datavorous/entrance-exam-dataset huggingface dataset:
What is water potential?

Options:
A The energy of water molecules in random motion
B The concentration of water in a system
C The movement of water from high energy to low energy system
D A concept fundamental to understanding water movement

The correct answer is:
A concept fundamental to understanding water movement

Solution:
Water potential (Ψw) is a concept fundamental to understanding water movement.

Topic: Biology, Transport in Plants, NEET
What is $\int \frac{x e^{x} d x}{(x+1)^{2}}$ equal to?

Options:
A $(x+1)^{2} e^{x}+c$
B $(x+1) e^{x}+c$
C $\frac{e^{x}}{x+1}+c$
D $\frac{e^{x}}{(x+1)^{2}}+c$
where $c$ is the constant integration.

The correct answer is:
$\frac{e^{x}}{x+1}+c$

Solution:
$\int \frac{x \mathrm{e}^{x}}{(1+x)^{2}} d x$
$=\int \mathrm{e}^{x}\left[\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right] \mathrm{d} x$
$=\frac{e^{x}}{1+x}+C \quad\left\{\int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)\right\}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2015 (Phase 1)
What is $\int\left(x^{2}+1\right)^{\frac{5}{2}} x d x$ equal to?

Options:
A $\left(x^{2}+1\right)^{\frac{7}{2}}+c$
B $\frac{2}{7}\left(x^{2}+1\right)^{\frac{7}{2}}+c$
C $\frac{1}{7}\left(x^{2}+1\right)^{\frac{7}{2}}+c$
D None of the above
where $\mathrm{c}$ is a constant of integration.

The correct answer is:
$\frac{1}{7}\left(x^{2}+1\right)^{\frac{7}{2}}+c$

Solution:
Let $\mathrm{I}=\int \mathrm{x}\left(\mathrm{x}^{2}+1\right)^{5 / 2} \mathrm{dx}$
Put $\mathrm{x}^{2}+1=\mathrm{t}$
$2 \mathrm{xdx}=\mathrm{dt}$
$\mathrm{xdx}=\frac{\mathrm{dt}}{2}$
$\therefore \mathrm{I}=\int(\mathrm{t})^{5 / 2} \frac{\mathrm{dt}}{2}=\frac{1}{2}\left(\frac{\mathrm{t}^{7 / 2}}{7 / 2}\right)+\mathrm{c}=\frac{1}{7}\left(\mathrm{x}^{2}+\mathrm{l}\right)^{7 / 2}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2012 (Phase 1)
What is $\int \frac{x^{4}+1}{x^{2}+1} d x$ equal to?

Options:
A $\frac{x^{3}}{3}-x+4 \tan ^{-1} x+c$
B $\frac{x^{3}}{3}+x+4 \tan ^{-1} x+c$
C $\frac{x^{3}}{3}-x+2 \tan ^{-1} x+c$
D $\frac{x^{3}}{3}-x-4 \tan ^{-1} x+c$
Where 'c' is a constant of integration.

The correct answer is:
$\frac{x^{3}}{3}-x+2 \tan ^{-1} x+c$

Solution:
$\int \frac{x^{4}+1}{x^{2}+1} d x=\int \frac{x^{4}+2-1}{x^{2}+1} d x=\int\left(\frac{x^{4}-1}{x^{2}+1}+\frac{2}{x^{2}+1}\right) d x$
$=\int\left[\frac{\left(x^{2}-1\right)\left(x^{2}+1\right)}{x^{2}+1}+\frac{2}{x^{2}+1}\right] d x$
$=\int\left(x^{2}-1+\frac{2}{x^{2}+1}\right) d x=\frac{x^{3}}{3}-x+2 \tan ^{-1} x+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2010 (Phase 1)
What is $\int \frac{d x}{\sqrt{x^{2}+a^{2}}}$ equal to?

Options:
A $\ln \left|\frac{x+\sqrt{x^{2}+a^{2}}}{a}\right|+c$
B $\ln \left|\frac{x-\sqrt{x^{2}+a^{2}}}{a}\right|+c$
C $\ln \left|\frac{x^{2}+\sqrt{x^{2}+a^{2}}}{a}\right|+c$
D None of these
where $c$ is the constant of integration.

The correct answer is:
$\ln \left|\frac{x+\sqrt{x^{2}+a^{2}}}{a}\right|+c$

Solution:
$\int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}$
Let $\mathrm{x}=\mathrm{a} \tan \mathrm{u}$
$\mathrm{dx}=\mathrm{a} \sec ^{2} \mathrm{u} \mathrm{du}$
$=\int \frac{\mathrm{a} \sec ^{2} \mathrm{u} \mathrm{du}}{\sqrt{\mathrm{a}^{2} \tan ^{2} \mathrm{u}+\mathrm{a}^{2}}}$
$=\mathrm{a} \int \frac{\sec ^{2} \mathrm{u} \mathrm{du}}{\sqrt{\mathrm{a}^{2}\left(1+\tan ^{2} \mathrm{u}\right)}} \Rightarrow \frac{\mathrm{a}}{\mathrm{a}} \int \frac{\sec ^{2} \mathrm{u} \mathrm{du}}{\sec \mathrm{u}}$
$=\int \frac{\sec ^{2} \mathrm{u} \mathrm{du}}{\sec \mathrm{u}}=\int \sec \mathrm{u} \mathrm{du}$
$=\ln [\tan (\mathrm{u})+\sec (\mathrm{u})]+\mathrm{c} \quad\left[\because \int \sec \mathrm{x} \mathrm{d} \mathrm{x}=\tan \mathrm{x}\right]$
$=\ln \left[\frac{\mathrm{x}}{\sqrt{\mathrm{a}^{2}}}+\sqrt{1+\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}}\right]+\mathrm{c}$
$=\ln \left[\frac{x}{a}+\frac{\sqrt{a^{2}+x^{2}}}{a}\right]+c$
$=\ln \left[\frac{x+\sqrt{x^{2}+a^{2}}}{a}\right]+c$
$\therefore \quad$ Option (a) is correct.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2015 (Phase 1)
What is $x \rightarrow \frac{\pi}{2} f(x)$ equal to ?
Consider the function $f(x)=\frac{1-\sin x}{(\pi-2 x)^{2}}$
Where $x \neq \frac{\pi}{2}$ and $f\left(\frac{\pi}{2}\right)=\lambda$

Options:
A 1
B $1 / 2$
C $1 / 4$
D $1 / 8$

The correct answer is:
$1 / 8$

Solution:
$\lim _{x \rightarrow \frac{\pi}{2}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2} \frac{1-\sin \mathrm{x}}{(\pi-2 \mathrm{x})^{2}}}$
$=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{-\cos \mathrm{x}}{(\pi-2 \mathrm{x})(-2)}$
$=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{\cos \mathrm{x}}{4(\pi-2 \mathrm{x})}$
and $=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{-\sin \mathrm{x}}{4(-2)}=\lim _{\mathrm{x} \rightarrow \frac{\pi}{2}} \frac{\sin \mathrm{x}}{8}$
$=\frac{1}{8} \cdot \sin \frac{\pi}{2}=\frac{1}{8} \times 1=\frac{1}{8}$

Topic: Mathematics, Functions, NDA, NDA 2014 (Phase 1)
What is $\int \frac{x^{4}-1}{x^{2} \sqrt{x^{4}+x^{2}+1}} d x$ equal to?

Options:
A $\sqrt{\frac{x^{4}+x^{2}+1}{x}}+c$
B $\sqrt{x^{4}+2-\frac{1}{x^{2}}}+c$
C $\sqrt{\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}+1}+\mathrm{c}$
D $\sqrt{\frac{x^{4}-x^{2}+1}{x}}+c$

The correct answer is:
$\sqrt{\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}+1}+\mathrm{c}$

Solution:
Take option (a)
$I_{1}=\sqrt{\frac{x^{4}+x^{3}+1}{x}}+C$
$\frac{d I_{1}}{d x}=\frac{d}{d x}\left[\left(x^{3}+x^{2}+x^{-1}\right)^{1 / 2}+C\right]$
$\frac{d I_{2}}{d x}=\frac{1}{2}\left(x^{3}+x^{2}+x^{-1}\right)^{-1 / 2}\left(3 x^{2}+2 x-x^{-2}\right)$
$=\frac{1}{2}\left[\frac{3 x^{2}+2 x-\frac{1}{x^{2}}}{\sqrt{x^{3}+x^{2}+\frac{1}{x}}}\right]$
$\frac{d I_{2}}{d x}=\frac{1}{2}\left[\frac{3 x^{4}+2 x^{3}-1}{\frac{x^{2}}{\sqrt{x}} \sqrt{x^{4}+x^{3}+1}}\right]$
Take option (b):
$I_{2}=\sqrt{x^{4}+2-\frac{1}{x^{2}}}+C$
$\frac{d I_{2}}{d x}=\frac{1}{2}\left[x^{4}+2-x^{-2}\right]^{-1 / 2}\left[4 x^{3}+0+2 x^{-3}\right]$
$\quad=\frac{1}{2}\left[\frac{4 x^{3}+\frac{2}{x^{3}}}{\sqrt{x^{4}+2-\frac{1}{x^{2}}}}\right]=\frac{2 x^{6}+1}{\frac{x^{3}}{x} \sqrt{x^{6}+2 x^{2}-1}}$
$\frac{d I_{2}}{d x}=\frac{2 x^{6}+1}{x^{2} \sqrt{x^{6}+2 x-1}}$
Take option (c):
$I_{3}=\sqrt{x^{2}+x^{-2}+1}+C$
$\frac{d I_{3}}{d x}=\frac{1}{2}\left[x^{2}+x^{-2}+1\right]^{-1 / 2}\left[2 x-2 x^{-3}+0\right]$
$=\frac{1}{2}\left[\frac{2 x-\frac{2}{x^{3}}}{\sqrt{x^{2}+\frac{1}{x^{2}}+1}}\right]$
$=\frac{1}{2}\left[\frac{2\left(x^{4}-1\right)}{x^{3} \sqrt{\frac{x^{4}+1+x^{2}}{x^{2}}}}\right]$
$\frac{d I_{3}}{d x}=\frac{x^{4}-1}{x^{2} \sqrt{x^{4}+x^{2}+1}}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2016 (Phase 2)
What is $\int\left(\mathrm{x}^{3}+\sin \mathrm{x}\right) \mathrm{d} \mathrm{x}$ equal to

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

The correct answer is:
0

Solution:
Let $\mathrm{I}=\int_{-\mathrm{a}}^{\mathrm{a}}\left(\mathrm{x}^{3}+\sin \mathrm{x}\right) \mathrm{d} \mathrm{x}$
Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\sin \mathrm{x}$
$\mathrm{f}(-\mathrm{x})=(-\mathrm{x})^{3}+\sin (-\mathrm{x})=-\mathrm{x}^{3}-\sin \mathrm{x}$
$=-\left(\mathrm{x}^{3}+\sin \mathrm{x}\right)=-\mathrm{f}(\mathrm{x})$
Since $\mathrm{f}(\mathrm{x})$ is an odd function
$\therefore \quad \int_{-a}^{a} f(x) d x=0$
Since $\mathrm{f}(\mathrm{x})$ is an odd function
$\therefore \quad \int_{-a}^{a} f(x) d x=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2013 (Phase 1)
What is $\int(x \cos x+\sin x) d x$ equal to?

Options:
A $x \sin x+c$
B $x \cos x+c$
C $-x \sin x+c$
D $-x \cos x+c$
Where $c$ is an arbitrary constant.

The correct answer is:
$x \sin x+c$

Solution:
$\int(\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}) \mathrm{dx}=\int \mathrm{x} \cos \mathrm{x} \mathrm{dx}+\int \sin \mathrm{x} \mathrm{d} \mathrm{x}$
$=\mathrm{x} \sin \mathrm{x}-\int \sin \mathrm{x} \mathrm{dx}+\int \sin \mathrm{x} \mathrm{dx}=\mathrm{x} \sin \mathrm{x}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2013 (Phase 2)
What is $\int[\mathrm{x}] \mathrm{dx}+\int^{\mathrm{b}}[-\mathrm{x}] \mathrm{dx}$ equal to, where $[.]$ is the greatest $\mathrm{a} \quad \mathrm{a}$
integer function?

Options:
A $\mathrm{b}-\mathrm{a}$
B $a-b$
C 0
D $2(\mathrm{~b}-\mathrm{a})$

The correct answer is:
$a-b$

Solution:
$\int_{a}^{b}[x] d x+\int_{a}^{b}[-x] d x$
$\quad=\int_{a}^{b}([x]+[-x]) d x=\int_{a}^{b}(-1) d x=a-b$

Topic: Mathematics, Definite Integration, NDA, NDA 2018 (Phase 2)
What is $\quad$ | $\mathrm{x}$ $\mathrm{dx}$ equal to ?

Options:
A 2
B 1
C 0
D $-1$

The correct answer is:
0

Solution:
$\begin{array}{l}\text { Let } \mathrm{I}=\int_{-1}^{1} \mathrm{x}|\mathrm{x}| \mathrm{dx}=\int_{-1}^{0} \mathrm{x}(-\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{0}^{1} \mathrm{x}(\mathrm{x}) \mathrm{d} \mathrm{x} \\ \left(\because|\mathrm{x}|=\left\{\begin{array}{ccc}\mathrm{x} & \text { if } & \mathrm{x} \geq 0 \\ -\mathrm{x} & \text { if } & \mathrm{x} < 0\end{array}\right)\right. \\ = & -\int_{-1}^{0} \mathrm{x}^{2} \mathrm{~d} \mathrm{x}+\int_{0}^{1} \mathrm{x}^{2} \mathrm{dx} \\ = & \left.\frac{-\mathrm{x}^{3}}{3}\right|_{-1} ^{0}+\left.\frac{\mathrm{x}^{3}}{3}\right|_{0} ^{1}=-\left(0+\frac{1}{3}\right)+\frac{1}{3}=0\end{array}$

Topic: Mathematics, Definite Integration, NDA, NDA 2012 (Phase 2)
What is $\int \frac{\left(x^{\mathrm{e}-1}+\mathrm{e}^{\mathrm{x}-1}\right) \mathrm{dx}}{\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}}$ equal to?

Options:
A $\frac{x^{2}}{2}+c$
B $\operatorname{In}(x+e)+c$
C $\operatorname{In}\left(\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c}$
D $\frac{1}{\mathrm{e}} \operatorname{In}\left(\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c}$

The correct answer is:
$\frac{1}{\mathrm{e}} \operatorname{In}\left(\mathrm{x}^{\mathrm{e}}+\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c}$

Solution:
$\int \frac{\left(x^{e-1}+e^{x-1}\right) d x}{x^{e}+e^{x}}$
Put $x^{e}+e^{x}=t \Rightarrow e x^{e-1}+e^{x}=\frac{d t}{d x}$
$\Rightarrow\left(e x^{e-1}+e^{x}\right) d x=d t$
$\therefore \int \frac{\left(x^{e-1}+e^{x-1}\right) d x}{x^{e}+e^{x}}=\frac{1}{e} \int \frac{e x^{e-1}+e^{x}}{x^{e}+e^{x}} \cdot d x$
$=\frac{1}{e} \int \frac{d t}{t}=\frac{1}{e} \cdot \ln t+c$
$=\frac{1}{e} \cdot \ln \left(x^{e}+e^{x}\right)+c$

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

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

The correct answer is:
$2 e^{\sqrt{x}}(x-2 \sqrt{x}+2)+c$

Solution:
Let $I=\int \sqrt{x} e^{\sqrt{x}} d x$
Put, $\sqrt{x}=t$
Differentiate both side w.r.t (t)
$\Rightarrow \quad \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow d x=2 t d t$
$I=\int t e^{t} 2 t d t=2 \int t^{2} e^{t} d t$
By parts, let first function $=t^{2}$ and second function $=e^{t .}$
$=2\left[t^{2} e^{t}-\int 2 t e^{t} d t\right]=2\left[t^{2} e^{t}-2\left\{t e^{t}-\int e^{t} d t\right\}\right]$
$=2\left[t^{2} e^{t}-2 t e^{t}+2 e^{t}\right]+c$
where $c$ is a constant of integration.
$=2\left[x e^{\sqrt{x}}-2 \sqrt{x} e^{\sqrt{x}}+2 e^{\sqrt{x}}\right]+c$
$=2 e^{\sqrt{x}}(x-2 \sqrt{x}+2)+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2010 (Phase 2)
What is $\int \frac{\mathrm{dx}}{\mathrm{x} \ln \mathrm{x}}$ equal to?

Options:
A $\ln (\ln x)+c$
B $\ln x+c$
C $(\ln x)^{2}+c$
D None of the above

The correct answer is:
$\ln (\ln x)+c$

Solution:
Let $\mathrm{I}=\int \frac{\mathrm{dx}}{\mathrm{x} \ell \mathrm{n} \mathrm{x}}$
Put $\ell \mathrm{n} \mathrm{x}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$
$\therefore \quad \mathrm{I}=\int \frac{1}{\mathrm{t}} \mathrm{dt}=\ell \mathrm{nt}+\mathrm{c}=\ell \mathrm{n}(\ln \mathrm{x})+\mathrm{c}$
where $\mathrm{c}$ is the constant of integration.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2013 (Phase 1)
What is $\int \frac{\ell \mathrm{n} \mathrm{x}}{\mathrm{x}} \mathrm{d} \mathrm{x}$ equal to?

Options:
A $\frac{(\ell \mathrm{n} \mathrm{x})^{2}}{2}+\mathrm{c}$
B $\frac{(\ell \mathrm{n} \mathrm{x})}{2}+\mathrm{c}$
C $(\ln x)^{2}+\mathrm{c}$
D None of the above

The correct answer is:
$\frac{(\ell \mathrm{n} \mathrm{x})^{2}}{2}+\mathrm{c}$

Solution:
Let $\mathrm{I}=\int \frac{\ell \mathrm{nx}}{\mathrm{x}} \mathrm{dx}$
Put $\ell \mathrm{n} \mathrm{x}=\mathrm{t}$
$\frac{1}{x} \mathrm{dx}=\mathrm{dt}$
$\therefore \mathrm{I}=\int \mathrm{t} \mathrm{dt}=\frac{\mathrm{t}^{2}}{2}+\mathrm{c}$
where $\mathrm{c}$ is the constant of integration.
$=\frac{(\ln \mathrm{x})^{2}}{2}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2012 (Phase 2)
What is $\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x$ equal to?

Options:
A $\frac{\cos \sqrt{x}}{2}+C$
B $2 \cos \sqrt{x}+C$
C $\frac{-\cos \sqrt{x}}{2}+C$
D $-2 \cos \sqrt{x}+C$

The correct answer is:
$-2 \cos \sqrt{x}+C$

Solution:
Let $I=\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x$
Put $\sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow \frac{1}{\sqrt{x}} d x=2 d t$
$\therefore \quad I=2 \int \sin t d t=-2 \cos t+C=-2 \cos \sqrt{x}+C$
where ' $C^{\prime}$ is a constant of integration.

Topic: Mathematics, Indefinite Integration, NDA, NDA 2011 (Phase 1)
What is $\int \frac{\mathrm{dx}}{\mathrm{x}\left(\mathrm{x}^{7}+1\right)}$ equal to?

Options:
A $\frac{1}{2} \ln \left|\frac{x^{7}-1}{x^{7}+1}\right|+c$
B $\frac{1}{7} \ln \left|\frac{x^{7}+1}{x^{7}}\right|+c$
C $\ln \left|\frac{x^{7}-1}{7 x}\right|+c$
D $\frac{1}{7} \ln \left|\frac{x^{7}}{x^{7}+1}\right|+c$

The correct answer is:
$\frac{1}{7} \ln \left|\frac{x^{7}}{x^{7}+1}\right|+c$

Solution:
$\int \frac{\mathrm{d} x}{x\left(x^{7}+1\right)}=\int \frac{x^{6}}{x^{7}\left(x^{7}+1\right)} \cdot \mathrm{dx}$
Let $x^{7}=t$
$\Rightarrow 7 x^{6} \cdot \mathrm{dx}=\mathrm{dt} \Rightarrow x^{6} \mathrm{dx}=\frac{\mathrm{dt}}{7}$
Then $\int \frac{x^{6} \mathrm{~d} x}{x^{7}\left(x^{7}+1\right)}=\frac{1}{7} \int \frac{d t}{t(t+1)}$
$=\frac{1}{7}\left[\int_{t}^{1} \frac{d t}{t+1} d t\right]$
$=\frac{1}{7}[\ell n|t|-\ell n|t+1|]+c$
$=\frac{1}{7} \ell n\left|\frac{t}{t+1}\right|+c$
$=\frac{1}{7} \ell n\left|\frac{x^{7}}{x^{7}+1}\right|+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2017 (Phase 1)
What is $\left[\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right]$ equal to ?

Options:
A $[a x+h y+g z h+b+f \quad g+f+c]$
B $\left[\begin{array}{lll}a & h & g \\ h x & b y & f z \\ g & f & c\end{array}\right]$
C $\left[\begin{array}{l}a x+h y+g z \\ h x+b y+f z \\ g x+f y+c z\end{array}\right]$
D $[a x+h y+g z \quad h x+b y+f z \quad g x+f y+c z]$

The correct answer is:
$[a x+h y+g z \quad h x+b y+f z \quad g x+f y+c z]$

Solution:
$\left[\begin{array}{lll}x & y & z\end{array}\right]\left[\begin{array}{lll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right]$
$=[a x+h y+g z \quad h x+b y+f z \quad g x+f y+c z]$

Topic: Mathematics, Matrices, NDA, NDA 2016 (Phase 2)
What is $\int \mathrm{xe}^{\mathrm{x}} \mathrm{dx}$ equal to

Options:
A 1
B $-1$
C 0
D $\mathrm{e}$

The correct answer is:
1

Solution:
Let $I=\int_{0}^{1} x e^{x} d x=x e^{x}-\int_{0}^{1} 1 . e^{x} d x=\left[x e^{x}-e^{x}\right]_{0}^{1}$
$=(e-e)-[0-1]=1$

Topic: Mathematics, Definite Integration, NDA, NDA 2013 (Phase 1)
What is/are the functional groups of ranitidine?

Options:
A primary amine
B secondary amine
C nitro
D All of the above

The correct answer is:
All of the above

Solution:
Correct Option is : 4
All of the above

Topic: Chemistry, Chemistry in Everyday Life, NEET
What is/are the solutions of the trigonometric
equation cosec $x+\cot x=\sqrt{3}$, where $0 < x < 2 \pi$ ?

Options:
A $\frac{5 \pi}{3}$ only
B $\frac{\pi}{3}$ only
C $\pi$ only
D $\pi, \frac{\pi}{3}, \frac{5 \pi}{3}$

The correct answer is:
$\frac{\pi}{3}$ only

Solution:
Given equation:
$\operatorname{cosec} x+\cot x=\sqrt{3}$
$\operatorname{cosec} x-\cot x=\frac{1}{\sqrt{3}}$
$\Rightarrow \quad 2 \operatorname{cosec} x=\sqrt{3}+\frac{1}{\sqrt{3}}=\frac{4}{\sqrt{3}}$
$\Rightarrow \operatorname{cosec} x=\frac{2}{\sqrt{3}}$
$\Rightarrow \sin x=\frac{\sqrt{3}}{2}$
Possible values of $x=\frac{\pi}{3}, \frac{2 \pi}{3}$ $\therefore$ Option (b) is correct.

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2018 (Phase 2)
What ix the alea enclused between the cul ves $y^{2}-12 x$ and the lines $x=0$ and $y=6 ?$

Options:
A 2 sq unit
B 4 sq unit
C 6 sq unit
D 8 sq unit

The correct answer is:
6 sq unit

Solution:
Equation of given curve is $y^{2}=12 x$ At $y=6,36=12 x \Rightarrow \quad x=3$
$\therefore$ Required area $=\int_{0}^{3}\left(y_{1}-y_{2}\right) d x$ where $y_{1}$ represents line and $y_{2}$ represents the curve.
$=\int_{0}^{3}(6-\sqrt{12 x}) d x=[6 x]_{0}^{3}-\sqrt{12}\left[\frac{2 x^{3 / 2}}{3}\right]_{0}^{3}$
$=[6 \times 3]-\frac{\sqrt{12} \times 2 \times \sqrt{27}}{3}=18-12=6$ sq unit

Topic: Mathematics, Area Under Curves, NDA, NDA 2010 (Phase 1)
What measures should be taken for public hygiene to prevent the spread of infectious diseases?

Options:
A Avoid close contact with infected persons
B Consume clean food and water
C Proper disposal of waste and excreta
D Spraying insecticides in ditches, drainage areas and swamps

The correct answer is:
Proper disposal of waste and excreta

Solution:
Public hygiene measures include proper disposal of waste and excreta, periodic cleaning and disinfection of water reservoirs, pools, cesspools, and tanks, and observing standard practices of hygiene in public catering.

Topic: Biology, Human Health and Diseases, NEET
What optical phenomenon is described in the paragraph as the scattering of light in all directions by colloidal particles?

Options:
A Diffraction
B Dispersion
C Tyndall effect
D Refraction

The correct answer is:
Tyndall effect

Solution:
The optical phenomenon described in the paragraph as the scattering of light in all directions by colloidal particles is called the Tyndall effect.

Topic: Chemistry, Surface Chemistry, NEET
What organization provides recommendations for the nomenclature of coordination entities?

Options:
A United Nations
B World Health Organization (WHO)
C International Union of Pure and Applied Chemistry (IUPA
C)
D European Space Agency (ES
A)

The correct answer is:
International Union of Pure and Applied Chemistry (IUPA
C)

Solution:
The recommendations for the nomenclature of coordination entities are based on the guidelines provided by the International Union of Pure and Applied Chemistry (IUPA

Topic: Chemistry, Coordination Compounds, NEET
What practical significance does the concept of surroundings hold in thermodynamics?

Options:
A It defines the boundaries of the universe
B It helps in understanding the behavior of the system
C It isolates the system from external influences
D It determines the speed of energy transformations

The correct answer is:
It isolates the system from external influences

Solution:
The surroundings in thermodynamics serve the practical purpose of isolating the system from external influences.

Topic: Chemistry, Thermodynamics (C), NEET
What product is formed when secondary alcohols are oxidised using chromic anhydride (CrO3)?

Options:
A Carboxylic acid
B Aldehyde
C Ketone
D Alkene

The correct answer is:
Ketone

Solution:
Secondary alcohols are oxidised to form ketones when using chromic anhydride (CrO3).

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
What property of alkali metals makes them strong reducing agents?

Options:
A High ionization enthalpy
B Low sublimation enthalpy
C High hydration enthalpy
D Low electronegativity

The correct answer is:
High hydration enthalpy

Solution:
The high reducing power of alkali metals is primarily attributed to their high hydration enthalpy. This is because the small size of alkali metal ions, such as lithium (Li+), results in strong electrostatic attraction with water molecules, leading to a highly negative standard electrode potential (E0) and strong reducing properties.

Topic: Chemistry, s Block Elements, NEET
What property of amines is used for the distinction of primary, secondary, and tertiary amines using benzenesulfonyl chloride?

Options:
A Solubility in water
B Boiling point
C Acidic character
D Basicity

The correct answer is:
Acidic character

Solution:
The reaction of amines with benzene sulphonyl chloride helps distinguish their acidic character.

Topic: Chemistry, Amines, NEET
What recommendation has the IUPAC made to avoid naming disputes for new elements?

Options:
A The IUPAC recommends that new elements should be named after famous scientists.
B The IUPAC recommends that new elements should not be named until their discovery is proven.
C The IUPAC recommends that new elements should be named by the first laboratory to synthesize them.
D The IUPAC recommends that new elements should be named after the laboratory where they were discovered.

The correct answer is:
The IUPAC recommends that new elements should not be named until their discovery is proven.

Solution:
To avoid problems, the IUPAC has made a recommendation that until a new element’s discovery is proved, and its name is officially recognised, a systematic nomenclature be derived directly from the atomic number of the element using the numerical roots for 0 and numbers 1-9.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
What $\int_{\sec x^{\circ}} d x$ is equal to?

Options:
A $\log \left(\sec x^{\circ}+\tan x^{\circ}\right)+\mathrm{c}$
B $\frac{\pi \log \tan \left(\frac{\pi}{4}+\frac{\pi}{2}\right)}{180^{\circ}}+c$
C $\frac{180^{\circ} \log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}{\pi}+c$
D $\frac{180^{\circ} \log \tan \left(\frac{\pi}{4}+\frac{x}{360^{\circ}}\right)}{\pi}+c$

The correct answer is:
$\log \left(\sec x^{\circ}+\tan x^{\circ}\right)+\mathrm{c}$

Solution:
$\int \sec x^{\circ} \cdot d x=\int \frac{\sec x^{\circ} \cdot\left(\sec x^{\circ}+\tan x^{\circ}\right)}{\sec x^{\circ}+\tan x^{\circ}} \cdot d x$
Let $u=\sec x^{\circ}+\tan x^{\circ}$
$\Rightarrow \quad \frac{d u}{d x}=\sec x^{\circ}+\tan x^{\circ}+\sec ^{2} x^{\circ}$
$\Rightarrow \quad d u=\left(\sec x^{\circ} \times \tan x^{\circ}+\sec ^{2} x^{\circ}\right) \cdot d x$
$\therefore \quad \int \frac{\sec x^{\circ} \cdot\left(\sec x^{\circ}+\tan x^{\circ}\right)}{\sec x^{\circ}+\tan x^{\circ}} d x$
$=\int \frac{\sec ^{2} x^{\circ}+\sec x^{\circ} \cdot \tan x^{\circ}}{\sec x^{\circ}+\tan x^{\circ}} \cdot d x$
$\quad=\int \frac{d u}{u}=\log (u)+\mathrm{C}=\log \left(\sec x^{\circ}+\tan x^{\circ}\right)+\mathrm{C}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2009 (Phase 1)
What should be the height of a flag where a 20 feet long ladder reaches 20 feet below the flag (The angle of elevation of the top of the flag at the foot of the ladder is $60^{\circ}$ )?

Options:
A 20 feet
B 30 feet
C 40 feet
D $\begin{array}{l}\text { } 20 \sqrt{2} \text { feet } & {\underline{}}\end{array}$

The correct answer is:
30 feet

Solution:
$\begin{array}{ll}\text { } & \text { } \operatorname{In} \Delta \mathrm{ABD} \text { , }\end{array}$
$\tan 60^{\circ}=\frac{B D}{A B}$
$\Rightarrow \sqrt{3}=\frac{\mathrm{h}}{\mathrm{AB}}$
$\Rightarrow \mathrm{AB}=\frac{\mathrm{h}}{\sqrt{3}}$
$\Rightarrow \mathrm{AB}=\frac{\mathrm{h}}{3} \sqrt{3}$
Now, in $\triangle \mathrm{ABC}$
$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$
$\Rightarrow 20^{2}=\left(\frac{\mathrm{h}}{\sqrt{3}}\right)^{2}+(\mathrm{h}-20)^{2}$
$\Rightarrow \mathrm{h}^{2}+3 \mathrm{~h}^{2}-120 \mathrm{~h}=0$
$\Rightarrow 4 \mathrm{~h}^{2}-120 \mathrm{~h}=0$
$\Rightarrow \mathrm{h}(\mathrm{h}-30)=0$
$\mathrm{h}=0$ or 30
$\mathrm{h}=0$ not possible $\Rightarrow \mathrm{h}=30 \mathrm{ft}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2007 (Phase 2)
What should be the length of a closed pipe to produce resonance with sound wave
of wavelength $62 \mathrm{~cm}$, in fundamental mode?
[Neglect end correction]

Options:
A $31 \mathrm{~cm}$
B $15 \cdot 5 \mathrm{~cm}$
C $20 \cdot 6 \mathrm{~cm}$
D $46 \cdot 5 \mathrm{~cm}$

The correct answer is:
$15 \cdot 5 \mathrm{~cm}$

Solution:
(C)
In a closed pipe in fundamental mode the length of the pipe is $\frac{\lambda}{4}$. $\therefore \ell=\frac{62}{4}=15.5 \mathrm{~cm}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (19 Oct Shift 1)