datavorous/entrance-exam-dataset huggingface dataset:
What is $\sqrt{\frac{1+\omega^{2}}{1+\omega}}$ equal to, where $\omega$ is the cube root of unity?

Options:
A 1
B $\omega$
C $\omega^{2}$
D $i\omega$, where $i=\sqrt{-1}$

The correct answer is:
$\omega$

Solution:
$\quad X=\sqrt{\frac{1+\omega^{2}}{1+\omega}} \quad\left(\because 1+\omega+\omega^{2}=0\right.$ and $\left.\omega^{3}=1\right)$
$\Rightarrow X=\sqrt{\frac{-\omega}{-\omega^{2}}}=\sqrt{\frac{1}{\omega}}=\sqrt{\frac{\omega^{3}}{\omega}}=\sqrt{\omega^{2}}=\omega$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 2)
What is $\int_{-1}^{1}\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \frac{1}{\mathrm{x}}\right)\right\} \mathrm{dx}$ equal to?

Options:
A 0
B $-\frac{\pi}{4}$
C $-\frac{\pi}{2}$
D $\frac{\pi}{2}$

The correct answer is:
$-\frac{\pi}{2}$

Solution:
$\begin{aligned} & \int_{-1}^{1}\left(\frac{d}{d x} \tan ^{-1} \frac{1}{x}\right) d x \\ &=\int_{-1}^{1}\left(\frac{d}{d x} \cot ^{-1} x\right) d x \\ &=\int_{-1}^{1} \frac{-1}{\left(1+x^{2}\right)} d x=-2 \int_{0}^{1} \frac{1}{1+x^{2}} d x \\ &=-2\left[\tan ^{-1} x\right]_{0}^{1}=-2 \times \frac{\pi}{4}=-\frac{\pi}{2} \end{aligned}$

Topic: Mathematics, Definite Integration, NDA, NDA 2018 (Phase 2)
What is $\int_{1}^{2} \ell n x d x$ equal to ?

Options:
A $\ln 2$
B 1
C $\ln \left(\frac{4}{e}\right)$
D $\ln \left(\frac{e}{4}\right)$

The correct answer is:
$\ln \left(\frac{4}{e}\right)$

Solution:
$\int_{1}^{2} \ln x \mathrm{~d} \mathrm{x}=[\mathrm{x} \ln \mathrm{x}-\mathrm{x}]_{1}^{2}$
$=2 \ln 2-2-\ln \mathrm{l}+1$
$=\ln 4-\ln \mathrm{e}=\ln \left(\frac{4}{\mathrm{e}}\right)$

Topic: Mathematics, Definite Integration, NDA, NDA 2013 (Phase 2)
What is $\int_{1}^{3}\left|1-x^{4}\right| d x$ equal to?

Options:
A $-232 / 5$
B $-116 / 5$
C $116 / 5$
D $232 / 5$

The correct answer is:
$232 / 5$

Solution:
$\begin{aligned} I &=\int_{1}^{3}\left|1-x^{4}\right| d x \\ I &=\int_{1}^{3}-\left(1-x^{4}\right) d x \quad\left(:-|x|=\left\{\begin{array}{ll}x, & x \geq 0 \\ -x, & x < 0\end{array}\right)\right.\\ I &=\int_{1}^{3}\left(x^{4}-1\right) d x \Rightarrow I=\left[\frac{x^{5}}{5}-x\right]_{1}^{3} \\ I &=\left(\frac{3^{5}}{5}-3\right)-\left(\frac{1^{5}}{5}-1\right) \Rightarrow I=\frac{232}{5} \end{aligned}$

Topic: Mathematics, Definite Integration, NDA, NDA 2016 (Phase 2)
What is $\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}$ equal to?

Options:
A $2 \tan x$
B $2 \operatorname{cosec} x$
C $2 \cos x$
D $2 \sin x$

The correct answer is:
$2 \operatorname{cosec} x$

Solution:
$\quad \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{\sin ^{2} x+1+\cos ^{2} x+2 \cos x}{(1+\cos x)(\sin x)}$
$=\frac{2+2 \cos x}{(1+\cos x)(\sin x)}=\frac{2(1+\cos x)}{(1+\cos x)(\sin x)}$
$=\frac{2}{\sin x}=2 \operatorname{cosec} x$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
What is $\int \frac{1}{1+e^{x}} d x$ equal to?

Options:
A $x-\log x+c$
B $x-\log (\tan x)+c$
C $x-\log \left(1+e^{x}\right)+c$
D $\log \left(1+e^{x}\right)+c$
where $\mathrm{c}$ is a constant of integration.

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

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

Topic: Mathematics, Indefinite Integration, NDA, NDA 2010 (Phase 1)
What is $\int_{1}^{\mathrm{e}} \mathrm{x} \ln \mathrm{x} \mathrm{dx}$ equal to?

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

The correct answer is:
$\frac{\mathrm{e}^{2}+1}{4}$

Solution:
$\int_{1}^{e} x \cdot \ell n x \cdot d x=\left[\ell n x \cdot \frac{x^{2}}{2}\right]_{1}^{e}-\int_{1}^{e} \frac{1}{x} \cdot x^{2} \cdot \mathrm{dx}$
$=\left(\frac{\mathrm{e}^{2}}{2}-0\right)-\frac{1}{2}\left(\frac{1}{2}\right)\left(x^{2}\right)_{1}^{e}=\frac{\mathrm{e}^{2}}{2}-\frac{1}{4}\left(\mathrm{e}^{2}-1\right)=\frac{2 \mathrm{e}^{2}-\mathrm{e}^{2}+1}{4}$
$=\frac{\mathrm{e}^{2}+1}{4}$

Topic: Mathematics, Definite Integration, NDA, NDA 2018 (Phase 1)
What is $\frac{(1+i)^{4 n+5}}{(1-i)^{4 n+3}}$ equal to, where $n$ is a natural number

Options:
A 2
B 2i
C -2
D i

The correct answer is:
2

Solution:
$\quad$ Given $\frac{(1+\mathrm{i})^{4 \mathrm{n}+5}}{(1-\mathrm{i})^{4 \mathrm{n}+3}}$
$=\frac{(1+\mathrm{i})^{4 \mathrm{n}+3} \cdot(1+\mathrm{i})^{2}}{(1-\mathrm{i})^{4 \mathrm{n}+3}}=\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{4 \mathrm{n}+3} \cdot(1+\mathrm{i})^{2}$
$=\left[\frac{(1+\mathrm{i})(1+\mathrm{i})}{(1-\mathrm{i})(1+\mathrm{i})}\right]^{4 \mathrm{n}+3} \cdot\left(1+\mathrm{i}^{2}+2 \mathrm{i}\right)$
$=\left[\frac{1+\mathrm{i}^{2}+2 \mathrm{i}}{1+1}\right]^{4 \mathrm{n}+3} \cdot 2 \mathrm{i}=(\mathrm{i})^{4 \mathrm{n}+3} \cdot 2 \mathrm{i}=2(\mathrm{i})^{4 \mathrm{n}+4}$
$=2 \cdot\left(\mathrm{i}^{4(\mathrm{n}+1)}\right)=2$

Topic: Mathematics, Complex Number, NDA, NDA 2014 (Phase 2)
What is $\int \frac{\log x}{(1+\log x)^{2}} d x$ equal to?

Options:
A $\frac{1}{(1+\log x)^{3}}+c$
B $\frac{1}{(1+\log x)^{2}}+c$
C $\frac{x}{(1+\log x)}+c$
D $\frac{x}{(1+\log x)^{2}}+c$
Where $c$ is a constant.

The correct answer is:
$\frac{x}{(1+\log x)}+c$

Solution:
Let $I=\int \frac{\log x}{(1+\log x)^{2}} d x$
Put $\log x=t \Rightarrow \frac{1}{x} d x=d t$
$\Rightarrow d x=x \cdot d t \Rightarrow d x=e^{t} \cdot d t \quad\left(\because x=e^{t}\right)$
$I=\int \frac{e^{t} t}{(1+t)^{2}} d t=\int \frac{e^{t} \cdot(t+1-1)}{(1+t)^{2}} d t$
$=\int \frac{e^{t}(1+t)}{(1+t)^{2}} d t-\int \frac{e^{t}}{(1+t)^{2}} d t=\int \frac{e^{t}}{1+t} d t-\int \frac{e^{t}}{(1+t)^{2}} d t$
$=\frac{e^{t}}{1+t}-\int-e^{t} \frac{1}{(1+t)^{2}} d t-\int \frac{e^{t}}{(1+t)^{2}} d t$
$=\frac{\mathrm{e}^{\mathrm{t}}}{1+t}+\int^{e^{t}} \frac{1}{(1+t)^{2}} d t-\int \frac{e^{t}}{(1+t)^{2}} d t$
$=\frac{e^{t}}{1+t}+c=\frac{x}{1+\log x}+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2009 (Phase 2)
What is $\frac{1+\sin A}{1-\sin A}-\frac{1-\sin A}{1+\sin A}$ equal to?

Options:
A $\sec A-\tan A$
B $2 \sec A \cdot \tan A$
C $4 \sec A \cdot \tan A$
D $4 \operatorname{cosec} A \cdot \cot A$

The correct answer is:
$4 \sec A \cdot \tan A$

Solution:
$\frac{1+\sin A}{1-\sin A}-\frac{1-\sin A}{1+\sin A}$
$=\frac{(1+\sin \mathrm{A})^{2}-(1-\sin \mathrm{A})^{2}}{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}=\frac{4 \sin \mathrm{A}}{\cos ^{2} \mathrm{~A}}$
$=\frac{4 \sin \mathrm{A}}{\cos \mathrm{A}} \cdot \frac{1}{\cos \mathrm{A}}=4 \sec \mathrm{A} \cdot \tan \mathrm{A}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 1)
What is $\left(1-\sin ^{2} \theta\right)\left(1+\tan ^{2} \theta\right)$ equal to ?

Options:
A $\sin ^{2} \theta$
B $\cos ^{2} \theta$
C $\tan ^{2} \theta$
D 1

The correct answer is:
1

Solution:
$\begin{aligned}\left(1-\sin ^{2} \theta\right)\left(1+\tan ^{2} \theta\right) &=\cos ^{2} \theta \cdot \sec ^{2} \theta \\=& \cos ^{2} \theta \cdot \frac{1}{\cos ^{2} \theta}=1 \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 2)
What is $\sqrt{1+\sin 2 \theta}$ equal to ?

Options:
A $\cos \theta-\sin \theta$
B $\cos \theta+\sin \theta$
C $2 \cos \theta+\sin \theta$
D $\cos \theta+2 \sin \theta$

The correct answer is:
$\cos \theta+\sin \theta$

Solution:
Consider, $\sqrt{1+\sin 2 \theta}$
$=\sqrt{\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta}$
$=\sqrt{(\sin \theta+\cos \theta)^{2}}=\sin \theta+\cos \theta$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 2)
What is $\frac{2 \tan \theta}{1+\tan ^{2} \theta}$ equal to?

Options:
A $\cos 2 \theta$
B $\tan 2 \theta$
C $\sin 2 \theta$
D $\operatorname{cosec} 2 \theta$

The correct answer is:
$\sin 2 \theta$

Solution:
$\frac{2 \tan \theta}{1+\tan ^{2} \theta}=\frac{2 \tan \theta}{\sec ^{2} \theta}$
$=2 \tan \theta \cdot \cos ^{2} \theta$
$=2 \sin \theta \cdot \cos \theta=\sin 2 \theta$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 2)
What is $\frac{2}{1+\tan ^{2} \frac{\mathrm{x}}{2}}$ equal to:

Options:
A $\sin x \cdot \cos x$
B $\tan x$
C $\sin x$
D $\cos x$

The correct answer is:
$\cos x$

Solution:
$\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}=\frac{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}}$
$=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}=\cos x$
$\left(\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
What is $\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}$ equal to?

Options:
A $\sin \theta-\cos \theta$
B $\sin \theta+\cos \theta$
C $2 \sin \theta$
D $2 \cos \theta$

The correct answer is:
$\sin \theta+\cos \theta$

Solution:
$\frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}$
$=\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}}$
$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}+\frac{\sin ^{2} \theta}{\sin \theta-\cos \theta}$
$=\frac{\cos ^{2} \theta}{\cos \theta-\sin \theta}-\frac{\sin ^{2} \theta}{\cos \theta-\sin \theta}$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta-\sin \theta}=\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta-\sin \theta}$
$=\cos \theta+\sin \theta$
$\therefore \quad$ Option (b) is correct.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2015 (Phase 2)
What is $(1+\tan \alpha \tan \beta)^{2}+(\tan \alpha-\tan \beta)^{2}-\sec ^{2} \alpha \sec ^{2} \beta$
equal to

Options:
A 0
B 1
C 2
D 4

The correct answer is:
0

Solution:
\begin{aligned}
&(1+\tan \alpha \tan \beta)^{2}+(\tan \alpha-\tan \beta)^{2}-\sec ^{2} \alpha \sec ^{2} \beta \\
&=1+\tan ^{2} \alpha \tan ^{2} \beta+2 \tan \alpha \tan \beta+\tan ^{2} \alpha+\tan ^{2} \beta \\
& \quad-2 \tan \alpha \tan \beta-\sec ^{2} \alpha \sec ^{2} \beta \\
&=1+\tan ^{2} \alpha \tan ^{2} \beta+\tan ^{2} \alpha+\tan ^{2} \beta-\sec ^{2} \alpha \sec ^{2} \beta \\
&=\left(1+\tan ^{2} \alpha\right)\left(1+\tan ^{2} \beta\right)-\sec ^{2} \alpha \sec ^{2} \beta \\
&=\sec ^{2} \alpha \sec ^{2} \beta-\sec ^{2} \alpha \sec ^{2} \beta \\
&=0 \\

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
What is $\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 A}}}$ equal to?

Options:
A $\cos A$
B $\cos (2 A)$
C $2 \cos (A / 2)$
D $\sqrt{2 \cos A}$

The correct answer is:
$2 \cos (A / 2)$

Solution:
$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 A}}}$
$=\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 4 A)}}}$
$=\sqrt{2+\sqrt{2+2 \cos 2 A}}=\sqrt{2+\sqrt{2(1+\cos 2 A)}}$
$\left(\because 1+\cos 4 \mathrm{~A}=2 \cos ^{2} 2 \mathrm{~A}\right)$
$=\sqrt{2+2 \cos A}=\sqrt{2(1+\cos A)}$
$\left(\because 1+\cos 2 \mathrm{~A}=2 \cos ^{2} \mathrm{~A}\right)$
$=2 \cos \left(\frac{A}{2}\right) \quad\left(\because 1+\cos \mathrm{A}=2 \cos ^{2}\left(\frac{\mathrm{A}}{2}\right)\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2008 (Phase 2)
What is $\int_{-\pi / 2}^{\pi / 2}|\sin x| \mathrm{d} x$ equal to?

Options:
A 2
B 1
C $\pi$
D 0

The correct answer is:
2

Solution:
Let $\mathrm{I}=\int_{-\pi / 2}^{\pi / 2}|\sin \mathrm{x}| \mathrm{dx}$
Consider, $|\sin x|=\left\{\begin{array}{l}\sin x, 0 < x < \frac{\pi}{2} \\ -\sin x,-\frac{\pi}{2} < x < 0\end{array}\right.$
$I=\int_{-\pi / 2}^{0}-\sin x d x+\int_{0}^{\pi / 2} \sin x d x$
$=-[-\cos x]_{-\pi / 2}^{0}+\left.(-\cos x)\right|_{0} ^{\pi / 2}$
$=(\cos 0-\cos (-\pi / 2))+(-\cos \pi / 2-(-\cos 0))$
$=(1-0)+(-0+1)=2 .$

Topic: Mathematics, Definite Integration, NDA, NDA 2012 (Phase 1)
What is
$\int_{-2}^{2} x d x-\int_{-2}^{2}[x] d x$
equal to, where $[\cdot]$ is the greatest integer function?

Options:
A 0
B 1
C 2
D 4

The correct answer is:
2

Solution:
$\int_{-2}^{2} x d x-\int_{-2}^{2}[x] d x$
$=\left[\frac{x^{2}}{2}\right]_{-2}^{2}-\int_{-2}^{-1}[x] d x-\int_{-1}^{0}[x] d x-\int_{0}^{1}[x] d x-\int_{1}^{2}[x] d x$
$=\frac{1}{2}[4-4]-(-2)-(-1)-0-(1)$
$=2+1-1=2$

Topic: Mathematics, Definite Integration, NDA, NDA 2016 (Phase 1)
What is $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x d x$ equal to?

Options:
A 0
B 2
C $-2$
D $\pi$

The correct answer is:
2

Solution:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x d x=2 \int_{0}^{\frac{\pi}{2}} x \sin x d x$
$\{\mathrm{x} \sin \mathrm{x}$ is an even function\}
$=2[-x \cos x+\sin x]_{0}^{\frac{\pi}{2}}=2$

Topic: Mathematics, Definite Integration, NDA, NDA 2014 (Phase 1)
What is $\int_{2}^{8}|x-5| \mathrm{dx}$ equal to?

Options:
A 2
B 3
C 4
D 9

The correct answer is:
9

Solution:
$\int_{2}^{8}|x-5| d x=2 \times \frac{1}{2} \times 3 \times 3=9$

Topic: Mathematics, Definite Integration, NDA, NDA 2018 (Phase 2)
What is $\int \frac{\mathrm{dx}}{2^{\mathrm{x}}-1}$ equal to?

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

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

Solution:
$\int \frac{\mathrm{dx}}{2^{\mathrm{x}}-1}=\int \frac{\mathrm{dx}}{\frac{1}{2^{-\mathrm{x}}}}$
$=\int \frac{2^{-\mathrm{x}}}{1-2^{-\mathrm{x}}} \cdot \mathrm{dx}$
Let $1-2^{-\mathrm{x}}=\mathrm{t}$
$\Rightarrow 2^{-\mathrm{x}} \cdot \log 2=\frac{\mathrm{dt}}{\mathrm{dx}} \Rightarrow 2^{-\mathrm{x}}=\frac{1}{\log 2} \cdot \frac{\mathrm{dt}}{\mathrm{d} \mathrm{x}} \Rightarrow 2^{-\mathrm{x}} \cdot \mathrm{d} \mathrm{x}=\frac{\mathrm{dt}}{\log 2}$
$\quad \therefore \int \frac{2^{-\mathrm{x}}}{1-2^{-\mathrm{x}}} \mathrm{dx}=\frac{1}{\log 2} \int \frac{\mathrm{dt}}{\mathrm{t}}=\frac{1}{\log 2}(\log \mathrm{t})+\mathrm{c}$
$=\frac{1}{\log 2}\left(\log \left(1-2^{-\mathrm{x}}\right)\right)+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2018 (Phase 1)
What is $(\sqrt{3}+i) /(1+\sqrt{3} i)$ equal to ?

Options:
A $1+i$
B $1-i$
C $\sqrt{3}(1-i) / 2$
D $(\sqrt{3}-i) / 2$

The correct answer is:
$(\sqrt{3}-i) / 2$

Solution:
$\frac{\sqrt{3}+i}{1+\sqrt{3} i}=\frac{(\sqrt{3}+i)(1-\sqrt{3} i)}{(1+\sqrt{3} i)(1-\sqrt{3} i)}$
$=\frac{\sqrt{3}-3 i+i+\sqrt{3}}{1+3}=\frac{2 \sqrt{3}-2 i}{4}=\frac{(\sqrt{3}-i)}{2}$

Topic: Mathematics, Complex Number, NDA, NDA 2009 (Phase 1)
What is $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^{6}$ equal to ?

Options:
A -1
B 0
C 1
D 2

The correct answer is:
1

Solution:
Consider $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)=\frac{(\sqrt{3}+i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)}$
$=\frac{(\sqrt{3}+i)^{2}}{(\sqrt{3})^{2}-(i)^{2}}=\frac{3+i^{2}+2 \sqrt{3} i}{3-i^{2}}$
$=\frac{2+2 \sqrt{3} i}{4}=\frac{1+\sqrt{3} i}{2}=-\omega^{2} \quad\left(\because i^{2}=-1\right.$
Now, $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^{6}=\left(-\omega^{2}\right)^{6}=\omega^{12}=1\left(\because \omega^{3}=1\right)$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 2)
What is $\int_{-\pi / 4}^{\pi / 4} \tan ^{3} x d x$ equal to ?

Options:
A $\sqrt{3}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D 0

The correct answer is:
0

Solution:
We know
$\int_{-a}^{a} f(x) d x=\left\{\begin{array}{l}0 \text { if } f(x) \text { is odd. } \\ 2 \int_{0}^{a} f(x) d x \text { if } f(x) \text { is even }\end{array}\right.$
Since, $\tan ^{3} x$ is an odd function
$\therefore \quad \int_{-\pi / 4}^{\pi / 4} \tan ^{3} x d x=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2010 (Phase 1)
What is $\int \frac{d x}{\sqrt{4+x^{2}}}$ equal to?

Options:
A $\ln \left|\sqrt{4+x^{2}}+x\right|+c$
B $\ln \left|\sqrt{4+x^{2}}-x\right|+c$
C $\sin ^{-1}\left(\frac{x}{2}\right)+c$
D None of these

The correct answer is:
$\ln \left|\sqrt{4+x^{2}}+x\right|+c$

Solution:
$\int \frac{\mathrm{dx}}{\sqrt{4+\mathrm{x}^{2}}}=\int \frac{\mathrm{dx}}{\sqrt{2^{2}+\mathrm{x}^{2}}}=\ln \left|\sqrt{4+\mathrm{x}^{2}}+\mathrm{x}\right|+\mathrm{C}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2013 (Phase 2)
What is $\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\sin ^{5} x \cos ^{3} x}{x^{4}} d x$ is equal to?

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{4}$
C $\frac{\pi}{8}$
D $\underline{0}$

The correct answer is:
$\underline{0}$

Solution:
Let $f(x)=\frac{\sin ^{5} x \cos ^{3} x}{x^{4}}$
$f(-x)=\frac{\sin ^{5}(-x) \cos ^{3}(-x)}{(-x)^{4}}$
$=\frac{-\sin ^{5} x \cos ^{3} x}{x^{4}}=-f(x)$
$\Rightarrow f(x)$ is an odd function.
Hence, $\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\sin ^{5} x \cos ^{3} x}{x^{4}} d x=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2013 (Phase 1)
What is $0.9+0.09+0.009+\ldots$ equal to?

Options:
A 1
B $1.01$
C $1.001$
D $1.1$

The correct answer is:
1

Solution:
$\begin{aligned} & \mathrm{S}=0.9+0.09+0.009+\ldots \\ &=9(0.1+0.01+0.001+\ldots) \\ &=9\left[\frac{0.1}{1-0.1}\right]=9 \times \frac{0.1}{0.9}=1 \end{aligned}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2013 (Phase 2)
What is ${ }^{47} \mathrm{C}_{4}+{ }^{51} \mathrm{C}_{3}+\sum_{\mathrm{j}=2}^{5}{ }^{52-\mathrm{j}} \mathrm{C}_{3}$ equal to?

Options:
A ${ }^{52} \mathrm{C}_{4}$
B ${ }^{51} \mathrm{C}_{5}$
C ${ }^{53} \mathrm{C}_{4}$
D ${ }^{52} \mathrm{C}_{5}$

The correct answer is:
${ }^{52} \mathrm{C}_{4}$

Solution:
$\begin{array}{l}{ }^{47} C_{4}+{ }^{51} C_{3}+{ }^{50} C_{3}+{ }^{49} C_{3}+{ }^{48} C_{3}+{ }^{47} C_{3} \\
= { }^{47} C_{3}+{ }^{47} C_{4}+{ }^{48} C_{3}+{ }^{49} C_{3}+{ }^{50} C_{3}+{ }^{51} C_{3}\end{array}$
${ }^{48} C_{4}+{ }^{48} C_{3}+{ }^{49} C_{3}+{ }^{50} C_{3}+{ }^{51} C_{3}$
$\left({ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}\right)$
$={ }^{49} C_{4}+{ }^{49} C_{3}+{ }^{50} C_{3}+{ }^{51} C_{3}$
$={ }^{50} C_{4}+{ }^{50} C_{3}+{ }^{51} C_{3}$
$={ }^{51} C_{4}+{ }^{51} C_{3}$
$={ }^{52} C_{4}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2016 (Phase 2)
What is $\omega^{100}+\omega^{200}+\omega^{300}$ equal to, where $\omega$ is the cube root of unity?

Options:
A 1
B $3 \omega$
C $3 \omega^{2}$
D 0

The correct answer is:
0

Solution:
$$
\begin{aligned}
\omega^{100}+\omega^{200}+\omega^{300} &=\left(\omega^{99} \cdot \omega\right)+\left(\omega^{100}\right)^{2}+\left(\omega^{3}\right)^{100} \\
&=\left(\omega^{99} \cdot \omega\right)+\left(\omega^{99} \cdot \omega\right)^{2}+\left(\omega^{3}\right)^{100} \\
\because \omega^{3}=1 \quad \omega^{99}=\left(\omega^{3}\right)^{33}=1^{33}=1 \\
\Rightarrow \omega^{100}+\omega^{200}+\omega^{300}=(1 \cdot \omega)+(1 \cdot \omega)^{2}+1^{100} & \\
=\omega+\omega^{2}+1=1+\omega+\omega^{2}=0
\end{aligned}
$$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 2)
What is $(1001)_{2}$ equal to ?

Options:
A $(5)_{10}$
B $(4)_{10}$
C $(17)_{10}$
D $(11)_{10}$

The correct answer is:
$(4)_{10}$

Solution:
$(1001)_{2}=\left(2^{3} \times 1+2^{2} \times 0+2^{1} \times 0+2^{0} \times 1\right)_{10}$
$(8+1)_{10}=(9)_{10}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 2)
What is $(1111)_{2}+(1001)_{2}-(1010)_{2}$ equal to?

Options:
A $(111)_{2}$
B $(1100)_{2}$
C $(1110)_{2}$
D $(1010)_{2}$

The correct answer is:
$(1110)_{2}$

Solution:
Since, $(1111)_{2}=1 \times 2^{3}+1 \times 2^{2}+1 \times 2^{1}+1 \times 2^{0}$
$=8+4+2+1=15$
$(1001)_{2}=1 \times 2^{3}+0 \times 2^{2}+0 \times 2^{1}+1 \times 2^{0}=8+1=9$
and $(1010)_{3}=1 \times 2^{3}+0 \times 2^{2}+1 \times 2^{1}+0 \times 2^{0}=8+2=10$
$(1111)_{2}+(1001)_{2}-(1010)_{2}=15+9-10=14$
$(14)_{10}=(1110)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2010 (Phase 2)
What is (1000000001) $_{2}-(0.0101)$, equal to?

Options:
A $(512.6775)_{10}$
B $(512.6875)_{10}$
C $(512.0909)_{10}$
D $(512.6975)_{10}$

The correct answer is:
$(512.6875)_{10}$

Solution:
$\quad(1000000001)_{2}$
$=1 \times 2^{0}+0 \times 2^{1}+0 \times 2^{2}+\ldots \ldots+1 \times 2^{9}$
$=1+0+0+\ldots \ldots .+512$
$=(513)_{00}$
$(0.0101)_{2}=0 \times 2^{-1}+1 \times 2^{-2}+0 \times 2^{-3}+1 \times 2^{-4}$
$=\frac{1}{4}+\frac{1}{16}=\frac{5}{16}=(0.3125)_{10}$
$(1000000001)_{2}-(0.0101)_{2}=513-0.312$
$=(512.6875)_{\mathrm{s}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2015 (Phase 2)
What is $\left|\begin{array}{ccc}-\mathrm{a}^{2} & \mathrm{ab} & \mathrm{ac} \\ \mathrm{ab} & -\mathrm{b}^{2} & \mathrm{bc} \\ \mathrm{ac} & \mathrm{bc} & -\mathrm{c}^{2}\end{array}\right|$ equal to?

Options:
A $4 \mathrm{abc}$
B $4 a^{2} b c$
C $4 a^{2} b^{2} c^{2}$
D $-4 a^{2} b^{2} c^{2}$

The correct answer is:
$4 a^{2} b^{2} c^{2}$

Solution:
Consider $\left|\begin{array}{ccc}-\mathrm{a}^{2} & \mathrm{ab} & \mathrm{ac} \\ \mathrm{ab} & -\mathrm{b}^{2} & \mathrm{bc} \\ \mathrm{ac} & \mathrm{bc} & -\mathrm{c}^{2}\end{array}\right|$
Take out $^{\circ} \mathrm{a}^{\prime},{ }^{\prime} \mathrm{b}^{\prime}$ and ' $\mathrm{c}$ ' common from $\mathrm{R}_{1}, \mathrm{R}_{2}$ and $\mathrm{R}_{3}$
respectively.
$=a b c\left|\begin{array}{ccc}-a & b & c \\ a & -b & c \\ a & b & -c\end{array}\right|$
Now, take out 'a', 'b' and 'c' common from $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ respectively.
$=\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{1}, \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}+\mathrm{R}_{1}$
$=a^{2} b^{2} c^{2}\left|\begin{array}{ccc}-1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0\end{array}\right|=a^{2} b^{2} c^{2}[(-1)(-4)]$
$=4 a^{2} b^{2} c^{2}$

Topic: Mathematics, Determinants, NDA, NDA 2012 (Phase 1)
What is $\int \frac{\mathrm{dx}}{\mathrm{a}^{2} \sin ^{2} \mathrm{x}+\mathrm{b}^{2} \cos ^{2} \mathrm{x}}$ equal to?

Options:
A $\mathrm{c}+\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\operatorname{atan} \mathrm{x}}{\mathrm{b}}\right)$
B $\mathrm{c}-\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\operatorname{btan} \mathrm{x}}{\mathrm{a}}\right)$
C $\mathrm{c}+\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\operatorname{btan} \mathrm{x}}{\mathrm{a}}\right)$
D None of these

The correct answer is:
$\mathrm{c}+\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\operatorname{atan} \mathrm{x}}{\mathrm{b}}\right)$

Solution:
$\mathrm{I}=\frac{1}{a^{2}} \int \frac{\sec ^{2} x}{\tan ^{2} x+\left(\frac{b}{a}\right)^{2}}$
$=\frac{1}{a^{2}} \times \frac{a}{b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)+c$
$=c+\frac{1}{a b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2018 (Phase 2)
What is $(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})$ equal to?

Options:
A $\overrightarrow{0}$
B $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$
C $2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$
D $|\vec{a}|^{2}-|\vec{b}|^{2}$

The correct answer is:
$2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$

Solution:
$\begin{aligned} &(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b}) \\ &=\vec{a} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{a}-\vec{b} \times \vec{b} \\ &=0+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}-0 \\ &=2(\vec{a} \times \vec{b}) \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 2)
What is $\int_{a}^{b} \frac{\log x}{x} d x$ equal to?

Options:
A $(1 / 2) \log (a b) \cdot \log (b / a)$
B $\log \mathrm{b} / \log \mathrm{a}$
C $\log (\mathrm{b} / \mathrm{a})$
D $(1 / 2) \log [(a+b) / a b]$

The correct answer is:
$(1 / 2) \log (a b) \cdot \log (b / a)$

Solution:
Let the given integer be
$$
\mathrm{I}=\int_{\mathrm{a}}^{b} \frac{\log \mathrm{x}}{\mathrm{x}} \mathrm{dx}
$$
Put $\log x=\operatorname{tand} \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}$ when $\mathrm{x}=\mathrm{a}, \mathrm{t}=\log \mathrm{a}$ and if
$x=b, t=\log b$
$\therefore \quad I=\int_{\log a}^{\log b} t d t=\left[\frac{t^{2}}{2}\right]_{a}^{b}$
$\quad=\frac{1}{2}\left[(\log b)^{2}-(\log a)^{2}\right]$
$\quad=\frac{1}{2}[(\log b+\log a)(\log b-\log a)]$
$\quad=\frac{1}{2} \log (a b) \log \left(\frac{b}{a}\right)$

Topic: Mathematics, Definite Integration, NDA, NDA 2017 (Phase 2)