datavorous/entrance-exam-dataset huggingface dataset:
What is the value of $\tan 15^{\circ} \cdot \tan 195^{\circ} ?$

Options:
A $7-4 \sqrt{3}$
B $7+4 \sqrt{3}$
C $7+2 \sqrt{3}$
D $7+6 \sqrt{3}$

The correct answer is:
$7-4 \sqrt{3}$

Solution:
Consider $\tan 15^{\circ}$ tan $195^{\circ}$ $=\tan 15^{\circ} \tan \left(180+15^{\circ}\right)$
$=\tan 15^{\circ} \tan 15^{\circ} \quad(\because \tan (180+\theta)=\tan \theta)$
$\begin{aligned} &=\left(\tan 15^{\circ}\right)^{2} \\ &=(2-\sqrt{3})^{2} \quad\left(\because \tan 15^{\circ}=2-\sqrt{3}\right) \\ &=4+3-4 \sqrt{3}=7-4 \sqrt{3} \\ \text { 109. (b) } & \frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{\sin ^{2} x+1+\cos ^{2} x+2 \cos x}{(1+\cos x)(\sin x)} \\ &=\frac{2+2 \cos x}{(1+\cos x)(\sin x)}=\frac{2(1+\cos x)}{(1+\cos x)(\sin x)} \\ &=\frac{2}{\sin x}=2 \operatorname{cosec} x \\ & \Rightarrow 3 \sin A-4 \sin ^{3} A=1 \\ & \Rightarrow 4 \sin ^{3} A-3 \sin A+1=0 \\ & \Rightarrow(\sin A+1)\left(4 \sin ^{2} A-4 \sin A+1\right)=0 \\ & \Rightarrow(\sin A+1)(2 \sin A-1)^{2}=0 \\ & \Rightarrow \sin A=-1 \text { or } \frac{1}{2} \end{aligned}$
Hence, $\sin A$ can take two distinct values.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
What is the value of
$\tan ^{-1}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)-\tan ^{-1}\left(\frac{\mathrm{m}-\mathrm{n}}{\mathrm{m}+\mathrm{n}}\right) ?$

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

The correct answer is:
$\frac{\pi}{4}$

Solution:
The given expression is:
$\tan ^{-1}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)-\tan ^{-1}\left(\frac{\mathrm{m}-\mathrm{n}}{\mathrm{m}+\mathrm{n}}\right)$
$=\tan ^{-1}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)-\tan ^{-1}\left(\frac{1-\frac{\mathrm{n}}{\mathrm{m}}}{1+\frac{\mathrm{n}}{\mathrm{m}}}\right)$
$=\tan ^{-1}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)-\tan ^{-1}(1)+\tan ^{-1}\left(\frac{\mathrm{n}}{\mathrm{m}}\right)$
$=\tan ^{-1}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)+\cot ^{-1}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)-\frac{\pi}{4}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2007 (Phase 2)
What is the value of $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ} ?$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Given expression is $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$
$=\tan 9^{\circ}-\tan 27^{\circ}-\tan \left(90^{\circ}-27^{\circ}\right)+\tan \left(90^{\circ}-9^{\circ}\right)$
$=\tan 9^{\circ}-\tan 27^{\circ}-\cot 27^{\circ}+\cot 9^{\circ}$
$=\left(\tan 9^{\circ}+\cot 9^{\circ}\right)-\left(\tan 27^{\circ}+\cot 27^{\circ}\right)$
$=\frac{1}{\sin 9^{\circ} \cos 9^{\circ}}-\frac{1}{\sin 27^{\circ} \cos 27^{\circ}}$
$=\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}$
$=\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin \left(90^{\circ}-36^{\circ}\right)}$
$=\frac{2}{\sin 18^{\circ}}-\frac{2}{\cos 36^{\circ}}$
$=2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right]=8 \times \frac{2}{4}=4$
(By putting value of $\sin 18^{\circ}$ and $\cos 36^{\circ}$ )

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)
What is the value of tan $105^{\circ}$ ?

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

The correct answer is:
$\frac{\sqrt{3}+1}{1-\sqrt{3}}$

Solution:
Consider tan $\left(105^{\circ}\right)=\tan \left(60+45^{\circ}\right)$
$=\frac{\tan 60^{\circ}+\tan 45^{\circ}}{1-\tan 60^{\circ} \tan 45^{\circ}}$
$=\frac{\sqrt{3}+1}{1-\sqrt{3}} \quad\left(\because \tan 60^{\circ}=\sqrt{3}\right.$ and $\left.\tan 45^{\circ}=1\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
What is the value of $\tan 15^{\circ}+\cot 15^{\circ} ?

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

The correct answer is:
4

Solution:
$\quad$ Consider $\tan 15^{\circ}+\cot 15^{\circ}=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}+\frac{\cos 15^{\circ}}{\sin 15^{\circ}}$
$=\frac{\sin ^{2} 15^{\circ}+\cos ^{2} 15^{\circ}}{\cos 15^{\circ} \sin 15^{\circ}}=\frac{2 \times 1}{2 \cos 15^{\circ} \sin 15^{\circ}}$
$\begin{aligned} &\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right) \\=\frac{2}{\sin 30^{\circ}}=4\left(\because \sin 30^{\circ}=\frac{1}{2}\right) & \end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
What is the value of \tan 18^{\circ} ?

Options:
A $\frac{\sqrt{5}-1}{\sqrt{10+2 \sqrt{5}}}$
B $\frac{\sqrt{5}-1}{\sqrt{10+\sqrt{5}}}$
C $\frac{\sqrt{10+2 \sqrt{5}}}{\sqrt{5}-1}$
D $\frac{\sqrt{10+\sqrt{5}}}{\sqrt{5}-1}$

The correct answer is:
$\frac{\sqrt{5}-1}{\sqrt{10+2 \sqrt{5}}}$

Solution:
$\tan 18^{\circ}=\frac{\sin 18^{\circ}}{\cos 18^{\circ}}=\frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2 \sqrt{5}}}{4}}=\frac{\sqrt{5}-1}{\sqrt{10+2 \sqrt{5}}}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 1)
What is the value of \tan \left(-1575^{\circ}\right) ?

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

The correct answer is:
1

Solution:
$\tan \left(-1575^{\circ}\right)=-\tan \left(4 \times 360^{\circ}+135^{\circ}\right)$
$=-\tan 135^{\circ}=-\tan \left(90^{\circ}+45^{\circ}\right)=\cot 45^{\circ}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
What is the value of $\tan \left(\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z\right)-\cot \left(\cot ^{-1} x+\cot ^{-1} y+\cot ^{-1} z\right) ?$

Options:
A 0
B $2(\mathrm{x}+\mathrm{y}+\mathrm{z})$
C $\frac{3 \pi}{2}$
D $\frac{3 \pi}{2}+\mathrm{x}+\mathrm{y}+\mathrm{z}$

The correct answer is:
0

Solution:
$\tan \left(\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z\right)$
$-\cot \left(\cot ^{-1} x+\cot ^{-1} y+\cot ^{-1} z\right)$
$=\tan \left(\tan ^{-1} x+\tan ^{-1} \mathrm{y}+\tan ^{-1} z\right)$
$-\cot \left(\frac{\pi}{2}-\tan ^{-1} x+\frac{\pi}{2}-\tan ^{-1} y+\frac{\pi}{2}-\tan ^{-1} z\right)$
$\quad\left(\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right)$
$=\tan \left(\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z\right)$
$=\tan \left(\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z\right)$
$-\tan \left(\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z\right)=0$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2006 (Phase 1)
What is the value of the determinant
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\mathrm{xyz} & 1 \\ 1 & 1 & 1+\mathrm{xyz}\end{array}\right| ?$

Options:
A $1+x+y+z$
B $2 x y z$
C $\mathrm{x}^{2} \mathrm{y}^{2} \mathrm{z}^{2}$
D $2 x^{2} y^{2} z^{2}$

The correct answer is:
$\mathrm{x}^{2} \mathrm{y}^{2} \mathrm{z}^{2}$

Solution:
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x y z & 1 \\ 1 & 1 & 1+x y z\end{array}\right| \begin{array}{l}c_{2} \rightarrow c_{2}-c_{1} \\ c_{3} \rightarrow c_{3}-c_{1}\end{array}$
$=\left|\begin{array}{ccc}1 & 0 & 0 \\ 1 & x y z & 0 \\ 1 & 0 & x y z\end{array}\right|=1\left(x^{2} y^{2} z^{2}-0\right)=x^{2} y^{2} z^{2}$

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 1)
What is the value of the determinant
$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|$ ?

Options:
A 0
B $a b c$
C $a b+b c+c a$
D $a b c(a+b+c)$

The correct answer is:
0

Solution:
$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & a c & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|$
Applying $C_{3} \rightarrow C_{2}+C_{3}$
$\left|\begin{array}{lll}1 & \text { bc } & a b+b c+a c \\ 1 & a c & a b+b c+a c \\ 1 & a b & a b+b c+a c\end{array}\right|$
$=(a b+b c+a c)\left|\begin{array}{lll}1 & b c & 1 \\ 1 & a c & 1 \\ 1 & a b & 1\end{array}\right|$
$=(a b+b c+a c) \times 0=0$

Topic: Mathematics, Determinants, NDA, NDA 2013 (Phase 2)
What is the value of the determinant
$\left|\begin{array}{lll}a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c\end{array}\right| ?$

Options:
A $a^{3}+b^{3}+c^{3}$
B $3 b c$
C $a^{3}+b^{3}+c^{3}-3 a b c$
D 0

The correct answer is:
$a^{3}+b^{3}+c^{3}-3 a b c$

Solution:
Given $\begin{array}{rrr}\left|\begin{array}{rrr}a-b & b+c & a \\ b-c & c+a & b \\ c-a & a+b & c\end{array}\right| \\ C_{1} \rightarrow C_{1}-C_{3} & \end{array}$
$=\left|\begin{array}{ccc}-b & b+c & a \\ -c & c+a & b \\ -a & a+b & c\end{array}\right|$
$\begin{aligned} & C_{2} \rightarrow C_{2}+C_{1} \\=&\left|\begin{array}{lll}-b & c & a \\ -c & a & b \\ -a & b & c\end{array}\right|=(-1)\left|\begin{array}{lll}b & c & a \\ c & a & b \\ a & b & c\end{array}\right| \\=&-1\left[b\left(a c-b^{2}\right)-c\left(c^{2}-a b\right)+a\left(b c-a^{2}\right)\right] \\=&-\left[a b c-b^{3}-c^{3}+a b c+a b c-a^{3}\right] \\=& a^{3}+b^{3}+c^{3}-3 a b c . \end{aligned}$

Topic: Mathematics, Determinants, NDA, NDA 2011 (Phase 2)
What is the value of the determinant
$\left|\begin{array}{ccc}x+1 & x+2 & x+4 \\ x+3 & x+5 & x+8 \\ x+7 & x+10 & x+14\end{array}\right|$

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

The correct answer is:
$-2$

Solution:
Let $\Delta=\left|\begin{array}{ccc}x+1 & x+2 & x+4 \\ x+3 & x+5 & x+8 \\ x+7 & x+10 & x+14\end{array}\right|$
By applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$
$\quad=\left|\begin{array}{lll}x+1 & 1 & 3 \\ x+3 & 2 & 5 \\ x+7 & 3 & 7\end{array}\right|$
By applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}$, we
get
$\begin{array}{l}
=\left|\begin{array}{ccc}
x+1 & 1 & 3 \\
2 & 1 & 2 \\
4 & 1 & 2
\end{array}\right| \\
=(x+1)(0)-1(4-8)+3(2-4) \\
=4-6=-2
\end{array}$

Topic: Mathematics, Determinants, NDA, NDA 2011 (Phase 1)
What is the value of the integral $\int_{-1}^{1}|x| d x ? \quad$

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

The correct answer is:
1

Solution:
Let $\mathrm{I}=\int_{-1}^{1}|\mathrm{x}| \mathrm{d} \mathrm{x}=-\int_{-1}^{0} \mathrm{xd} \mathrm{x}+\int_{0}^{1} \mathrm{x} \mathrm{d} \mathrm{x}$
Since, $|\mathrm{x}|=\left\{\begin{array}{ccc}\mathrm{x} & \text { if } & \mathrm{x} \geq 0 \\ -\mathrm{x} & \text { if } & \mathrm{x} < 0\end{array}\right\}$
therefore $|x|=-x$ when $x$ lies, between $-1$ and 0 . and $|x|=x$ when $x$ lies between 0 and 1 . $=-\left[\frac{\mathrm{x}^{2}}{2}\right]^{0}+\left[\frac{\mathrm{x}^{2}}{2}\right]^{1}=-\left[-\frac{1}{2}\right]+\left[\frac{1}{2}\right]=1$

Topic: Mathematics, Definite Integration, NDA, NDA 2010 (Phase 2)
What is the value of the minor of the element 9 in the determinant $\left|\begin{array}{lll}10 & 19 & 2 \\ 0 & 13 & 1 \\ 9 & 24 & 2\end{array}\right| ?$

Options:
A $-9$
B $-7$
C 7
D 0

The correct answer is:
$-7$

Solution:
Minor of element $9=\left|\begin{array}{cc}19 & 2 \\ 13 & 1\end{array}\right|=19-26=-7$

Topic: Mathematics, Determinants, NDA, NDA 2013 (Phase 1)
What is the value of the sum $\quad[2018-I]$
$\sum_{n=2}^{11}\left(i^{n}+i^{n+1}\right)$, where $i=\sqrt{-1} ?$

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

The correct answer is:
$-2 \mathrm{i}$

Solution:
$\sum_{n=2}^{11}\left(i^{n}+i^{n+1}\right)$
We know, $\mathrm{i}+i^{2}+i^{3}+i^{4}=i-1-i+1=0$
Also, $i^{3}+i^{4}+i^{5}+i^{6}=0$
The sum of 4 consecutive powers of 1 is always 0 .
$\therefore \sum_{n=2}^{11}\left(i^{n}+i^{n+1}\right)=i^{2}+i^{3}+i^{3}+i^{4}+i^{4}+i^{5}+i^{5}$
$+i^{6}+i^{6}+i^{7}+i^{7}+i^{8}+i^{8}+i^{9}+i^{9}+i^{10}+i^{10}$
$+i^{11}+i^{11}+i^{12}$
$=\left(i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+i^{10}+i^{11}\right)+$
$\quad\left(i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+i^{10}+i^{11}+i^{12}\right)$
$=\left(i^{2}+i^{3}+0\right)+\left(i^{3}+i^{4}+0\right)$
$=i^{2}+i^{3}+i^{3}+i^{4}$
$=i^{2}+2 i^{3}+i^{4}$
$=-1+2(-i)+1$
$=-2 i$

Topic: Mathematics, Complex Number, NDA, NDA 2018 (Phase 1)
What is the value of $\int \frac{\mathrm{d} \mathrm{x}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}+\mathrm{b}^{2}\right)} ?$

Options:
A $\int \frac{\left[\left\{\tan ^{-1}(x / a)\right\} / a-\left\{\tan ^{-1}(x / b)\right\} / b\right]}{\left(a^{2}+b^{2}\right)}+c$
B $\int \frac{\left[\left\{\tan ^{-1}(x / a)\right\} / a+\left\{\tan ^{-1}(x / b)\right\} / b\right]}{\left(a^{2}+b^{2}\right)}+c$
C $\int \frac{\left[\left\{\tan ^{-1}(x / a)\right\} / a+\left\{\tan ^{-1}(x / b)\right\} / b\right]}{\left(b^{2}-a^{2}\right)}+c$
D $\int \frac{\left[\left\{\tan ^{-1}(x / a)\right\} / a+\left\{\tan ^{-1}(x / b)\right\} / b\right]}{\left(b^{2}-a^{2}\right)}+c$

The correct answer is:
$\int \frac{\left[\left\{\tan ^{-1}(x / a)\right\} / a+\left\{\tan ^{-1}(x / b)\right\} / b\right]}{\left(b^{2}-a^{2}\right)}+c$

Solution:
The given integral is $\int \frac{\mathrm{dx}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}+\mathrm{b}^{2}\right)}$
Breaking the expression under integral into partial fraction $\frac{1}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}$
$=\left(\frac{1}{\left(x^{2}+a^{2}\right)}-\frac{1}{\left(x^{2}+b^{2}\right)}\right) \times \frac{1}{b^{2}-a^{2}}$
The given integral is
$\frac{1}{\left(b^{2}-a^{2}\right)} \int\left(\frac{1}{\left(x^{2}+a^{2}\right)}-\frac{1}{\left(x^{2}+b^{2}\right)}\right) d x$
$=\frac{1}{\left(b^{2}-a^{2}\right)} \int\left[\frac{1}{x^{2}+a^{2}} d x-\int \frac{1}{x^{2}+b^{2}} d x\right]$
$=\frac{1}{\left(b^{2}-a^{2}\right)}\left\{\frac{\tan ^{-1}\left(\frac{x}{a}\right)}{a}-\frac{\tan ^{-1}\left(\frac{x}{b}\right)}{b}\right\}+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2007 (Phase 1)
What is the value of $x$ at the intersection of $y=\frac{8}{\left(x^{2}+4\right)}$
and $x+y=2$

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

The correct answer is:
0

Solution:
Given equations are
$y=\frac{8}{x^{2}+4}$ and $x+y=2$
Putting value of $y$ from 1 st equation into second
equation
$x+\frac{8}{x^{2}+4}=2$
$\Rightarrow x^{3}+4 x+8=2 x^{2}+8$
$\Rightarrow x^{3}-2 x^{2}+4 x=0$
$\Rightarrow x\left(x^{2}-2 x+4\right)=0$
$\mathrm{x}=0$
[The other valueof $x$ is not real $]$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 1)
What is the value of $x$ satisfying the equation
$6\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x} ?$

Options:
A $a / 2$
B $a / 3$
C 0
D $a / 4$

The correct answer is:
$a / 3$

Solution:
Consider $16\left(\frac{a-x}{a+x}\right)^{3}=\frac{a+x}{a-x}$
$\Rightarrow\left(\frac{a-x}{a+x}\right)^{3} \times\left(\frac{a-x}{a+x}\right)=\frac{1}{16}$
$\Rightarrow\left(\frac{a-x}{a+x}\right)^{4}=\left(\frac{1}{2}\right)^{4}$
$\Rightarrow \frac{a-x}{a+x}=\frac{1}{2}$
$\Rightarrow 2 a-2 x=a+x$
$\Rightarrow a=3 x$
$\Rightarrow x=\frac{a}{3}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 1)
What is the value of $x$ that satisfies the equation $\cos ^{-1} x=2 \sin ^{-1} x ?$

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

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

Solution:
Given that $\cos ^{-1} \mathrm{x}=2 \sin ^{-1} \mathrm{x}$
$\Rightarrow \frac{\pi}{2}-\sin ^{-1} x=2 \sin ^{-1} x$
$\Rightarrow \frac{\pi}{2}=3 \sin ^{-1} \mathrm{x}$
$\Rightarrow \sin ^{-1} x=\frac{\pi}{6}$
So, $x=\sin \frac{\pi}{6}=\frac{1}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2006 (Phase 1)
What is the value of $\int(\sqrt{x}+x)^{-1} \mathrm{dx}?$

Options:
A $\operatorname{In}(x+\sqrt{x})+c$
B $2 \ln (1+\sqrt{x})+c$
C $2 \ln (x+\sqrt{x})+c$
D $2 \ln (x-\sqrt{x})+c$

The correct answer is:
$2 \ln (1+\sqrt{x})+c$

Solution:
$\int(x+\sqrt{x})^{-1} d x=\int \frac{1}{(x+\sqrt{x})} d x$
$=\int \frac{1}{(\sqrt{x} \cdot \sqrt{x}+\sqrt{x})} \cdot d x$
Let $\sqrt{x}+1=t$
then, $\frac{1}{2 \sqrt{x}} d x=d t \Rightarrow \frac{1}{\sqrt{x}} d x=2 d t$
$\therefore \int \frac{1}{\sqrt{x}(\sqrt{x}+1)} d x=\int \frac{2 d t}{t}=2 \log t+c$
$=2 \log (1+\sqrt{x})+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2007 (Phase 1)
What is the variance of the firs
ILnatural number:

Options:
A 10
B 12
C 11
D 13

The correct answer is:
10

Solution:
Variance of 11 natural numbers $=\frac{11^{2}-1}{12}=10$

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 2)
What is the vector equally inclined to the vectors $\hat{\mathbf{i}}+3 \hat{\mathbf{j}}$
and $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}$ ?

Options:
A $\hat{\mathbf{i}}+\hat{\mathbf{j}}$
B $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}$
C $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}$
D None of these

The correct answer is:
$\hat{\mathbf{i}}+\hat{\mathbf{j}}$

Solution:
Let the required vector be $\hat{i}+\hat{j}$
Since the vector $\hat{i}+\hat{j}$ is equally inclined to the vectors
$\hat{i}+3 \hat{j}$ and $3 \hat{i}+\hat{j}$ therefore
Angle $\mathrm{b} / \mathrm{w} \hat{i}+\hat{j}$ and $\hat{i}+3 \hat{j}=\theta_{1}$ is equal to angle
between $\hat{i}+\hat{j}$ and $3 \hat{i}+\hat{j}=\theta_{2}$
$\therefore \quad$ Angle between $\hat{i}+\hat{j}$ and $\hat{i}+3 \hat{j}$
$=\cos ^{-1}\left[\frac{(1)(1)+(1)(3)}{\sqrt{(1)^{2}+(1)^{2}} \sqrt{(1)^{2}+(3)^{2}}}\right]$
$=\cos ^{-1}\left[\frac{1+3}{\sqrt{2} \sqrt{10}}\right]=\cos ^{-1}\left[\frac{4}{\sqrt{2} \sqrt{10}}\right]$
$=\cos ^{-1}\left[\frac{2}{\sqrt{5}}\right]$ and
angle between $\hat{i}+\hat{j}$ and $(3 \hat{i}+\hat{j})$
$=\cos ^{-1}\left|\frac{1+3}{\sqrt{10} \sqrt{2}}\right|$
$=\cos ^{-1}\left(\frac{4}{\sqrt{2} \sqrt{10}}\right)=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)$
Hence required vector is $\hat{i}+\hat{j}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2010 (Phase 2)
What is the vector in the $x y$ -plane through origin and perpendicular to the vector $\vec{r}=a \hat{i}+b \hat{j}$ and of the same length?

Options:
A $-a \hat{i}-b \vec{b}$
B $a \hat{i}-b \hat{j}$
C $-a \hat{i}+b \hat{j}$
D $\quad b \hat{i}-a \hat{j}$

The correct answer is:
$\quad b \hat{i}-a \hat{j}$

Solution:
Let $\vec{r}_{1}=b \hat{i}-a \hat{j}$ be the required vector
Given, $\vec{r}=a \vec{i}+b \vec{j}$
Now, $\vec{r}_{1} \cdot \vec{r}=(b \hat{i}-a \hat{j}) \cdot(a \hat{i}+b \hat{j})$
$=a b-a b=0$
Hence, option (d) is the correct answer

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 2)
What is the vector perpendicular to both the vectors
$\hat{i}-\hat{j}$ and $\hat{\mathrm{i}} ?$

Options:
A $\hat{i}$
B $-\hat{j}$
C $\hat{j}$
D $\hat{\mathrm{k}}$

The correct answer is:
$\hat{\mathrm{k}}$

Solution:
$\quad(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \times(\hat{\mathrm{j}})$
$\quad=\hat{\mathrm{i}} \times \hat{\mathrm{i}}-\hat{\mathrm{j}} \times \hat{\mathrm{j}}=\hat{\mathrm{k}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 2)
What is the vector whose magnitude is 3 , and is perpendicular
to $\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\hat{\mathrm{j}}+\hat{\mathrm{k}} ?$

Options:
A $3(\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}-\overrightarrow{\mathrm{k}})$
B $\sqrt{3}(\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})$
C $\sqrt{3}(\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})$
D $3(\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})$

The correct answer is:
$\sqrt{3}(\overrightarrow{\mathrm{i}}-\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}})$

Solution:
Let the vector be $\mathrm{r}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$. Since, $\overrightarrow{\mathrm{r}}$ and $\hat{\mathrm{i}}+\hat{\mathrm{j}}$
are perpendicular to each other.
Hence, $\overrightarrow{\mathrm{r}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})=0 \Rightarrow \mathrm{x}+\mathrm{y}=0$
also $\overrightarrow{\mathrm{r}}$ and $\hat{\mathrm{j}}+\hat{\mathrm{k}}$ are perpendicular to each other. so,
$\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{j}}+\hat{\mathrm{k}})=0$
So, $\mathrm{y}+\mathrm{z}=0$
(ii) and $x^{2}+y^{2}+z^{2}=9$
$\Rightarrow \quad\left(-y^{2}\right)+y^{2}+(-y)^{2}=9$
$\Rightarrow \quad 3 y^{2}=9$
$\Rightarrow \quad y=\pm \sqrt{3}$
$\therefore \quad \mathrm{x}=\mp \sqrt{3} \quad[$ from $(\mathrm{i})$
and $\mathrm{z}=\pm \sqrt{3}$
[from (ii)]
So, vector is $\sqrt{3}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
What is the water potential of pure water at standard temperatures when it is not under any pressure?

Options:
A 10
B 0
C -10
D None of the above

The correct answer is:
0

Solution:
By convention, the water potential of pure water at standard temperatures, which is not under any pressure, is taken to be zero.

Topic: Biology, Transport in Plants, NEET
What is the $\mathrm{x}$ -coordinate of the point on the curve $\mathrm{f}(\mathrm{x})=$ $\sqrt{\mathrm{x}}(7 \mathrm{x}-6)$, where the tangent is parallel to $\mathrm{x}$ -axis?

Options:
A $-\frac{1}{3}$
B $\frac{2}{7}$
C $\frac{6}{7}$
D $\frac{1}{2}$

The correct answer is:
$\frac{2}{7}$

Solution:
$\begin{aligned} \mathrm{f}(\mathrm{x}) &=\sqrt{\mathrm{x}}(7 \mathrm{x}-6)=7 \mathrm{x}^{3 / 2}-6 \mathrm{x}^{1 / 2} \\ \mathrm{f}^{\prime}(\mathrm{x}) &=7 \times \frac{3}{2} \mathrm{x}^{1 / 2}-6 \times \frac{1}{2} \mathrm{x}^{-1 / 2} \end{aligned}$
When tangent is parallel to $\mathrm{x}$ axis $\mathrm{f}^{\prime}(\mathrm{x})=0$ or, $\frac{21}{2} \mathrm{x}^{1 / 2}-3 \mathrm{x}^{-1 / 2}=0$
$\frac{21}{2} \sqrt{x}=\frac{3}{\sqrt{x}}$
or, $7 x=2 \Rightarrow x=\frac{2}{7}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 1)
What is true about earthworm, scorpion, cockroach, and silverfish?

Options:
A They all posses dorsal heart
B None of them in aquatic
C They all belong to the same phylum
D They all have joined paired appendages

The correct answer is:
They all posses dorsal heart

Solution:
Earthworm belonging to phylum Annelida, scorpion, silverfish, and cockroach (belonging to phylum Arthropoda) all have dorsal heart.

Topic: Biology, Body Fluids and Circulation, NEET
What is true about ecosystem?

Options:
A Primary consumers are least dependent upon producers
B Primary consumers out-number producers
C Producers are more than primary consumers
D Secondary consumers are the largest and most powerful

The correct answer is:
Producers are more than primary consumers

Solution:
In an ecosystem producers (green plants) are always more than consumers (herbivores, carnivores etc.)

Topic: Biology, Ecosystem, NEET
What is true about Nereis, scorpion, cockroach and silver fish?

Options:
A they all possess dorsal heart
B none of them is aquatic
C they all belong to the same phylum
D they all have jointed paired appendages.

The correct answer is:
they all possess dorsal heart

Solution:
(Ref to Table)
$\begin{array}{|l|l|l|l|l|}
\hline & \text{Nereis} & \text{Scorpion} & \text{Cockroach} & \text{Silver fish} \\
\hline \text{Dorsal heart} & \text{Present} & \text{Present} & \text{Present} & \text{Present} \\
\hline \text{Habitat} & \begin{array}{l}
\text{Sand or rocks in intertidal and} \\
\text{shallow marine waters.}
\end{array} & \text{under logs, rocks, boards and clutter.} & \text{damp but warm places} & \text{dark, damp areas} \\
\hline \text{Phylum} & \text{Annelida} & \text{Arthropoda} & \text{Arthropoda} & \text{Arthropoda} \\
\hline \text{Presence of jointed appendages} & \text{Absent} & \text{Present} & \text{Present} & \text{Present} \\
\hline
\end{array}$
Related Theory
Scorpions are found on every continent except Antarctica, in habitats ranging from tropical rainforests to grasslands and deserts. As aduits, most scorpions are nocturnal and solitary, usually staying in the same territory throughout their lives. Many live in burrows they dig or claim and defend from other wildlife.

Topic: Biology, Animal Kingdom, NEET, NEET 2007
What is true about the Euro II norms?

Options:
A It stipulates to control sulphur at 350 ppm in diesel and 150 ppm in petrol
B It stipulates to reduce sulphur levels to 50 ppm in petrol and diesel
C It stipulates to reduce sulphur levels to 200 ppm in diesel and petrol
D It stipulates to reduce sulphur levels to 200 ppm in diesel and 100 ppm in petrol

The correct answer is:
It stipulates to control sulphur at 350 ppm in diesel and 150 ppm in petrol

Solution:
Standard norms to keep check on air pollution is Euro II that stipulates that vehicular sulphur emission should be at 350ppm for diesel and 150ppm for petrol. Aromatic hydrocarbons should contain at 42% of the concerned fuel whether diesel or petrol.

Topic: Biology, Environmental Issues, NEET