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. $g(x) > 0$
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