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$\lim_{\text{y} \rightarrow 4}\text{f}(\text{y}) =\text{y}^ 2+6\text{y}$
$f(4) = 4^2 + 6(4)$
$f(4) = 16 + 24$
$f(4) = 40$
View full question & answer→MCQ 511 Mark
$ \lim_\limits{\text{x}→0} \frac{| \sin \text{x}|}{\text{x}}$ is equal to:
View full question & answer→MCQ 521 Mark
if $\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100},$then $f'(1)$ is equal to:
- A
$\frac{1}{100}$
- ✓
$100$
- C
$50$
- D
$0$
Answer$\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}$
Differentiate both the sides with respect to $x$, we get
$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}\Big(1+\text{x}+\frac{\text{x}^2}{2}+\dots+\frac{\text{x}^{100}}{100}\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2}\Big)+\dots+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^{100}}{100}\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\dots+\frac{1}{100}\frac{\text{d}}{\text{dx}}(\text{x}^{100})$
$=0+1+\frac{1}{2}\times\text{2x}+\dots+\frac{1}{100}\times100\text{x}^{99}$
$=1+\text{x}+\text{x}^2+\dots+\text{x}^{99}$
Putting $x = 1,$ we get
$\text{f}'(\text{x})=1+1+1+\dots+1 (100$ terms$)$
$=100$
View full question & answer→MCQ 531 Mark
What is the value of What is the value of $\lim_{\text{x} \rightarrow \infty}\frac{\text{x}^2-9}{\text{x}^2-3\text{x} +2}$
AnswerSince it is of the form $ \frac{\infty }{\infty }$
we use $L$ 'Hospital' s rule and differentiate the numerator and denominator
$\text{L}=\lim_{\text{x} \rightarrow \infty}\frac{\text{x}^2-9}{\text{x}^2-3\text{x} +2}$
On differentiating once, we get $\text{L}=\lim_{\text{x} \rightarrow \infty}\frac{2\text{x}}{2\text{x}}$
Which is equal to, $\lim_{\text{x} \rightarrow \infty}1 = 1.$
View full question & answer→MCQ 541 Mark
What is the value of $\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \tan \text{x})$ at $\text{x} = 0?$
AnswerWe need to use product rule in both the terms to get the answer.
$\frac{\text{d}}{\text{dx}} (\text{f.g}) = \text{g}.\frac{\text{d}}{\text{dx}} (\text{f})+ (\text{f}).\frac{\text{dy}}{\text{dx}} (\text{g})$
Here $f = e^x$ and $g = \tan x$
$ \frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \tan \text{x}) = \tan \text{x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}}) + \text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}}(\tan \text{x})$
$\frac{d}{dx} (\text{e}^{\text{x}} \tan \text{x}) = \tan \text{e}^{\text{x}} + \text{e}^{\text{x}}.\sec2\text {x}$
At $x = 0$ we get,
$= \tan 0.\text{e}0 + \text{e}0.\sec20$
$= 0.(1) + 1.(1)$
$= 1$
View full question & answer→MCQ 551 Mark
$\lim_\limits{\text{n} \rightarrow \infty}\frac{\text{n}(2\text{n}+1)2}{(\text{n}+2)(\text{n}2+3\text{n}−1)}$ is equal to:
Answer$=\lim_\limits{\text{n} \rightarrow \infty}\frac{\text{n}(2\text{n}+1)2}{(\text{n}+2)(\text{n}2+3\text{n}−1)}$
$ = \displaystyle \lim_{\text{n}\to\infty}{\displaystyle \frac {\left(2+\Large \frac{1}{\text{n}} \right)^2}{\left(1+\Large \frac{2}{\text{n}} \right)\left(1+\Large \frac{3}{\text{n}} - \Large \frac{1}{\text{n}^2} \right)} }$
$=\text{n}$
$ = \displaystyle \frac{(2+0)^2}{(1+0)(1+0+0)}$
View full question & answer→MCQ 561 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 571 Mark
If $\text{L}=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}\sin\text{x}-\sin^2\text{x}}{\tan^3\text{x}}$ is finite, then the value of $L$ is:
View full question & answer→MCQ 581 Mark
What is the value of $\lim_{\text{y} \rightarrow \frac{\pi}{2}}\frac{\text{sin}}{\text{x}} ?$
- ✓
$\frac{2}{π}$
- B
$\frac{π}{2}$
- C
$1$
- D
$0$
AnswerCorrect option: A. $\frac{2}{π}$
$ \sin \frac{π}{2} = 1$
$\lim_{\text{y} \rightarrow \frac{\pi}{2}}\frac{\text{sin}}{\text{x}}=\frac{\sin\frac{\pi}{2}}{\frac{pi}{2}}$
$=\frac{1}{\frac{\pi}{2}}$
$ = \frac{2}{π}$
View full question & answer→MCQ 591 Mark
Choose the correct answer. If $\text{f}(\text{x})=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is equal to:
- A
$1$
- B
$\frac{1}{2}$
- C
$\frac{1}{\sqrt{2}}$
- ✓
$0$
AnswerGiven that $\text{f}(\text{x})=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\text{x}}}-\frac{1}{2\text{x}^{\frac{3}{2}}}$
$\big(\frac{\text{dy}}{\text{dx}}\big)=\frac{1}{2}-\frac{1}{2}=0$
View full question & answer→MCQ 601 Mark
What is the value of the limit $ \text{f(x)} = \text{x}2+\sqrt{2\text{x}}\sqrt{\text{x}2}−4\text{x}$ if $x$ approaches infinity?
AnswerThis is of the form $\infty ,$ therefore we use $L$’Hospital’s,
rule and differentiate the numerator and denominator.
View full question & answer→MCQ 611 Mark
if $\text{f}(\text{x})=\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1,$ then $f'(1)$ is equal to
- ✓
$5050$
- B
$5049$
- C
$5051$
- D
$50051$
AnswerCorrect option: A. $5050$
$\text{f}(\text{x})=\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1$
Differentiate both the sides with respect to $x,$ we get
$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}^{100}+\text{x}^{99}+\dots+\text{x}+1)$
$=\frac{\text{d}}{\text{dx}}(\text{x}^{100})+\frac{\text{d}}{\text{dx}}(\text{x}^{99})+\dots+\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}(1)$
$=100\text{x}^{99}+99\text{x}^{98}+\dots+\text{2x}+1+0$
$=100\text{x}^{99}+99\text{x}^{98}+\dots+\text{2x}+1$
Putting $x = 1,$ we get
$\text{f}'(\text{x})=100+99+98+\dots+2+1$
$=\frac{100(100+1)}{2}\ \Big(\text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big)$
$=50\times101$
$=5050$
View full question & answer→MCQ 621 Mark
$ \lim\limits_{\text{x}→3}\ 2\text{x}^2−3\text{x}−5 =$
Answer$ \lim\limits_{\text{x}→3}\ 2\text{x}^2−3\text{x}−5 $
$ = 2(3)^2 - 3(3) - 5$
$= 18 - 9 - 5$
$= 4$
View full question & answer→MCQ 631 Mark
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. $5050$
$ f(x)=x^{100}+x^{99}+\ldots+x+1 $
$ f(x)=100 x^{99}+99 x^{98}+\ldots+1+0 $
$ f(1)=100(1)^{99}+99(1)^{98}+\ldots+1 $
$= 100 + 99 + …. + 1$
This is an $AP$ with common difference $-1, a = 100, n = 100$ and $l = 1.$
So, the sum of this $AP = \Big(\frac{100}{2}\Big)[100 + 1]$
$= 50(101)$
$= 5050$
Therefore, $f(1) = 5050$
View full question & answer→MCQ 641 Mark
What is the value of $\text{ddx} (\sin x^3 \cos x^2)$?
- A
$ 3 x^2 \cos x^2 \cos x^3+2 x \sin x^3 \sin x^2 $
- ✓
$ 3 x^2 \cos 2 \cos x^3-2 x \sin x^3 \sin x^2 $
- C
$ 2 x \cos x^2 \cos x^3-2 x \sin x^3 \sin x^2 $
- D
$ 2 x \cos x^2 \cos x^3+3 x^2 \sin x^3 \sin x^2 $
AnswerCorrect option: B. $ 3 x^2 \cos 2 \cos x^3-2 x \sin x^3 \sin x^2 $
We follow product rule $\frac{\text{d}}{\text{dx}}(\text{f}.\text{g}.)=\text{g}.\frac{\text{d}}{\text{dx}}(\text{f})+\text{f}.\frac{\text{dy}}{\text{dx}}(\text{g})$
Here $ \text{f} = \sin x^3$ and $g = \cos x^2$
$\frac{\text{d}}{\text{dx}} (\text{f}) = 3x^2 \cos x^3$
$\frac{\text{d}}{\text{dx}} (\text{g}) = -2\text{x}\ \sin \text{x}^2$
We now substitute this in our main equation,
$=\cos x^2.3x^2 \cos x^3 + \sin x^3.(-2x \sin x^2)$
$=3x^2 \cos x^2 \cos x^3 – 2x \sin x^3 \sin x2$
View full question & answer→MCQ 651 Mark
If $ \mathop {\lim }\limits_{\text{x} \to 0} {\left( {\cos \text{x} + \text{a}\sin \text{bx}} \right)^{\frac{1}{\text{x}}}} = {\text{e}^2}$ then the possible values of $a\ \&\ \text{amp};$ bare$:′a′\ \&\ ′b′$ are:
- ✓
$a = 1, b = 2$
- B
$a = 2, b = 1$
- C
$a = 3, b = 2$
- D
AnswerCorrect option: A. $a = 1, b = 2$
$ \mathop {\lim }\limits_{\text{x} \to 0} {\left( {\cos \text{x} + \text{a}\sin \text{bx}} \right)^{\frac{1}{\text{x}}}}$
so its limit will be $e^k$,
where $\text{k}=\lim\limits_{\text{x}\to0}\frac{1}{\text{x}}(\text{cos x}+\text{asinbx}-1)$
$=\lim\limits_{\text{x}\to0}\frac{-\sin\text{x}+\text{abcosbx}}{1}$
$=\text{ab}$
$=2$
Hence all possible combination of $aa$ and $bb$ are possible whose product is $2$
View full question & answer→MCQ 661 Mark
$ \lim\limits_{\text{x}\rightarrow \infty } [\text{x}-1]$ - where $[.]$ is greatest integer function, is equal to:
View full question & answer→MCQ 671 Mark
Choose the correct answer. If $\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^{2}}{2}+.....+\frac{\text{x}^{100}}{100}$ then $f'(1)$ is equal to:
- A
$\frac{1}{100}$
- ✓
$100$
- C
- D
$0$
AnswerGiven $\text{f}(\text{x})=1+\text{x}+\frac{\text{x}^{2}}{2}+.....+\frac{\text{x}^{100}}{100}$
$\text{f}(\text{x})=1+\frac{2\text{x}}{2}+......+\frac{100\text{x}}{100}$
$\therefore \text{f'}(1)=1+11+....+1=100$
View full question & answer→MCQ 681 Mark
$f\ f(x) = 3\cos x,$ then $f ¢(x)$ at $ \text{x} = \frac{Π}{2}$ = is:
View full question & answer→MCQ 691 Mark
What is the number of critical points of $ \text{f(x)} =\frac{|x^2 - 1|}{x^2}$?
AnswerClearly $f(x)$ is not differentiable at $x = 1$
and $x = -1$ And $x = 0$ is not a critical point not in the domain.
Therefore $1$ and $-1$ are critical points.
Thus, there are $2$ critical points.
View full question & answer→MCQ 701 Mark
What is the derivative of $ \lim\limits_{\text{x}\rightarrow \infty }\Big({\text{xsinx} (\frac{2}{\text{x}})}\Big)? $
View full question & answer→MCQ 711 Mark
Identify the value of $\lim\limits_{\text{x} \rightarrow 2} \text{x}^2 - 5\text{x} + 6$
AnswerLet $\lim\limits_{\text{x} \rightarrow 2} \text{x}^2 - 5\text{x} + 6$ This is not an indeterminate form
Therefore, $\text{L}=(2)2−5(2)+6$
$\Rightarrow \text{L}=0$.
View full question & answer→MCQ 721 Mark
Let $f(x) = x - [x] \in R,$ then $\text{f}\Big(\frac{1}{2}\Big)$ is:
- A
$ \frac{3}{2}$
- ✓
$1$
- C
$0$
- D
$-1$
AnswerGiven,
$f(x) = x - [x]$
$f(x) = 1 - 0 \{[x] =$ integer less than or equal to $x\}$
$\text{f}\Big(\frac{1}{2}\Big)=1$
View full question & answer→MCQ 731 Mark
$\text{f}(\text{x})=\frac{3\text{x}^2+\text{ax}+\text{a}+1}{\text{x}^2+\text{x}-2}$ and $\lim_\limits{\text{x} \rightarrow -2}\text{f}(\text{x})$ exists. Then the value of $(a - 4)$ is?
Answer$\text{f}(\text{x})=\frac{3\text{x}^2+\text{ax}+\text{a}+1}{\text{x}^2+\text{x}-2}$
As $0 x \rightarrow -2, D^r\rightarrow 0$.
Hence, as $x \rightarrow -2, N^r\rightarrow 0$.
Therefore, $12 - 2a + a + 1 = 0$ or $a = 13$
Hence, option A is correct.
View full question & answer→MCQ 741 Mark
Choose the correct answer. If $\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$ then $f'(1)$ is equal to:
- ✓
$\frac{5}{4}$
- B
$\frac{4}{5}$
- C
$1$
- D
$0$
AnswerCorrect option: A. $\frac{5}{4}$
Given that $\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$
$\therefore\text{f}(\text{x})=\frac{1}{2}\bigg[\frac{\sqrt{\text{x}.}1-(\text{x}-4).\frac{1}{2\sqrt{\text{x}}}}{\text{x}}\bigg]$
$=\frac{1}{2}\Big[\frac{2\text{x}-\text{x}+4}{2\sqrt{\text{x}.\text{x}}}\Big]$
$=\frac{1}{2}\Bigg[\frac{\text{x}+4}{2(\text{x})^{\frac{3}{2}}}\Bigg]$
$\therefore\text{f}(\text{x})\ \text{x}=1=\frac{1}{2}\Big[\frac{1+4}{2\times1}\Big]=\frac{5}{4}$
View full question & answer→MCQ 751 Mark
The derivative of $f(x) = \sin^2 x$ is:
- A
$\cos 2x$
- B
$\tan 2x$
- ✓
$\sin 2x$
- D
$\text{cosec}\ 2x$
AnswerCorrect option: C. $\sin 2x$
View full question & answer→MCQ 761 Mark
Choose the correct answer. If $\text{f}(\text{x})=\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}$ for some constant, $a,$ then $f'(a)$ is equal to:
AnswerGiven $\text{f}(\text{x})=\frac{\text{x}^{\text{n}}-\text{a}^{\text{n}}}{\text{x}-\text{a}}$
$\text{f'}(\text{x})=\frac{(\text{x}-\text{a})(\text{n.}\text{x}^{\text{n-1}}-(\text{x}^{\text{n}-\text{a}^{\text{n}}})1}{(\text{x}-\text{a})^{2}}$
So, $\text{f}(\text{a})=\frac{0}{0} =$ Does not exist.
View full question & answer→MCQ 771 Mark
Evaluate: $\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{x}^{2}+4\text{x}+4}{2\text{x}-1}:$
AnswerGiven, $\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{x}^{2}+4\text{x}+4}{2\text{x}-1}:$
Substituting $x = 1$ we get
$\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{x}^{2}+4\text{x}+4}{2\text{x}-1}=$
$=\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{(1)}^{2}+4\text{(1)}+4}{2\text{(1)}-1}$
$=\displaystyle \lim_{\text{x}\rightarrow 1}\frac{2\text{}+4\text{}+4}{2\text{}-1}$
$=10$
View full question & answer→MCQ 781 Mark
If $\text{f(x)}\frac{\sin(\text{x}+9)}{\cos\text{x}}$ then $f(x)$ at $x = 0$ is:
- ✓
$\cos 9$
- B
$\sin 9$
- C
$0$
- D
$1$
AnswerCorrect option: A. $\cos 9$
View full question & answer→MCQ 791 Mark
What is the value of the $\lim_\limits{\text{x} \rightarrow 5}\frac{32\text{x}+1}{\text{x}^2-5\text{x}}?$
AnswerUse $L’$ Hospital’s Rule, and differentiate the numerator and denominator.
$ \lim_\limits{\text{x} \rightarrow 5}\frac{32\text{x}+1}{\text{x}^2-5\text{x}}$
$ =\frac{32}{5}$
$= 6.4$
View full question & answer→MCQ 801 Mark
$ \lim_\limits{\text{x}\rightarrow 1} \sqrt{x +1) (2x - 3)} \sqrt{2\times 3 + x -3}$ is:
AnswerCorrect option: B. $\frac{1}{10}$
View full question & answer→MCQ 811 Mark
If $\lim\limits_{\text{x}→5} \frac{\text{Xk -5K}}{\text{x} -5} = 500$ then $k$ is equal to:
View full question & answer→MCQ 821 Mark
$\lim_\limits{\text{x} \rightarrow \text{a}}\frac{1}{(\text{x}-\text{a})^{2\text{n}-1}}(n\ \epsilon\ N)$ equals:
- A
$ \infty$
- B
$ -\infty$
- C
$0$
- ✓
AnswerLeft hand limit is$\lim_\limits{\text{x} \rightarrow \text{a}}\frac{1}{(\text{x}-\text{a})^{2\text{n}-1}}=-\infty$
And Right hand limit is $\lim_\limits{\text{x} \rightarrow \text{a}}\frac{1}{(\text{x}-\text{a})^{2\text{n}-1}}=+\infty$
$\text{L.H.L.}\neq \text{R.H.L.}$
Therefore, the given limit does not exist.
Hence, the option $D$ is correct.
View full question & answer→MCQ 831 Mark
What is the value of $ \lim_\limits{\text{y} \rightarrow 2}\big(1+\frac{1}{\text{n}}\big)^\text{n}\text{y}^2-4\text{y}-2?$
Answer$y^2- 4 = (y - 2)(y + 2)$ Therefore the fraction becomes,
$(y + 2)$ As $y$ tends to $2,$ the fraction becomes $4$
View full question & answer→MCQ 841 Mark
If $f′ (0) = 0$ and $f(x)$ is a differentiable and increasing function,then $\lim\text{x}\rightarrow0 \ \frac{\text{x},\text{f(x)}^2}{\text{f(x)}}$:
- ✓
- B
May not exist as left hand limit may not exist
- C
May not exist as left hand limit may not exist
- D
Right hand limit is always zero
View full question & answer→MCQ 851 Mark
What is the value of the $\lim_{\text{x} \rightarrow 5}\frac{32\text{x}+1}{\text{x} ^2-5\text{x}} ?$
AnswerUse $L$ ’Hospital’ s Rule,
and differentiate the numerator and denominator.
$\lim_{\text{x} \rightarrow 5}\frac{32\text{x}+1}{\text{x} ^2-5\text{x}} ?$
$=\frac{32}{5}$
$= 6.4$
View full question & answer→MCQ 861 Mark
What is the value of $\text{limy}_{\text{y}\rightarrow\infty}\frac{2}{\text{y}} ?$
AnswerAny number divided by infinity gives us $0.$
Here, since the number $2$ is divided by $y,$
as $y$ approaches infinity, we get $0.$
View full question & answer→MCQ 871 Mark
The coefficient of $y$ in the expansion of $\Big({y^2} + \frac{\text{c}}{\text{y}}\Big)5$ is
AnswerCorrect option: B. $10c^2$
Given, binomial expression is $\Big({y^2} + \frac{\text{c}}{\text{y}})5$
Now, $T_{r+1}={ }^5 C_r \times\left(y^2\right)^{5-r} \times\left(\frac{c}{y}\right)^r$
$={ }^5 C_r \times y^{10-3 r} \times C^r$
Now, $10-3 r=1$
$\Rightarrow 3 r=9 $
$\Rightarrow r=3$
So, the coefficient of $y={ }^5 \mathrm{C}_3 \times \mathrm{c}^3=10 \mathrm{c}^3$
View full question & answer→MCQ 881 Mark
If the third term in the binomial expansion of $(1 + x)^m$ is $ \frac{-1}{8}\text{x}^2$ then the rational value of $m$ is:
- A
$2$
- ✓
$\frac{1}{2}$
- C
$3$
- D
$4$
AnswerCorrect option: B. $\frac{1}{2}$
$(1 + x)^m = 1 + mx + \frac{\text{m}(\text{m - 1)}}2\text{x}^2 + ........$
Now, $ \frac{\text{m}(\text{m - 1)}}2\text{x}^2$ = $ \frac{1}{8}\text{x}^2$
$\Rightarrow \frac{\text{m}(\text{m - 1)}}2\text{x}^2$ = $ \frac{-1}{8}\text{x}$
$\Rightarrow 4m^2 - 4m = -1$
$\Rightarrow 4m^2- 4m + 1 = 0$
$\Rightarrow (2m - 1)^2 = 0$
$\Rightarrow 2m - 1 = 0$
$\Rightarrow m = \frac{1}{2}$
View full question & answer→MCQ 891 Mark
is $ \text{f(x)} = \displaystyle \frac {\text{x}^2+6\text{x}}{\sin \text{x}}$ then $\lim_\limits{\text{x} \rightarrow 0} \text{f(x)=}$
View full question & answer→MCQ 901 Mark
What is the value of $\lim y \rightarrow 2y^2- 4y - 2\ ?$
AnswerExplanation: $y^2- 4 = (y - 2)(y + 2)$
herefore the fraction becomes, $(y + 2)$
As y tends to $2$, the fraction becomes $4$
View full question & answer→MCQ 911 Mark
$\lim_\limits{\text{n} \rightarrow \infty}\frac{\text{n}(2\text{n}+1)2}{(\text{n}+2)(\text{n}2+3\text{n}−1)}$ is equal to:
Answer$=\lim_\limits{\text{n} \rightarrow \infty}\frac{\text{n}(2\text{n}+1)2}{(\text{n}+2)(\text{n}2+3\text{n}−1)}$
$= \displaystyle \lim_{\text{n}\to\infty}{\displaystyle \frac {\left(2+\Large \frac{1}{\text{n}} \right)^2}{\left(1+\Large \frac{2}{\text{n}} \right)\left(1+\Large \frac{3}{\text{n}} - \Large \frac{1}{\text{n}^2} \right)} }=\text{n}$
$= \displaystyle \frac{(2+0)^2}{(1+0)(1+0+0)}$
View full question & answer→MCQ 921 Mark
If,$ \text{y} = (\sin^{-1}\text{x})^2 $, then what is the value of $(1 - x^2)y - xy + 4?$
AnswerWe have, $ \text{y} = (\sin^{-1}\text{x})^2………..(1)$
Differentiating with respect to $x,$
we get, $ \text{y} = \frac{(\sin^{-1}\text{x})^2}{1-\text{x}^2} 1/2$ or,
Squaring both sides,
$(1 - x^2 )(y)^2 = 4(\sin - 1x)^2$ From $(1),$
$(1 - x^2 )( y)^2 = 4y$ Differentiating with respect to $x,$
$4 = 2 + 4 = 6$
View full question & answer→MCQ 931 Mark
Consider the differential equation $\frac{\text{dy}}{\text{dx}}=\cos\text{x}$ Then we observe that:
- A
$\text{y}=\sin\text{x}$
- B
$\text{y}=\sin\text{x}+2$
- C
$\text{y}=\sin\text{x}-\frac{1}{2}$
- ✓
$\text{y}=\sin\text{x}+\text{c}$
AnswerCorrect option: D. $\text{y}=\sin\text{x}+\text{c}$
View full question & answer→MCQ 941 Mark
What is the value of $\frac{\text{d}}{\text{dx}}\text{(ex sinx + ex cos x)}?$
- A
$0$
- B
$\text{2 cosx}$
- C
$2e^x.\sin x$
- ✓
$2e^x.\cos x$
AnswerCorrect option: D. $2e^x.\cos x$
We need to use product rule in both the terms to get the answer.
$=\frac{d}{dx}(\text{f.g})=\text{g}\frac{d}{dx}(\text{f})+\text{f}\frac{d}{dx}(\text{fg})$
$=\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{x}) = ({\text{e}^{\text{x}},.\frac{\text{d}}{\text{dx}}} (\sin\text{x}) + \text{sin x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}})) + (\text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}} (\cos \text{x}) + \cos \text{x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}}))$
$=\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{x}) =(\text{e}^{\text{x}}.\cos \text{x} + \sin \text{x} . \text{e}^{\text{x}}) + (\text{e}^{\text{x}}.(-\sin \text{x}) + \cos \text{x}.\text{e}^{\text{x}})$
$= \frac{\text{d}}{\text{dx}} (\text{e}^{\text {x}} \sin \text{x} + \text{e}^{\text {x}} \cos \text{x}) = \text{e}^{\text {x}}.\cos \text{x} + \sin \text{x} . \text{e}^{\text {x}} – \text{ex}.\sin \text{x} + \cos \text{x}.\text{e}^{\text {x}}$
$= \frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{x}) = 2\text{e}^{\text{x}}.\text{cos} \text{x}$
View full question & answer→MCQ 951 Mark
Choose the correct answer. If $f(x) = 1 - x + x^2- x^3+ ....-x^{99}+ x^{100}$, then $f'(1)$ is equal to:
AnswerGiven that $f(x) = 1 - x + x^2- x^3+ .......-x^{99}+ x^{100}$
$f'(x) = -1 + 2x - 3x^2+ ... -99.x^{98}+ 100.x^{99}$
$f'(x) = -1 + 2 - 3 + ....-99 + 100$
$=(- 1 - 3 - 5 ....-99) + (2 + 4 + 6 + ....100)$
$=\frac{50}{2}\big[2\times1+(50-1)(-2)\big]+\frac{50}{2}\big[2\times2(50-1)2\big] $
$=25\big[-11+102\big]$
$=25\times2$
$=50$
View full question & answer→MCQ 961 Mark
The derivative of $x^2 \cos x$ is:
AnswerCorrect option: B. $2x \cos x - x^2 \sin x$
$\frac{ \text{d}}{\text{dx}(x^2 \text{cos x})}$
Using the formula $ \frac{\text{d}}{\text{dx} [\text{f(x) g(x)}]} = \text{f}(\text{x}) \Big[\frac{\text{d}}{\text{dx} \text{g}(\text{x})}\Big] + \text{g(x)} \Big[\frac{\text{d}}{\text{dx} \text{f(x})}\Big]$
$= \frac{\text{d}}{\text{dx}(\text{x}^2 \cos \text{x})}$
$= \text{x}^2 \Big[\frac{\text{d}}{\text{dx} (\cos \text{x})}\Big] + \cos x \Big[\frac{\text{d}}{\text{dx } \text{x}^2}\Big]$
$ = \text{x}^2(-\sin \text{x}) + \cos\text{x}(2\text{x})$
$ = 2\text{x} \cos \text{x} – \text{x}2 \sin \text{x}$
View full question & answer→MCQ 971 Mark
What is the value of the limit $\text{f}(\text{x}) = \frac{\text{sin}^2\text{x}+2\sqrt{\text{sinx}}}{\text{x}^2−4\text{x}}$ if $x$ approaches $0?$
- A
$\frac{1}{\sqrt{2}}$
- B
$\frac{-1}{\sqrt{2}}$
- ✓
$\frac{-1}{2\sqrt{2}}$
- D
$\frac{-1}{\sqrt{-2}}$
AnswerCorrect option: C. $\frac{-1}{2\sqrt{2}}$
This is of the form $\frac{0}{0}$,
therefore we use $L$ ’Hospital’ s rule and differentiate the numerator and
denominator.
$ =\lim_\limits{a \rightarrow b} \frac{\text{2sin}\text{x cos}+\cos\text{x}\sqrt{\text{2}}}{\text{2x}−4\text{x}}$
$= \frac{0+\sqrt{2}}{-4}$
$=\frac{-1}{2\sqrt{2}}$
View full question & answer→MCQ 981 Mark
Let $ 3\text{f(x)} - 2{\text{f}(\frac{1}{\text{x}}) = \text{x}}$ then $f(2)$ is equal to:
- A
$ \frac{2}{7}$
- ✓
$ \frac{1}{2}$
- C
$2$
- D
$7$
AnswerCorrect option: B. $ \frac{1}{2}$
View full question & answer→MCQ 991 Mark
If $y = 5x^2 + 8x$ find $\frac{\text{dy}}{\text{dx}}$
- ✓
$10x + 8$
- B
$5x + 8$
- C
$10x^2 + 8x$
- D
AnswerCorrect option: A. $10x + 8$
View full question & answer→MCQ 1001 Mark
lf $f(x)$ is a quadratic expression which is positive for all real vaues of $x$ and $g(x) = f(x) + f′(x) + f′′(x)$ then for any real value of $x:$
- A
$g(x) < 0$
- ✓
$g(x) > 0$
- C
$g(x) = 0$
- D
$g(x) > 0$
AnswerCorrect option: B. $g(x) > 0$
View full question & answer→MCQ 1011 Mark
$\lim \tan\text{x} = \text{x}\rightarrow \frac{\pi }{2}$
- A
$1$
- B
$0$
- C
$\frac{1}{\pi}$
- ✓
Answer$\text{L.H.L}.=\lim \tan\text{x}=+\infty \ \text{x}\rightarrow \Big(\frac{\pi}{2}\Big)^-$
$\text{R.H.L}.=\lim \tan\text{x}=-\infty \ \text{x}\rightarrow \Big(\frac{\pi}{2}\Big)^+$
Clearly left hand $ \text{limit} \neq$ right hand limit.
Hence given limit does not exist.
View full question & answer→MCQ 1021 Mark
Evaluate $\underset{\text{x}\, \rightarrow\,3}{\lim}\, (4\text{x}^2\, +\, 3)$
Answer$ =\displaystyle \lim _{\text{x}\rightarrow 3 }{ \left( 4{ \text{x} }^{ 2 }+3 \right) }$
$=4{ \left( 3 \right) }^{ 2 }+3$
$=36+3$
$=39$
View full question & answer→MCQ 1031 Mark
$\lim_\limits{\text{x}→0} \sin\text{x} (\sqrt{\text{x} + 1} - \sqrt{(1- \text{x})}$ is:
View full question & answer→MCQ 1041 Mark
The limit of $ \left[\frac{1}{\text{x}^2}+\frac{(2013)^\text{x}}{\text{e}^\text{x}-1}-\frac{1}{\text{e}^\text{x}-1}\right]\text{as} \text{ x}\rightarrow 0:$
AnswerCorrect option: A. Approaches $+\infty$
View full question & answer→MCQ 1051 Mark
lf $f(x) = 2x - 3, a = 2, l = 1f(x) = 2x - 3, a = 2, l = 1$ and $\epsilon = 0.001$ then $δ > 0$ satisfying $0<|x - a|< δ, ∣ f(x) - l ∣ < \epsilon $, is:
- A
$0.0050$
- ✓
$0.0005$
- C
$0.001$
- D
AnswerCorrect option: B. $0.0005$
$|f(x) - l|< 0.001 = \epsilon$
$\Rightarrow |2x - 3 - 1| (x) - l∣ < 0.001$
$\Rightarrow -0.001 < 2 x - 4 < 0.001$
$\Rightarrow -0.0005 < x - 2 < 0.0005$
$\Rightarrow ∣x - 2∣ < 0.0005$
$\Rightarrow ∣x - a∣ < 0.0005 = δ$
Hence, $δ = 0.0005 > 0$
View full question & answer→MCQ 1061 Mark
$\lim_\limits{\text{x} \rightarrow 2}\Bigg(\frac{\sqrt{1-\text{cos}{2(\text{x}-2)}}}{\text{x}-2}\Bigg):$
Answer$\lim_\limits{\text{t} \rightarrow 0}\frac{\sqrt{1-\cos2\text{t}}}{\text{t}}$
Clearly $\text{R.H.L}. = \sqrt{2}$
$\text{L.H.L.} = -\sqrt{2}$
Since $\text{R.H.L}. \neq \text{L.H.L.}$
So, limit does not exist.
Hence, option $A$ is correct.
View full question & answer→MCQ 1071 Mark
$\lim_\limits{\text{x} \rightarrow 1}(1+\sin\pi)π\text{x}:$
- ✓
$\pi$
- B
${\pi }^{ 2 }$
- C
${\pi }^{ 3 }$
- D
Answer$=\lim_\limits{\text{x} \rightarrow 1}(1+\sin\pi)π\text{x}$
$= (1+\sin \pi(1))$
$=\pi(1+0)$
$= \pi$
View full question & answer→MCQ 1081 Mark
$\lim_\limits{\text{x} \rightarrow \text{a}}\frac{\text{x}-\text{a}}{|\text{x}-\text{a}|}=$
AnswerUsing,
$\lim_\limits{\text{x} \rightarrow 0}|\text{x}|=-\text{x}$
$\lim_\limits{\text{x} \rightarrow 0}|\text{x}|=+\text{x}$
we get $\lim_\limits{\text{x} \rightarrow \text{a}}-\frac{\text{x}-\text{a}}{-(\text{x}-\text{a})}=-1$
$\lim_\limits{\text{x} \rightarrow \text{a}}+\frac{\text{x}-\text{a}}{-(\text{x}-\text{a})}=-1$
Since, $\text{LHL}$ is not equal to $\text{RHL}$,
hence the limit does not exist.
View full question & answer→MCQ 1091 Mark
$\lim_\limits{\text{x} \rightarrow 0}\frac{\sin|\text{x}|}{\text{x}}$ is equal to:
Answer$=\lim_\limits{\text{x} \rightarrow 0}\frac{\sin|\text{x}|}{\text{x}}$
$\text{ LHL} =-1,\text{RHL}=1$
Limit does not exist.
View full question & answer→MCQ 1101 Mark
If $\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}},$ then $f'(1)$ is:
- ✓
$\frac{5}{4}$
- B
$\frac{4}{5}$
- C
$1$
- D
$0$
AnswerCorrect option: A. $\frac{5}{4}$
$\text{f}(\text{x})=\frac{\text{x}-4}{2\sqrt{\text{x}}}$
$=\frac{1}{2}\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}$
$=\frac{1}{2}\text{x}^{\frac{1}{2}}-\text{2x}^{-\frac{1}{2}}$
Differentiate both the sides with respect to $x,$ we get
$\text{f}'(\text{x})=\frac{1}{2}\times\frac{1}{2}\text{x}^{\frac{1}{2}-1}-2\times\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}\ [\text{f}(\text{x})=\text{x}^\text{n}$
$\Rightarrow\text{f}'(\text{x})=\text{nx}^{\text{n}-1}]$
$\Rightarrow\text{f}'(\text{x})=\frac{1}{4}\text{x}^{-\frac{1}{2}}+\text{x}^{-\frac{3}{2}}$
$\therefore\text{f}'(\text{1})=\frac{1}{4}\times1+1=\frac{5}{4}$
View full question & answer→MCQ 1111 Mark
Mark the correct alternative in each of the following : If $\text{f(x)}=\text{x}\sin\text{x},$ then $\text{f}'\Big(\frac{\text{x}}{2}\Big)=$
- A
$0$
- ✓
$1$
- C
$-1$
- D
$\frac{1}{2}$
Answer$\text{f(x)}=\text{x}\sin\text{x}$
Differentiating both sides with respect to $x,$ we get
$\text{f}'\text{(x)}=\text{x}\times\frac{\text{d}}{\text{dx}}(\sin\text{x})+\sin\text{x}\times\frac{\text{d}}{\text{dx}}\text{(x)} ($Product rule$)$
$=\text{x}\times\cos\text{x}+\sin\text{x}\times1$
$=\text{x}\cos\text{x}+\sin\text{x}$
Putting $\text{x}=\frac\pi{2},$ we get
$=\frac{\pi}{2}\times0+1$
$=1$
Hence, the correct answer is option $(b)$
View full question & answer→MCQ 1121 Mark
Evaluate the following limit : $ \displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin^2 3\text{x}}{\text{x}^2}$
Answer$\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin^2 3\text{x}}{\text{x}^2}$
$=\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{x}} \times\frac{\sin 3\text{x}}{\text{x}}$
$= \displaystyle\lim_{\text{x} \rightarrow 0} 3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg) \times3\Bigg(\frac{\sin 3\text{x}}{\text{3x}}\Bigg)$
$=3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}\times3\displaystyle\lim_{\text{x} \rightarrow 0} \frac{\sin 3\text{x}}{\text{3x}}$
$ = 3\times3$
$=9$
View full question & answer→MCQ 1131 Mark
Choose the correct answer. If $\text{f}(\text{x})=\text{x}-[\text{x}],\in\text{R}$ then $\text{f}'\big(\frac{1}{2}\big)$ is equal to:
- A
$\frac{3}{2}$
- ✓
$1$
- C
$0$
- D
$-1$
AnswerGiven $f(x) = x - [x]$
we have ti first check for differentiability of $f(x)$ at $\text{x}=\frac{1}{2}$
$\therefore \text{Lf}'\Big(\frac{1}{2}\Big)=\text{LHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}-\text{h}\big)-\big[\frac{1}{2}-\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}-\text{h}-0-\frac{1}{2}+0}{-\text{h}}$
$=\frac{-\text{h}}{-\text{h}}$
$=1$
$\therefore \text{Rf}'\Big(\frac{1}{2}\Big)=\text{RHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}+\text{h}\big)-\big[\frac{1}{2}+\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}+\text{h}-1-\frac{1}{2}+1}{\text{h}}$
$=\frac{\text{h}}{\text{h}}$
$=1$
Since, $\text{LHD = RHD}$
$\text{f}'\big(\frac{1}{2}\big)=1$
View full question & answer→MCQ 1141 Mark
Differentiate with respect to $x^4+ 3x^2− 2x:$
- ✓
$4x^3+ 6x − 2$
- B
$4x^3+ 6x − 3$
- C
$4x^4+ 6x − 2$
- D
AnswerCorrect option: A. $4x^3+ 6x − 2$
View full question & answer→MCQ 1151 Mark
What is the value of $ \lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}^2\sec\text{x}}{\sin\text{x}}?$
Answer$\lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}^2\sec\text{x}}{\sin\text{x}}\times\ \lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}}{\cos\text{x}} $$$
$= 1 \times 0$
$= 0$
View full question & answer→MCQ 1161 Mark
If $\lim\limits_{\text{x}\to \text{a}}\frac{\text{x}^5-\text{a}^5}{\text{x - a}}=80$ then the value of $aa$ is:
AnswerGiven,$ \lim\limits_{\text{x}\to \text{a}}\frac{\text{x}^5-\text{a}^5}{\text{x - a}}=80$
or, $ 5\text{a}^4=80 [$Using direct formula$]$ or, $\text{a}^4=16$ or, $\text{a}=2.$
View full question & answer→MCQ 1171 Mark
if $\text{f(x)} = \begin{vmatrix} \cos \text{x}& \text{x} & 1\\ 2\sin \text{x} & \text{x}^{2} & 2\text{x}\ \\ \tan \text{x} & \text{x} & 1\end{vmatrix},$ then $\displaystyle \lim_{\text{x}\rightarrow 0} \dfrac {\text{f}(\text{x})}{\text{x}}.$
AnswerCorrect option: A. Exists and is equal to $-2$
View full question & answer→MCQ 1181 Mark
Evaluate : $\displaystyle\lim_{\text{x}\rightarrow 2} \dfrac{\text{x}^2-4}{\text{x}+3}:$
AnswerUsing direct substitution, we obtain,$ =\displaystyle\lim_{\text{x}\rightarrow 2} \dfrac{\text{x}^2-4}{\text{x}+3}$
$ =\dfrac{4-4}{2+3}$
$=0$
View full question & answer→MCQ 1191 Mark
$\lim_\limits{\text{x} \rightarrow 0}\frac{2\text{x}^2+3\text{x}+4}{2}=$
AnswerAs their is not any $x$ term in the denominator,
we can directly substitute the value of $x$ as $0.$
Thus, we have $\lim_\limits{\text{x} \rightarrow 0}\frac{2\text{x}^2+3\text{x}+4}{2}$
$=\frac { 2.0+3.0+4 }{ 2 }$
$=\frac{4}{2}$
$=2$
View full question & answer→MCQ 1201 Mark
Evaluate: $\displaystyle \lim_{\text{x} \rightarrow 0}{\left( \frac{\text{a}^\text{x} + \text{b}^\text{x} + \text{c}^\text{x}}{3} \right)^{\frac{2}{\text{x}}}}$
AnswerCorrect option: D. $(\text{abc})^{\frac{2}{3}}$
$ \displaystyle \lim _{ \text{x} \rightarrow 0 }{ { \left( \frac { { \text{a} }^{ \text{x} }+{ \text{b} }^{ \text{x} }+{ \text{c} }^{ \text{x} } }{ 3 } \right) }^{ \frac{ 2 }{ \text{x} } } } ={ \left( \frac { 3 }{ 3 } \right) }^{\frac { 2 }{ 0 } }$
$ =1\infty \text{ form}={ \text{e} }^{ \displaystyle \lim _{ \text{x}\rightarrow 0 }{ { \left( \cfrac { { \text{a} }^{ \text{x} }+{ \text{b} }^{ \text{x} }+{ \text{c} }^{ \text{x} } }{ 3 } -1 \right) }^{\frac{ 2 }{ \text{x} } } } }$
$ =\text{e}\frac{2}{3}(\log \text{a}+\log \text{b}+\log \text{c})$
$= (\text{abc})^{\frac{2}{3}}$
View full question & answer→MCQ 1211 Mark
What is the number of critical points of $\text{f}(\text{x}) = \frac{|\text{x}^2 - 1|}{ \text{x}^2} ?$
Answer$f(x)$ is not differentiable at $x = 1$ and $x = -1$
And $x = 0$ is not a critical point not in the domain.
Therefore $1$ and $-1$ are critical points.
Thus, there are $2$ critical points.
View full question & answer→MCQ 1221 Mark
Derivative of the function $f(x) = (x - 1) (x - 2)$ is:
- A
$2x + 3$
- B
$3x - 2$
- C
$3x + 2$
- ✓
$2x - 3$
AnswerCorrect option: D. $2x - 3$
View full question & answer→MCQ 1231 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}$ is equal to:
- A
$1$
- ✓
$\frac{\text{m}}{\text{n}}$
- C
$\frac{-\text{m}}{\text{n}}$
- D
$\text{m}^{2}\text{n}^{2}$
AnswerCorrect option: B. $\frac{\text{m}}{\text{n}}$
Given $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}=\lim\limits_{\text{x} \rightarrow 1}\frac{\frac{\text{x}^{\text{m}-(1)^\text{m}}}{\text{x}-1}}{\frac{\text{x}^{\text{n}-(1)^{\text{n}}}}{\text{x}-1}}$
$=\frac{\text{m}(1)^{\text{m}-1}}{\text{n}(1)^{\text{n}-1}}=\frac{\text{m}}{\text{n}}$
$=\frac{\text{m}}{\text{n}}$
View full question & answer→MCQ 1241 Mark
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{t}^\text{x}-\text{e}\sin\text{x}}{2(\text{x}-\sin\text{x)}}=$
- ✓
$-\frac{1}{2}$
- B
$\frac{1}{2}$
- C
$1$
- D
$\frac{3}{2}$
AnswerCorrect option: A. $-\frac{1}{2}$
View full question & answer→MCQ 1251 Mark
Evaluate : $ \displaystyle \underset{\text{x}\rightarrow 2}{\lim} \text{x}^2-5\text{x}+6$
Answer$ = \displaystyle \underset{\text{x}\rightarrow 2}{\lim} \text{x}^2-5\text{x}+6$
$ ={ 2 }^{ 2 }-5\times 2+6$
$= 4 - 10 + 6$
$= 0$
View full question & answer→MCQ 1261 Mark
If $\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}},$ then $\frac{\text{dy}}{\text{dx}}=$
- ✓
$-\frac{4\text{x}}{(\text{x}^2-1)^2}$
- B
$-\frac{4\text{x}}{\text{x}^2-1}$
- C
$\frac{1-\text{x}^2}{\text{4x}}$
- D
$\frac{4\text{x}}{\text{x}^2-1}$
AnswerCorrect option: A. $-\frac{4\text{x}}{(\text{x}^2-1)^2}$
$\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}}$
$=\frac{\text{x}^2+1}{\text{x}^2-1}$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)}{(\text{x}^2-1)^2} ($Quotient rule$)$
$=\frac{(\text{x}^2-1)(\text{2x}+0)-(\text{x}^2+1)(\text{2x}-0)}{(\text{x}^2-1)^2}$
$=\frac{\text{2x}^3-\text{2x}-\text{2x}^3-\text{2x}}{(\text{x}^2-1)^2}$
$=\frac{-\text{4x}}{(\text{x}^2-1)^2}$
View full question & answer→MCQ 1271 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{1-\cos4\theta}{1-\cos6\theta}$ is equal to:
- ✓
$\frac{4}{9}$
- B
$\frac{1}{2}$
- C
$\frac{-1}{2}$
- D
$-1$
AnswerCorrect option: A. $\frac{4}{9}$
Given $\lim\limits_{\theta \rightarrow 0}\frac{1-\cos4\theta}{1-\cos6\theta}=\lim\limits_{\theta \rightarrow 0}\frac{2\sin^{2}2\theta}{2\sin^{2}3\theta}$
$=\lim\limits_{\theta \rightarrow 0}\frac{\sin^{2}2\theta}{\sin^{2}3\theta}=\lim\limits_{\theta \rightarrow 0}\Big[\frac{\sin2\theta}{\sin3\theta}\Big]^{2}$
$=\lim\limits_{\theta \rightarrow 0}\bigg[\frac{\frac{\sin2\theta}{2\theta}\times2\theta}{\frac{\sin3\theta}{2\theta}\times3\theta}\bigg]$
$=\Big[\frac{2\theta}{2\theta}\Big]^{2}$
$=\Big(\frac{2}{3}\Big)^{2}$
$=\frac{4}{9}$
View full question & answer→MCQ 1281 Mark
The value of $ \lim\limits_{\text{x} \rightarrow 3^{+}} \dfrac{|\text{x}-3|}{\text{x}-3}$ equals:
Answerfor $ \text{x}=30^+,$
$ \text{ x}-3 > 0$
Let $\text{L}=\displaystyle \lim_{\text{x}-3^+}\dfrac {|\text{x}-3|}{\text{x}-3}$
$=\displaystyle \lim_{\text{x}\rightarrow 3^+}$
$=\dfrac{(\text{x}-3)}{(\text{x}-3)}$
$=\lim\limits_{\text{x}\rightarrow 3^+}(1)$
$=1$
View full question & answer→MCQ 1291 Mark
$\lim_\limits{\text{x} \rightarrow 1}{1−\text{x}+[\text{x}+1]+[1−\text{x}]}$, where $[x]$ denotes greatest integer function, is
AnswerSubstitute $x = 1 + t$
$ \text{L.H.S} \lim_\limits{\text{t}\rightarrow o^{-}} (-\text{t}+[2+\text{t}]+[-\text{t}])$
$= 0 + 1 + 0 = 1$
$ \text{R.H.S} \lim_\limits{\text{t}\rightarrow o^{-}} (-\text{t}+[2+\text{t}]+[-\text{t}])$
$= 0 + 2 - 1 = 1$
$\text{L.H.S = R.H.S}$
View full question & answer→MCQ 1301 Mark
$\lim\limits_{\text{n}\rightarrow \infty } \frac{{\text{np}\ \text{sin}^2(\text{n}!)}}{\text{n}+1}, 0 < p < 1$ is equal to:
View full question & answer→MCQ 1311 Mark
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin7\text{x}}{\sin3\text{x}}$ equals:
- ✓
$\frac{7}{3}$
- B
$\frac{10}{3}$
- C
$\frac{14}{3}$
- D
$\frac{1}{3}$
AnswerCorrect option: A. $\frac{7}{3}$
View full question & answer→MCQ 1321 Mark
What is the value of $\text{ddx} (\sin x \tan x)?$
- ✓
$\sin x + \tan x \sec x$
- B
$\cos x + \tan x \sec x$
- C
$\sin x + \tan x$
- D
$\sin x + \tan x \sec^2x$
AnswerCorrect option: A. $\sin x + \tan x \sec x$
We follow product rule $\frac{\text{d}}{\text{dx}}(\text{f}.\text{g})=\text{g.}\frac{\text{d}}{\text{dx}}(\text{f})+(\text{f})\frac{\text{d}}{\text{dx}}(\text{f}.\text{g})$
Here, $f = \sin x$ and $g = \tan x$
$\frac{\text{d}}{\text{dx}} (\sin x \tan x) = \cos x \tan x + \sec^2 x \sin x$
$\frac{\text{d}}{\text{dx}} (\sin x \tan x) = \sin x + \tan x \sec x$
View full question & answer→MCQ 1331 Mark
What is the derivative of $f(x) = x|x|?$
- A
$|x| + x$
- B
$2x$
- ✓
$2|x|$
- D
$-2|x|$
AnswerCorrect option: C. $2|x|$
View full question & answer→MCQ 1341 Mark
If $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points on the curve $y = ax^2+ bx + c$, then as perLagrange’s mean value theorem whichof the following is correct?
- A
At least one point $C(x_3, y_3)$ where the tangent will be intersecting the chord $AB$
- B
At least one point $C(x_3, y_3)$ where the tangent will be overlapping to the chord $AB$
- C
At least two points where the tangent will be parallel to the chord $AB$
- ✓
At least one point $C(x_3, y_3)$ where the tangent will be parallel to the chord $AB$
AnswerCorrect option: D. At least one point $C(x_3, y_3)$ where the tangent will be parallel to the chord $AB$
Here, $y = f(x) = ax^2+ bx + c$
As $f(x)$ is a polynomial function, it is continuous and differentiable for all $x.$
So, according to geometrical interpretation of mean value theorem there,
will be at least one point $C(x_3, y_3)$ between $A(x_1, y_1)$ and $B(x_2, y_2)$ where,
tangent will be parallel chord $AB.$
View full question & answer→MCQ 1351 Mark
if $f(x) = x, x < 0: f(x) = 0, x = 0; f(x) = x, x > 0,$ then $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})$ is equal to:
AnswerGiven: $\text{f}(\text{x})= \left\{ \begin{matrix} \text{x} \\ 0 \\ \text{x}^ 2 \end{matrix}\begin{matrix}\quad \text{x}<0 \\\quad \text{x}=0 \\ \quad \text{x}>0 \end{matrix} \right\}$
$ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})=?$
Sol: left hand $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x}) x = 0$ right hand limit $\rightarrow \lim_\limits{\text{x} \rightarrow 0}\text{f}(\text{x})=\lim_\limits{\text{x} \rightarrow0},\text{x}=0 $
right hand limit → $ \lim_\limits{\text{x}\rightarrow 0}\text{f}(\text{x})=\lim_\limits{\text{x}\rightarrow 0}\text{x}^2=0$
$\text{LHL = RHL f(0) = 0}$
$\text{LHL = RHL = f(0) = 0}$
View full question & answer→MCQ 1361 Mark
If $n$ is a positive integer, then ($\sqrt{3} + 1)^{2n+1} + (\sqrt{3} - 1)^{2n+1}$ is:
AnswerSince $n$ is a positive integer, assume $n = 1$
$(\sqrt{3}+1)^3 + (\sqrt{3}−1)^3$
$ = {3\sqrt{3} + 1 + 3\sqrt{3}(\sqrt{3} + 1)} + {3\sqrt{3} - 1 - 3\sqrt{3}(\sqrt{3} - 1)}$
$ = 3\sqrt{3} + 1 + 9 + 3\sqrt{3} + 3\sqrt{3}- 1 - 9 + 3\sqrt{3}$
$= 12\sqrt{3}$, which is an irrational number.
View full question & answer→MCQ 1371 Mark
$\lim_\limits{\text{x} \rightarrow \infty}\frac{2\sqrt{\text{x}}+3\sqrt[3]{\text{x}}+4\sqrt[4]{\text{x}}+...+\text{n}\sqrt[\text{n}]{\text{x}}}{\sqrt{(2\text{x}-3}+{\sqrt[3]{(2\text{x}-3)}+{\sqrt[4]{(2\text{x}-3)}+...+}{\sqrt[\text{n}]{(2\text{x}-3)}}}}$ is equal to:
AnswerCorrect option: C. $\sqrt{2}$
$ \lim_\limits{\text{x} \rightarrow \infty}\frac{2\sqrt{\text{x}}+3\sqrt[3]{\text{x}}+4\sqrt[4]{\text{x}}+...+\text{n}\sqrt[\text{n}]{\text{x}}}{\sqrt{(2\text{x}-3}+{\sqrt[3]{(2\text{x}-3)}+{\sqrt[4]{(2\text{x}-3)}+...+}{\sqrt[\text{n}]{(2\text{x}-3)}}}}$
Dividing numerator and denominator by$ \sqrt{\text{x}}$
$=\frac{2}{\sqrt{2}}$
$=\sqrt{2}$
View full question & answer→MCQ 1381 Mark
Find the value of $\lim\limits_{\text{x} \rightarrow 0} \frac{2\text{x}^2 + 3\text{x} + 4}{2}$
AnswerLet $ \lim\limits_{\text{x} \rightarrow 0} \frac{2\text{x}^2 + 3\text{x} + 4}{2}$ This is not an indeterminate form,
Therefore, $\text{L}=\lim\limits_{\text{x}\rightarrow 0}\dfrac {2(0)+3(0)+4}{2}=\dfrac {4}{2}\text{L}=2$
View full question & answer→MCQ 1391 Mark
$\lim\limits_{X\rightarrow 0} 2\sin\frac{23\text{x}}{x^2}$ is equal to:
View full question & answer→MCQ 1401 Mark
The coefficient of $x^n$ in the expansion of $(1 - 2x + 3x^2- 4x^3+ ........)^{-n}$ is:
AnswerCorrect option: B. $ \frac{(2\text{n})!}{(\text{n}!)^2}$
We have,
$(1 - 2x + 3x^2 - 4x^3 + ........)^{-n}$
$= \{(1 + x)^{-2}\}^{-n}$
$= (1 + x)^{2n}$
So, the coefficient of $x^nC_3= ^{2n}C_n$
$=\frac{(2\text{n})!}{(\text{n}!)^2}$
View full question & answer→MCQ 1411 Mark
Choose the correct answer.
If $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$ where $[.]$ denotes the greatest integer function. then $\lim\limits_{\text{x} \rightarrow 0}\text{f}(\text{x})$ is equal to :
AnswerGiven $\begin{cases}\frac{\sin[\text{x}]}{[\text{x}]} & \text{x}\neq0\\0, & [\text{x}]=0\end{cases}$
$\text{L}.\text{H}.\text{H}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0-\text{h}]}{[0-\text{h}]} $
$=\lim\limits_{\text{h} \rightarrow 0}\frac{-\sin[\text{-h}]}{[-\text{h}]}=-1$
$\text{R}.\text{H}.\text{L}=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin[\text{x}]}{[\text{x}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[0+\text{h}]}{[0+\text{h}]}=\lim\limits_{\text{h} \rightarrow 0}\frac{\sin[\text{h}]}{[\text{h}]}=1$
$\text{L}.\text{H}.\text{L}\neq\text{R}.\text{H}.\text{L}$
So, the limit does not exist.
View full question & answer→MCQ 1421 Mark
Let $f(x) = (x - a) (x - b) (x - c), a < b < c.$ Then $f(x) = 0$ has two roots. At which interval does these roots belongs?
- A
Both the roots in $(a, b)$
- ✓
At least one root in $(a, b)$ and at least one root in $(b, c)$
- C
Both the roots in $(b, c)$
- D
Neither in $(a, b)$ nor in $(b, c)$
AnswerCorrect option: B. At least one root in $(a, b)$ and at least one root in $(b, c)$
$f(x)$ being a polynomial is continuous and differentiable for all real values of $x.$
We also have $f(a) = f(b) = f(c).$
If we apply Rolle’s theorem to $f(x)$ in $[a, b]$ and $[b, c]$ we will observe that $f(x) = 0$
will have at least one root in $(a, b)$ and at least one root in $(b, c).$
But $f(x)$ is a polynomial of degree two,
so that $f(x) = 0$
can’t have more than two roots.
It implies that exactly one root of $f(x) = 0$
will lie in $(a, b)$ and exactly one root of $f(x) = 0$ will lie in $(b, c).$
Let $y = f(x)$ be a polynomial function of degree $n.$
If $f(x) = 0$ has real roots only,
then $f(x) = 0, f(x) = 0, … , f^{n-1}(x) = 0$ will have real roots.
It is in fact the general version of above mentioned application,
because if $f(x) = 0$ have all real roots, then between two consecutive roots of $f(x) = 0,$
exactly one root of $f(x) = 0$ will lie.
View full question & answer→MCQ 1431 Mark
$ \lim_\limits{\text{x} \rightarrow 1}\frac{(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{n}})}{[(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{}})]^2},n \in N,$ equals:
AnswerCorrect option: B. $^{2\text{n}}{\text{C}}_\text{n}$
$\lim_\limits{\text{x} \rightarrow 1}\frac{(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{n}})}{[(1-\text{x})(1-\text{x}^2)...(1-\text{x}^{2\text{}})]^2}$
$\lim_\limits{\text{x} \rightarrow 1}\frac{\frac{(1-\text{x})}{(1-\text{x})}\frac{(1-\text{x}^2)}{(1-\text{x})}...\frac{(1-\text{x}^{2\text{n}})}{(1-\text{x})}}{\Big[\frac{(1-\text{x})}{(1-\text{x})}\frac{(1-\text{x}^2)}{(1-\text{x})}...\frac{(1-\text{x}^{\text{n}})}{(1-\text{x})}\Big]^2}$
$=\frac{1\times2\times3\times\ldots(2\text{n})}{(1\times2\times3\ldots \text{n})^2} = \frac{(2n)!}{\text{n}!\text{n}!}={}^{2\text{n}}\text{C}_\text{n}$
Hence, option $B$ is correct.
View full question & answer→MCQ 1441 Mark
$f(x) = x − 1+ x − 3$ then $f (2) =:$
View full question & answer→MCQ 1451 Mark
Choose the correct answer. If $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is equal to:
AnswerGiven $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\frac{-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{-(\sin\text{x}+\cos\text{x})^{2}(\sin\text{x}+\cos\text{x})}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{\sin^{2}\text{x}+\cos^{2}\text{x}+2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^{2}}$
$=\frac{-2}{(\sin\text{x}-\cos\text{x})^{2}}$
$\therefore \Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-2}{(\sin\text{0}-\cos0)^{2}}=\frac{-2}{(-1)^{2}}=2$
View full question & answer→MCQ 1461 Mark
$\lim x\rightarrow 1x−1xn−1$ is equal to
Answer$= {\lim _\limits{\text{x}\to 1}}\frac{{{\text{x}}^{\text{n}}}-1}{\text{x}-1}$
$ =\lim_\limits{\text{x}\to 1}\frac{\left( \text{x}-1 \right)\left( {{\text{x}}^{\text{n}-1}}+{{\text{x}}^{\text{n} -2}}+.....+\text{x}+1 \right)}{\text{x}-1}$
$ =\lim_\limits{\text{x}\to 1}\sum\limits_{\text{i}=0}^{\text{n}-1}{{{\text{x}}^{\text{i}}}}$
$ =\sum\limits_{\text{i}=0}^{\text{n}-1}{{{1}^{\text{i}}}}$
$ =\sum\limits_{\text{i}=0}^{\text{n}-1}{{{1}^{\text{}}}}$
$ = \text{n}$
Hence, this is the answer.
View full question & answer→MCQ 1471 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 1481 Mark
$\lim\limits_{θ→0} \sin\text{m}^2\frac{θ}{θ}$ is equal to:
View full question & answer→MCQ 1491 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}$ is:
AnswerGiven $\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\sec^{2}\text{x}-2}{\tan\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{1+\tan^{2}\text{x}-2}{\tan\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{\tan\text{x}-1}{\tan\text{x}-1}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}\frac{(\tan\text{x}+1)(\tan\text{x}-1)}{(\tan\text{x}-1)}$
$=\lim\limits_{\text{x} \rightarrow{\frac{\pi}{4}}}(\tan\text{x}+1)$
$=\tan\frac{\pi}{4}+1$
$=1+1$
$=2$
View full question & answer→MCQ 1501 Mark
Given that $f(x)$ is a differentiable function of $x$ and that $f(x) f(y) = f(x) + f(y) + f(xy) −2$ and that $f(2) = 5.$Then $f(3)$ is equal to?
View full question & answer→MCQ 1511 Mark
Find the derivative of ${e^x}^2$:
- A
${e^x}^2$
- B
$2x$
- C
${2e^x}^2$
- ✓
${2xe^x}^2$
AnswerCorrect option: D. ${2xe^x}^2$
We apply chain rule. First we differentiate $x^2$.
$\frac{\text{d}}{\text{dx}} (\text{x}^2) = 2\text{x}$
Now, we know that $\frac{\text{d}}{\text{dx}} (\text{e}^x) = \text{e}^\text{x}$
We differentiate ${e^x}^2$ in the same manner and then multiply with the derivative of $x^2$
$\frac{\text{d}}{\text{dx}} (\text{e}^\text{x}) = 2\text{x}\text{e}^\text{x}$
View full question & answer→MCQ 1521 Mark
What is the value of $\lim_\limits{\text{x}\rightarrow 0}\frac{\text{x}\tan\text{x}}{\cot\text{x}} ?$
- ✓
$0$
- B
$1$
- C
$2$
- D
$\frac{1}{2}$
Answer$\lim_\limits{\text{x}\rightarrow 0}\frac{\text{x}\tan\text{x}}{\cot\text{x}}=\lim_\limits{\text{x}\rightarrow 0}\frac{\text{x}\frac{\sin\text{x}}{\cos\text{x}}}{\frac{\cos\text{x}}{\sin\text{x}}}$
$\lim_\limits{\text{x} \rightarrow 0}\text{x}= 0$
View full question & answer→MCQ 1531 Mark
$\lim\limits_{\text{x}\rightarrow\infty}\sin\text{x}$ equals:
View full question & answer→MCQ 1541 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$ is:
AnswerGiven $\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\sqrt{\text{x}+1}-\sqrt{1-\text{x}}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\big(\sqrt{\text{x}+1}-\sqrt{1-\text{x}}\big)\big(\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big)}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}\big[\sqrt{\text{x}+1}\sqrt{1-\text{x}}\big] }{\text{x}+1-1+\text{x}}$
$=\frac{1}{2}\cdot\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}\big[\sqrt{\text{x}+1}+\sqrt{1-\text{x}}\big]$
Taking limit, we get
$=\frac{1}{2}\times1\times\big[\sqrt{0+1}+\sqrt{0-1}\big]$
$=\frac{1}{2}\times1\times2$
$=1$
View full question & answer→MCQ 1551 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow \pi}\frac{\sin\text{x}}{\text{x}-\pi}$ is:
AnswerGiven, $\lim\limits_{\text{x} \rightarrow \pi}\frac{\sin\text{x}}{\text{x}-\pi}$
$=\lim\limits_{\text{x} \rightarrow\pi}\frac{\sin(\pi)-\text{x}}{-(\pi-\text{x})}$
$=-1$
View full question & answer→MCQ 1561 Mark
$\mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{\tan 3\text{x}}}$
- A
$-\frac{5}{3}$
- ✓
$\frac{5}{3}$
- C
$-\frac{7}{3}$
- D
AnswerCorrect option: B. $\frac{5}{3}$
$\mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{\tan 3\text{x}}}$
$=\mathop {\lim }\limits_{\text{x} \to 0} \frac{{\sin 5\text{x}}}{{5\text{x}}}\times\frac{\text{3x}}{\tan\text{x}}\times\frac{5}{3}$
we know that $ =\displaystyle \lim_{\text{x}\rightarrow 0}\frac {\sin 5\text{x}}{5\text{x}}=1$
$=\mathop {\lim }\limits_{\text{x} \to 0}\times\frac{\text{3x}}{\tan\text{x}}=1$
$= \text{L}=1\times 1\times \dfrac {5}{3}$
$= \displaystyle \lim_{\text{x}\rightarrow 0}\frac {\sin 5\text{x}}{5\text{x}}=\frac{5}{3}$
View full question & answer→MCQ 1571 Mark
What is the value of $ \lim_\limits{\text{y} \rightarrow \infty}\frac{2}{\text{y}}?$
AnswerAny number divided by infinity gives us $0.$
Here, since the number $2$ is divided by $y,$ as $y$ approaches infinity, we get $0.$
View full question & answer→MCQ 1581 Mark
Evaluate the following limit $\lim_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\text{x}^2}:$
View full question & answer→MCQ 1591 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$ is:
- A
$\frac{1}{10}$
- ✓
$-\frac{1}{10}$
- C
$1$
- D
AnswerCorrect option: B. $-\frac{1}{10}$
Given $\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$
$=\lim\limits_{\text{x} \rightarrow1}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{\text{x}\big(2\text{x}+3\big)-1\big(2\text{x}+3\big)}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{\big(\sqrt{\text{x}}-1\big)(\big(2\text{x}-3\big)}{\big(\text{x}-1\big)\big(2\text{x}+3\big)}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{2\text{x}-3}{\big(\sqrt{\text{x}}+1\big)\big(2\text{x}+3\big)}$
Taking limit, we get
$=\frac{2(1)-3}{\big(\sqrt{\text{x}}+1\big)\big(2\times1+3\big)}$
$=\frac{-1}{2\times5}$
$=\frac{-1}{10}$
View full question & answer→MCQ 1601 Mark
If $\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is
Answer$\text{y}=\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}$
$=\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}$
Differentiate both the sides with respect to $x$, we get
$\frac{1}{2}\text{x}^{\frac{1}{2}-1}+\Big(-\frac{1}{2}\Big)\text{x}^{-\frac{1}{2}-1}$
$\frac{1}{2}\text{x}^{-\frac{1}{2}}-\Big(\frac{1}{2}\Big)\text{x}^{-\frac{3}{2}}$
Putting $x = 1,$ we get
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=1}=\frac{1}{2}\times1-\frac{1}{2}\times1=0$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 1$ is $0.$
View full question & answer→MCQ 1611 Mark
$\lim_{x\rightarrow 0} x \sin \frac{1}{\text{x}}$ is equal to:
AnswerWe know that,
$= \lim_\limits{\text{x}\rightarrow 0} \text{x} = 0$
And
$= -1 \leq \sin \frac{1}{\text{x}} \leq 1$
By Sandwich theorem,
$=\lim_\limits{\text{x}\rightarrow 0} \text{x} \sin \frac{1}{\text{x}}$
$=0$
View full question & answer→MCQ 1621 Mark
What is the value of $\lim_\limits{\text{x} \rightarrow 3}\frac{\text{x}^2-9}{\text{x}-3}?$
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.
View full question & answer→MCQ 1631 Mark
Find the derivative of $ e^{x^2} $:
- A
$ e^{x^2} $
- B
$2^x$
- C
$ 2 e^{x^2} $
- ✓
$2xe^{x^2}$
AnswerCorrect option: D. $2xe^{x^2}$
We apply chain rule.
First we differentiate $x^2$.
$\frac{\text{d}}{\text{dx}} (x^2 ) = 2\text{x}$
Now, we know that $\frac{\text{d}}{\text{dx}}(e^x ) = e^x$
We differentiate $ex^2$ in the same manner and then,
multiply with the derivative of $\frac{{\text{x}}^2 \text{d}}{\text{dx} (\text{e}^{x^2})} = \text{2xe}^{x^2}.$
View full question & answer→MCQ 1641 Mark
Choose the correct answer.
If $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$ then the quadeatic equation whose roots are $\lim\limits_{\text{x} \rightarrow 2^{-}}\text{f}(\text{x})$ and $\lim\limits_{\text{x} \rightarrow 2^{+}}\text{f}(\text{x})$ is:
- A
$\text{x}^{2}-6\text{x}+9=0$
- B
$\text{x}^{2}-7\text{x}+8=0$
- C
$\text{x}^{2}+14\text{x}+49=0$
- ✓
$\text{x}^{2}-10\text{x}+21=0$
AnswerCorrect option: D. $\text{x}^{2}-10\text{x}+21=0$
Given $\text{f}(\text{x})=\begin{cases}\text{x}^{2}-1 & 0<\text{x}<2\\2\text{x}+3, & 2\geq\text{3}<3\end{cases}$
$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{-}}(\text{x}^{2}-1)$
$ \lim\limits_{\text{h} \rightarrow 0}[(2-\text{x})^{2}-1]$
$=\lim\limits_{\text{h} \rightarrow 0}(4+\text{h}^{2}-4\text{h}-1)$
$ =\lim\limits_{\text{h} \rightarrow 0}(\text{h}^{2}-4\text{h}+3)=3$
$\therefore\ \lim\limits_{\text{x} \rightarrow 2}\text{f}(\text{x})=\lim\limits_{\text{x} \rightarrow 2^{+}}(2\text{x}+3)$
$ =\lim\limits_{\text{h} \rightarrow 0}[2(2+\text{h})+3]$
$=7$
Therefore, the quadratic equation whose roots are $3$ and $7$ is $\text{x}^{2}-10\text{x}+21=0$
View full question & answer→MCQ 1651 Mark
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}$ is equal to:
- ✓
$2$
- B
$\frac{3}{2}$
- C
$-\frac{3}{2}$
- D
$1$
AnswerGiven $\lim\limits_{\text{x} \rightarrow \pi}\frac{\text{x}^{2}\cos\text{x}}{1-\cos\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{\text{x}^{2}\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\frac{\text{x}^{2}}{4}\times4\cos\text{x}}{2\sin^{2}\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\frac{\text{x}}{2}\Big)\cdot2\cos\text{x}}{\sin^{2}\frac{\text{x}}{2}}$
$=\lim\limits_{\text{x} \rightarrow 0}\bigg(\frac{\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\bigg)\cdot2\cos\text{x}$
$=2\cos\text{x}$
$=2\times1$
$=2$
View full question & answer→MCQ 1661 Mark
If $\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}},$ then $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is:
Answer$\text{y}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}-\cos\text{x}}$
Differentiate both the sides with respect to $x,$ we get
$=\frac{(\sin\text{x}-\cos\text{x})(\cos\text{x}-\sin\text{x})-(\sin\text{x}+\cos\text{x})(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-(\cos^2\text{x}+\sin^2\text{x}-2\cos\text{x}\sin\text{x})-(\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x})}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-1+2\cos\text{x}\sin\text{x}-1-2\sin\text{x}\cos\text{x}}{(\sin\text{x}-\cos\text{x})^2}$
$=\frac{-2}{(\sin\text{x}-\cos\text{x})^2}$
Putting $x = 0$ is $-2$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=0}$
$=\frac{-2}{(\sin0-\cos0)}$
$=\frac{-2}{(0-1)^2}$
$=-2$
Thus, $\frac{\text{dy}}{\text{dx}}$ at $x = 0$ is $-2$
View full question & answer→MCQ 1671 Mark
If $\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots,$ then $\frac{\text{dy}}{\text{dx}}=$
- A
$y + 1$
- B
$y - 1$
- ✓
$y$
- D
$\text{y}^2$
Answer$\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\text{y}=1+\frac{\text{x}}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\Big)$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{1!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{2!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^3}{3!}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^4}{4!}\Big)+\dots$
$=\frac{\text{d}}{\text{dx}}(1)+\frac{1}{1!}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{2!}\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{1}{3 !}\frac{\text{d}}{\text{dx}}(\text{x}^3)+\frac{1}{4!}\frac{\text{d}}{\text{dx}}(\text{x}^4)+\dots$
$=0+\frac{1}{1!}\times1+\frac{1}{2!}\times2\text{x}+\frac{1}{3!}\times3\text{x}^2+\frac{1}{4!}\times4\text{x}^3+\dots$
$=1+\frac{1}{1!}+\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}+\dots\ \Big[\frac{\text{n}}{\text{n}!}=\frac{1}{(\text{n}-1)!}\Big]$
$=\text{y}$
View full question & answer→MCQ 1681 Mark
$\lim_\limits{\text{x} \rightarrow\infty}\frac{\sqrt{\text{x}^2+1}-^3\sqrt{\text{x}^2+1}}{^4\sqrt{\text{x}^4+1}-^5\sqrt{\text{x}^4+1}}$ is equal to:
Answer$\lim_\limits{\text{x} \rightarrow\infty}\frac{\sqrt{\text{x}^2+1}-^3\sqrt{\text{x}^2+1}}{^4\sqrt{\text{x}^4+1}-^5\sqrt{\text{x}^4+1}}$
$=\displaystyle \lim_{\text{x}\rightarrow \infty}\frac {\sqrt {1+1/\text{x}^2}-\sqrt [3]{1/\text{x}+1/\text{x}^3}}{\sqrt [4]{1+1/\text{x}^4}-\sqrt [5]{1/\text{x}-1/\text{x}^5}}$
$ =\frac{1-0}{1-0}$
$=1$
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