that the function is continuous at $x = 0,$ is:
AnswerThe value of the function $f$ at $x = 0,$ so that it is continuous at $x = 0$ is $0.$
View full question & answer→MCQ 1081 Mark
If $f(x)=4 x^8$, then :
- A
$\text{f}\ '\Big(\frac{1}{2}\Big)=\text{f}'\Big(-\frac{1}{2}\Big)$
- B
$\text{f}\Big(\frac{1}{2}\Big)=-\text{f}'\Big(-\frac{1}{2}\Big)$
- ✓
$\text{f}\Big(-\frac{1}{2}\Big)=\text{f}\Big(-\frac{1}{2}\Big)$
- D
$\text{f}\Big(\frac{1}{2}\Big)=\text{f}'\Big(-\frac{1}{2}\Big)$
AnswerCorrect option: C. $\text{f}\Big(-\frac{1}{2}\Big)=\text{f}\Big(-\frac{1}{2}\Big)$
$f(x)=4 x^8,$
$\Rightarrow\text{f}\Big(\frac{1}{2}\Big)=4\Big(\frac{1}{2}\Big)^8=\frac{4}{256}=\frac{1}{64}$
$\Rightarrow\text{f}\Big(-\frac{1}{2}\Big)=4\Big(-\frac{1}{2}\Big)^8=\frac{4}{256}=\frac{1}{64}$
$\Rightarrow\text{f}\Big(-\frac{1}{2}\Big)=\text{f}\Big(-\frac{1}{2}\Big)$
View full question & answer→MCQ 1091 Mark
If $\text{f(x)}=|\log_\text{e}|\text{x}\|,$ then :
- A
$f(x)$ is continuous and differentiable for all $x$ in its domain.
- ✓
$f(x)$ is continuous for all for all $\times $ in its domain but not differentiable at $\text{x}=\pm1$
- C
$f(x)$ is neither continuous nor differentiable at $\text{x}=\pm1$
- D
AnswerCorrect option: B. $f(x)$ is continuous for all for all $\times $ in its domain but not differentiable at $\text{x}=\pm1$
We have,
$\text{f(x)}=|\log_\text{e}|\text{x}\|$
We know that log function is defined for posirive value.
Here, $|x$| is positive for all non zero $x.$
Therefore, domian of function is $R - {0}$
And we know that logarithmic function continuous in its domain.
Therefore, $|\log_\text{e}|\text{x}||$ is continuous in its domain.
We will check the differentiability at its critical points.
$|\log_\text{e}|\text{x}||=\begin{cases}\log_\text{e}(-\text{x}) & -\infty<\text{x<-1}\\-\log_\text{e}(-\text{x}) &-1<\text{x}<0\\-\log_\text{e}(\text{x})&0<\text{x}<1\\\log_\text{e}(\text{x})&1<\text{x}<\infty\end{cases}$
$(\text{LHL}$ at $x = -1) =\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$
$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\log_\text{e}(-\text{x})-0}{\text{x}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[-(-1-\text{h})]}{-1-\text{h}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{-\text{h}}$
$=-1$
$\text{(RHL}$ at $x = -1) =\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$
$=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{-\log_\text{e}(-\text{x})-0}{\text{x}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[-(-1+\text{h})]}{-1+\text{h}+1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}(1-\text{h})}{\text{h}}$
$=-\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$
$=-1\times-1=1$
Here, $\text{LHL}\neq\text{RHL}$
Therefore, the given function is not differentiable at $x = -1.$
$\text{(LHL}$ at $x = 1) =\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{-\log_\text{e}(\text{x})-0}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[(1-\text{h})]}{1-\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$
$=-1$
$\text{(RHL}$ at $x = 1) =\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-(1)}$
$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\log_\text{e}(\text{x})-0}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[(1+\text{h})]}{1+\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{\text{h}}$
$=1$
Here, $\text{LHL}\neq\text{RHL}$
Therefore, the given function is not differentiable at $x =1$.
Therefore, given function is continuous for all $x$ in its domain but not differentiable at $\text{x}=\pm1.$
View full question & answer→MCQ 1101 Mark
If $\text{y}^2=\text{ax}^2+\text{bx}+\text{c},$ then $\text{y}^3\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is :
AnswerCorrect option: C. a function of $y$ only
$\text{y}^2=\text{ax}^2+\text{bx}+\text{c}$
$\frac{\text{dy}}{\text{dx}}=2\text{ax}+\text{b}$
$\frac{\text{d}^2\text{y}}{\text{d}^2}=2\text{a}$
$=\text{y}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ay}^3$
$=$ A function of $y$ only
View full question & answer→MCQ 1111 Mark
The set of points where the function $f(x) = x |x|$ is differentiable is :
- ✓
$(-\infty,\infty)-\infty,\infty$
- B
$(-\infty,0)\cup(0,\infty)-\infty,0\cup0,\infty$
- C
$(0,\infty)0,\infty$
- D
$[0,\infty]0,\infty$
AnswerCorrect option: A. $(-\infty,\infty)-\infty,\infty$
We have $,\text{f(x)}=\text{x}|\text{x}|$
$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}^2, & \text{x} < 0\\0 ,& \text{x}= 0\\\text{x}^2,&\text{x}>0\end{cases}$
When, $x < 0,$ we have
$f(x)=-x^2$ which being a polynomial function is continuous and differentable in $(-\infty,0)$
When, $x > 0$, we have
$f(x)=-x^2$ which being a polynomial function is continuous and differentable in $(0,\infty,)$
Thus possible point of non $-$ differentiability of $f(x)$ is $x = 0$
Now, $\text{LHL}\ ($at $x = 0) =\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{x}^2-0}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-(-\text{h})^2}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}$
$=0$
And $\text{RHL}\ ($at $x = 0) =\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{x}^2-0}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}$
$=0$
$\therefore \text{LHL}\ ($at $x = 0) = \text{RHL}\ ($at $x = 0)$
So, $f(x)$ is also differentiable at $x = 0$
i.e. $f(x)$ is differentiable in $(-\infty,\infty).$
View full question & answer→MCQ 1121 Mark
if $\text{y}=\text{e}^{{\tan}}\ \text{x},$ then $(\cos^2\text{x})\text{y}_2=$
- A
$(1-\sin2\text{x})\text{y}_1$
- B
$-(1+\sin2\text{x})\text{y}_1$
- ✓
$(1+\sin2\text{x})\text{y}_1$
- D
AnswerCorrect option: C. $(1+\sin2\text{x})\text{y}_1$
$\text{y}=\text{e}^{{\tan}}\text{x},$
$\text{y}_1=\text{sec}^2\text{xe}^{\tan}\text{x}$
$\Rightarrow\cos^2\text{xy}_1=\text{e}^{\tan}\text{x}$
again differentiating $\text{ w.r.t. x},$ we get
$\cos^2\text{xy}_2-2\cos\text{x}\sin\text{xy}_1=\sec^2\text{xe}^{\tan}\text{x}$
$\Rightarrow\cos^2\text{xy}_2=\text{y}_1\sin2\text{x}+\text{y}_1$
View full question & answer→MCQ 1131 Mark
The derivative of $\sec^{-1}\Big(\frac{1}{2\text{x}^2+1}\Big) \ \text{w.r.t.}\ \sqrt{1+3\text{x}}$ at $\text{x}=\frac{-1}{3}$
- ✓
- B
$0$
- C
$\frac{1}{2}$
- D
$\frac{1}{3}$
AnswerPut, $\text{u}=\sec^{-1}\Big(\frac{1}{2\text{x}^2+1}\Big)$ and $\text{v}=\sqrt{1+3\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{1}{2\sqrt{1+3\text{x}}}\times3$
But at $\text{x}=\frac{-1}{3}\frac{\text{dv}}{\text{dx}}$ does not exist
Hence, derivative of $\sec^{-1}\Big(\frac{1}{2\text{x}^2+1}\Big)$
With respect to $\sqrt{1+3\text{x}}$ does not exist.
View full question & answer→MCQ 1141 Mark
If $\text{f(x)}=\begin{cases}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ and $f(x)$ is continous at $x = 0,$ then the value of $k$ is :
AnswerCorrect option: B. $a + b$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})-\log(1-\text{bx})}{\text{x}}=\text{k}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{x}}-\frac{\log(1-\text{bx})}{\text{x}}=\text{k}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{ax})}{\text{ax}}\times\text{a} -\frac{\log(1-\text{bx})}{\text{-bx}}\times(-\text{b})=\text{k}$
$\text{a}+\text{b}=\text{k}$
View full question & answer→MCQ 1151 Mark
Choose the correct answers from the given four options : The derivative of $\cos^{-1}(2\text{x}^2-1)\ \text{w.r.t.} \cos^{-1}\text{x}$ is :
AnswerLet $\text{u}=\cos^{-1}(2\text{x}^2-1)$ and $\text{v}=\cos^{-1}\text{x}$
$\text{u}=\cos^{-1}(2\text{x}^2-1)$
$=\cos^{-1}(2\cos^2\text{v}-1)=\cos^{-1}(\cos2\text{v})=2\text{v}$
$\therefore\ \frac{\text{du}}{\text{dv}}=2$
View full question & answer→MCQ 1161 Mark
If $\text{y}=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\tan\frac{\text{x}}{2}\Big),\text{a} > \text{b} > 0,$ then :
- A
$\text{y}_1=\frac{-1}{\text{a}+\text{b}\cos\text{x}}$
- ✓
$\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$
- C
$\text{y}_1=\frac{1}{\text{a}-\text{b}\cos\text{x}}$
- D
$\text{y}_2=\frac{-\text{b}\sin\text{x}}{(\text{a}-\text{b}\cos\text{x})^2}$
AnswerCorrect option: B. $\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$
View full question & answer→MCQ 1171 Mark
If $\text{x}=\text{at}^2,\text{y}=2\text{at}$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}=$
AnswerCorrect option: D. $-\frac{1}{2\text{at}^3}$
$\text{x}=\text{at}^2,\text{y}=2\text{at}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\text{a}}{2\text{at}}$
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{d}}{\text{dt}}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\frac{\text{dx}}{\text{dt}}}$
$=\frac{\frac{-1}{\text{t}^2}}{2\text{at}}=\frac{-1}{2\text{at}^3}$
View full question & answer→MCQ 1181 Mark
Let $\text{f(x)}=|\cos\text{x}|.$ Then,
- A
$f(x)$ is everywhere differentiable.
- B
$f(x)$ is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$
- ✓
$f(x)$ is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$
- D
AnswerCorrect option: C. $f(x)$ is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$
$\text{f}(\text{x)} = |\cos\text{x}|$
Given function is trigonometric function.
$\Rightarrow $ Hence, it is continuous.
Function is not differentiable at odd multiples of $\frac{\pi}{2}.$
$\Rightarrow f(x)$ is not differentiable at $\text{x} = (2\text{n} + 1) \frac{\pi}{2}$
View full question & answer→MCQ 1191 Mark
The function $\text{f(x)}=\frac{\sin(\text{x}|\text{x}-\pi|)}{4+|\text{x}|^2},$ where $[.]$ denotes the greatest integer function, is :
- ✓
Continuous as well as differentiable for all $\text{x}\in\text{R}$
- B
Continuous for all $x$ but differentiable at some $x$
- C
Differentiable for all $x$ but not continuous at some $x$
- D
AnswerCorrect option: A. Continuous as well as differentiable for all $\text{x}\in\text{R}$
Here,
$\text{f(x)}=\frac{\sin(\text{x}|\text{x}-\pi|)}{4+|\text{x}|^2}$
Since, we know that $\pi(\text{x}-\pi)=\text{n}\pi$ and $\sin\text{n}\ \pi=0.$
$\because4+\text{x}[\text{x}]^2\neq0$
$\therefore\text{f(x)}=0$ for all $x$
Thus, $f(x)$ is a constant function and it is continuous and differentible everywhere.
View full question & answer→MCQ 1201 Mark
The function $\text{f(x)}=|\cos\text{x}|$ is :
- A
Everywhere continuous and differentiable.
- ✓
Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
- C
Neither continuous nor differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
- D
AnswerCorrect option: B. Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
As $\cos x$ is even function it is continuous everywhere but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$
$\cos\Big[(2\text{n}+1)\frac{\pi}{2}=\cos\Big(\text{n}\pi+\frac{\pi}{2}\Big)\Big]=-\sin\text{n}\ \pi$
For $n$ as an integer
$\Rightarrow\sin\text{n}\ \pi=0$
For $n$ as rational
$\Rightarrow\sin\text{n}\ \pi=-1$
View full question & answer→MCQ 1211 Mark
For the curve $\sqrt{\text{x}}+\sqrt{\text{y}}=1,\frac{\text{dy}}{\text{dx}}$ at $\Big(\frac{1}{4},\frac{1}{4}\Big)$ is :
- A
$\frac{1}{2}$
- B
$1$
- ✓
$-1$
- D
$0$
AnswerWe have, $\sqrt{\text{x}}+\sqrt{\text{y}}=1$
Differentiating with respect to $x,$ we get,
$\frac{1}{2\sqrt{\text{x}}}+\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{1}{2\sqrt{\text{x}}}\times\frac{2\sqrt{\text{y}}}{1}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\sqrt{\text{y}}}{\sqrt{\text{x}}}$
Now, $\Big[\frac{\text{dy}}{\text{dx}}\Big]_{\Big(\frac{1}{4},\frac{1}{4}\Big)}=-\frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-1$
View full question & answer→MCQ 1221 Mark
Let $\text{f(x)}=|\sin\text{x}|.$ then,
- A
$f(x)$ is everywhere differentiable.
- ✓
$f(x)$ is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$
- C
$f(x)$ is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$
- D
AnswerCorrect option: B. $f(x)$ is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$
$\text{f(x)}=|\sin\text{x}|$
Given function is continuous and differentiable on $(2\text{n}\pi,(2\text{n}+1)\pi)$
But not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$
As $\sin\text{n}\pi=0$ for $\text{n}\in\text{Z}.$
View full question & answer→MCQ 1231 Mark
The value of a for which the function $\text{f(x)}=\begin{cases}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}},&\text{x}\neq0\\12(\log4)^3,&\text{x}=0\end{cases}$ may be continuous at $x = 0$ is :
Answer$\lim\limits_{\text{x}\rightarrow0}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^2}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^3}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{(4^\text{x}-1}{\text{x}^3}\Big)}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}}\text{a}\text{x}\ {\times}\frac{\frac{1}{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}}{\frac{\text{x}^3}{3}}\text{x}^3=12(\log4)^3$
$3(\log4)^3=12(\log4)^3$
$3\text{a}=12$
$\text{a}=12$
Note : The question is incorrect, so it has been modified.
View full question & answer→MCQ 1241 Mark
The value of $c$ in Lagrange's mean value theorem for the function $f(x) = x(x - 2)$ when $\text{x}\in[1,2]$ is :
- A
$1$
- B
$\frac{1}{2}$
- C
$\frac{2}{3}$
- ✓
$\frac{3}{2}$
AnswerCorrect option: D. $\frac{3}{2}$
We have
$f(x) = x(x - 2)$
It can be rewritten as $f(x)=x^2-2 x$
We know that a polynomial function is everywhere continuous and differentiable.
Since, $f(x)$ is polynomial, it is continuous on $[1, 2]$ and differentiable on $[1, 2].$
So, there must exist at least one real number $\text{c}\in(1,2)$ such that
$\text{f}\ '(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$=\frac{\text{f}(2)-\text{f}(1)}{1}$
Now, $f(x)=x^2-2 x$
$\Rightarrow f^{\prime}(x)=2 x-2$
and $f(1)=-1, f(2)=d$
$\therefore\ \text{f}\ '(\text{x})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow\text{f}(\text{x})=\frac{0+1}{1}$
$\Rightarrow2\text{x}-2=1$
$\Rightarrow\text{x}=\frac{3}{2}$
$\therefore\ \text{c}=\frac{3}{2}\in(1,2)$
View full question & answer→MCQ 1251 Mark
If $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}=$
- A
$\frac{4\text{x}^3}{1-\text{x}^4}$
- ✓
$-\frac{4\text{x}}{1-\text{x}^4}$
- C
$\frac{1}{4-\text{x}^4}$
- D
$-\frac{4\text{x}^3}{1-\text{x}^4}$
AnswerCorrect option: B. $-\frac{4\text{x}}{1-\text{x}^4}$
We have, $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\frac{1-\text{x}^2}{1+\text{x}^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\text{x}^2}{1-\text{x}^2}\Big[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1-\text{x}^2}\Big[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{(1+\text{x}^2)}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}}{1-\text{x}^4}$
View full question & answer→MCQ 1261 Mark
If $\text{f(x)}=\frac{1}{1-\text{x}},$ then the set of points discontinuity of the function $f(f(f(x)))$ is :
- A
$\{1\}$
- ✓
$\{0,1\}$
- C
$\{-1, 1\}$
- D
AnswerCorrect option: B. $\{0,1\}$
Given, $\text{f}\text{(x)}=\frac{1}{1-\text{x}}$
Clearly, $\text{f}:\text{R}-\big\{1\big\}\rightarrow\text{R}$
Now,
$\text{f}\big(\text{f}\text{(x)}\big)=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\Bigg(\frac{1}{1-\big(\frac{1}{1-\text{x}}\big)}\Bigg)$
$=\Big(\frac{1-\text{x}}{-\text{x}}\Big)=\Big(\frac{\text{x}-1}{\text{x}}\Big)$
$\therefore\text{f}\text{o}\text{f}:\text{R}-\big\{0,1\big\}\rightarrow\text{R}$
Now,
$\text{f}\big(\text{f}\text{(x)}\big)=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\Bigg(\frac{1}{1-\big(\frac{1}{1-\text{x}}\big)}\Bigg)=\text{x}$
$\therefore\ \text{f}\text{o}\text{f}:\text{R}-\big\{0,1\big\}\rightarrow\text{R}$
Thus, $f(f(f(x)))$ is not defind at $x = 0, 1$
Hence, $f(f(f(x)))$ is discontinuous at $\{0, 1\}$
View full question & answer→MCQ 1271 Mark
Choose the correct answers from the given four options : If $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}$ is equal to :
- A
$\frac{4\text{x}^3}{1-\text{x}^4}$
- ✓
$\frac{-4\text{x}}{1-\text{x}^4}$
- C
$\frac{1}{4-\text{x}^4}$
- D
$\frac{-4\text{x}^3}{1-\text{x}^4}$
AnswerCorrect option: B. $\frac{-4\text{x}}{1-\text{x}^4}$
We have, $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\frac{1-\text{x}^2}{1+\text{x}^2}}\cdot\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$=\frac{(1+\text{x}^2)}{(1-\text{x}^2)}\cdot\frac{(1+\text{x}^2)\cdot(-2\text{x})-(1-\text{x}^2)\cdot2\text{x}}{(1+\text{x}^2)^2}$
$=\frac{-2\text{x}[1+\text{x}^2+1-\text{x}^2]}{(1-\text{x}^2)\cdot(1+\text{x}^2)}=\frac{-4\text{x}}{1-\text{x}^4}$
View full question & answer→MCQ 1281 Mark
The value of $c$ in Rolle's theorem for the function $f(x)=x^3-3 x$ in the interval $\big[0,\sqrt3\big]$ is :
- ✓
$1$
- B
$-1$
- C
$\frac{3}{2}$
- D
$\frac{1}{3}$
Answer$f(x)=x^3-3 x$ in the interval $\big[0,\sqrt3\big]$
$\text{f}(0)=0$ and $\text{f}\big(\sqrt3\big)=0$
$f^{\prime}(x)=3 x^2-3$
$f^{\prime}(c)=3 c^2-3$
$f^{\prime}(c)=0$
$3 c^2-3=0$
$3 c^2=3$
$c^2=1$
$\text{c}=\pm1$
$\text{x}\in\big[0,\sqrt3\big]$
Hence, $x = 1$
View full question & answer→MCQ 1291 Mark
If $y=a x^2+b$, then $\frac{d y}{d x}$ at $x=2$ is equal to :
AnswerGiven that, $y=a x^2+b$
Then, $\frac{\text{dy}}{\text{dx}} = 2\text{ax}$
At $x = 2, \frac{\text{dy}}{\text{dx}}=2(\text{a})(2)=4\text{a}$
View full question & answer→MCQ 1301 Mark
Let $3\sin(\text{xy})+4\cos(\text{xy})=5,$ then $\frac{\text{dy}}{\text{dx}}=$
- ✓
$-\frac{\text{y}}{\text{x}}$
- B
$\frac{3\sin(\text{xy})+4\cos(\text{xy})}{3\cos(\text{xy})-4\sin(\text{xy})}$
- C
$\frac{3\cos(\text{xy})+4\sin(\text{xy})}{4\cos(\text{xy})-3\sin(\text{xy})}$
- D
AnswerCorrect option: A. $-\frac{\text{y}}{\text{x}}$
We have, $3\sin(\text{xy})+4\cos(\text{xy})=5$
$\Rightarrow3\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]-4\sin(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]=0$
$\Rightarrow\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big]\big[3\cos(\text{xy})-4\sin(\text{xy})\big]=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=-\text{y}$
$\therefore\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
View full question & answer→MCQ 1311 Mark
If $\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{bx}}\Big)^\text{x}$ then $\text{x}^3\text{y}_2=$
AnswerCorrect option: A. $(\text{xy}_1-\text{y})^2$
$\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{b}}\Big)^\text{x}$
$\text{y}=\text{x}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))$
$\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\text{x}\Big(\frac{1}{\text{x}}-\frac{\text{b}}{\text{a}+\text{bx}}\Big)$
$=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+1-\frac{\text{bx}}{\text{a}+\text{bx}}$
$(\text{a}+\text{bx})\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))(\text{a}+\text{bx})+\text{a}$
Again differentiating $\text{w.r.t. x},$ we get
$(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))$
$+(\text{a}+\text{bx})\Big(\frac{1}{\text{x}}\frac{\text{b}}{\text{a}+\text{bx}}\Big)$
$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\frac{\text{a}}{\text{x}}$
$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\frac{\text{by}}{\text{x}}+\frac{\text{a}}{\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} =\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\Big(\frac{\text{y}}{\text{x}}+\frac{\text{a}}{\text{a}+\text{bx}}\Big)$
$=\frac{(\text{a}+\text{by})(\text{a}+\text{bx})-\text{b}(\text{ay}+\text{bxy}+\text{ax})}{\text{x}(\text{a}+\text{b})^2}$
$=\frac{\text{a}^2+\text{abx}+\text{aby}+\text{b}^2\text{xy}-\text{bay}-\text{b}^2\text{xy}-\text{abx}}{\text{x}(\text{a}+\text{bx})^2}$
$=\frac{\text{a}^2}{\text{x}(\text{a}+\text{bx})^2}$
$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}$
$(\text{xy}_1-\text{y})=\frac{\text{ax}}{\text{a}+\text{bx}}$
$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}=(\text{xy}_1-\text{y})^2$
View full question & answer→MCQ 1321 Mark
If $\text{f}(\text{x})=\text{e}^{\text{x}}\sin\text{x}$ in $[0,\pi],$ then $c$ in Rolle's theorem is :
- A
$\frac{\pi}{6}$
- B
$\frac{\pi}{4}$
- C
$\frac{\pi}{2}$
- ✓
$\frac{3\pi}{4}$
AnswerCorrect option: D. $\frac{3\pi}{4}$
$\text{f}(\text{x})=\text{e}^{\text{x}}\sin\text{x}$
$\text{f}\ '(\text{x})=\text{e}^{\text{x}}\cos\text{x}+\text{e}^{\text{x}}\sin\text{x}$
$\text{f}\ '(\text{c})=0$
$\text{e}^\text{c}(\cos\text{c}+\sin\text{c})=0$
$\cos\text{c}+\sin\text{c}=0$
$\cos\text{c}=-\sin\text{c}$
$\tan\text{c}=-1$
$\text{c}=\frac{3\pi}{4}\in(0,\pi)$
View full question & answer→MCQ 1331 Mark
If $\text{f(x)}=\begin{cases}\frac{\sin(\cos\text{x})-\cos\text{x}}{(\pi-2\text{x})^2},&\text{x}\neq\frac{\pi}{2}\\\text{k},&\text{x}=\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then $k$ is equal to :
- ✓
$0$
- B
$\frac{1}{2}$
- C
$1$
- D
$-1$
Answer$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}}\frac{\sin(\cos\text{x})-\cos\text{x}}{(\pi-2\text{x})^2}$
$\text{x}\rightarrow\frac{\pi}{2}+\text{h}$
$\text{x}\rightarrow\frac{\pi}{2}$
$\Rightarrow\text{h}\rightarrow0$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\Big(\cos\big(\frac{\pi}{2}+\text{h}\big)\Big)-\cos\big(\frac{\pi}{2}+\text{h}\big)}{\Big(\pi-2\big(\frac{\pi}{2}+\text{h}\big)\Big)^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin(-\sin\text{h})+\sin\text{h}}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}-\sin(\sin\text{h})}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{4\text{h}^2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\sin\big(\frac{\text{h}-\sin\text{h}}{2}\big)}{\frac{\text{h}-\sin\text{h}}{2}}\times\frac{\text{h}-\sin\text{h}}{2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times\frac{\text{h}-\sin\text{h}}{2}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{2\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{4\text{h}^2}\times(\text{h}-\sin\text{h})$
$\text{f}\Big(\frac{\pi}{2}\Big)=\lim\limits_{\text{h}\rightarrow0}\frac{1}{4}\frac{\cos\big(\frac{\text{h}+\sin\text{h}}{2}\big)}{\text{h}^2}\times(\text{h}-\sin\text{h})=0$
$\text{k}=0$
View full question & answer→MCQ 1341 Mark
If $\text{f(x)}=|\text{x}-\text{a}|\ \phi\ (\text{x}),$ where $\phi(\text{x})$ is continuous function, then :
- A
$\text{f}\ '(\text{a}^+)=\phi(\text{a})$
- ✓
$\text{f}\ '(\text{a}^-)=-\phi(\text{a})$
- C
$\text{f}\ '(\text{a}^+)=\text{f}'(\text{a}^-)$
- D
AnswerCorrect option: B. $\text{f}\ '(\text{a}^-)=-\phi(\text{a})$
Given that $\text{f(x)}=|\text{x}-\text{a}|\ \phi\ (\text{x}),$ where $\phi(\text{x})$ continuous function.
$|\text{x}-\text{a}|$
$\Rightarrow\text{x}-\text{a}$ if $\text{x}-\text{a} > 0$
$|\text{x}-\text{a}|$
$\Rightarrow-(\text{x}-\text{a})$ if $\text{x}-\text{a} < 0$
By definition of continuity,
$\text{f}'(\text{a})= \lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{a}+\text{h})-\text{f(a)}}{\text{h}}$
Hence, $\text{f}(\text{a}^+)=\phi(\text{x})$ and $\text{f}'(\text{a}^-)=-\phi(\text{x})$
View full question & answer→MCQ 1351 Mark
The value of $f(0),$ so that the function $\text{f(x)}=\frac{(27-2\text{x})^\frac{1}{3}-3}{9-3(243+5\text{x})^\frac{1}{5}}$ is continuous, is given by :
- A
$\frac{2}{3}$
- B
$6$
- ✓
$2$
- D
$4$
AnswerFor $f(x)$ to be continuous at $x = 0,$ we must have
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\text{f}(0)$
$\Rightarrow\text{f}(0)=\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\frac{(27-2\text{x})^\frac{1}{3}-3}{9-3(243+5\text{x})^\frac{1}{5}}$
$\Rightarrow\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{3\Big(243^{\frac{1}{5}}-(243+5\text{x})^\frac{1}{5}\Big)}$
$=\frac{1}{3}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{\text{x}}}{\frac{\bigg(243^\frac{1}{5}-(243+5\text{x})^\frac{1}{5}\bigg)}{\text{x}}}$
$=\frac{-1}{3}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{\text{x}}}{\frac{\bigg((243+5\text{x})^{\frac{1}{5}}-243\frac{1}{5}\bigg)}{\text{x}}}$
$=\frac{2}{15}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{-2\text{x}}}{\frac{\bigg((243+5\text{x})^\frac{1}{5}-243^\frac{1}{5}\bigg)}{5\text{x}}}$
$=\frac{2}{15}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(27-2\text{x})^\frac{1}{3}-27^\frac{1}{3}}{27-2\text{x}-27}}{\frac{\bigg((243+5\text{x})^\frac{1}{5}{-243^{\frac{1}{5}}\bigg)}}{243+5\text{x}-243}}$
$=\frac{2}{15}\times\frac{\frac{1}{3}\times27^{\frac{-2}{3}}}{\frac{1}{5}\times243^\frac{-4}{5}}$
$=\frac{2}{15}\times\frac{\frac{1}{3}\times\frac{1}{27^{\frac{-2}{3}}}}{\frac{1}{5}\times\frac{1}{243^\frac{-4}{5}}}$
$=2$
View full question & answer→MCQ 1361 Mark
Rolle's theorem is applicable in case of $\phi(\text{x})=\text{a}^{\sin\text{x}},\text{a}>\text{a}$ in :
AnswerCorrect option: B. Any interval $[0,\pi]$
$\phi(\text{x})$ is continuous and differentiable function then using statement of Roll's theorem $f(a) = f(b).$
Hence, here $\sin0=0$ also $\sin\pi=0.$
View full question & answer→MCQ 1371 Mark
If $\text{y}=\sin(\text{m}\sin^{-1}\text{x}),$ then $(1-\text{x}^2)\text{y}_2-\text{xy}_1$ is equal to :
AnswerCorrect option: C. $-m^2 y$
Here,$\text{y}=\sin(\text{m}\sin^{-1}\text{x}),$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{mx}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)^\frac{3}{2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xm}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)\times\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xy}_1}{(1-\text{x}^2)}$
$\Rightarrow(1-\text{x}^2)\text{y}_2=-\text{ym}^2+\text{xy}_1$
$\Rightarrow(1-\text{x}^2)\text{y}_2-\text{xy}_1=-\text{m}^2\text{y}$
View full question & answer→MCQ 1381 Mark
If $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$ Then $f(x)$ is continuous at $\text{x}=\frac{\pi}{2},$ if :
- A
$\text{a}=\frac{1}{3},\text{ b}=2$
- ✓
$\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
- C
$\text{a}=\frac{2}{3},\text{ b}=\frac{8}{3}$
- D
AnswerCorrect option: B. $\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
Given, $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x} < \frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x } > \frac{\pi}{2}\end{cases}$
We have
$\Big(\text{LHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{1-\sin^2\big(\frac{\pi}{2}-\text{h}\big)}{3\cos^2\big(\frac{\pi}{2}-\text{h}\big)}\Bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{1-\cos^2\text{h}}{3\sin^2\text{h}}\Big)$
$=\frac{1}{3}\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin^2\text{h}}{\sin^2\text{h}}\Big)$
$=\frac{1}{3}$
$\Big(\text{RHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\text{b}\big[1-\sin\big(\frac{\pi}{2}+\text{h})\big]}{\big[\text{x}-2\big(\frac{\pi}{2}+\text{h}\big)\big]^2}\Bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{b}(1-\cos\text{h})}{[-2\text{h}]^2}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^{2}\frac{\text{h}}{2}}{4\text{h}^2}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^2\frac{\text{h}}{2}}{16\frac{\text{h}^2}{4}}\bigg)$
$=\frac{\text{b}}{8}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2$
$=\frac{\text{b}}{8}\times1$
$=\frac{\text{b}}{8}$
Also, $\text{f}\big(\frac{\pi}{2}\big)=\text{a}$
If $f(x)$ is continuous at $\text{x}=\frac{\pi}{2},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\text{f}\big(\frac{\pi}{2}\big)$
$\Rightarrow\frac{1}{3}=\frac{\text{b}}{8}=\text{a}$
$\Rightarrow\text{a}=\frac{1}{3}$ and $\text{b}=\frac{8}{3}$
View full question & answer→MCQ 1391 Mark
If $\text{y}=\text{ax}^{\text{n+1}}+\text{bx}^{-\text{n}}$ Then $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2} =$
- A
$n(n - 1) y$
- ✓
$n(n + 1) y$
- C
$ny$
- D
$n^2 y$
AnswerCorrect option: B. $n(n + 1) y$
Here
$\text{y}=\text{ax}^{\text{n}+1}+\text{bx}^{\text{-n}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^\text{n}-\text{bn}\text{x}^{-\text{n}-1}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}$
$\therefore\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{x}^2\{\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}\}$
$=\text{n}(\text{n}+1)(\text{ax}^{\text{n}+1}+\text{b x}^{-\text{n}})$
$=\text{n}(\text{n}+1)\text{y}$
View full question & answer→MCQ 1401 Mark
The points of discontinuity of the function $\text{f(x)}=\begin{cases}2\sqrt{\text{x}},&0\leq\text{x}\leq1\\4-2\text{x},&1<\text{x}<\frac{5}{2}\\2\text{x}-7,&\frac{5}{2}\leq\text{x}\leq4\end{cases}$ is $($are$) :$
- A
$\text{x}=1,\text{x}=\frac{5}{2}$
- ✓
$\text{x}=\frac{5}{2}$
- C
$\text{x}=1,\frac{5}{2},4$
- D
$\text{x}=0,4$
AnswerCorrect option: B. $\text{x}=\frac{5}{2}$
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}2\sqrt{\text{x}}=2$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}4-2\text{x}=2$
Function is continuous at $x = 1$
$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}4-2\text{x}=-1$
$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}2\text{x}-7=-2$
Function is discontinuous at $\text{x}=\frac{5}{2}$
View full question & answer→MCQ 1411 Mark
Let $\text{f(x)}=\begin{cases}\frac{1}{|\text{x}|} & \text{for |x|}\geq1\\\text{ax}^2+\text{b} & \text{for |x|}<1\end{cases}$ if $f(x)$ is continuous and differentiable at any point, then :
- A
$\text{a}=\frac{1}{2},\text{b}=-\frac{3}{2}$
- ✓
$\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$
- C
$\text{a}=1,\text{b}=-1$
- D
AnswerCorrect option: B. $\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$
Given function is continuous at $x = 1.$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{1}{\text{x}}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{ax}^2+\text{b}$
$\Rightarrow1=\text{a}+\text{b}\ \dots(1)$
Function is derivable at $x = 1.$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\text{f}(1+\text{h}-\text{f}(1))}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{f}(0+\text{h}-\text{f}(1))}{\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\frac{1}{1+\text{h}}+1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{a}(1+\text{h})^2-\text{a}}{\text{h}}$
$\Rightarrow-1=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{h}(2\text{a}+\text{h})}{\text{h}}$
$\Rightarrow2\text{a}=-1$
$\Rightarrow\text{a}=\frac{-1}{2}$
$\text{a}+\text{b}=1\ ($From $(1))$
$\frac{-1}{2}+\text{a}=1$
$\Rightarrow\text{b}=\frac{3}{2}$
View full question & answer→MCQ 1421 Mark
Choose the correct answers from the given four options : The value of $c$ in Rolle’s theorem for the function $f(x) = x^3 - 3x$ in the interval $\big[0,\sqrt{3}\big]$ is :
- ✓
$1$
- B
$-1$
- C
$\frac{3}{2}$
- D
$\frac{1}{3}$
AnswerConsider, $\text{f(x)}=\text{x}^3-3\text{x}$
$\Rightarrow\ \text{f(x)}=3\text{x}^2-3$
$\Rightarrow\ \text{f}'(\text{c})=3\text{c}^2-3$
Now, $f(c) = 0$
$\Rightarrow\ 3\text{c}^2-3=0$
$\Rightarrow\ \text{c}^2=\frac{3}{3}=1$
$\Rightarrow\ \text{c}=\pm1$ where $1\in\big(0,\sqrt{3}\big)$
View full question & answer→MCQ 1431 Mark
Choose the correct answers from the given four options : If $x=t^2, y=t^3$, then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is :
- A
$\frac{3}{2}$
- ✓
$\frac{3}{4\text{t}}$
- C
$\frac{3}{2\text{t}}$
- D
$\frac{3}{2\text{t}}$
AnswerCorrect option: B. $\frac{3}{4\text{t}}$
We are given, $x=t^2$ and $y=t^3$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=2\text{t}$ and $\frac{\text{dy}}{\text{dt}}=3\text{t}^2$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{3\text{t}^2}{2\text{t}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{3}{2}\text{t}$
$\Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{3}{2}\cdot\frac{\text{dt}}{\text{dx}}$
$=\frac{3}{2}\cdot\frac{1}{2\text{t}}$
$\begin{bmatrix}\because\frac{\text{dx}}{\text{dt}}=2\text{t}\\\Rightarrow\frac{\text{dt}}{\text{dx}}=\frac{1}{2\text{t}}\end{bmatrix}$
$=\frac{3}{4\text{t}}$
View full question & answer→MCQ 1441 Mark
If $\text{y}=\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}},$ then $\frac{\text{dy}}{\text{dx}}$ is equal to :
Answer$\text{y}=\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}}$
$\text{y}=\frac{1}{1+\frac{\text{x}^\text{a}}{\text{x}^\text{b}}+\frac{\text{x}^\text{c}}{\text{x}^\text{b}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{c}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{a}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}}}$
$\text{y}=\frac{\text{x}^\text{b}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$
$\text{y}=\frac{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$
$\text{y}=1$
$\frac{\text{dy}}{\text{dx}}=0$
View full question & answer→MCQ 1451 Mark
Choose the correct answers from the given four options : Let $\text{f(x)}=|\sin\text{x}|.$ Then :
- A
$f$ is everywhere differentiable.
- ✓
$f$ is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$
- C
$f$ is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$
- D
AnswerCorrect option: B. $f$ is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$
Let $\text{u(x)}=\sin\text{x}$ and $\text{v(x)}=|\text{x}|$
$\therefore\ \text{f(x)}=\text{vou(x)}=\text{v}[\text{u(x)}]$
Since, $u(x)$ and $v(x)$ both are continuous functions.
Hence, $f(x) = \text{vou(x)}$ is also a continuous function but $v(x)$ is not differentiable at $x = 0.$
So, $f(x)$ is not differentiable where $\sin\text{x}=0$
$\Rightarrow\ \text{x}=\text{n}\pi,\text{n}\in\text{Z}$
Hence, $f(x)$ is continuous everywhere but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$
View full question & answer→MCQ 1461 Mark
If $\text{y}=\tan^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big),$ then $\frac{\text{dy}}{\text{dx}}$ is equals to :
AnswerWe have, $\text{y}=\tan^{-1}\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2}\times$
$\bigg[\frac{(\cos\text{x}-\sin\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x}+\cos\text{x})-(\sin\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\cos\text{x}-\sin\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2} \times$
$\bigg[\frac{(\cos\text{x}-\sin\text{x})(\cos\text{x}-\sin\text{x})-(\sin\text{x}+\cos\text{x})(-\sin\text{x}-\cos\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2} \times$
$\bigg[\frac{(\cos\text{x}-\sin\text{x})(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\cos\text{x})(-\sin\text{x}+\cos\text{x})}{(\cos\text{x}-\sin\text{x})^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x}-\sin\text{x})^2}{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2}\times\frac{(\cos\text{x}-\sin\text{x})^2+(\sin\text{x}+\cos\text{x})^2}{(\cos\text{x}-\sin\text{x})^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→MCQ 1471 Mark
$\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}-\sqrt{1-\text{px}}}{\text{x}},&\text{if }0\leq\text{x}<0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{x}\leq1\end{cases}$ is continuous in the interval $[-1, 1],$ then $p$ is equal to :
- A
$-1$
- ✓
$-\frac{1}{2}$
- C
$\frac{1}{2}$
- D
$1$
AnswerCorrect option: B. $-\frac{1}{2}$
Given, $\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}-\sqrt{1-\text{px}}}{\text{x}},&\text{if }0\leq\text{x}<0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{x}\leq1\end{cases}$
If $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\text{f(0)}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}}{-\text{h}}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(1-\text{ph}-1-\text{ph})}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(-2\text{ph})}{-\text{h}\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{(2\text{p})}{\big(\sqrt{1-\text{ph}}-\sqrt{1+\text{ph}}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\Big(\frac{(2\text{p})}{(2)}\Big)=\Big(\frac{1}{-2}\Big)$
$\Rightarrow\text{p}=\frac{-1}{2}$
View full question & answer→MCQ 1481 Mark
If $\text{f(x)}=\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)}+....+\frac{\text{x}^2}{(1+\text{x}^2)}+....,$ then at $x = 0, f(x) :$
- A
- ✓
- C
Is continuous but not differentiable.
- D
Answer$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Big(\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\Big(1+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\bigg(\frac{1}{1-\frac{1}{1+\text{x}^2}}\bigg)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}(1+\text{x}^2)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}=1$
But, $f(0)=0$
$\text{f}(0)\neq\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$
Function is discontinuous.
View full question & answer→MCQ 1491 Mark
Choose the correct answers from the given four options : If $\text{f(x)}=\begin{cases}\text{mx}+1,&\text{if x}\leq\frac{\pi}{2}\\\sin\text{x}+\text{n},&\text{if x}>\frac{\pi}{2}\end{cases},$ is continuous at $\text{x}=\frac{\pi}{2},$ then :
- A
$\text{m}=1,\text{n}=0$
- B
$\text{m}=\frac{\text{n}\pi}{2}+1$
- ✓
$\text{n}=\frac{\text{m}\pi}{2}$
- D
$\text{m}=\text{n}=\frac{\pi}{2}$
AnswerCorrect option: C. $\text{n}=\frac{\text{m}\pi}{2}$
We have, $\text{f(x)}=\begin{cases}\text{mx}+1,&\text{if x}\leq\frac{\pi}{2}\\\sin\text{x}+\text{n},&\text{if x}>\frac{\pi}{2}\end{cases},$ is continuous at $\text{x}=\frac{\pi}{2}$
$\therefore\ \text{L.H.L}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}(\text{mx}+1)$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{m}\Big(\frac{\pi}{2}-\text{h}\Big)+1\bigg]=\frac{\text{m}\pi}{2}+1$
and $\text{R.H.L}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}(\sin\text{x}+\text{n})$ $\lim\limits_{\text{h}\rightarrow0}\bigg[\sin\Big(\frac{\pi}{2}+\text{h}\Big)+\text{n}\bigg]$
$\lim\limits_{\text{h}\rightarrow0}[\cos\text{h}+\text{n}]=1+\text{n}$
We must have $\text{L.H.L = R.H.L}$
$\Rightarrow\ \text{m}\frac{\pi}{2}+1=\text{n}+1$
$\therefore\ \text{n}=\text{m}\frac{\pi}{2}$
View full question & answer→MCQ 1501 Mark
If $\text{y}=\text{a}\sin\text{mx}+\text{b}\cos\text{mx},$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is equal to :
- ✓
$-m^2 y$
- B
$m^2 y$
- C
$-my$
- D
$my$
AnswerCorrect option: A. $-m^2 y$
$\text{y}=\text{a}\sin\text{mx}+\text{b}\cos\text{mx}$
$\frac{\text{dy}}{\text{dx}}=\text{am}\cos\text{mx}-\text{bm}\sin\text{mx}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{am}^2\sin\text{mx}-\text{bm}^2\cos\text{mx}=-\text{m}^2\text{y}$
View full question & answer→