MCQ 11 Mark
If $x$ is very large, then $\frac{2\text{x}}{1+\text{x}}\text{is:}$
- A
Close to $0$
- B
- C
Lie between $2$ and $3$
- ✓
Close to $2$
AnswerCorrect option: D. Close to $2$
View full question & answer→MCQ 21 Mark
If $ \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}2\mathrm{x}+\mathrm{b}(\mathrm{x}<\alpha)\\\mathrm{x}+\mathrm{d}(\mathrm{\text{x}}\geq\alpha)\end{array}\right.$ is such that $ \lim_\limits{\text{x} \rightarrow \text{a}}\text{f}(\text{x}=\text{L}),$ then $L.$
- ✓
$2d - b$
- B
$b - db$
- C
$d + bd$
- D
$b- 2d$
AnswerCorrect option: A. $2d - b$
$\lim _{h \rightarrow 0^{-}} f(x)=2(\alpha-h)+b=2 \alpha+b=L \ldots \ldots \ldots \ldots(1) .$
$\lim _{\mathrm{h} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x})=(\alpha+\mathrm{h})+\mathrm{d}=\mathrm{L} \alpha=\mathrm{L}-\mathrm{d} \ldots \ldots \ldots \ldots \ldots (2).$
Substituting value of euation $(2)$ in $(1),$ we get
$2(\mathrm{~L}-\mathrm{d})+\mathrm{b}=\mathrm{L}$
$ \mathrm{~L}=2 \mathrm{~d}-\mathrm{b}$
Hence, option $A$ is correct.
View full question & answer→MCQ 31 Mark
The fourth term in the expansion $(x - 2y)^{12}$ is:
- A
$ -1670 x^9 \times y^3 $
- B
$ -7160 x^9 \times y^3 $
- ✓
$ -1760 x^9 \times y^3 $
- D
$ -1607 x^9 \times y^3 $
AnswerCorrect option: C. $ -1760 x^9 \times y^3 $
$4^{th}$ term in $(x-2 y)^{12}=T_4$
$ =T_3+1 $
$ ={ }^{12} C_3(x)^{12-3} \times(-2 y)^3$
$={ }^{12} C_3 x^9 \times\left(-8 y^3\right)$
$ = {\frac{(12 \times 11 \times 10)}{(3 \times 2 \times 1)} \times \text{x}^9 \times (-8\text{y}^³)}$
$= - (2 \times 11 \times 10 \times 8) \times x^9\times y^3$
$= -1760 x^9\times y^3$
View full question & answer→MCQ 41 Mark
Is Rolle’s theorem valid for $f(x) = x^2 - 3x + 4$ in the interval $[1, 2]\ ?$
View full question & answer→MCQ 51 Mark
If $\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100},$then $f'(1)$ equals
Answer$\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100}$
Differentiate both the sides with respect to $x,$ we get
$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100})$
$=\frac{\text{d}}{\text{dx}}(1)-\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{\text{d}}{\text{dx}}(\text{x}^3)+\dots-\frac{\text{d}}{\text{dx}}(\text{x}^{99})+\frac{\text{d}}{\text{dx}}(\text{x}^{100})$
$=0-1+\text{2x}-\text{3x}^2+\dots-\text{99x}^{98}+100\text{x}^{99}$
Putting $x = 1,$ we get
$\text{f}'(1)=-1+2-3+\dots-99+100$
$=(-1+2)+(-3+4)+(-5+6)+\dots+(-99+100)$
$=1+1+1+\dots+1(50$terms$)$
$=50$
View full question & answer→MCQ 61 Mark
$x\rightarrow 0 \lim \Bigg(\dfrac{(1+x)^{2}}{e^{x}}\Bigg)^\dfrac{4}{\sin x}(ex(1+x)^2)\ \sin x^4$ is:
- A
$ \text{e}^2$
- ✓
$ \text{e}^4$
- C
$ \text{e}^8$
- D
AnswerCorrect option: B. $ \text{e}^4$
$ \lim\limits_{\text{x} \to 0}\left(\frac{(1+\text{x})^2}{\text{e}^{\text{x}}}\right)^{\frac{4}{\sin \text{x}}}=\lim\limits_{\text{x} \to 0}\frac{\left(\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\frac{\text{x}}{\sin \text{x}}}\right)^8}{\text{e}^{\frac{4\text{x}}{\sin \text{x}}}}$
We have
$ \lim\limits_{\text{x} \to 0}\frac{\sin \text{x}}{\text{x}}=1$
and $ \lim\limits_{\text{x} \to 0}(1+\text{x})^{\frac{1}{\text{x}}}=$
both the limits of the numerator and denominator exists,and the limit of the numerator is non$-$vanishing,
$=\lim_\limits{\text{x} \rightarrow 0}\frac{\left(\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}}{\sin \text{x}}}\right)^8}{\lim_\limits{\text{x} \rightarrow 0}\text{e}^{\frac{4\text{x}}{\sin \text{x}}}}$
[Using division property of limits]
$=\frac{\left(\lim\limits_{\text{x}\to 0}\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\left(\lim\limits_{\text{x}\to 0}\frac{\text{x}}{\sin \text{x}}\right)}\right)^8}{\text{e}^{\left(\lim\limits_{\text{x}\to 0}\dfrac{4\text{x}}{\sin \text{x}}\right)}} [$Using limit property$]$
$= \text{e}^4$
View full question & answer→MCQ 71 Mark
What is the value of $\lim_\limits{\text{x} \rightarrow 0}\frac{\text{e}^\text{X}(\sin^2\text{x})}{\text{x}^3}?$
Answer$\lim_\limits{\text{x} \rightarrow 0}\frac{\sin^2}{\text{x}^2}\times\lim_\limits{\text{x} \rightarrow 0}\frac{\text{e}^x}{\text{x}}$
We apply L’Hospital’s rule and differentiate numerator and denominator.
$1\times\lim_\limits{\text{x} \rightarrow 0}\frac{\text{e}^x}{\text{1}}$
$= 1$
View full question & answer→MCQ 81 Mark
In the binomial expansion of $(a + b)^n$, the coefficient of fourth and thirteenth terms are equal to each other, then the value of $n$ is:
AnswerGiven, in the binomial expansion of $(a + b)^n$,
the coefficient of fourth and thirteenth terms are equal to each other
$\Rightarrow {^nC_3}={^nC_{12}}$
This is possible when $n = 15$
Because ${ }^{15} C_{13}={ }^{15} C_{12}$
View full question & answer→MCQ 91 Mark
Let $\text{f}(\text{x})=\text{x}-[\text{x}],\text{x}\in\text{R},$ then $\text{f}'\Big(\frac{1}{2}\Big)$ is:
- A
$\frac{3}{2}$
- ✓
$1$
- C
$0$
- D
$-1$
AnswerGiven: $\text{f}(\text{x})=\text{x}-[\text{x}],\text{x}\in\text{R}$
Now,
For $0\le\text{x}<1,[\text{x}]=0$
$\therefore\text{f}(\text{x})=\text{x}-0=\text{x},\forall\text{x}\in[0,1)$
Differentiate with respect to $x,$ we get
$\text{f}'(\text{x})=1,\forall\text{x}\in[0,1)$
$\therefore\text{f}'\Big(\frac{1}{2}\Big)=1$
View full question & answer→MCQ 101 Mark
Choose the correct answer. If $f(x) = x{100}+ x^{99}....... + x + 1$, then $f'(1)$ is equal to:
- ✓
$5050$
- B
$5049$
- C
$5051$
- D
$50051$
AnswerCorrect option: A. $5050$
Given $f(x) = x^{100}+ x^{99}+ .... + x + 1$
$f'(x) = 100.x^{100}+ 99.x^{98}+ .... + 1$
S0, $f'(1) = 100 + 99 + 98 + ..... + 1$
$=\frac{100}{2}[2\times100+(100-1)(-1)]$
$=50[200-99]$
$=50\times101$
$=5050$
View full question & answer→MCQ 111 Mark
If $f(x) = 2\sin x - 3x^4 + 8$, then $f ¢(x)$ is:
- A
$2\sin x - 12x^3$
- ✓
$2\cos x - 12x^3$
- C
$2\cos x + 12x^3$
- D
$2\sin x + 12x^3$
AnswerCorrect option: B. $2\cos x - 12x^3$
View full question & answer→MCQ 121 Mark
If $ \text{f(x)} = \text{x} \sin\text{x},$ then $ \text{f}\Big(\frac{Π}{2}\Big)$ is equal to:
- A
$0$
- ✓
$1$
- C
$1$
- D
$\frac{1}{2}$
View full question & answer→MCQ 131 Mark
What is the value of $\lim_{\text{y} \rightarrow 0}(32\text{x}^2 \text{cosec} ^2 4\text{x}) ?$
Answerhe limit can be written as,
$\lim_{\text{x} \rightarrow 0}\frac{32\text{x}^2}{\sin^2 4\text{x}}$
$2\times\lim_\limits{\text{x} \rightarrow 0}\frac{4\text{x}}{\sin 4\text{x}}\times\ \lim_\limits{\text{x} \rightarrow 0}\frac{4\text{x}}{\sin 4\text{x}}$
$= 2 \times 1 \times 1$
$= 2$
View full question & answer→MCQ 141 Mark
lf $\lim_\limits{\text{x}\rightarrow \text{a}}\text{f}(\text{x})=\text{L},$ then for eac $ \epsilon > 0$, there exists $ δ>0$ so that:
- A
$0<∣\text{x−a}∣<δ\Rightarrow ∣\text{f(x)}−L∣\geq \epsilon$
- ✓
$0<∣\text{x−a}∣<δ\Rightarrow ∣\text{f(x)}−\text{L}∣<\epsilon$
- C
$ \text{a} < \text{x} < \text{a}+\delta\Rightarrow \text{f(x)} −\text{L}<\epsilon$
- D
$ a-\delta < x < a\Rightarrow |f(x)- L|<\epsilon$
AnswerCorrect option: B. $0<∣\text{x−a}∣<δ\Rightarrow ∣\text{f(x)}−\text{L}∣<\epsilon$
It is fundamental concept that,
for limit of a function $f(x)$ to exist at any point aa there exists a real number $δ>0,$
such that $0 < |x - a|< δ,$
for which $|f(x) - L| < \epsilon$ ,
where $\epsilon > 0$
View full question & answer→MCQ 151 Mark
If $\text{f(x)}=\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}},$ then $\text{f}'\text{(a)}$ is:
- A
$1$
- B
$0$
- C
$\frac{1}{2}$
- ✓
$\text{dose not exist}$
AnswerCorrect option: D. $\text{dose not exist}$
Given: $\text{f(x)}=\frac{\text{x}^\text{n}-\text{a}^\text{n}}{\text{x}-\text{a}}$
Now, $f(x)$ is not difined at $x = a.$
Therefore, $f(x)$ is not differentiable at $x = a.$
So, $f'(a)$ dose not exist.
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 161 Mark
Choose the correct answer. $ \lim\limits_{\text{x} \rightarrow 0}\frac{\tan2\text{x}-\text{x}}{3\text{x}-\sin\text{x}}$ is equal to:
- A
$2$
- ✓
$\frac{1}{2}$
- C
$-\frac{1}{2}$
- D
$\frac{1}{4}$
AnswerCorrect option: B. $\frac{1}{2}$
Given $ \lim\limits_{\text{x} \rightarrow 0}\frac{\tan2\text{x}-\text{x}}{3\text{x}-\sin\text{x}} =\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}\big[\frac{\tan2\text{x}}{\text{x}}-1\big]}{\text{x}\big[3-\frac{\sin\text{x}}{\text{x}}\big]}$
$ =\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\tan2\text{x}}{2\text{x}}\times2-1}{3-\frac{\sin\text{x}}{\text{x}}}$
$=\frac{1.2-1}{3-1}$
$=\frac{2-1}{2}$
$=\frac{1}{2}$
View full question & answer→MCQ 171 Mark
The value of $ \displaystyle \lim _{ \text{x}\rightarrow \text{a} }{ \frac { \sqrt { \text{x-b} } -\sqrt {\text{ a-b} } }{ { \text{x} }^{ 2 }-{ \text{a} }^{ 2 } } } \text{(a > b)}:$
AnswerCorrect option: D. $ \dfrac {1}{\text{4a}\sqrt {\text{a}-\text{b}}}$
$= \displaystyle \lim _{ \text{x}\rightarrow \text{a} }{ \frac { \sqrt { \text{x-b} } -\sqrt {\text{ a-b} } }{ { \text{x} }^{ 2 }-{ \text{a} }^{ 2 } } } \text{(a > b)}:$
This is the $ \frac{0}{0}$ form.
Apply $L-$hospital rule
$= \lim_\limits{\text{x}\to \text{a}}\dfrac{\dfrac{1}{2\sqrt {\text{x}-b}}-0}{2\text{x}-0}$
$\lim_\limits{\text{x}\to \text{a}}\frac{1}{4\text{x}\sqrt {\text{x}-\text{b}}}$
$=\dfrac{1}{4\text{a}\sqrt{\text{a}-\text{b}}}$
Hence, this is the answer.
View full question & answer→MCQ 181 Mark
$ \lim_\limits{\text{x} \rightarrow 0}\frac{3\sin(2\text{x}^2)}{\text{x}^2}= A$ then the value of $A$ is:
Answer$ \lim_\limits{\text{x} \rightarrow 0}\frac{3\sin(2\text{x}^2)}{\text{x}^2}=$
$ \lim_\limits{\text{x} \rightarrow 0}\frac{2\ *\ 3\sin(2\text{x}^2)}{\text{2x}^2}=6$
View full question & answer→MCQ 191 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}$ is equal to:
- ✓
$\text{n}$
- B
$1 $
- C
$-\text{n}$
- D
$0$
AnswerCorrect option: A. $\text{n}$
Given $\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{(1+\text{x})-(1)}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{1+\text{x}-(2)}=\text{n}(1)^{\text{n}-1}$
$=\text{n}$
View full question & answer→MCQ 201 Mark
What is the number of critical points for $ \text{f}(\text{x}) = \max(\sin\text{x}, \cos\text{x})$ for $x$ belonging to $ (0, 2π)?$
AnswerWe know that in the range of $ (0, 2\pi )$ the graph of $\sin x$ and $\cos x$ intersects each other in three points.
And we know that these points of intersection are only the critical points
Thus, there are $3$ critical points.
View full question & answer→MCQ 211 Mark
lf $f′(x) = g(x)$ and $g′(x) = −f(x)$ for all $x$ and $f(2) = 4 = g(2),$ then $f^2(24)+g^2(24)$ is:
View full question & answer→MCQ 221 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\text{cosec}-\cot\text{x}}{\text{x}}$ is equal to:
- A
$-\frac{1}{2}$
- B
$1$
- ✓
$\frac{1}{2}$
- D
$-1$
AnswerCorrect option: C. $\frac{1}{2}$
Given $\lim\limits_{\text{x} \rightarrow 0}\frac{\text{cosec}\text{x}-\cot\text{x}}{\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}{\text{x}}$
$ =\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos\text{x}}{\text{x}\sin\text{x}}$
$=\frac{2\sin^{2}\frac{\text{x}}{2}}{\text{x}\cdot\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow1}\frac{\sin\frac{\text{x}}{2}}{\text{x}\cos\frac{\text{x}}{2}} $
$=\frac{\tan\frac{\text{x}}{2}}{\text{x}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\tan\frac{\text{x}}{2}}{2\times\frac{\text{x}}{2}}$
$=\frac{1}{2}\times1$
$=\frac{1}{2}$
View full question & answer→MCQ 231 Mark
Evaluate: $\lim_{\text{n} \rightarrow \infty} \dfrac{\text{n}!}{(\text{n}+1)!-\text{n}!}$
AnswerWe have,$\lim\limits_{\text{n} \rightarrow \infty} \dfrac{\text{n}!}{(\text{n}+1)!-\text{n}!}$
$=\lim\limits_{\text{n} \rightarrow \infty} \dfrac{\text{n}!}{(\text{n}+1)\text{n}!-\text{n}!}$
$ =\lim\limits_{\text{n}\rightarrow \infty} \dfrac{1}{\text{n}+1-1}$
$ =\lim\limits_{\text{n}\rightarrow \infty}\frac{1}{\text{n}}=0$
View full question & answer→MCQ 241 Mark
The greatest coefficient in the expansion of $(1 + x)^{10}$ is:
- A
$10!(5!)$
- ✓
$ \frac{10}{(5!)^2}$
- C
$10!(5! \times 4!)^2$
- D
AnswerCorrect option: B. $ \frac{10}{(5!)^2}$
The coefficient of $x^r$ in the expansion of $(1+x)^{10}$ is ${ }^{10} \mathrm{C}_{\mathrm{r}}$ and ${ }^{10} \mathrm{C}_{\mathrm{r}}$ is maximum for
$r=10=5$
Hence, the greatest coefficient $={ }^{10} \mathrm{C}_5$
$=\frac{10}{(5!)^2}$
View full question & answer→MCQ 251 Mark
What is the value of the limit $\text{f}(\text{x}) = {x}^2+2\text{x}\sqrt{\text{x}^2−4\text{x}}$ if $x$ approaches infinity?
- ✓
$0$
- B
$2$
- C
$\frac{1}{2}$
- D
$4$
AnswerThis is of the form $ \frac{\infty }{\infty }$,
therefore we use $L$ ’Hospital’ s rule and,
differentiate the numerator and denominator.
View full question & answer→MCQ 261 Mark
$\lim_\limits{\text{x} \rightarrow -1\frac{\text{x}^2}{\text{x}-1}}=...........$
- A
$0$
- B
$1$
- ✓
$ \frac{-1}{2}$
- D
$-2$
AnswerCorrect option: C. $ \frac{-1}{2}$
$=\lim_\limits{\text{x} \rightarrow -1\frac{\text{x}^2}{\text{x}-1}}$
Substituting $x = -1$
we get, $ \displaystyle \lim _{ \text{x}\rightarrow -1 } \frac{\text{x}^2}{\text{x}-1}= \frac { { (-1) }^2 1 }{ -1-1 } = -\frac {1}{2}$
View full question & answer→MCQ 271 Mark
In the expansion of $(a + b)^n$, if $n$ is odd then the number of middle term is:
- A
$0$
- B
$1$
- ✓
$2$
- D
More than $2$
AnswerIn the expansion of $(a + b)^n$,
if $n$ is odd then there are two middle terms which are
${(n + 1)/2}^{th}$ term and ${(n+1)/2 + 1}^{th}$ term.
View full question & answer→MCQ 281 Mark
What is the number of critical points for $f(x) = \text{max}(\sin x, \cos x)$ for $x$ belonging to $(0, 2\pi )?$
AnswerWe know that in the range of $(0, 2\pi )$ the graph of
$\sin x$ and $\cos x$ intersects each other in three points.
And we know that these points of intersection are only the critical points
Thus, there are $3$ critical points.
View full question & answer→MCQ 291 Mark
Derivative of the function $f(x) = 7x^{-3}$ is:
- A
$ 21 x^{-4}$
- ✓
$-21 x^{-4} $
- C
$21 x^4 $
- D
$ -21 x^4 $
AnswerCorrect option: B. $-21 x^{-4} $
View full question & answer→MCQ 301 Mark
Choose the correct answer. If $\text{y}=\frac{1+\frac{1}{\text{x}^{2}}}{1-\frac{1}{\text{x}^{2}}}$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
- ✓
$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
- B
$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
- C
$\frac{1-\text{x}^{2}}{4\text{x}}$
- D
$\frac{4\text{x}}{\text{x}^{2}-1}$
AnswerCorrect option: A. $\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
Given, $\text{y}=\frac{1+\frac{1}{\text{x}^{2}}}{1-\frac{1}{\text{x}^{2}}}$
$\Rightarrow\text{y}=\frac{\text{x}^{2}+1}{\text{x}^{2}-1}$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^{2}-1).2\text{x}-(\text{x}^{2}+1).2\text{x}}{(\text{x}^{2}-1)^{2}}$
$=\frac{2\text{x}(\text{x}^{2}-1-\text{x}^{2}-1)}{(\text{x}^{2}-1)^{2}}$
$=\frac{2\text{x}(-2)}{(\text{x}^{2}-1)^{2}}$
$=\frac{-4\text{x}}{(\text{x}^{2}-1)^2{}}$
View full question & answer→MCQ 311 Mark
Let $f:(a, b) \rightarrow R$ be a differentiable function. Which of the following statements is true:
- ✓
$\displaystyle \lim_{\text{x} \rightarrow \text{a}}\text{f(x)}=\infty \Longrightarrow \lim_{\text{x} \rightarrow \text{a}} |\text{f(x)}|=\infty $
- B
$\displaystyle \lim_{\text{x} \rightarrow \text{a}}\text{f(y)}=\infty \Longrightarrow \lim_{\text{x} \rightarrow \text{a}} |\text{f(y)}|=\infty $
- C
$\displaystyle \lim_{\text{x} \rightarrow \text{a}}\text{f(y)}=\infty \Longrightarrow \lim_{\text{x} \rightarrow \text{a}} |\text{f(y)}|=\infty \pi$
- D
$\displaystyle \lim_{\text{b} \rightarrow \text{a}}\text{f(y)}=\infty \Longrightarrow \lim_{\text{x} \rightarrow \text{a}} |\text{f(y)}|=\infty \pi$
AnswerCorrect option: A. $\displaystyle \lim_{\text{x} \rightarrow \text{a}}\text{f(x)}=\infty \Longrightarrow \lim_{\text{x} \rightarrow \text{a}} |\text{f(x)}|=\infty $
$f : (a, b) \rightarrow R$ is differentiable.
If $ \lim _\limits{ \text{x}\rightarrow \text{a} }{ \text{f}(\text{x} )}$
View full question & answer→MCQ 321 Mark
What is the value of $\lim_{\text{x} \rightarrow 3}\frac{\text{x}^2-9}{\text{x}-3}$
AnswerWhen $x$ tends to $3,$ both the numerator and,
the denominator become $0$ and it becomes of the form, $0$.
Therefore, we use $L’$Hospital’s rule,
which states the we differentiate the numerator and the denominator,
until a definite answer is reached.
On differentiating once we get,
$\lim_{\text{x} \rightarrow 3}\frac{2\text{x}}{1}$
Since, this not an indeterminate form now, we can substitute the value of $x.$
$= 2 \times 3$
$= 6$
View full question & answer→MCQ 331 Mark
if a differentiable function $f$ defined for $x > 0$ satisfies the relation $f(x^2) = x^3, x > 0,$ then what is the value of $f(4) ?$
View full question & answer→MCQ 341 Mark
Evaluate: $ \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\sin \text{x}+\cos \text{x}}{\sin \text{x}-\cos\text{x}}$
Answer$ \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\sin \text{x}+\cos \text{x}}{\sin \text{x}-\cos\text{x}}$
Substituting $x = 0,$ we get
$= \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\sin \text{x}+\cos \text{x}}{\sin \text{x}-\cos\text{x}}$
$= \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\sin \text{0}+\cos \text{0}}{\sin \text{0}-\cos\text{0}}$
$ = \displaystyle \lim_{\text{x}\rightarrow 0}\frac{\ \text{0}+\text{1}}{0-1}$
View full question & answer→MCQ 351 Mark
What is the value $ \lim_{\text{x} \rightarrow 4}\frac{\text{x}^2-2\text{x}-8}{\text{x}-4}$ :
AnswerThe denominator becomes $0,$ as $x$ approaches $4.$
$ \lim_{\text{x} \rightarrow 4}\frac{\text{x}^2-2\text{x}-8}{\text{x}-4}$
Here, if we factorize the numerator we get
$ \lim_{\text{x} \rightarrow 4}\frac{(\text{x}-4) (\text{x}+2)}{\text{x}-4}$
We can now cancel out $(x - 4)$ from both the numerator and denominator.
We get, $ \lim_{\text{x} \rightarrow 4}(\text{x}+2)=6$
View full question & answer→MCQ 361 Mark
The value of the limit $\lim _\limits{ \text{x}\rightarrow 1 }{ \frac { \sin { \left( { \text{e} }^{ \text{x}-1 }-1 \right) } }{ \log { \text{x} } } }$ is:
Answer$\lim _\limits{ \text{x}\rightarrow 1 }{ \frac { \sin { \left( { \text{e} }^{ \text{x}-1 }-1 \right) } }{ \log { \text{x} } } }$
$ =\lim _\limits{ \text{h}\rightarrow 0 }{ \frac { \sin { \left( { \text{e} }^{\text{ h }}-1 \right) } }{ \log { \left( 1+\text{h} \right) } } }$
$ =\displaystyle\lim _{\text{ h}\rightarrow 0 }{ \frac { \sin { \left( {\text{ e} }^{\text{ h} }-1 \right) } }{ \left( { \text{e} }^{ \text{h} }-1 \right) } } \times \frac { \left( {\text{ e} }^{ \text{h} }-1 \right) }{ \log { \left( 1+\text{h} \right) } }$
$ =1\times\lim _\limits{ \text{h}\rightarrow 0 }{ \frac { \left( \text{h}+\frac { {\text{ h} }^{ 2 } }{ 2! } +\dots \right) }{ \left(\text{ h}-\frac { {]\text{ h} }^{ 2 } }{ 2 } +\dots \infty \right) } }$
$= 1 \times 1$
$= 1$
View full question & answer→MCQ 371 Mark
The value of $n$ in the expansion of $(a + b)^n$ if the first three terms of the expansion are $729, 7290$ and $30375,$ respectively is:
AnswerGiven that the first three terms of the expansion are $729, 7290$ and $30375$ respectively.
Now $T_1={ }^n C_0 \times a^{n-0} \times b^0=729 $
$\Rightarrow a^n=729 \ldots \ldots \ldots \ldots .1 $
$T_2={ }^n C_1 \times a^{n-1} \times b^1=7290 $
$\Rightarrow n $
$a^{n-1} \times b=7290 \ldots \ldots .2 $
$T_3={ }^n C_2 \times a^{n-2} \times b^2=30375 $
$\Rightarrow n \frac{(n-1)}{2} $
$a^{n-2} \times b^2=30375 \ldots \ldots .3 $
Now equation $2,$ equation $1$
$n=\text{a}^{\text{n}-1}\times\text{b}=\frac{7290}{729}$
$\Rightarrow \frac{\text{n}\times \text{b}}{\text{n}} = 10 ....... 4$
Now equation $3$, equation $2$
$\Rightarrow \text{n}\frac{(\text{n}-1)}{2}$
$\text{an}-2 \times \frac{\text{b}^2}{\text{n}}$
$=\text{a}^{\text{n}-1}\times\text{b}=\frac{30375}{7290}$
$\Rightarrow \text{n}\frac{(\text{n}-1)}{\text{2a}}=\frac{30375\times2}{7290}$
$\Rightarrow \text{n}\frac{(\text{n}-1)}{\text{a}}=\frac{30375\times2}{7290}$
$\Rightarrow \text{n}\frac{(\text{n}-1)}{\text{a}}-\frac{\text{b}}{\text{a}} = \frac{60750}{7290}$
$\Rightarrow 10 - \frac{\text{b}}{\text{a}} = \frac{60750}{729}$
$(60750$ and $7290$ is divided by $10)$
$\Rightarrow 10 - \frac{\text{b}}{\text{a}} = \frac{25}{3} (6075$ and $729$ is divided by $243)$
$\Rightarrow 10 -\frac{25}{3} = \frac{\text{b}}{\text{a}}$
$\Rightarrow \frac{(30-25)}{3} = \frac{\text{b}}{\text{a}}$
$\Rightarrow \frac{5}{3} =\frac{\text{b}}{\text{a}} $
$\Rightarrow \frac{\text{b}}{\text{a}} = \frac{5}{3}\ \dots\dots (5)$
Put this value in equation $4,$ we get
$\text{n} \times \frac{5}{3} = 10$
$\Rightarrow 5\text{n} = 30$
$\Rightarrow \text{n} = \frac{30}{5}$
$\Rightarrow \text{n} = 6$
So, the value of $n$ is $6.$
View full question & answer→MCQ 381 Mark
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{ae}^\text{x}+\text{b}\cos\text{x}+\text{c.e}^\text{x}}{\sin^2\text{x}}=4$ then $\text{ b:}$
View full question & answer→MCQ 391 Mark
$\lim_\limits{\text{x}\rightarrow 4} \frac{|x-4 |}{x - 4}$ is equal to:
View full question & answer→MCQ 401 Mark
If $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is:
AnswerCorrect option: A. $\cos9$
$\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\cos\text{x})\frac{\text{d}}{\text{dx}}\sin(\text{x}+9)-\sin(\text{x}+9)\frac{\text{d}}{\text{dx}}(\cos\text{x})}{\cos^2\text{x}} ($Quotient rule$)$
$=\frac{(\cos\text{x})(\cos(\text{x}+9))-(\sin(\text{x}+9))(-\sin\text{x})}{\cos^2\text{x}}$
$=\frac{(\cos\text{x})(\cos(\text{x}+9))+(\sin(\text{x}+9))(\sin\text{x})}{\cos^2\text{x}}$
$=\frac{\cos(\text{x}+9-\text{x})}{\cos^2\text{x}}$
$=\frac{\cos9}{\cos^2\text{x}}$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is $\cos9$
View full question & answer→MCQ 411 Mark
Choose the correct answer. If $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is equal to:
AnswerCorrect option: A. $\cos9$
Given $\text{y}=\frac{\sin(\text{x}+9)}{\cos\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}.\cos(\text{x}+9)-\sin(\text{x}+9)(-\sin\text{x})}{\cos^{2}\text{x}}$
$=\frac{\cos\text{x}\cos(\text{x}+9)+\sin\text{x}\sin(\text{x}+9)}{\cos^{2}\text{x}}$
$=\frac{\cos(\text{x}+9-\text{x})}{\cos^{2}\text{x}}=\frac{\cos9}{\cos^{2}\text{x}}$
$=\frac{\cos9}{(1)^{2}}$
$=\cos9$
View full question & answer→MCQ 421 Mark
What is the value of $(x + y)^2$ y if $x = et \sin t$ and $y = et \cos t\ ?$
- A
$12(y + y)$
- B
$2(y - y)$
- C
$2(xy + y)$
- ✓
$2(xy - y)$
AnswerCorrect option: D. $2(xy - y)$
Since, $x = et \sin t$ and $y = et \cos t$ Therefore,
$\frac{\text{dx}}{\text{dt}} = e^t \sin t+ e^t \cos t = y + x$ And,
$\frac{\text{dx}}{\text{dt}} = et \cos t - et \sin t = y - x $So,
$=\text{y}= \frac{\text{dx}}{\text{dt}} $
$=\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}$
$= \frac{(\text{y} - \text{x})}{(\text{y} + \text{x})}$
Thus, $\text{y}= \frac{[(\text{x} + \text{y})(\text{y} - 1) - (\text{y} - \text{x})(\text{y} + 1)]}{(\text{x} + \text{y})^2}$
Or, $(x + y)^2 y = (x + y - y + x)y - x - y + x$
$= 2xy - 2y$
$= 2(xy - y) 10$.
If, $ y = (\sin-1x)^2$ , then what is the value $0.$
View full question & answer→MCQ 431 Mark
Is Rolle’s theorem valid for $f(x) = x^2- 3x + 4$ in the interval $[1, 2]?$
AnswerObviously, $f(x)$ is continuous at $[1, 2]$
And, $f(x)$ differentiable at $[1, 2]$
Also, $f(1) = f(2) = 2$
Now, $f(x) = 0$
$\Rightarrow 2x - 3 = 0$
$\Rightarrow \text{x} = \frac{3}{2}$
Thus, $x$ belongs to $[1, 2]$
Hence, it is verified.
View full question & answer→MCQ 441 Mark
If $ {\text{z}_\text{r}} = \cos \frac{{\text{r}\alpha }}{{{\text{n}^2}}} + \text{i}\sin \frac{{\text{r}\alpha }}{{{\text{n}^2}}}$ where $r = 1, 2, 3, ....n$ then $ \mathop {\lim }\limits_{\text{n} \to \infty } \left( {{\text{z}_1}.{\text{z}_2}.....{\text{z}_\text{n}}} \right)$ is equal to:
- A
$ \cos \frac{\alpha }{2} $
- B
$ \sin \frac{\alpha }{2} $
- C
${{\text{e}^{\text{i}\alpha }}}$
- ✓
$ \sqrt {{\text{e}^{\text{i}\alpha }}}$
AnswerCorrect option: D. $ \sqrt {{\text{e}^{\text{i}\alpha }}}$
View full question & answer→MCQ 451 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow 0}\frac{|\sin\text{x}|}{\text{x}}$ is equal to:
AnswerGiven $\lim\limits_{\text{x} \rightarrow 0}\frac{|\sin\text{x}|}{\text{x}}$
$\text{L}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{-\sin\text{x}}{\text{x}}=-1$
$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}=1$
$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$
So, the limit does not exist.
View full question & answer→MCQ 461 Mark
if n is a positive ineger then $2^{3n}n - 7n - 1$ is divisible by:
AnswerGiven, $2^{3 n}-7 n-1=2^{3 \times n}-7 n-1 $
$ =8^n-7 n-1 $
$ =(1+7)^n-7 n-1 $
$ =\left\{{ }^n C_0+{ }^n C_1 {~7}+{ }^n C_2 {~7}^2+\ldots \ldots \ldots+{ }^n C_n {~7}^n\right\}-7 n-1 $
$ =\left\{1+7 n+{ }^n C_2 {~7}^2+\ldots \ldots \ldots+{ }^n C_n {~7}^n\right\}-7 n-1 $
$ ={ }^n C_2 {~7}^2+\ldots \ldots \ldots+{ }^n C_n {~7}^n $
$ =49\left({ }^n C_2+\ldots \ldots \ldots+{ }^n C_n {~7}^{n-2}\right)$
which is divisible by $49$
So, $2^{3n} - 7n - 1$ is divisible by $49$
View full question & answer→MCQ 471 Mark
If $f(x) = |4x - x^2- 3|$ when $x\ €\ [0, 4],$ then, which of the following is correct?
- A
$x = 1$ is the global maximum
- B
$x = 2$ is the global maximum
- ✓
$x = 3$ is the global maximum
- D
$x = 0$ is the global maximum
AnswerCorrect option: C. $x = 3$ is the global maximum
Clearly, $x = 1, 3$ are the points of global minimum $($as the values being equal$).$
And, $x = 0, 4$ are the points of global maximum $($as the values being equal$).$
And, $x = 2$, is the point of local maximum $($as the values being equal$).$
Thus, $x = 3$ is the global maximum.
View full question & answer→MCQ 481 Mark
$\lim\limits_{\text{x}\rightarrow1}(1+\cos\pi)\cot^2\pi\text{x}:$
- A
$1$
- B
$-1$
- ✓
$\frac{1}{2}$
- D
$0$
AnswerCorrect option: C. $\frac{1}{2}$
View full question & answer→MCQ 491 Mark
$(1.1)^{10000}$ is $.......... 1000:$
AnswerGiven, $ (1.1)^{10000 }= (1 + 0.1)^{10000}$
$ 10000\text{C}_0 + 10000\text{C}_1 \times (0.1) + 10000\text{C}_2 \times (0.1)^2 +$ other $+ve$ terms
$= 1 + 10000 \times (0.1) +$ other $+ ve$ terms
$= 1 + 1000 +$ other $+ ve$ terms
$> 1000$
So, $(1.1)^{10000}$ is greater than $1000$
View full question & answer→MCQ 501 Mark
What is the value of $\lim_{\text{y} \rightarrow 4}\text{f}(\text{y}) ?$ It is given that $f(y) = y^2 + 6y (y ≥ 2)$ and $f(y) = 0(y < 2).$
Answer$\lim_{\text{y} \rightarrow 4}\text{f}(\text{y}) =\text{y}^ 2+6\text{y}$
$f(4) = 4^2 + 6(4)$
$f(4) = 16 + 24$
$f(4) = 40$
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