datavorous/entrance-exam-dataset huggingface dataset:
If $y=\sin \left(m \sin ^{-1} x\right)$, what is the value of $d^{2} y / d x^{2}$ at $x=0$?

Options:
A $m$
B $m^{2}$
C $m^{2}+2$
D None of these

The correct answer is:
None of these

Solution:
$y=\sin \left(m \sin ^{-1} x\right)$
Then, $\frac{d y}{d x}=\cos \left(m \sin ^{-1} x\right) \frac{m}{\sqrt{1-x^{2}}}$
$\therefore \frac{d^{2} y}{d x^{2}}=\cos \left(m \sin ^{-1} x\right) \cdot m\left\{\frac{-1}{2} \cdot \frac{(-2 x)}{\left(1-x^{2}\right)^{3 / 2}}\right\}$
$+\frac{m}{\sqrt{1-x^{2}}}\left\{-\sin \left(m \sin ^{-1} x\right)\right\} \cdot \frac{m}{\sqrt{1-x^{2}}}$
$=\frac{m}{\sqrt{1-x^{2}}}\left[\frac{x}{\left(1-x^{2}\right)} \cos \left(m \sin ^{-1} x\right)\right]$
$\left.-\frac{m}{\sqrt{1-x^{2}}} \sin \left(m \sin ^{-1} x\right)\right]$
Now, $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is $m[0-0]=0 \quad\left(\because \sin ^{-1} 0=0\right)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 2)
If $\mathrm{y}=\sin (\ell \mathrm{nx})$, then which one of the following is correct?

Options:
A $\frac{d^{2} y}{d x^{2}}+y=0$
B $\frac{d^{2} y}{d x^{2}}=0$
C $\mathrm{x}^{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$
D $x^{2} \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+y=0$

The correct answer is:
$\mathrm{x}^{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$

Solution:
(c) $y=\sin (\log x)$
$\Rightarrow \frac{d y}{d x}=\frac{\cos (\log x)}{x}$
$\Rightarrow \quad x\left(\frac{d y}{d x}\right)=\cos (\log x)$
Again differentiating, $x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\frac{\sin \log x}{x}=\frac{-y}{x}$
$\Rightarrow \quad x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 2)
If $\mathrm{y}=|\sin \mathrm{x}| \mathrm{x} \mid$, then what is the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=\frac{\pi}{6} ?$

Options:
A $\frac{2^{-\frac{\pi}{6}}(6 \ln 2-\sqrt{3} \pi)}{6}$
B $\frac{2^{\frac{\pi}{6}}(6 \ln 2+\sqrt{3} \pi)}{6}$
C $\frac{2^{-\frac{\pi}{6}}(6 \ln 2+\sqrt{3} \pi)}{6}$
D $\frac{2^{\frac{\pi}{6}}(6 \ln 2-\sqrt{3} \pi)}{6}$

The correct answer is:
$\frac{2^{-\frac{\pi}{6}}(6 \ln 2-\sqrt{3} \pi)}{6}$

Solution:
$\begin{aligned} y &=(-\sin x)^{-x} \\ & \frac{d y}{d x}=(-\sin )^{-x}\left[\frac{x}{\sin x}(-\cos x)+\log (-\sin x) \cdot(-1)\right] \\ &\left[\frac{d y}{d x}\right] x=-\frac{\pi}{6}=\left(\frac{1}{2}\right)^{\pi / 6}\left[-\frac{\pi}{6} \sqrt{3}-\log \frac{1}{2}\right] \\ &=2^{-\pi / 6}\left(\frac{6 \log 2-\sqrt{3} \pi}{6}\right) \end{aligned}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2018 (Phase 2)
If $\mathrm{y}=\tan ^{-1}\left(\frac{5-2 \tan \sqrt{\mathrm{x}}}{2+5 \tan \sqrt{\mathrm{x}}}\right)$, then what is $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to?

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

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

Solution:
$\begin{aligned} & y=\tan ^{-1}\left(\frac{5-2 \tan \sqrt{x}}{2+5 \tan \sqrt{x}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{5}{2}-\tan \sqrt{x}}{1+\left(\frac{5}{2}\right) \tan \sqrt{x}}\right) \\ &=\tan ^{-1} \frac{5}{2}-\tan ^{-1} \tan \sqrt{x} \\ &=\tan ^{-1} \frac{5}{2}-\sqrt{x} \\ \therefore & \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}} \end{aligned}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2018 (Phase 2)
If $\mathrm{y}=\frac{\mathrm{x}+1}{\mathrm{x}-1}$, then what is $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to?

Options:
A $\frac{-2}{x-1}$
B $\frac{-2}{(x-1)^{2}}$
C $\frac{2}{(x-1)^{2}}$
D $\frac{2}{(x-1)}$

The correct answer is:
$\frac{-2}{(x-1)^{2}}$

Solution:
Let $\mathrm{y}=\frac{\mathrm{x}+1}{\mathrm{x}-1}$
Differentiate both the side w.r.t ' $x$ '
$\begin{aligned} \frac{\mathrm{dy}}{\mathrm{dx}} &=\frac{(\mathrm{x}-1)(1)-(\mathrm{x}+1) 1}{(\mathrm{x}-1)^{2}} \\ &=\frac{\mathrm{x}-1-\mathrm{x}-1}{(\mathrm{x}-1)^{2}}=\frac{-2}{(\mathrm{x}-1)^{2}} \end{aligned}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 1)
If $y=x+e^{x}$, then, what is $\frac{d^{2} x}{d y^{2}}$ equal to?

Options:
A $\mathrm{e}^{\mathrm{x}}$
B $-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{3}}$
C $-\frac{e^{x}}{\left(1+e^{x}\right)}$
D $-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{2}}$

The correct answer is:
$-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{3}}$

Solution:
Given that $y=x+e^{x}$ Differentiating w. r. t. $\mathrm{x}$
$\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=1+\mathrm{e}^{\mathrm{x}}$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{1+\mathrm{e}^{\mathrm{x}}}$
Differentiating w. r.t. $\mathrm{y}$
$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=-\frac{(1)\left(\mathrm{e}^{\mathrm{x}}\right)}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{2}} \cdot \frac{\mathrm{dx}}{\mathrm{dy}}$
$=-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{2}} \cdot \frac{1}{\left(1+\mathrm{e}^{\mathrm{x}}\right)}=-\frac{\mathrm{e}^{\mathrm{x}}}{\left(1+\mathrm{e}^{\mathrm{x}}\right)^{3}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 1)
If $y=x \ln x+x e^{x}$, then what is the value of $\frac{d y}{d x}$ at $x=1$?

Options:
A $1+\mathrm{e}$
B $1-\mathrm{e}$
C $1+2 \mathrm{e}$
D None of these

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

Solution:
$y=x \ln x+x e^{x}$
After differentiating both sides with respect to $\mathrm{x}$,
we get $\frac{d y}{d x}=x \cdot \frac{1}{x}+\log x+x e^{x}+e^{x}$
or $=1+\log x+x e^{x}+e^{x}$
Therefore $\left(\frac{d y}{d x}\right)_{x=1}=1+\log 1+1 . e^{1}+e^{1}=1+2 e$
$[\because \log 1=0]$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2014 (Phase 2)
If $y=x+x^{2}+x^{3}+\ldots$ up to infinite terms where $x < 1$, then which one of the following is correct?

Options:
A $\mathrm{x}=\frac{\mathrm{y}}{1+\mathrm{y}}$
B $\mathrm{x}=\frac{\mathrm{y}}{1-\mathrm{y}}$
C $\mathrm{x}=\frac{1+\mathrm{y}}{\mathrm{y}}$
D $\mathrm{x}=\frac{1-\mathrm{y}}{\mathrm{y}}$

The correct answer is:
$\mathrm{x}=\frac{\mathrm{y}}{1+\mathrm{y}}$

Solution:
$\mathrm{y}=\mathrm{x}+\mathrm{x}^{2}+\mathrm{x}^{3}+\ldots$ upto infinite terms. $\mathrm{x}+\mathrm{x}^{2}+\mathrm{x}^{3}+\ldots$ is a G.P. with $\mathrm{a}=\mathrm{x}$ and $\mathrm{r}=\mathrm{x}$
$\therefore \mathrm{y}=\frac{\mathrm{x}}{1-\mathrm{x}}$
$\Rightarrow \mathrm{y}-\mathrm{xy}=\mathrm{x}$
$\Rightarrow \mathrm{x}+\mathrm{xy}=\mathrm{y}$
$\Rightarrow \mathrm{x}(1+\mathrm{y})=\mathrm{y}$
$\Rightarrow \mathrm{x}=\frac{\mathrm{y}}{1+\mathrm{y}}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2017 (Phase 2)
If $y=x^{x}$, what is $\frac{d y}{d x}$ at $x=1$ equal to?

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

The correct answer is:
1

Solution:
Let $\mathrm{y}=\mathrm{x}^{\mathrm{x}}$
Take $\log$ on both the sides $\Rightarrow \ell \mathrm{ny}=\mathrm{x} \cdot \ln \mathrm{x}$
$\Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=1+\ell n x \quad \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\ell n e+\ell n x$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y} \ell \mathrm{n} \mathrm{ex}==\left(\mathrm{x}^{\mathrm{x}}\right) \ln \mathrm{ex}$
$\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=1}=1 . \quad$ $\ell n\mathrm{e}=1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 1)
If $\left|\begin{array}{lll}y & x & y+z \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$, then which one of the following is
correct?

Options:
A Either $x+y=z$ or $x=y$
B Either $x+y=-z$ or $x=z$
C Either $x+z=y$ or $z=y$
D Either $z+y=x$ or $x=y$

The correct answer is:
Either $x+y=-z$ or $x=z$

Solution:
Given, $\left|\begin{array}{ccc}y & x & y+z \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$
$\Rightarrow\left|\begin{array}{ccc}x+y+ & x+y+z & 2(x+y+z) \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$
$\Rightarrow(x+y+z)\left|\begin{array}{ccc}1 & 1 & 2 \\ z & y & x+y \\ x & z & z+x\end{array}\right|=0$
Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-2 \mathrm{C}_{1}$
$\Rightarrow(x+y+z)\left|\begin{array}{ccc}1 & 0 & 0 \\ z & z-y & x+y-2 z \\ x & z-x & z-x\end{array}\right|=0$
$\Rightarrow(x+y+z)\left|\begin{array}{cc}z-y & x+y-2 z \\ z-x & z-x\end{array}\right|=0$
$\Rightarrow(x+y+z)(z-x)(z-y-x-y+2 z)=0$
$\Rightarrow x+y=-z$ or $z=x$

Topic: Mathematics, Determinants, NDA, NDA 2009 (Phase 1)
If \( y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c\end{array}\right| \), then \( \frac{d y}{d x} \) is equal to

Options:
A \( \left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right| \)
B \( \left|\begin{array}{ccc}l & m & n \\ f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ a & b & c\end{array}\right| \)
C \( \left|\begin{array}{lll}f^{\prime}(x) & l & a \\ g^{\prime}(x) & m & b \\ h^{\prime}(x) & n & c\end{array}\right| \)
D \( \left|\begin{array}{ccc}l & m & n \\ a & b & c \\ f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x)\end{array}\right| \)

The correct answer is:
\( \left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right| \)

Solution:
Given that, \( y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c\end{array}\right| \)
So,
\( \frac{d y}{d x}=\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right|+\left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ 0 & 0 & 0 \\ a & b & c\end{array}\right|+ \)
\( \left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ l & m & n \\ 0 & 0 & 0\end{array}\right| \)
\( =\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right| \)

Topic: Mathematics, Vector Algebra, KCET, KCET 2017
If you want to measure the intelligence of a group of students, which one of the following measures will be more suitable?

Options:
A Arithmetic mean
B Mode
C Median
D Geometric mean

The correct answer is:
Mode

Solution:
To measure the intelligence of a group of students mode will be more suitable

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 1)
If $\mathrm{z}_{1}$ and $\mathrm{z}_{2}$ are complex numbers with $\left|\mathrm{z}_{1}\right|=\left|\mathrm{z}_{2}\right|$, then which of the following is/are correct?
1- $z_{1}=z_{2}$
2- Real part of $\mathrm{z}_{1}=$ Real part of $\mathrm{z}_{2}$
3- Imaginary part of $z_{1}=$ Imaginary part of $z_{2}$ Select the correct answer using the code given below:

Options:
A 1 only
B 2 only
C 3 only
D None

The correct answer is:
None

Solution:
Let $Z_{1}=a_{1}+i b_{1}$ $\mathrm{Z}_{2}=\mathrm{a}_{2}+\mathrm{ib}_{2}$
$\left|\mathrm{Z}_{1}\right|=\left|\mathrm{Z}_{2}\right|$
$\sqrt{\left(\mathrm{a}_{1}\right)^{2}+\left(\mathrm{b}_{1}\right)^{2}}=\sqrt{\left(\mathrm{a}_{2}\right)^{2}+\left(\mathrm{b}_{2}\right)^{2}}$
It is true for many values of $\mathrm{a}_{1}, \mathrm{a}_{2} \& \mathrm{~b}_{1}, \mathrm{~b}_{2}$. So $\mathrm{a}_{1}$ must not equal to $\mathrm{a}_{2}$, and $\mathrm{b}_{1}$ must not equal to $\mathrm{b}_{2}$.

Topic: Mathematics, Complex Number, NDA, NDA 2015 (Phase 2)
If $z=1+\cos \frac{\pi}{5}+i \sin \frac{\pi}{5}$, then what is $|z|$ equal to ?

Options:
A $2 \cos \frac{\pi}{5}$
B $2 \sin \frac{\pi}{5}$
C $2 \cos \frac{\pi}{10}$
D $2 \sin \frac{\pi}{10}$

The correct answer is:
$2 \cos \frac{\pi}{10}$

Solution:
$\begin{aligned}|z| &=\sqrt{1^{2}+\cos ^{2} \frac{\pi}{5}+2 \cos \frac{\pi}{5}+\sin ^{2} \frac{\pi}{5}} \\ &=\sqrt{1^{2}+1+2 \cos \frac{\pi}{5}}=\sqrt{2\left(1+\cos \frac{\pi}{5}\right)} \\ &=\sqrt{2\left(1+2 \cos ^{2} \frac{\pi}{10}-1\right)}=\sqrt{2\left(2 \cos ^{2} \frac{\pi}{10}\right)} \\ &=2 \cos \frac{\pi}{10} \end{aligned}$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 1)
If $z=1+i \tan \alpha$ where $\pi < \alpha < \frac{3 \pi}{2}$, then what is $|z|$ equal to?

Options:
A $\sec \alpha$
B $-\sec \alpha$
C $\sec ^{2} \alpha$
D $-\sec ^{2} \alpha$

The correct answer is:
$-\sec \alpha$

Solution:
Given : $z=1+i \tan \alpha, \pi < \alpha < \frac{3 \pi}{2}$
$|z|=\sqrt{(1)^{2}+(\tan \alpha)^{2}}=\sqrt{\sec ^{2} \alpha}$
$|z|=\sec \alpha .$
Since, $\pi < \alpha < \frac{3 \pi}{2}$
$\therefore$ Sec $\alpha$ lies in III $^{\mathrm{rd}}$ quadrant
In $3^{\mathrm{rd}}$ quadrant, $\mathrm{sec} \alpha$ is negative.

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 2)
If $z=1+i$, then what is the inverse of $z^{2}$ ?

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

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

Solution:
Given that $z=1+i$ $\Rightarrow z^{2}=(1+i)^{2}=1+i^{2}+2 i=1-1+2 i=2 i$
Inverse of $z^{2}=\frac{1}{2 i}=-\frac{i^{2}}{2 \mathrm{i}}\left[\right.$ Since $\left.i^{2}=-1 ;-\mathrm{i}^{2}=1\right]$
$=-\frac{\mathrm{i}}{2}$

Topic: Mathematics, Complex Number, NDA, NDA 2006 (Phase 1)
If $z_{1}, z_{2}$ are any two complex numbers such that $\left|z_{1}+z_{2}\right|=$ $\left|z_{1}\right|+\left|z_{2}\right|$, which one of the following is correct ?

Options:
A $\mathrm{z}_{1}=\alpha \mathrm{z}_{2}$ with $\alpha \in \mathrm{R}$
B $\mathrm{z}_{1} \geq 0$ or $\mathrm{z}_{2} \geq 0$
C $\quad z_{1}=\alpha z_{2}$ with $\alpha>0$
D $\left|z_{1}\right|=\left|z_{2}\right|$

The correct answer is:
$\mathrm{z}_{1}=\alpha \mathrm{z}_{2}$ with $\alpha \in \mathrm{R}$

Solution:
Let $z_{1}=r_{1}(\cos \theta+i \sin \theta)$
Squaring both the sides $\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}+2 \mathrm{r}_{1} \mathrm{r}_{2} \cos (\theta-\phi)=\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)^{2}$
$=\mathrm{r}_{1}{ }^{2}+\mathrm{r}_{2}{ }^{2}+2 \mathrm{r}_{1} \mathrm{r}_{2}$
or, $2 \mathrm{r}_{1} \mathrm{r}_{2} \cos (\theta-\phi)=2 \mathrm{r}_{1} \mathrm{r}_{2}$
or, $\cos (\theta-\phi)=1$
$\theta-\phi=0 \Rightarrow \theta=\phi$
$\operatorname{Arg}\left(z_{1}\right)=\operatorname{Arg}\left(z_{2}\right)$ so, $z_{1}=\alpha z_{2}$
where $\alpha \in \mathrm{R}$

Topic: Mathematics, Complex Number, NDA, NDA 2006 (Phase 1)
If $\mathrm{z}=\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)^{107}+\left(\frac{\sqrt{3}}{2}-\frac{\mathrm{i}}{2}\right)^{107}$, then what is the

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

The correct answer is:
0

Solution:
$z=\left[\frac{\sqrt{3}}{2}+\frac{i}{2}\right]^{107}+\left[\frac{\sqrt{3}}{2}-\frac{i}{2}\right]^{107} .$
$\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \& \sin \frac{\pi}{6}=\frac{1}{2}$
$\Rightarrow \quad \mathrm{z}=\left[\cos \frac{\pi}{6}+\mathrm{i} \sin \frac{\pi}{6}\right]^{107}+\left[\cos \frac{\pi}{6}-\mathrm{i} \sin \frac{\pi}{6}\right]^{107}$
$\mathrm{Also},(\cos \theta+\mathrm{i} \sin \theta)^{\mathrm{n}}=\cos n \theta+\mathrm{i} \sin \mathrm{n} \theta$
$\Rightarrow \quad \mathrm{z}=\cos \frac{107 \pi}{6}+\mathrm{i} \sin \frac{107 \pi}{6}+\cos \frac{107 \pi}{6}-\mathrm{i} \sin \frac{107 \mathrm{t}}{6} .$
$\Rightarrow \quad \operatorname{Im}(\mathrm{z})=0$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 2)
If $z=\frac{1+2 i}{2-i}-\frac{2-i}{1+2 i}$, then what is the value of $z^{2}+z \bar{z}$ ?
$(i=\sqrt{-1})$

Options:
A 0
B -1
C 1
D 8

The correct answer is:
0

Solution:
$z=\frac{1+2 i}{2-i}-\frac{2-i}{1+2 i}$
$z=\frac{(1+2 i)^{2}-(2-i)^{2}}{(2-i)(1+2 i)}=\frac{1+4 i^{2}+4 i-4-i^{2}+4 i}{2+4 i-i-2 i^{2}}$
$=\frac{-3-4+8 i+1}{4+3 i}=\frac{-6+8 i}{4+3 i}=\frac{(-6+8 i)(4-3 i)}{16+9}$
$=\frac{-24+18 i+32 i-24 i^{2}}{25}=\frac{50 i}{25}=2 i$
Consider,
$z^{2}+z \bar{z}=(2 i)^{2}+(2 i)(-2 i)=4 i^{2}-4 i^{2}=0$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 2)
If $z=\frac{-2(1+2 i)}{3+i}$ where $i=\sqrt{-1}$, then argument
$\theta(-\pi < \theta \leq \pi)$ of $z$ is

Options:
A $\frac{3 \pi}{4}$
B $\frac{\pi}{4}$
C $\frac{5 \pi}{6}$
D $-\frac{3 \pi}{4}$

The correct answer is:
$-\frac{3 \pi}{4}$

Solution:
$\begin{aligned} & z=\frac{-2(1+2 i)}{3+i} \\ &=\frac{-2-4 i}{3+i}=\frac{-2-4 i}{3+i} \times \frac{3-i}{3-i} \\ &=\frac{-6+2 i-12 i+4 i^{2}}{10} \\ &=\frac{-6-10 i-4}{10}=\frac{-10-10 \mathrm{i}}{10}=-1-i \\ & z=-1-i=r(\cos \theta+i \sin \theta) \end{aligned}$
On comparing real and imaginary part on both sides, we get $r \cos \theta=-1$
$r \sin \theta=-1$
On dividing eq. (ii) by (i), we get
$\frac{\mathrm{r} \sin \theta}{\mathrm{r} \cos \theta}=\frac{-1}{-1}$
$\tan \theta=1=\tan \frac{\pi}{4}$
$\Rightarrow \quad \theta=\frac{\pi}{4}$
$\therefore \quad \theta=\frac{\pi}{4}$
$\therefore$ Option (b) is correct.

Topic: Mathematics, Complex Number, NDA, NDA 2015 (Phase 1)
If $|z+5| \leq 3$, then the maximum value of $|z+1|$ is

Options:
A 0
B 4
C 6
D 10

The correct answer is:
6

Solution:
$|z+4| \leq 3$
$|z+1|=|z+4+(-3)|$
We know, $\left|z_{1}+z_{2}\right| \leq\left|z_{1}\right|+\left|z_{2}\right|$
So, $|z+4+(-3)| \leq|z+4|+|-3|$
$\leq|3|+|-3|$
$\leq 6$
So, maximum value $=6$.

Topic: Mathematics, Complex Number, NDA, NDA 2017 (Phase 1)
If $z=f o f(x)=x^{2}$ where $f(x)=x^{2}$, then what is $\frac{d z}{d x}$ equal to?

Options:
A $x^{3}$
B $2 x^{3}$
C $4 x^{3}$
D $4 x^{2}$

The correct answer is:
$4 x^{3}$

Solution:
$\mathrm{z}=$ fof $(\mathrm{x})=\mathrm{f}\left(\mathrm{x}^{2}\right)=\mathrm{x}^{4}$
$\frac{\mathrm{dz}}{\mathrm{dx}}=4 \mathrm{x}^{3}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2014 (Phase 1)
If $z$ is a complex number such that $z+z^{-1}=1$, then what is the value of $z^{99}+z^{-99} ?$

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

The correct answer is:
-2

Solution:
Given, $z+z^{-1}=1 \Rightarrow z+\frac{1}{z}=1$
$\Rightarrow z^{2}-z+1=0 \Rightarrow z=-\omega$ and $-\omega^{2}$
when $z=-\omega$
$\therefore z^{99}+z^{-99}=(-\omega)^{99}+(-\omega)^{-99}=-1-1=-2$
when $z=-\omega^{2}$
$\therefore z^{99}+z^{-99}=\left(-\omega^{2}\right)^{99}+\left(-\omega^{2}\right)^{-99}$
$\quad=-1-1=-2$

Topic: Mathematics, Complex Number, NDA, NDA 2009 (Phase 2)
If $\mathrm{z}=\mathrm{x}+\mathrm{iy}=\left(\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}}\right)^{-25}$, where $\mathrm{i}=\sqrt{-1}$, then what is
the fundamental amplitude of $\frac{\mathrm{z}-\sqrt{2}}{\mathrm{z}-\mathrm{i} \sqrt{2}} ?$

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

The correct answer is:
$\pi$

Solution:
$\begin{aligned} z &=x+i y \\ &=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)^{-25} \\ &=\left[\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right]^{-25} \end{aligned}$
$\because(\operatorname{Cos} \pi-\mathrm{i} \sin \pi)^{\mathrm{n}}=\operatorname{Cos} \mathrm{n} \pi-\mathrm{i} \operatorname{Sin} \mathrm{n} \pi$
$\left[\begin{array}{l}\cos (-\theta)=\cos \theta \\ \sin (-\theta)=-\sin \theta\end{array}\right]$
$\begin{array}{l}\mathrm{z}=\left[\cos \left(\frac{25 \pi}{4}\right)+i \sin \left(\frac{25 \pi}{4}\right)\right] \\ = & {\left[\cos \left(6 \pi+\frac{\pi}{4}\right)+i \sin \left(6 \pi+\frac{\pi}{4}\right)\right]} \\ =\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \\ =\frac{1+i}{\sqrt{2}} \\ \frac{z-\sqrt{2}}{z-i \sqrt{2}}=\frac{(1+i-2) \sqrt{2}}{\sqrt{2}(1+i-2 i)} \\ =\frac{-1+i}{1-i}=-1 \\ \text { Amplitude }=\tan ^{-1} \frac{b}{a}=\tan ^{-1}\left(\frac{0}{-1}\right)=\tan ^{-1} 0 \\ =\tan ^{-1}(\tan \pi)=\pi\end{array}$
Hence fundamental amplitude of $\left(\frac{z-\sqrt{2}}{z-i \sqrt{2}}\right)$ is $\pi$

Topic: Mathematics, Complex Number, NDA, NDA 2016 (Phase 1)
If $\left|z-\frac{4}{z}\right|=2$, then the maximum value of $|z|$ is equal to

Options:
A $1+\sqrt{3}$
B $1+\sqrt{5}$
C $1-\sqrt{5}$
D $\sqrt{5}-1$

The correct answer is:
$1+\sqrt{5}$

Solution:
$\left|\mathrm{z}-\frac{4}{\mathrm{z}}\right|=2 .$
We know $|\mathrm{a}-\mathrm{b}| \geq|\mathrm{a}|-|\mathrm{b}|$
$\therefore\left|z-\frac{4}{z}\right| \geq|z|-\left|\frac{4}{z}\right|$
$\Rightarrow 2 \geq|z|-\frac{4}{|z|} \Rightarrow|z|^{2}-2|z|-4 \leq 0$
$\Rightarrow|z|=\frac{2 \pm \sqrt{4-4(1)(-4)}}{2(1)}=\frac{2 \pm \sqrt{20}}{2}=\frac{2 \pm 2 \sqrt{5}}{2}=1 \pm \sqrt{5}$

Topic: Mathematics, Complex Number, NDA, NDA 2017 (Phase 2)
If $z=-\bar{z}$, then which one of the following is correct?

Options:
A real part of $z$ is zero. $[2012-I]$
B The imaginary part of $\mathrm{z}$ is zero.
C The real part of $z$ is equal to imaginary
D The sum of real and imaginary parts of z is z.

The correct answer is:
real part of $z$ is zero. $[2012-I]$

Solution:
Let z $=\mathrm{x}+\mathrm{y}$ then $\overline{\mathrm{z}}=\mathrm{x}-\mathrm{iy}$
Now $z=-\bar{z}$
$\Rightarrow(\mathrm{x}+\mathrm{iy})=-(\mathrm{x}-\mathrm{iy}) \Rightarrow \mathrm{x}+\mathrm{iy}=-\mathrm{x}+\mathrm{iy}$
$\Rightarrow 2 \mathrm{x}=0 \Rightarrow \operatorname{Re}(\mathrm{z})=0$

Topic: Mathematics, Complex Number, NDA, NDA 2012 (Phase 1)
If $|z+\bar{z}|=|z-\bar{z}|$, then the locus of $z$ is:

Options:
A A pair of straight lines
B A line
C A set of four straight lines
D A circle

The correct answer is:
A pair of straight lines

Solution:
Let $z=x+$ iy, $\bar{z}=x-i y$
$|\mathrm{z}+\overline{\mathrm{z}}|=|\mathrm{z}-\overline{\mathrm{z}}|$
$|(\mathrm{x}+\mathrm{iy})+(\mathrm{x}-\mathrm{iy})|=|(\mathrm{x}+\mathrm{iy})-(\mathrm{x}-\mathrm{iy})|$
$|2 \mathrm{x}|=|2 \mathrm{y}|$
$\mathrm{x}=\pm \mathrm{y}$

Topic: Mathematics, Complex Number, NDA, NDA 2014 (Phase 1)
If \(z=\left(\begin{array}{ccc}1 & 1+2 i & -5 i \\ 1-2 i & -3 & 5+3 i \\ 5 i & 5-3 i & 7\end{array}\right)\) then \((i=\sqrt{-1})\)

Options:
A \(z\) is purely real
B \(z\) is purely imaginary
C \(z+\bar{z}=0\)
D \((z-\bar{z})\) i is purely imaginary

The correct answer is:
\(z\) is purely real

Solution:
\(\begin{aligned}
& \text { Hints : } z=\left|\begin{array}{ccc}
1 & 1+2 i & -5 i \\
1-2 i & -3 & 5+3 i \\
5 i & 5-3 i & 7
\end{array}\right|=1(-21-64)-((1-2 i)(7(1+2 i)+5 i(5-3 i)))+5 i((1+2 i)(5+3 i)-15 i) \\
& \text { = Real } \\
&
\end{aligned}\)

Topic: Mathematics, Determinants, WBJEE, WBJEE 2011
IfA is an invertible matrix of order $\mathrm{n}$ and $\mathrm{k}$ is any positive real number, then the value of $[\operatorname{det}(\mathrm{kA})]^{-1}$ det $\mathrm{A}$ is

Options:
A $\mathrm{k}^{-\mathrm{n}}$
B $\mathrm{k}^{-1}$
C $\mathrm{k}^{\mathrm{n}}$
D nk

The correct answer is:
$\mathrm{k}^{-\mathrm{n}}$

Solution:
If A is a matrix that is invertible then det- $(\mathrm{kA})$ will be $\mathrm{k}^{\mathrm{n}} \cdot \operatorname{det}(\mathrm{A})$, where $\mathrm{n}$ is the order.
$\therefore [\operatorname{det}(\mathrm{KA})]^{-1} \operatorname{det}(\mathrm{A})$
$=\left[(\mathrm{K})^{\mathrm{n}} \times \operatorname{det}(\mathrm{A})\right]^{-1} \cdot \operatorname{det}(\mathrm{A})$
$=\mathrm{K}^{-\mathrm{n}} \times \frac{1}{\operatorname{det}(\mathrm{A})} \times \operatorname{det}(\mathrm{A})$
$=\mathrm{K}^{-\mathrm{n}}$

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 2)
IfABCD is a cyclic quadrilateral then what is $\sin A+\sin B-\sin C-\sin D$ equal to?

Options:
A 0
B 1
C 2
D $2(\sin A+\sin B)$

The correct answer is:
0

Solution:
We know in cyclic quadrilateral, ABCD $\mathrm{A}+\mathrm{C}=180^{\circ}, \mathrm{B}+\mathrm{D}=180^{\circ}$
$\therefore \mathrm{A}=180^{\circ}-\mathrm{C}, \mathrm{B}=180^{\circ}-\mathrm{D}$
$\sin \mathrm{A}+\sin \mathrm{B}-\sin \mathrm{C}-\sin \mathrm{D}$
$=\sin \left(180^{\circ}-\mathrm{C}\right)+\sin \left(180^{\circ}-\mathrm{D}\right)-\sin \mathrm{C}-\sin \mathrm{D}$
$=\sin \mathrm{C}+\sin \mathrm{D}-\sin \mathrm{C}-\sin \mathrm{D}=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
Iff: $\mathrm{R} \rightarrow \mathrm{R}^{+}$ such that $\mathrm{f}(\mathrm{x})=(1 / 3)^{\mathrm{x}}$, then what is the value $\operatorname{off}^{-1}(\mathrm{x}) ?$

Options:
A $(1 / 3)^{\mathrm{x}}$
B $3^{x}$
C $\log _{1 / 3} x$
D $\log _{x}(1 / 3)$

The correct answer is:
$\log _{1 / 3} x$

Solution:
Given function is $\mathrm{f}(\mathrm{x})=\left(\frac{1}{3}\right)^{\mathrm{x}}$
Let $f(x)=y$, so, $y=\left(\frac{1}{3}\right)^{x}$
Taking $\log _{1 / 3}$ on both sides $\Rightarrow \mathrm{x} \times \log _{1 / 3}\left(\frac{1}{3}\right)=\log _{(1 / 3)} \mathrm{y}$
$\mathrm{x}=\log _{(1 / 3)} \mathrm{y}$
$\Rightarrow \mathrm{f}^{-1}(\mathrm{x})=\log _{(1 / 3)} \mathrm{x}$

Topic: Mathematics, Functions, NDA, NDA 2006 (Phase 2)