datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{x}, \frac{3}{2}, \mathrm{z}$ are in $\mathrm{AP} ; \mathrm{x}, 3, \mathrm{z}$ are in GP; then which one of the following will be in HP?

Options:
A $\mathrm{x}, 6, \mathrm{z}$
B $\mathrm{x}, 4, \mathrm{z}$
C $\mathrm{x}, 2, \mathrm{z}$
D $x, 1, z$

The correct answer is:
$\mathrm{x}, 6, \mathrm{z}$

Solution:
$\mathrm{x}, \frac{3}{2}, \mathrm{z}$ are in A.P
If $a, b, c$ are in $A \cdot P 2 b=a+c$
$\therefore 2\left(\frac{3}{2}\right)=x+z$
$\Rightarrow 3=x+z$
$\ldots .(1)$
$\mathrm{x}, 3, \mathrm{z}$ are in G.P. If $a, b, c$ are in G.P. $b^{2}=a c$ $\therefore 3^{2}=\mathrm{xz}$
$\Rightarrow 9=\mathrm{xz}$...(2
If $\mathrm{x}, 6, \mathrm{z}$ are in $\mathrm{H.P.} \frac{2}{6}=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{z}}$
$\quad\left(\because \frac{2}{b}=\frac{1}{a}+\frac{1}{c}\right.$, if $a, b, c$ are is H.P. $)$
$\Rightarrow \frac{1}{3}=\frac{z+x}{x z}=\frac{3}{9}=\frac{1}{3} \quad($ from $(1)(2))$
L.HS. $=$ R.H.S.

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 1)
If $\mathrm{x}=2^{1 / 3}-2^{-1 / 3}$, then what is the value of $2 \mathrm{x}^{3}+6 \mathrm{x}$ ?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Given equation is: $\mathrm{x}=2^{1 / 3}-2^{-1 / 3}$ On cubing both sides, we get
$\begin{aligned} x^{3} &=\left(2^{1 / 3}-2^{-1 / 3}\right)^{3} \\ &=2-2^{-1}-3.2^{1 / 3} \cdot 2^{-1 / 3}\left(2^{1 / 3}-2^{-1 / 3}\right) \\ \Rightarrow & x^{3}=-3 x+2-\frac{1}{2} \Rightarrow x^{3}=-3 x+\frac{3}{2} \\ \Rightarrow & 2 x^{3}+6 x=3 \end{aligned}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 1)
If $\int x^{2} \ln x d x=\frac{x^{3}}{m} \ln x+\frac{x^{3}}{n}+c$, then what are the values of $m$ and $n$ respectively?

Options:
A $\frac{1}{3,}-\frac{1}{9}$
B $3,-9$
C 3,9
D 3,3
where $c$ is a constant of integration.

The correct answer is:
$3,-9$

Solution:
$\int x^{2} \ln x d x=\ln x \frac{x^{3}}{3}-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x$
$=\frac{x^{3}}{3} \ln x-\int \frac{x^{2}}{3} d x=\frac{x^{3}}{3} \ln x-\frac{1}{3} \cdot \frac{x^{3}}{3}+c$ ...(1)
$=\frac{x^{3}}{3} \ln x-\frac{x^{3}}{9}+c$
But $\int x^{2} \ln x d x=\frac{x^{3}}{m} \ln x+\frac{x^{3}}{n}+c$
On comparing with (1), we get $m=3$ and $n=-9$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2010 (Phase 1)
If $x^{2}+p x+4>0$ for all real values of $x$, then which one of the following is correct?

Options:
A $|\mathrm{p}| < 4$
B $|\mathrm{p}| \leq 4$
C $|p|>4$
D $|\mathrm{p}| \geq 4$

The correct answer is:
$|\mathrm{p}| \leq 4$

Solution:
$x^{2}-p x+4>0 \quad \forall$ real values of $x$. If $b^{2}-4 a c \leq 0$
$\Rightarrow p^{2}-4(1)(4) \leq 0$
$\Rightarrow p^{2} \leq 16 \Rightarrow|p| \leq 4$
Sol. $(137-138)$ :
$f(x)=\frac{27}{4}\left(x^{2 / 3}-x\right)$
$=\frac{27}{4}\left(\left(x^{2}\right)^{1 / 3}-x\right.$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 1)
If $x, 2 x+2,3 x+3$ are the first three terms of a GP, then what is its fourth term?

Options:
A $-27 / 2$
B $27 / 2$
C $-33 / 2$
D $33 / 2$

The correct answer is:
$-27 / 2$

Solution:
Since, $\mathrm{x}, 2 \mathrm{x}+2,3 \mathrm{x}+3$ are the terms of G.P
therefore $\frac{2 x+2}{x}=\frac{3 x+3}{2 x+2}$
$\Rightarrow(2 x+2)^{2}=x(3 x+3) \Rightarrow 4 x^{2}+4+8 x=3 x^{2}+3 x$
$\Rightarrow x^{2}+5 x+4=0$
$\Rightarrow x^{2}+4 x+x+4=0$
$\Rightarrow x(x+4)+1(x+4)=0$
$\Rightarrow \mathrm{x}=-1,-4$
Now, first term $\mathrm{a}=\mathrm{x}$
Second term, ar $=2(\mathrm{x}+1) \Rightarrow \mathrm{r}=\frac{2(x+1)}{x}$
Fourth term $=\mathrm{ar}^{3}=\mathrm{x}\left(\frac{2(x+1)}{x}\right)^{3}$
Put $x=-4$, we get
Fourth term $=-4\left(\frac{2(-4+1)}{-4}\right)^{3}=-4 \times\left(\frac{3}{2}\right)^{3}=-\frac{27}{2}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2009 (Phase 2)
If $x^{2}+y^{2}=1$, then what is $\frac{1+x+i y}{1+x-i y}$ equal to?

Options:
A $x-i y$
B $x+i y$
C $2 x$
D $-2 i y$

The correct answer is:
$x+i y$

Solution:
Consider $\frac{1+x+\text { iy }}{1+x-i y}=\frac{(1+x+i y)(1+x+i y)}{(1+x-i y)(1+x+i y)}$
(By Rationalizing)
$=\frac{(1+x)^{2}+i y(1+x)+i y(1+x)-y^{2}}{1+x^{2}+2 x+y^{2}}\left(\because i^{2}=-1\right)$
$=\frac{1+x^{2}+2 x-y^{2}+2 i y(1+x)}{2(1+x)}\left(\because x^{2}+y^{2}=1\right)$
$=\frac{1-y^{2}+2 x+x^{2}+2 i y(1+x)}{2(1+x)}$
$=\frac{2 x^{2}+2 x+2 i y(1+x)}{2(1+x)}=x+i y \quad\left(\because 1-y^{2}=x^{2}\right)$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 1)
If $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=1$, then what is the value of
$\left|\begin{array}{ccc}1 & z & -y \\ -z & 1 & x \\ y & -x & 1\end{array}\right|=?$

Options:
A 0
B 1
C 2
D $2-2 x y z$

The correct answer is:
2

Solution:
Let $D=\left|\begin{array}{ccc}1 & z & -y \\ -z & 1 & x \\ y & -x & 1\end{array}\right|$
Expanding along $\mathrm{R}_{1}$ $\Rightarrow \mathrm{D}=1\left|\begin{array}{cc}1 & \mathrm{x} \\ -\mathrm{x} & 1\end{array}\right|-\mathrm{z}\left|\begin{array}{cc}-\mathrm{z} & \mathrm{x} \\ \mathrm{y} & 1\end{array}\right|-\mathrm{y}\left|\begin{array}{cc}-\mathrm{z} & 1 \\ \mathrm{y} & -\mathrm{x}\end{array}\right|$
$=\left(1+x^{2}\right)-z(-z-x y)-y(x z-y)$
$=1+x^{2}+z^{2}+x y z-x y z+y^{2}$
$=1+x^{2}+y^{2}+z^{2}=1+1=2$

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 1)
If $x^{2}, y^{2}, z^{2}$ are in AP, then $y+z, z+x, x+y$ are in

Options:
A AP
B $\mathrm{HP}$
C GP
D None of these

The correct answer is:
AP

Solution:
Let $\mathrm{x}^{2}, \mathrm{y}^{2}, \mathrm{z}^{2}$ are in A.P $\Rightarrow \mathrm{y}^{2}-\mathrm{x}^{2}=\mathrm{z}^{2}-\mathrm{y}^{2}$
$2 y^{2}=x^{2}+z^{2}$
(a) Suppose $\mathrm{y}+\mathrm{z}, \mathrm{z}+\mathrm{x}$ and $\mathrm{x}+\mathrm{y}$ are in $\mathrm{A}-\mathrm{P}$
$\therefore(\mathrm{z}+\mathrm{x})-(\mathrm{y}+\mathrm{z})=(\mathrm{x}+\mathrm{y})-(\mathrm{z}+\mathrm{x})$
$2(z+x)=(y+z)+(x+y)$
$\Rightarrow 2 z+2 x=2 y+z+x \Rightarrow z+x=2 y$
$\Rightarrow \mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are in $\mathrm{AP}$. Which is true
(b) Let $\mathrm{y}+\mathrm{z}, \mathrm{z}+\mathrm{x}, \mathrm{x}+\mathrm{y}$ are in $\mathrm{HP}$.
$z+x=\frac{2(y+z)(x+y)}{y+z+x+y}$
$\Rightarrow \quad z+x=\frac{2(y+z)(x+y)}{2 y+z+x}$
$\Rightarrow \quad 2 y z+z^{2}+z x+2 x y+x z+x^{2}$
$\quad=2 y x+2 y^{2}+2 z x+2 y z$
$\Rightarrow z^{2}+x^{2}=2 y^{2}$
$\Rightarrow \mathrm{x}^{2}, \mathrm{y}^{2}$ and $\mathrm{z}^{2}$ are in AP. Which is true. Hence, $\mathrm{y}+\mathrm{z}, \mathrm{z}+\mathrm{x}$ and $\mathrm{x}+$ y are in A.P.

Topic: Mathematics, Sequences and Series, NDA, NDA 2009 (Phase 2)
If $x^{2} y^{b}=(x-y)^{a+b}$, then the value of $\frac{d y}{d x}-\frac{y}{x}$ is equal to

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

The correct answer is:
0

Solution:
$x^{a} y^{b}=(x-y)^{a+b}$
taking log both the sides.
$\log \left(x^{a} y^{b}\right)=\log (x-y)^{(a+b)}$
$a \log x+b \log y=(a+b) \log (x-y)$
differentiating both sides w.r.t 'x'.
$\frac{a}{x}+\frac{b}{y} \frac{d y}{d x}=\frac{(a+b)}{(x-y)}\left[1-\frac{d y}{d x}\right]$
$\frac{d y}{d x}\left[\frac{b}{y}+\frac{a+b}{x-y}\right]=\frac{a+b}{x-y}-\frac{a}{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{\mathrm{bx}-\mathrm{by}+\mathrm{ay}+\mathrm{by}}{\mathrm{y}(\mathrm{x}-\mathrm{y})}\right]=\frac{\mathrm{ax}+\mathrm{bx}-\mathrm{ax}+\mathrm{ay}}{\mathrm{x}(\mathrm{x}-\mathrm{y})}$
$\frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{\mathrm{bx}+\mathrm{ay}}{\mathrm{y}}\right]=\frac{\mathrm{bx}+\mathrm{ay}}{\mathrm{x}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}$
$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}}=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2015 (Phase 2)
If $\quad X=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right], B=\left[\begin{array}{cc}5 & 2 \\ -2 & 1\end{array}\right]$ and $A=\left[\begin{array}{cc}p & q \\ r & s\end{array}\right]$ satisfy
the equation $A X=B$, then the matrix $A$ is equal to

Options:
A $\left[\begin{array}{cc}-7 & 26 \\ 1 & -5\end{array}\right]$
B $\left[\begin{array}{ll}7 & 26 \\ 4 & 17\end{array}\right]$
C $\left[\begin{array}{ll}-7 & -4 \\ 26 & 13\end{array}\right]$
D $\left[\begin{array}{rr}-7 & 26 \\ -6 & 23\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}-7 & 26 \\ 1 & -5\end{array}\right]$

Solution:
$\quad \mathrm{X}=\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}5 & 2 \\ -2 & 1\end{array}\right]$ and $\mathrm{A}=\left[\begin{array}{ll}\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{s}\end{array}\right]$
Now, $\mathrm{AX}=\mathrm{B}$
$\left[\begin{array}{ll}\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{s}\end{array}\right]\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]=\left[\begin{array}{cc}5 & 2 \\ -2 & 1\end{array}\right]$
$\Rightarrow \quad\left[\begin{array}{ll}3 \mathrm{p}+\mathrm{q} & -4 \mathrm{p}-\mathrm{q} \\ 3 \mathrm{r}+\mathrm{s} & -4 \mathrm{r}-\mathrm{s}\end{array}\right]=\left[\begin{array}{cc}5 & 2 \\ -2 & 1\end{array}\right]$
$3 p+q=5$ ...(i)
$-4 p-q=2$ ...(ii)
$3 r+s=-2$ ...(iii)
$-4 r-s=1$ ...(iv)
From equations (i) and (ii), we get $-p=7$
$\therefore \mathrm{p}=-7$
$\begin{aligned} \Rightarrow \quad \mathrm{q} &=5-3(-7) \\ \mathrm{q} &=26 \end{aligned}$
From
(iii) and (iv),
$\begin{array}{ll} & -r=-1 \\ \therefore & \quad r=1 \\ \Rightarrow & s=-2-3=-5 \\ \therefore & s=-5\end{array}$
Hence, $\mathrm{A}=\left[\begin{array}{ll}\mathrm{p} & \mathrm{q} \\ \mathrm{r} & \mathrm{s}\end{array}\right]=\left[\begin{array}{cc}-7 & 26 \\ 1 & -5\end{array}\right]$
$\therefore$ Option (a) is correct.

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 1)
If $X=\left\{\left(4^{n}-3 n-1\right) \mid n \in N\right\}$ and $\mathrm{Y}=\{9(n-1) n \in N\}$, then
what is $X \cup Y$ equal to?

Options:
A $\mathrm{X}$
B $\mathrm{Y}$
C $\mathrm{N}$
D A null set

The correct answer is:
$\mathrm{Y}$

Solution:
Let $X=\left\{\left(4^{n}-3 n-1\right) \mid n \in N\right\}$
and $Y=\{9(n-1) \mid n \in N\}$
$\Rightarrow X=\{0,9,54, \ldots\}$ (By putting $n=1,2, \ldots .)$
and $Y=\{0,9,18,27,36,54, \ldots$ ?
(By putting $n=1,2, \ldots$ $\therefore X \cup Y=\{0,9,18,27,36,54, \ldots\}=Y$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
If $\mathrm{x}^{4}$ occurs in the rth term in the expansion of
$\left(\mathrm{x}^{4}+\frac{1}{\mathrm{x}^{3}}\right)^{15}$, then what is the value of r?

Options:
A 4
B 8
C 9
D 10

The correct answer is:
9

Solution:
In the expansion of $\left(x^{4}+\frac{1}{x^{3}}\right)^{15}$, let $T_{r}$ is the $n_{\text {th }}$
term
$\mathrm{T}_{\mathrm{r}}=15_{\mathrm{C}_{\mathrm{r}-1}}\left(\mathrm{x}^{4}\right)^{15-\mathrm{r}+1}\left(\frac{1}{\mathrm{x}^{3}}\right)^{\mathrm{r}-1}$
$=15_{\mathrm{C}_{\mathrm{r}-1}} \mathrm{x}^{64-4 \mathrm{r}-3 \mathrm{r}+3}=15_{\mathrm{C}_{\mathrm{r}-1}} \mathrm{x}^{67-7 \mathrm{r}}$
$\mathrm{x}^{4}$ occurs in this term
$\Rightarrow 4=67-7 \mathrm{r}$
$\Rightarrow 7 \mathrm{r}=63$
$\Rightarrow \quad \mathrm{r}=9$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2007 (Phase 2)
If $x^{1 / 3}+y^{1 / 3}+z^{1 / 3}=0$ then what is $(x+y+z)^{3}$ equal to?

Options:
A 1
B 3
C $3 \mathrm{xy}$
D 27

The correct answer is:
27

Solution:
$x^{1 / 3}+y^{1 / 3}+z^{1 / 3}=0$
Given formula, $(a+b)^{3}=a^{3}+b^{3}+3 a b(A+b)$
so, $x^{1 / 3}=-\left(y^{1 / 3}+z^{1 / 3}\right)$
Raising both the to the of cube $x=-\left\{y+z+3 y^{1 / 3} z^{1 / 3}\left(y^{1 / 3}+z^{1 / 3}\right)\right\}$
$=-\left\{y+z+3 y^{1 / 3} z^{1 / 3}\left(-x^{1 / 3}\right)\right\}=-\left\{y+z-3 x^{1 / 3} y^{1 / 3} z^{1 / 3}\right\}$
$=-y-z+3 x^{1 / 3} y^{1 / 3} z^{1 / 3} \Rightarrow x+y+z=3 x^{1 / 3} y^{1 / 3} z^{1 / 3}$
or $x+y+z=3(x y z)^{1 / 3}$
$\Rightarrow(x+y+z)^{3}=27 x y z$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 1)
If $x=(1101)_{2}$ and $y=(110)_{2}$, then what is the value of $x^{2}-y^{2}$ ?

Options:
A $\left(\begin{array}{llll}10 & 0 & 0 & 1 & 0 & 1\end{array}\right)_{2}$
B $\left(\begin{array}{llllll}1 & 0 & 0 & 0 & 0 & 1 & 0 & 1\end{array}\right)_{2}$
C $\left(\begin{array}{llllll}1 & 0 & 0 & 0 & 1 & 1 & 0 & 1\end{array}\right)_{2}$
D $\left(\begin{array}{llllll}1 & 0 & 0 & 1 & 0 & 1 & 0 & 1\end{array}\right)_{2}$

The correct answer is:
$\left(\begin{array}{llllll}1 & 0 & 0 & 0 & 0 & 1 & 0 & 1\end{array}\right)_{2}$

Solution:
Given, $x=(1101)_{2}=1 \times 2^{3}+1 \times 2^{2}+0 \times 2^{1}+1$
$=8+4+1=13$
and $\begin{aligned} y=(110)_{2}=& 1 \times 2^{2}+1 \times 2^{1}+0 \times 2^{0} \\=4+2=6 \end{aligned}$
$\therefore x^{2}-y^{2}=(13)^{2}-(6)^{2}=169-36=133$
Now, \begin{array}{l|l|l}
2 & 133 & \\
\hline 2 & 66 & 1 \\
\hline 2 & 33 & 0 \\
\hline 2 & 16 & 1 \\
\hline 2 & 8 & 0 \\
\hline 2 & 4 & 0 \\
\hline 2 & 2 & 0 \\
\hline & 1 & 0
\end{array}
$\therefore 133=(10000101)_{2}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 2)
If $\mathrm{x}=\mathrm{a}^{1 / 3}-\mathrm{a}^{-1 / 3}$, then what is $\mathrm{x}^{3}+3 \mathrm{x}$ equal to?

Options:
A zero
B $\mathrm{a}+\left(\frac{1}{\mathrm{a}}\right)$
C $\mathrm{a}-\left(\frac{1}{\mathrm{a}}\right)$
D $a^{3}+\left(\frac{1}{a^{3}}\right)$

The correct answer is:
$\mathrm{a}-\left(\frac{1}{\mathrm{a}}\right)$

Solution:
In the given equation, Given formula, $(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}$
$x=a^{1 / 3}-a^{-1 / 3}$
Raising both the sides to the power of cube $\begin{aligned} x^{3} &=a-3 a^{2 / 3} \cdot a^{-1 / 3}+3 a^{1 / 3} \times a^{-2 / 3}-a^{-1} \\ &=a-3 a^{1 / 3}+3 a^{-1 / 3}-a^{-1}=a-3\left(a^{1 / 3}-a^{-1 / 3}\right)-a^{-1} \end{aligned}$
$\Rightarrow x^{3}=a-3 x-a^{-1}$ or $x^{3}+3 x=a-\frac{1}{a}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 1)
If $(x+a)$ is a factor of both the quadratic polynomials $x^{2}+p x+q$ and $x^{2}+L x+m$, where $p, q, l$ and $m$ are constants, then which one of the following is correct?

Options:
A $a=(m-q) /(l-p)(l \neq p)$
B $a=(m+q) /(l+p)(l \neq-p)$
C $l=(m-q) /(a-p)(a \neq p)$
D $p=(m-q) /(a-l)(a \neq l)$

The correct answer is:
$a=(m-q) /(l-p)(l \neq p)$

Solution:
Given $(x+a)$ is a factor of quadratic polynomials $x^{2}+p x+q$ and $x^{2}+b x+m$
then $a^{2}-a p+q=0$
and $a^{2}-l a+m=0$
(i) $-($ ii $) \Rightarrow-a p+q+l a-m=0$
$\Rightarrow(I-p) a=m-q$
$\Rightarrow a=\frac{m-q}{l-p}(l \neq p)$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2009 (Phase 1)
If $\mathrm{x}=\mathrm{a} \sec \theta \cos \phi, \mathrm{y}=\mathrm{bsec} \theta \sin \phi, \mathrm{z}=\operatorname{ctan} \theta$, then what
is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}$ equal to?

Options:
A 1
B 0
C $-1$
D $\mathrm{a}^{2}+\mathrm{b}^{2}$

The correct answer is:
1

Solution:
As given:
$\mathrm{x}=\mathrm{a} \sec \theta \cos \phi, \mathrm{y}=\mathrm{b} \sec \theta \sin \phi, \mathrm{z}=\mathrm{c} \tan \theta$
So, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\frac{a^{2} \sec ^{2} \theta \cos ^{2} \phi}{a^{2}}$
$+\frac{b^{2} \sec ^{2} \theta \sin ^{2} \phi}{b^{2}}-\frac{c^{2} \tan ^{2} \theta}{c^{2}}$
$=\sec ^{2} \theta\left(\cos ^{2} \phi+\sin ^{2} \phi\right)-\tan ^{2} \theta=\sec ^{2} \theta-\tan ^{2} \theta=1$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 2)
If $X$ and $Y$ are any two non-empty sets, then what is $(X-Y)$ ' equal to?

Options:
A $X^{\prime}-Y^{\prime}$
B $X^{\prime} \cap Y$
C $X^{\prime} \cup Y$
D $X-Y$

The correct answer is:
$X^{\prime} \cup Y$

Solution:
We know
$X-Y=X \cap Y^{\prime}$
$(X-Y)^{\prime}=\left(X \cap Y^{\prime}\right)^{\prime}$
$=X^{\prime} \cup\left(Y^{\prime}\right)^{\prime}=X^{\prime} \cup Y$

Topic: Mathematics, Sets and Relations, NDA, NDA 2009 (Phase 1)
If $x$ and $y$ are positive and $x y>1$, then what is $\tan ^{-1} x+$ $\tan ^{-1} y$ equal to ?

Options:
A $\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
B $\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
C $\pi-\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
D $\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$

The correct answer is:
$\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

Solution:
$\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right]$, when $\mathrm{xy} < 1$.
And if $x < 0, y < 0$ and $x y>1$, then $\tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2014 (Phase 1)
If $X$ and $Y$ are the matrices of order $2 \times 2$ each and
$2 X-3 Y=\left[\begin{array}{cc}-7 & 0 \\ 7 & -13\end{array}\right]$ and $3 X+2 Y=\left[\begin{array}{ll}9 & 13 \\ 4 & 13\end{array}\right]$, then what is $Y$ equal to?

Options:
A $\left[\begin{array}{cc}1 & 3 \\ -2 & 1\end{array}\right]$
B $\left[\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right]$
C $\left[\begin{array}{cc}3 & 2 \\ -1 & 5\end{array}\right]$
D $\left[\begin{array}{cc}3 & 2 \\ 1 & -5\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}3 & 2 \\ -1 & 5\end{array}\right]$

Solution:
Let $X$ and $Y$ be two matrices of order $2 \times 2$ each.
Given, $2 X-3 Y=\left[\begin{array}{cc}-7 & 0 \\ 7 & -13\end{array}\right]$ ...(i)
and $3 X+2 Y=\left[\begin{array}{ll}9 & 13 \\ 4 & 13\end{array}\right]$ ...(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 2, we get
$6 X-9 Y=\left[\begin{array}{cc}-21 & 0 \\ 21 & -39\end{array}\right]$ ...(iii)
$6 X+4 Y=\left[\begin{array}{cc}18 & 26 \\ 8 & 26\end{array}\right]$ ...(iv)
On subtracting Eqs. (iii) from (iv), we get
$13 Y=\left[\begin{array}{cc}39 & 26 \\ -13 & 65\end{array}\right] \Rightarrow Y=\left[\begin{array}{cc}3 & 2 \\ -1 & 5\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2009 (Phase 2)
If $\bar{x}$ and $\bar{y}$ are the means of two distrubutions such that
$\bar{x} < \bar{y}$ and $\bar{z}$ is the mean of the combined distrubution, then which one of the following statements is correc

Options:
A $\quad \bar{x} < \bar{y} < \bar{z}$
B $\bar{x}>\bar{y}>\bar{z}$
C $\bar{z}=\frac{\bar{x}+\vec{\jmath}}{2}$
D $\bar{x} < \bar{z} < \bar{v}$

The correct answer is:
$\bar{x} < \bar{z} < \bar{v}$

Solution:
It is obvious that, $\bar{x} < \bar{z} < \bar{y}$.

Topic: Mathematics, Statistics, NDA, NDA 2014 (Phase 2)
If $x$ -axis is tangent to the circle $x^{2}+y^{2}+2 g x+2 f y+k=0$, then which one of the following is correct?

Options:
A $g^{2}=k$
B $g^{2}=f$
C $f^{2}=k$
D $f^{2}=g$

The correct answer is:
$g^{2}=k$

Solution:
Length of intercept on the $x$ -axis made by the circle
$$
x^{2}+y^{2}+2 g x+2 f y+k=0 \text { is } 2 \sqrt{g^{2}-k}
$$
Since, circle touches the $\mathrm{x}$ -axis therefore intercept on $\mathrm{x}=$ axis $=0$
$$
\begin{array}{l}
\therefore \quad \sqrt{g^{2}-k}=0 \\
\Rightarrow \quad g^{2}=k
\end{array}
$$

Topic: Mathematics, Circle, NDA, NDA 2009 (Phase 1)
If $x=\cos (2 \mathrm{t})$ and $y=\sin ^{2} \mathrm{t}$, then what is $\frac{d^{2} y}{d x^{2}}$ equal to?

Options:
A 0
B $\sin (2 \mathrm{t})$
C $-\cos (2 \mathrm{t})$
D $-\frac{1}{2}$

The correct answer is:
0

Solution:
Let $x=\cos 2 t$ and $y=\sin ^{2} t$
Differentiate both the functions w.r.t. ' $\mathrm{t}^{\prime}$
$\frac{\mathrm{dx}}{\mathrm{dt}}=-2 \sin 2 \mathrm{t}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}=2 \sin t \cos \mathrm{t}=\sin 2 \mathrm{t}$
$\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=\frac{\sin 2 \mathrm{t}}{-2 \sin 2 \mathrm{t}}=-\frac{1}{2}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
If $\mathrm{x}=\cos \mathrm{t}, \mathrm{y}=\sin \mathrm{t}$, then what is $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ equal to?

Options:
A $\mathrm{y}^{-3}$
B $\mathrm{y}^{3}$
C $-\mathrm{y}^{-3}$
D $-y^{3}$

The correct answer is:
$-\mathrm{y}^{-3}$

Solution:
Given that $x=\cos t, y=\sin t$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\sin \mathrm{t}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}=\cot \mathrm{t}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=-\frac{\cos \mathrm{t}}{\sin \mathrm{t}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\cot \mathrm{t}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=\operatorname{cosec}^{2} \mathrm{t} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\operatorname{cosec}^{2} \mathrm{t} \frac{1}{-\sin \mathrm{t}}=-\frac{1}{\sin ^{3} \mathrm{t}}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=-\mathrm{y}^{-3}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 1)
If $x \cos \theta+y \sin \theta=2$ is perpendicular to the line $x-y=3$, then what is one of the value of $\theta ?$

Options:
A $\pi / 6$
B $\pi / 4$
C $\pi / 2$
D $\pi / 3$

The correct answer is:
$\pi / 4$

Solution:
Consider a line
$x \cos \theta+y \sin \theta=2$
$\Rightarrow y \sin \theta=-x \cos \theta+2$
$\Rightarrow y=-x \frac{\cos \theta}{\sin \theta}+\frac{2}{\sin \theta}$
$\Rightarrow \mathrm{y}=-x \cot \theta+2 \operatorname{cosec} \theta$
On comparing this equation with $y=m x+c$ we get slope of line $x \cos \theta+y \sin \theta=2$ is $-\cot \theta$
Also, we have a line $x-y=3$ $\Rightarrow y=x-3$
slope of line $x-y=3$ is 1 . Since, both the lines are perpendicular to each other Product of their slopes $=-1$ $\Rightarrow(-\cot \theta)(1)=-1$
$\Rightarrow \cot \theta=1=\cot \frac{\pi}{4}$
$\Rightarrow \theta=\frac{\pi}{4}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
If $x \cos \theta+y \sin \theta=z$, then what is the value of $(x \sin \theta-y \cos \theta)^{2} ?$

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

The correct answer is:
$x^{2}+y^{2}-z^{2}$

Solution:
Here, $z=x \cos \theta+y \sin \theta$
$z^{2}=x^{2} \cos ^{2} \theta+y^{2} \sin ^{2} \theta+2 x y \sin \theta \cos \theta$
$\Rightarrow 2 x y \sin \theta \cos \theta=z^{2}-x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta$
Let, $L=(x \sin \theta-y \cos \theta)^{2}$
$\Rightarrow L=x^{2} \sin ^{2} \theta+y^{2} \cos ^{2} \theta-2 x y \sin \theta \cos \theta$
$\Rightarrow L=x^{2} \sin ^{2} \theta+y^{2} \cos ^{2} \theta-\left[z^{2}-x^{2} \cos ^{2} \theta-y^{2} \sin ^{2} \theta\right]$
$\Rightarrow L=x^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]+y^{2}\left[\sin ^{2} \theta+\cos ^{2} \theta\right]-z^{2}$
$\Rightarrow L=x^{2}+y^{2}-z^{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 2)
If $x d y=y(d x+y d y) ; y(1)=1$ and $y(x)>0$, then what is $y(-$ 3) equal to? $\quad$

Options:
A 3
B 2
C 1
D 0

The correct answer is:
3

Solution:
$x d y=y d x+y^{2} d y$
$\Rightarrow x \cdot \frac{d y}{d x}=y+y^{2} \cdot \frac{d y}{d x} \Rightarrow \frac{y-x \cdot \frac{d y}{d x}}{y^{2}}=\frac{-d y}{d x}$
Integrating both sides
$\frac{x}{y}=-y+c$
Given, $x=1, y=1$
$\Rightarrow \frac{1}{1}=-1+c \quad \Rightarrow c=2$
$\therefore \frac{\mathrm{x}}{\mathrm{y}}+\mathrm{y}=2$
$\Rightarrow x+y^{2}=2 y$
$\Rightarrow-3+y^{2}-2 y=0$
$\Rightarrow y^{2}-2 y-3=0$
$y=3,-1$
$\therefore y=3$

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 1)
If $\mathrm{X}$ follows a binomial distribution with parameters $\mathrm{n}=100$ and $\mathrm{p}=1 / 3$, then $\mathrm{P}(\mathrm{X}=\mathrm{r})$ is maximum when

Options:
A $\quad r=16$
B $\quad r=32$
C $\quad r=33$
D $\quad r=34$

The correct answer is:
$\quad r=33$

Solution:
$\mathrm{P}(\mathrm{X}=r)$ will be maximum when $r$ is mode. There are 2 cases.
(i) If $(n+1) \mathrm{P}$ is an integer, binomial distribution is bimodal and two modal values are $(n+1) \mathrm{P}$ and $(n+1) \mathrm{P}-1$
(ii) If $(n+1) \mathrm{P}$ is not an integer, then modal value is integral part of $(n+1) \mathrm{P}$.
Here, $(n+1) \mathrm{P}=(100+1) \frac{1}{3}=101 \times \frac{1}{3}=33.69$, which is
not integer. $\therefore \mathrm{P}(\mathrm{X}=r)$ is maximum at integral part of $33.69$ i.e., 33 .

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 1)