datavorous/entrance-exam-dataset huggingface dataset:
What is $\int_{e^{-1}}^{\mathrm{e}^{2}}\left|\frac{\ln x}{\mathrm{x}}\right| \mathrm{d} \mathrm{x}$ equal to?

Options:
A $\frac{3}{2}$
B $\frac{5}{2}$
C 3
D 4

The correct answer is:
$\frac{5}{2}$

Solution:
$\int_{x-1}^{e^{2}}\left|\frac{\log x}{x}\right| d x=\int_{s^{-1}}^{e^{0}} \frac{-\log x}{x} d x+\int_{\varepsilon^{0}}^{e^{2}} \frac{\log x}{x} d x$
$=\frac{-1}{2}\left[(\log \mathrm{x})^{2}\right]_{\mathrm{e}}^{\mathrm{e}^{0}}-\frac{1}{2}\left[(\log \mathrm{x})^{2}\right]_{e^{0}}^{\mathrm{c}^{2}}$
$=\frac{-1}{2}\left[0-\left(\log \mathrm{e}^{-1}\right)^{2}\right]+\frac{1}{2}\left[\left(\log \mathrm{e}^{2}\right)^{2}-0\right]$
$=\frac{-1}{2}(-1)+\frac{1}{2}(2)^{2}$
$=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}$

Topic: Mathematics, Definite Integration, NDA, NDA 2017 (Phase 1)
What is $\int e^{e^{x}} e^{x} d x$ equal to?

Options:
A $e^{e^{x}}+c$
B $2 e^{e^{x}}+c$
C $e^{e^{x}} e^{x}+c$
D $2 e^{e^{x}} e^{x}+c$

The correct answer is:
$e^{e^{x}}+c$

Solution:
$\mathrm{I}=\int \mathrm{e}^{\mathrm{e}^{\mathrm{x}}} \mathrm{e}^{\mathrm{x}} \mathrm{dx}$
Let $\mathrm{e}^{x}=\mathrm{y} \Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dy}$
$\mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{e}^{\mathrm{x}}}$
$\mathrm{I}=\int \mathrm{e}^{\mathrm{y}} \mathrm{e}^{\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{e}^{\mathrm{x}}}=\int \mathrm{e}^{\mathrm{y}} \mathrm{dy}=\mathrm{e}^{y}+\mathrm{c}$
$\mathrm{I}=\mathrm{e}^{\mathrm{e}^{\mathrm{x}}}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2013 (Phase 2)
What is $\int \mathrm{e}^{\operatorname{In}(\tan x)} \mathrm{dx}$ equal to?

Options:
A In $|\tan x|+c$
B $\operatorname{In}|\sec x|+c$
C $\tan x+\mathrm{c}$
D $\mathrm{e}^{\tan \mathrm{x}}+\mathrm{c}$
Where $c$ is the constant of integration.

The correct answer is:
$\operatorname{In}|\sec x|+c$

Solution:
$\int e^{\ln (\tan x)} d x=\int \tan x d x$
$=\ell n|\sec x|+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2018 (Phase 2)
What is $\int \mathrm{e}^{\ln \mathrm{x}} \mathrm{dx}$ equal to?

Options:
A $\mathrm{xe}^{\mathrm{fnx}}+\mathrm{c}$
B $-x e^{-i n x}+c$
C $x+c$
D $\frac{x^{2}}{2}+c$
where $\mathrm{c}$ is constant of integration.

The correct answer is:
$\frac{x^{2}}{2}+c$
where $\mathrm{c}$ is constant of integration.

Solution:
Let $\mathrm{I}=\int \mathrm{e}^{\ln \mathrm{x}} \mathrm{dx}=\int \mathrm{x} \mathrm{dx}=\frac{\mathrm{x}^{2}}{2}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2013 (Phase 1)
What is $\int e^{\ln x} \sin x d x$ equal to?

Options:
A $e^{\ln x}(\sin x-\cos x)+c$
B $(\sin x-x \cos x)+c$
C $(x \sin x+\cos x)+c$
D $(\sin x+x \cos x)-c$
Where 'c' is a constant of integration.

The correct answer is:
$(\sin x-x \cos x)+c$

Solution:
Let $\mathrm{I}=\int e^{\ln x} \sin x d x=\int x \sin x d x\left(\because e^{\log a}=a\right)$
$=-x \cos x+\int 1 \cdot \cos x d x=(\sin x-x \cos x)+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2010 (Phase 1)
What is $\int \mathrm{e}^{\sin \mathrm{x}} \frac{\mathrm{x} \cos ^{3} \mathrm{x}-\sin \mathrm{x}}{\cos ^{2} \mathrm{x}} \mathrm{dx}$ equal to?

Options:
A $(x+\sec x) e^{\sin x}+c$
B $(x-\sec x) e^{\sin x}+c$
C $(x+\tan x) e^{\sin x}+c$
D $(x-\tan x) e^{\sin x}+c$

The correct answer is:
$(x-\sec x) e^{\sin x}+c$

Solution:
Let us differentiate all the options one by one to get the expression in the question whose integral is to be found.
Here $x e^{\sin x}$ is the common term in all the options. So, let us differentiate it first.
Let $l=x e^{\sin x}$
$\Rightarrow \frac{d l}{d x}=e^{\sin x}[x \cos x+1]$
$\Rightarrow \frac{d l}{d x}=\frac{e^{\sin x}}{\cos ^{2} x}\left[x \cos ^{3} x+\cos ^{2} x\right]$
Let $m=\sec x e^{\sin x}$
$\Rightarrow \frac{d m}{d x}=\sec x e^{\sin x} \cdot \cos x+e^{\sin x} \sec x \tan x$
$\Rightarrow \frac{d m}{d x}=e^{\sin x}\left[1+\frac{\sin x}{\cos ^{2} x}\right]$
$\Rightarrow \frac{d m}{d x}=\frac{e^{\sin x}}{\cos ^{2} x}\left[\cos ^{2} x+\sin x\right]$
Differentiation of option (a) is
$=\frac{e^{\sin x}}{\cos ^{2} x}\left[x \cos ^{3} x+\cos ^{2} x+\cos ^{2} x+\sin x\right]$
$=\frac{e^{\sin x}}{\cos ^{2} x}\left[x \cos ^{3} x+2 \cos ^{2} x+\sin x\right]$
Differentiation of option (b) is
$=\frac{e^{\sin x}}{\cos ^{2} x}\left[x \cos ^{3} x+\cos ^{2} x-\cos ^{2} x-\sin x\right]$
$=\frac{e^{\sin x}}{\cos ^{2} x}\left[x \cos ^{3} x-\sin x\right]$
$\therefore$ Option (b) is correct.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2016 (Phase 2)
What is $\int\left(e^{x}+1\right)^{-1} d x$ equal to?

Options:
A $\ln \left(e^{x}+1\right)+c$
B $\ln \left(e^{-x}+1\right)+c$
C $-\ln \left(e^{-x}+1\right)+c$
D $-\left(e^{x}+1\right)+c$

The correct answer is:
$-\ln \left(e^{-x}+1\right)+c$

Solution:
Let $\mathrm{I}=\int\left(e^{x}+1\right)^{-1} d x=\int \frac{1}{e^{x}+1} d x=\int \frac{e^{-x}}{1+e^{-x}} d x$
Let $1+e^{-x}=t \Rightarrow-e^{-x} d x=d t$
$\therefore I=-\int \frac{1}{t} d t=-\log t+c=-\log \left(1+e^{-x}\right)+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2008 (Phase 2)
What is $\int e^{x \ln (a)} d x$ equal to?

Options:
A $\frac{a^{x}}{\ln (a)}+c$
B $\frac{e^{x}}{\ln (a)}+c$
C $\frac{e^{x}}{\ln (a e)}+c$
D $\frac{a e^{x}}{\ln (a)}+c$

The correct answer is:
$\frac{a^{x}}{\ln (a)}+c$

Solution:
$\int e^{x \ln (a)} \cdot d x=\int e^{x} \cdot d x=\frac{a^{x}}{\ln (a)}+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2019 (Phase 1)
What is $\int e^{x}\left(\sqrt{x}+\frac{1}{2 \sqrt{x}}\right) d x$ equal to

Options:
A $x e^{x}+C$
B $e^{x}(\sqrt{x})+C$
C $2 e^{x}(\sqrt{x})+C$
D $2 x e^{x}+C$
(where $\mathrm{C}$ is a constant of integration.)

The correct answer is:
$e^{x}(\sqrt{x})+C$

Solution:
Let $I=\int e^{x}\left(\sqrt{x}+\frac{1}{2 \sqrt{x}}\right) d x$
$=\int e^{x} \sqrt{x} d x+\int e^{x} \cdot \frac{1}{2 \sqrt{x}} d x$
$=e^{x} \cdot \sqrt{x}-\int e^{x} \cdot \frac{1}{2 \sqrt{x}} d x+\int e^{x} \cdot \frac{1}{2 \sqrt{x}} d x$
$=e^{x} \cdot \sqrt{x}+C$
where $C$ is constant of integration.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2011 (Phase 1)
What is equation of straight line pass through the point of
intersection of the line $\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{3}+\frac{y}{2}=1$, and parallel the $4 \mathrm{x}+5 \mathrm{y}-6=0 ?$

Options:
A $20 \mathrm{x}+25 \mathrm{y}-54=0$
B $25 x+20 y-54=0$
C $4 x+5 y-54=0$
D $4 x+5 y-45=0$

The correct answer is:
$20 \mathrm{x}+25 \mathrm{y}-54=0$

Solution:
Given equatiosns: $\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{3}+\frac{y}{2}=1$
Point of intersection $=\left(\frac{6}{5}, \frac{6}{5}\right)$
Let equation of line be $4 x+5 y+k=0$
Putting $\left(\frac{6}{5}, \frac{6}{5}\right), k=-\frac{54}{5}$
Equation of line is $20 x+25 y-54=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 2)
What is $f(2 x)$ equal to ?
Consider the function $f(x)=\frac{x-1}{x+1}$.

Options:
A $\frac{f(x)+1}{f(x)+3}$
B $\frac{f(x)+1}{3 f(x)+1}$
C $\frac{3 f(x)+1}{f(x)+3}$
D $\frac{f(x)+3}{3 f(x)+1}$

The correct answer is:
$\frac{3 f(x)+1}{f(x)+3}$

Solution:
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-1}{\mathrm{x}+1} \Rightarrow x=\frac{1+\mathrm{f}(\mathrm{x})}{1-\mathrm{f}(\mathrm{x})}$
$\Rightarrow \mathrm{f}(2 \mathrm{x})=\frac{2 \mathrm{x}-1}{2 \mathrm{x}+1} \Rightarrow \mathrm{f}(2 \mathrm{x})=\frac{2\left[\frac{[1+\mathrm{f}(\mathrm{x})]}{1-\mathrm{f}(\mathrm{x})}\right]-1}{2\left[\frac{[1+\mathrm{f}(\mathrm{x})]}{1-\mathrm{f}(\mathrm{x})}\right]+1}$
$\Rightarrow \mathrm{f}(2 \mathrm{x})=\frac{3 \mathrm{f}(\mathrm{x})+1}{\mathrm{f}(\mathrm{x})+3}$

Topic: Mathematics, Functions, NDA, NDA 2014 (Phase 2)
What is $f(f(x))$ equal to ?
Consider the function $f(x)=\frac{x-1}{x+1}$.

Options:
A x
B -x
C $-\frac{1}{x}$
D None of these

The correct answer is:
$-\frac{1}{x}$

Solution:
Given $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-1}{\mathrm{x}+1}$
$\Rightarrow \mathrm{f}[\mathrm{f}(\mathrm{x})]=\frac{\mathrm{f}(\mathrm{x})-1}{\mathrm{f}(\mathrm{x})+1}$
$\Rightarrow \mathrm{f}[\mathrm{f}(\mathrm{x})]=-\frac{1}{\mathrm{x}} \quad\left[\because \mathrm{x}=-\left\{\frac{\mathrm{f}(\mathrm{x})+1}{-\mathrm{f}(\mathrm{x})-1}\right\}\right]$

Topic: Mathematics, Functions, NDA, NDA 2014 (Phase 2)
What is Hinsberg reagent?

Options:

None

Solution:
Beneze sulphonyl chloride $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}\right)$ is as Hinsberg reagent. It is used to distinguish between primary, secondary and tertiary amine.

Topic: Chemistry, Amines
What is $\mathrm{i}^{1000}+\mathrm{i}^{1001}+\mathrm{i}^{1002}+\mathrm{i}^{1003}$ equal to (where
$\mathrm{i}=\sqrt{-1}$ ?

Options:
A 0
B i
C -i
D 1

The correct answer is:
0

Solution:
$\mathrm{i}^{1000}+\mathrm{i}^{1001}+\mathrm{i}^{1002}+\mathrm{i}^{1003}$
$=\mathrm{i}=\sqrt{-1}$
$=\mathrm{i}^{1000}\left(1+\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}\right)$
$\Rightarrow \mathrm{i}^{2}=-1$
$\begin{array}{ll}= & \mathrm{i}^{1000}(\not{\chi}+\dot{\lambda}-\not{\lambda}-\dot{\lambda}) \quad \begin{array}{l}\mathrm{i}^{3}=\mathrm{i}^{2} . \mathrm{i} \\ =-\mathrm{i}\end{array} \\ =0\end{array}$

Topic: Mathematics, Complex Number, NDA, NDA 2018 (Phase 1)
What is $\sqrt{-\mathrm{i}}$ where $\mathrm{i}=\sqrt{-1}$ equal to?

Options:
A $\pm \frac{1-\mathrm{i}}{\sqrt{2}}$
B $\pm \frac{1+\mathrm{i}}{\sqrt{2}}$
C $\pm \frac{1-\mathrm{i}}{2}$
D $\pm \frac{1+\mathrm{i}}{2}$

The correct answer is:
$\pm \frac{1-\mathrm{i}}{\sqrt{2}}$

Solution:
Consider $\sqrt{-\mathrm{i}}=\sqrt{\mathrm{e}^{-\mathrm{i} \pi / 2}}=\pm \mathrm{e}^{-\mathrm{i} \pi / 4}=\pm\left(\frac{1-\mathrm{i}}{\sqrt{2}}\right)$

Topic: Mathematics, Complex Number, NDA, NDA 2013 (Phase 1)
What is imbricate aestivation?

Options:
A When the margins of sepals or petals overlap one another but not in any particular direction
B When sepals or petals overlap one another in a particular direction
C When sepals or petals just touch one another at the margin, without overlapping
D When one margin of the appendage overlaps that of the next one and so on

The correct answer is:
When the margins of sepals or petals overlap one another but not in any particular direction

Solution:
Imbricate aestivation refers to a situation where the margins of the sepals or petals overlap one another, but not in any particular direction.

Topic: Biology, Morphology of Flowering Plants, NEET
What is $\frac{1}{\log _{2} \mathrm{~N}}+\frac{1}{\log _{3} \mathrm{~N}}+\frac{1}{\log _{4} \mathrm{~N}}+\ldots .+\frac{1}{\log _{100} \mathrm{~N}}$ equal
to $(\mathrm{N} \neq 1)$ ?

Options:
A $\frac{1}{\log _{1001} \mathrm{~N}}$
B $\frac{1}{\log _{991} N}$
C $\frac{99}{\log _{1001} \mathrm{~N}}$
D $\frac{99}{\log _{991} \mathrm{~N}}$

The correct answer is:
$\frac{1}{\log _{1001} \mathrm{~N}}$

Solution:
$\frac{1}{\log _{2} \mathrm{~N}}+\frac{1}{\log _{3} \mathrm{~N}}+\frac{1}{\log _{4} \mathrm{~N}}+\ldots .+\frac{1}{\log _{100} \mathrm{~N}}$
$=\frac{\frac{1}{\log N}}{\log _{2}}+\frac{1}{\frac{\log N}{\log _{3}}}+\frac{1}{\frac{\log N}{\log _{4}}}+\ldots .+\frac{1}{\frac{\log N}{\log _{1} \infty}}$
$\left(\because \log _{a} b=\frac{\log b}{\log a}\right)$
$=\frac{\log _{2}}{\log N}+\frac{\log _{3}}{\log N}+\frac{\log _{4}}{\log N}+\ldots .+\frac{\log _{100}}{\log N}$
$=\frac{\log _{2}+\log _{3}+\log _{4}+\ldots+\log _{100}}{\log \mathrm{N}}$
$=\frac{\log (2.3 .4 \ldots .100)}{\log \mathrm{N}}$
$=\frac{\log 100 !}{\log N}=\frac{1}{\frac{\log \mathrm{N}}{\log 100 !}}=\frac{1}{\log _{100 !} \mathrm{N}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Options:

None

Solution:
Lanthanoid Contraction : In the lanthanoids, the electrons are getting filled in the \(4 f\)-subshell. On moving from left to right, the nuclear charge increases and this increase is expected to be compensated by the increase in the magnitude of shielding effect by the \(4 f\) - electrons However, the \(f\)-electrons have very poor shielding effect. Consequently, the atomic and ionic radii decrease from left to right and this is knwon as lanthanoid contraction.
Consequences of lanthanoid Contraction
(a) Separation of Lanthanoids: All the lanthanoids have quite similar properties and due to this reason they are difficult to separate.
(b) Variation in basic strength of hydroxides:
Due to lanthanoid contraction, size of \(\mathrm{M}^{3+}\) ions decreases and thus there is a corresponding increase in the covalent character in \(\mathrm{M}-\mathrm{OH}\) bond. Thus basic character of oxides and hydroxides decreases from \(\mathrm{La}(\mathrm{OH})_3\) to \(\mathrm{Lu}(\mathrm{OH})_3\).
(c) Similarity in the atomic sizes of the elements of second and third transition series present in the same group. The difference in the value of atomic radii of \(\mathrm{Y}\) and La is quite large as compared to the difference in the value of \(\mathrm{Zr}\) and \(\mathrm{Hf}\). This is because of the lanthanoid contraction.
(d) Variation in standard reduciton potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential \(\left(\mathrm{E}^{\circ}\right)\) for the reduction process.
\(\mathrm{M}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{M}(a q)\)
(e) Variation in physical properties like melting point, boiling point, hardness etc.

Topic: Chemistry, d and f Block Elements
What is $\lim _{\theta \rightarrow 0} \frac{\sqrt{1-\cos \theta}}{\theta}$ equal to?

Options:
A $\sqrt{2}$
B $2 \sqrt{2}$
C $\frac{1}{\sqrt{2}}$
D $-\frac{1}{2 \sqrt{2}}$

The correct answer is:
$\frac{1}{\sqrt{2}}$

Solution:
$\begin{aligned} & \lim _{\theta \rightarrow 0} \frac{\sqrt{1-\cos \theta}}{\theta}=\lim _{\theta \rightarrow 0} \frac{\sqrt{2} \sin (\theta / 2)}{\theta} \\ &=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2018 (Phase 2)
What is $\lim _{h \rightarrow 0} \frac{\sqrt{2 x+3 h}-\sqrt{2 x}}{2 h}$ equal to?

Options:
A $\frac{1}{2 \sqrt{2 x}}$
B $\frac{3}{\sqrt{2 \mathrm{x}}}$
C $\frac{3}{2 \sqrt{2 x}}$
D $\frac{3}{4 \sqrt{2 x}}$

The correct answer is:
$\frac{3}{4 \sqrt{2 x}}$

Solution:
$\lim _{h \rightarrow 0} \frac{\sqrt{2 x+3 h}-\sqrt{2 x}}{2 h}$
Rationalise the numerator.
$\lim _{h \rightarrow 0}\left(\frac{\sqrt{2 x+3 h}-\sqrt{2 x}}{2 h} \times \frac{\sqrt{2 x+3 h}+\sqrt{2 x}}{\sqrt{2 x+3 h}+\sqrt{2 x}}\right)$
$=\lim _{h \rightarrow 0} \frac{2 x+3 h-2 x}{2 h(\sqrt{2 x+3 h}+\sqrt{2 x})}$
$=\frac{3}{2(\sqrt{2 x+0}+\sqrt{2 x})}=\frac{3}{4 \sqrt{2 x}}$

Topic: Mathematics, Limits, NDA, NDA 2018 (Phase 1)
What is $\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{1^{2}+2^{2}+3^{2}+\ldots n^{2}}$ equal to ?

Options:
A 5
B 2
C 1
D 0

The correct answer is:
0

Solution:
$\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{1^{2}+2^{2}+3^{2}+\ldots .+n^{2}}=\lim _{n \rightarrow \infty} \frac{\frac{n(n+1)}{2}}{\frac{n(n+1)(2 n+1)}{6}}$
$\therefore \lim _{n \rightarrow \infty} \frac{3}{2 n+1}=0$
Note: $1+2+3+\ldots .+n=\frac{n(n+1)}{2}$
$1^{2}+2^{2}+3^{2}+\ldots .+n^{2}=\frac{n(n+1)(2 n+1)}{6}$

Topic: Mathematics, Limits, NDA, NDA 2014 (Phase 2)
What is $\lim _{x \rightarrow 0} \frac{x}{\sqrt{1-\cos x}}$ equal to ?

Options:
A $\sqrt{2}$
B $-\sqrt{2}$
C $\frac{1}{\sqrt{2}}$
D Limit does not exist

The correct answer is:
Limit does not exist

Solution:
$\lim _{x \rightarrow 0} \frac{x}{\sqrt{1-\cos x}}=\lim _{x \rightarrow 0} \frac{x}{\sqrt{1-\left(1-2 \sin ^{2} \frac{x}{2}\right)}}$
$=\lim _{x \rightarrow 0} \frac{x}{\sqrt{2 \sin ^{2} \frac{x}{2}}}=\frac{1}{\sqrt{2}} \lim _{x \rightarrow 0} \frac{x}{\left|\sin \frac{x}{2}\right|}$
L.H.L $=f(0-0)=\lim _{h \rightarrow 0} \frac{x}{\left|\sin \frac{x}{2}\right|}$
$=-\frac{1}{\sqrt{2}} \lim _{x \rightarrow 0} \frac{2\left(\frac{h}{2}\right)}{\sin \frac{\mathrm{h}}{2}}$
$=-\frac{1}{\sqrt{2}} \times 2 \times 1 \quad\left(\because \lim _{\theta \rightarrow 0} \frac{\theta}{\sin \theta}=1\right)$
$=-\sqrt{2}$
RHL $=\mathrm{f}(0+0)=\lim _{h \rightarrow 0} \mathrm{f}(0+\mathrm{h})$
$=\frac{1}{\sqrt{2}} \lim _{\mathrm{h} \rightarrow 0} \frac{2\left(\frac{\mathrm{h}}{2}\right)}{\sin \frac{\mathrm{h}}{2}}=\frac{1}{\sqrt{2}} \times 2 \times 1$
$=\mathrm{LHL} \neq \mathrm{RHL}=\sqrt{2}$
Therefore limit does not exist.

Topic: Mathematics, Limits, NDA, NDA 2014 (Phase 1)
What is $\lim _{x \rightarrow 0} \frac{\sin 2 x+4 x}{2 x+\sin 4 x}$ equal to ?

Options:
A 0
B $\frac{1}{2}$
C 1
D 2

The correct answer is:
1

Solution:
$\lim _{x \rightarrow 0} \frac{\sin 2 x+4 x}{2 x+\sin 4 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x}{x}+4}{2+\frac{\sin 4 x}{x}}$
$=\lim _{x \rightarrow 0} \frac{2\left(\frac{\sin 2 x}{2 x}\right)+4}{2+4\left(\frac{\sin 4 x}{4 x}\right)}$
Applying limit, we get $\frac{2+4}{2+4}=1$

Topic: Mathematics, Limits, NDA, NDA 2013 (Phase 2)
What is $\lim _{x \rightarrow 0} e^{-\frac{1}{x^{2}}}$ equal to?

Options:
A 0
B 1
C $-1$
D Limit does not exist

The correct answer is:
0

Solution:
$\lim _{x \rightarrow 0} e^{-\frac{1}{x^{2}}}=e^{\frac{-1}{0}}=e^{-\infty}=0$

Topic: Mathematics, Limits, NDA, NDA 2016 (Phase 1)
What is $\lim _{x \rightarrow 0} f(x)$ equal to?
Consider the function $f(x)=\left\{\begin{array}{ccc}\frac{\alpha \cos x}{\pi-2 x} & \text { If } & x \neq \frac{\pi}{2} \\ 3 & \text { If } & x=\frac{\pi}{2}\end{array}\right.$
Which is continuous at $x=\frac{\pi}{2}$, where $\alpha$ is a constant.

Options:
A 0
B 3
C $\frac{3}{\pi}$
D $\frac{6}{\pi}$

The correct answer is:
$\frac{6}{\pi}$

Solution:
$\lim _{x \rightarrow 0} f(x)=\frac{\alpha \cos x}{\pi-2 x}$
$=\frac{\alpha(\cos 0)}{\pi-2(0)}=\frac{6}{\pi}$
$\therefore \quad$ Option $(\mathrm{d})$ is correct.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 1)
What is $\lim _{x \rightarrow 0^{-}} f(x)$ equal to?
Consider the function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{a}^{[\mathrm{x}]+\mathrm{x}}-1}{[\mathrm{x}]+\mathrm{x}}$ where $[\cdot]$ denotes the
greatest integer function.

Options:
A 0
B $\ln$ a
C $1-\mathrm{a}^{-1}$
D Limit does not exist

The correct answer is:
$1-\mathrm{a}^{-1}$

Solution:
$\begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{h \rightarrow 0}\left[\frac{a^{[0-h]+(0-h)}-1}{[0-h]+(0-h)}\right] \\ &=\lim _{h \rightarrow 0} \frac{a^{[-h]-h}-1}{[-h]+(-h)} \\ &=\lim _{h \rightarrow 0} \frac{a^{-1-h}-1}{-1-h} \\ &=\frac{a^{-1-0}-1}{-1-0}=\frac{a^{-1}-1}{-1} \\ &=\lim _{h \rightarrow 0^{-}} f(x)=\left(1-a^{-1}\right) \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2016 (Phase 1)
What is $\lim _{x \rightarrow 0^{+}} f(x)$ equal to?
Consider the function $\mathrm{f}(\mathrm{x})=\frac{\mathrm{a}^{[\mathrm{x}]+\mathrm{x}}-1}{[\mathrm{x}]+\mathrm{x}}$ where $[\cdot]$ denotes the
greatest integer function.

Options:
A 1
B $\ln$ a
C $\quad 1-\mathrm{a}^{-1}$
D Limit does not exist

The correct answer is:
$\ln$ a

Solution:
Given $f(x)=\frac{a^{[x]+x}-1}{[x]+x}$
$\therefore \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} \frac{a^{[0+h]+(0+h)}-1}{[0+h]+(0+h)}$
$=\lim _{h \rightarrow 0} \frac{a^{[h]+(h)}-1}{[h]+h}$
$=\lim _{h \rightarrow 0} \frac{\left(a^{h}-1\right)}{h}$
$=\log _{e} a$

Topic: Mathematics, Limits, NDA, NDA 2016 (Phase 1)
What is $\lim _{x \rightarrow 0} \frac{\tan x}{\sin 2 x}$ equal to?

Options:
A $\frac{1}{2}$
B 1
C 2
D Limit does not exist

The correct answer is:
$\frac{1}{2}$

Solution:
$\lim _{x \rightarrow 0} \frac{\tan x}{\sin 2 x}$
$\lim _{x \rightarrow 0} \frac{\frac{\tan x}{x}}{2\left(\frac{\sin 2 x}{2 x}\right)}=\frac{1}{2} .$

Topic: Mathematics, Limits, NDA, NDA 2018 (Phase 1)
What is $\lim _{x \rightarrow 0} \frac{1-\sqrt{1+x}}{x}$ equal to?

Options:
A 1/2
B -1/2
C 1
D -1

The correct answer is:
-1/2

Solution:
Consider $\lim _{x \rightarrow 0} \frac{1-\sqrt{1+x}}{x}$
$=\lim _{x \rightarrow 0} \frac{1-(1+x)}{x(1+\sqrt{1+x})}($ ByRationalizing $)$
$=\lim _{x \rightarrow 0} \frac{-x}{x(1+\sqrt{1+x})}=\frac{-1}{2}$

Topic: Mathematics, Limits, NDA, NDA 2013 (Phase 1)
What is $\lim _{x \rightarrow 0} \frac{1-\cos x}{x}$ equal to ?

Options:
A 0
B $\frac{1}{2}$
C 1
D 2

The correct answer is:
0

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{1-\cos x}{x}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{x} \\ &=2\left[\lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2} \times 2} \cdot \lim _{x \rightarrow 0} \sin \frac{x}{2}\right] \\ &=2\left[\frac{1}{2} \times 0\right]=0 \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2013 (Phase 2)
What is $\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}$ equal to ?

Options:
A 0
B $\frac{1}{2}$
C 1
D $-\frac{1}{2}$

The correct answer is:
$\frac{1}{2}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x} \\ &=\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x} \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}=\lim _{x \rightarrow 0} \frac{1+x-1}{x[\sqrt{1+x}+1]} \\ &=\lim _{x \rightarrow 0} \frac{1}{\sqrt{1+x}+1}=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2} \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2012 (Phase 2)
What is $\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}$ equal to ?

Options:
A 0
B 1
C n
D n-1

The correct answer is:
n

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x} \\ &=\lim _{x \rightarrow 0} \frac{{ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots .+{ }^{n} C_{n} x^{n}-1}{x} \\ &=\lim _{x \rightarrow 0} \frac{x\left({ }^{n} C_{1}+{ }^{n} C_{2} x+\ldots .+{ }^{n} C_{n} x^{n-1}\right)}{x} \\ &=\lim _{x \rightarrow 0}{ }^{n} C_{1}+{ }^{n} C_{2} x+\ldots .{ }^{n} C_{n} x^{n-1} \\ & \text { Put } x=0 \Rightarrow{ }^{n} C_{1}=n \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2014 (Phase 1)
What is $\lim _{x \rightarrow 0} \frac{2(1-\cos \mathrm{x})}{\mathrm{x}^{2}}$ equal to ?

Options:
A 0
B $1 / 2$
C $1 / 4$
D 1

The correct answer is:
1

Solution:
$\begin{aligned} \lim _{x \rightarrow 0} \frac{2(1-\cos x)}{x^{2}}=\lim _{x \rightarrow 0} \frac{2.2 \sin ^{2} \frac{x}{2}}{x^{2}} \\=& 4 \lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2} \times 2} \cdot \lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2} \times 2} \\=& \lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}} \cdot \lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}}=1 \times 1=1 \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2012 (Phase 2)
What is $\lim _{x \rightarrow 0} \frac{\mathrm{e}^{\mathrm{x}}-(1+\mathrm{x})}{\mathrm{x}^{2}}$ equal to?

Options:
A 0
B $\frac{1}{2}$
C 1
D 2

The correct answer is:
$\frac{1}{2}$

Solution:
$\lim _{x \rightarrow 0} \frac{e^{x}-(1+x)}{x^{2}}$
If we keep $\mathrm{x}=0$, it is $\frac{0}{0}$.
So, applying L'Hospital role.
$\lim _{x \rightarrow 0} \frac{e^{x}-(1+x)}{x^{2}}=\frac{1}{2} \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=\frac{1}{2} \times 1=\frac{1}{2}$

Topic: Mathematics, Limits, NDA, NDA 2017 (Phase 1)
What is $\lim _{x \rightarrow 0} x^{2} \sin \left(\frac{1}{x}\right)$ equal to?

Options:
A 0
B 1
C $1 / 2$
D Limit does not exist

The correct answer is:
0

Solution:
$\begin{aligned} \lim _{\mathrm{x} \rightarrow 0}\left(\mathrm{x}^{2}\right)\left[\sin \left(\frac{1}{\mathrm{x}}\right)\right] &=0 \times \lim _{\mathrm{x} \rightarrow 0} \sin \left(\frac{1}{\mathrm{x}}\right) \\ &=0 \times \text { finite quantity }=0 \end{aligned}$

Topic: Mathematics, Limits, NDA, NDA 2012 (Phase 1)
What is $\lim _{x \rightarrow 0} \frac{5^{x}-1}{x}$ equal to ?

Options:
A $\log _{\mathrm{e}} 5$
B $\log _{5} \mathrm{e}$
C 5
D 1

The correct answer is:
$\log _{\mathrm{e}} 5$

Solution:
$\because \lim _{x \rightarrow 0} \frac{\mathrm{a}^{\mathrm{x}}-1}{\mathrm{x}}=\operatorname{loe}_{\mathrm{e}} \mathrm{a}$
$\therefore \lim _{x \rightarrow 0} \frac{5^{x}-1}{x}=\log _{e} 5$
Hence, option (a) is correct

Topic: Mathematics, Limits, NDA, NDA 2014 (Phase 2)
What is $\operatorname{Lim}_{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$ equal to?

Options:
A $\ln (\mathrm{ab})$
B $\frac{\ln a}{\ln h}$
C $\ln \left(\frac{a}{b}\right)$
D $\ln \left(\frac{b}{a}\right)$

The correct answer is:
$\ln \left(\frac{a}{b}\right)$

Solution:
Consider
$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}=\lim _{x \rightarrow 0} \frac{a^{x} \log a-b^{x} \log b}{1}$
(Using L' Hospital's rule)
$=\log \mathrm{a}-\log \mathrm{b}=\log \frac{\mathrm{a}}{\mathrm{b}}$

Topic: Mathematics, Limits, NDA, NDA 2010 (Phase 2)