datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{x}_{\mathrm{i}}>0, \mathrm{y}_{\mathrm{i}}>0(\mathrm{i}=1,2,3, \ldots \ldots \mathrm{n})$ are the values of two variable $X$ and Y with geometric mean Pand Q respectively, then the geometric mean of $\frac{\mathrm{X}}{\mathrm{Y}}$ is

Options:
A $\frac{\mathrm{P}}{\mathrm{Q}}$
B antilog $\left(\frac{\mathrm{P}}{\mathrm{Q}}\right)$
C $\mathrm{n}(\log \mathrm{P}-\log \mathrm{Q})$
D $\mathrm{n}(\log \mathrm{P}+\log \mathrm{Q})$

The correct answer is:
antilog $\left(\frac{\mathrm{P}}{\mathrm{Q}}\right)$

Solution:
$\left(x_{1} x_{2} \ldots \ldots \ldots x_{n}\right)^{1 / n}=\mathrm{P}$
$\left(y_{1} y_{2} \ldots \ldots . y_{n}\right)^{1 / n}=\mathrm{Q}$
$\left(\frac{x_{1} \cdot x_{2} \ldots \ldots \ldots \ldots x_{n}}{y_{1} \cdot y_{2} \ldots \ldots \ldots . y_{n}}\right)^{1 / n}=\frac{\left(x_{1} \cdot x_{2} \ldots \ldots \ldots \ldots x_{n}\right)^{1 / n}}{\left(y_{1} y_{2} \ldots \ldots \ldots . y_{n}\right)^{1 / n}}=\frac{\mathrm{P}}{\mathrm{Q}}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2018 (Phase 2)
If $\sum\left(x_{i}-2\right)=110, \quad \sum^{n}\left(x_{i}-5\right)=20$, then what is the
mean?

Options:
A $11 / 2$
B $2 / 11$
C $17 / 3$
D $17 h$

The correct answer is:
$17 / 3$

Solution:
$\begin{aligned} & \sum_{t=1}^{n}\left(x_{i}-2\right)=110 \\ & \therefore x_{1}+x_{2}+\ldots+x_{n}-2 n=110 \\ & \Rightarrow x_{1}+x_{2}+\ldots+x_{n}=2 n+110 \\ & \text { and } \sum_{i=1}^{n}\left(x_{i}-5\right)=20 \\ & \Rightarrow x_{1}+x_{2}+\ldots+x_{n}-5 n=20 \\ & \Rightarrow x_{1}+x_{2}+\ldots+x_{n}=5 n+20 \end{aligned}$
From equations (i) and (ii), we get $5 n+20=2 n+110$
$\Rightarrow 3 n=90 \Rightarrow n=30$
Now, mean $=\frac{x_{1}+x_{2}+\ldots x_{n}}{n}$
$=\frac{5 \times 30+20}{30}=\frac{170}{30}=\frac{17}{3}$

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 1)
If $x+i y=\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|$, then what is $x-i y$ equal to?

Options:
A $3+i$
B $1+3 i$
C $3 i$
D 0

The correct answer is:
0

Solution:
Let $x+i y=\left|\begin{array}{ccc}6 i & -3 i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i\end{array}\right|$
$\Rightarrow \quad x+i y=6 i\left(3 i^{2}+3\right)+3 i(4 i+20)+1(12-60 i)$
$\quad=-12+60 i+12-60 i=0\left(\because i^{2}=-1\right)$
$\Rightarrow \quad x+i y=0$
$\Rightarrow \quad x=0$ and $y=0$
Hence, $x-i y=0-i(0)=0$

Topic: Mathematics, Determinants, NDA, NDA 2010 (Phase 2)
If $\mathrm{x} \hat{\mathbf{i}}+\mathrm{y} \hat{\mathbf{j}}+\mathrm{z} \hat{\mathbf{k}}$ is a unit vector and $\mathrm{x}: \mathrm{y}: \mathrm{z}=\sqrt{3}: 2: 3$, then what is the value of $z ?$

Options:
A $\frac{3}{16}$
B 3
C $\frac{3}{4}$
D 2

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

Solution:
Let $x \hat{i}+y \hat{j}+z \hat{k}$ is a unit vector.
$\therefore \quad x^{2}+y^{2}+z^{2}=1$
Given $x: y: z=\sqrt{3}: 2: 3$
$\Rightarrow x=\sqrt{3} k, y=2 k$ and $z=3 k$
$\therefore \quad(\sqrt{3} k)^{2}+(2 k)^{2}+(3 k)^{2}=1$
$\Rightarrow 3 k^{2}+4 k^{2}+9 k^{2}=1$
$\Rightarrow k^{2}=\frac{1}{16} \Rightarrow k=\frac{1}{4}$
Hence, $z=3 k=3 \times \frac{1}{4}=\frac{3}{4}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 1)
If X is a Poisson variate such that 2P(X=1)=5P(X=5)+2P(X=3), then the standard deviation of X is

Options:
A 4
B 2
C 12
D 2

The correct answer is:
2

Solution:
It is given that,

2P(X=1)=5P(X=5)+2P(X=3),

The formula for Poisson PX=r=e-λλrr!.

⇒2e-λλ1!=5e-λλ55!+2e-λλ33!

⇒2=λ424+λ23

⇒2=3λ4+24λ272

⇒144=3λ4+24λ2

⇒λ4+8λ2-48=0

Consider λ2=x,

⇒x2+8x-48=0

⇒(x+12)(x-4)=0

x=4 and x≠-12

The standard deviation is,

x=λ2=4 and λ=2

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
If $x$ is an integer and satisfies $9 < 4 x-1 \leq 19$, then $x$ is an element of which one of the following sets?

Options:
A $\{3,4\}$
B $\{2,3,4\}$
C $\{3,4,5\}$
D $\{2,3,4,5$

The correct answer is:
$\{3,4,5\}$

Solution:
Given, $9 < 4 x-1 \leq 19$
$\Rightarrow 9 < 4 x-1$ and $4 x-1 \leq 19$
$\Rightarrow 9+1 < 4 x$ and $4 x \leq 19+1$
$\Rightarrow x>\frac{5}{2}$ and $x \leq 5$
$\therefore x \in\{3,4,5\}$

Topic: Mathematics, Basic of Mathematics, NDA, NDA 2008 (Phase 2)
If x is anyreal number, then $\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{4}}$ belongs to which one of the following intervals?

Options:
A $(0,1)$
B $\left(0, \frac{1}{2}\right]$
C $\left(0, \frac{1}{2}\right)$
D $[0,1]$

The correct answer is:
$\left(0, \frac{1}{2}\right]$

Solution:
$\mathrm{y}=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{4}} \Rightarrow \mathrm{y} \geq 0 .$
Also, $\mathrm{y}=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{4}}=\frac{1}{\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}}$
$\Rightarrow \mathrm{y} \leq \frac{1}{2}$
$\therefore \frac{\mathrm{x}^{2}}{1+\mathrm{x}^{4}}$ belongs to $\left(0, \frac{1}{2}\right]$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 2)
If $\mathrm{X}$ is changed to $+\mathrm{hU}$ and $\mathrm{Y}$ to $\mathrm{b}+\mathrm{kV}$, then which one of the following is the correct relation between the regression coefficients $\mathrm{b}_{\mathrm{XY}}$ and $\mathrm{b}_{\mathrm{UV}}$ ?

Options:
A $\mathrm{hb}_{\mathrm{XY}}=\mathrm{k} \mathrm{b}_{\mathrm{UV}}$
B $\mathrm{k} \mathrm{b}_{\mathrm{XY}}=\mathrm{h} \mathrm{b}$
C $b_{X Y}=b_{U V}$
D $\mathrm{k}^{2} \mathrm{~b}_{\mathrm{XY}}=\mathrm{h}^{2} \mathrm{~b}_{\mathrm{UV}}$

The correct answer is:
$\mathrm{k} \mathrm{b}_{\mathrm{XY}}=\mathrm{h} \mathrm{b}$

Solution:
If $\mathrm{X}$ is changed to $\mathrm{a}+\mathrm{hU}$ and $\mathrm{Y}$ to $\mathrm{b}+\mathrm{k} \mathrm{V}$, then
$\mathbf{b}_{\mathrm{XY}}=\left|\frac{\mathrm{h}}{\mathrm{k}}\right| \mathrm{b}_{\mathrm{UV}}$
$\Rightarrow \mathrm{kb}_{\mathrm{XY}}=\mathrm{h} \cdot \mathrm{b}_{\mathrm{UV}}$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 1)
If $\mathrm{x}$ is real and $\mathrm{x}^{2}-3 \mathrm{x}+2>0, \mathrm{x}^{2}-3 \mathrm{x}-4 \leq 0$, then which on
of the following is correct?

Options:
A $\quad-1 \leq x \leq 4$
B $\quad 2 \leq x \leq 4$
C $-1 < x \leq 1$
D $-1 \leq x < 1$ or $2 < x \leq 4$

The correct answer is:
$-1 \leq x < 1$ or $2 < x \leq 4$

Solution:
Consider first: $\mathrm{x}^{2}-3 \mathrm{x}+2>0$ $\Rightarrow \quad(x-1)(x-2)>0$...(1)
$\Rightarrow \quad x < 1$ or $x>2$
and $x^{2}-3 x-4 \leq 0$
$\Rightarrow \quad(x-4)(x+1) \leq 0$
$\Rightarrow-1 \leq x \leq 4$...(2)
Combining $(1)$ and $(2)$ $-1 \leq x < 1$ or $2 < x \leq 4$
Drawing on number line:

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 1)
If x¯ is the mean of n observations x1,x2,…,xn then the mean of the absolute deviations of these observations from x¯ is

Options:
A the variance of the data
B the mean proportion of the data
C the standard deviation of the data
D the mean deviation of the data

The correct answer is:
the mean deviation of the data

Solution:
If x¯ is the mean of n observations x1,x2,…,xn then the mean of the absolute deviations of these observations from x¯ is the mean deviation of the data.

Here,

x¯=x1+x2+...+xnn ....1

Mean deviation

=x1-x¯+x2-x¯+...+xn-x¯n

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 1)
If $X$ is the number of tails in three tosses of a coin, then determine the standard deviation of $\mathrm{X}$.

Options:

None

Solution:
Let, random variable $\mathrm{X}$ is the number of tails in three tosses of the coin.
$$
\begin{aligned}
&\therefore \quad \mathrm{X}=0,1,2,3 \\
&\Rightarrow \quad \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} C_{\mathrm{x}}(\mathrm{p})^{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}} \\
&\text { where } \\
&\quad \mathrm{n}=3, \mathrm{p}=1 / 2, \mathrm{q}=1 / 2 \text { and } \mathrm{x}=0,1,2,3 .
\end{aligned}
$$
\begin{array}{|l|l|l|l|l|}
\hline \mathbf{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} \\
\hline \mathrm{P}(\mathrm{X}) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\
\mathrm{XP}(\mathrm{X}) & 0 & \frac{3}{8} & \frac{3}{4} & \frac{3}{8} \\
\mathrm{X}^2 \mathrm{P}(\mathrm{X}) & 0 & \frac{3}{8} & \frac{3}{2} & \frac{9}{8} \\
\hline
\end{array}
We know that, $\operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2$
Also $\mathrm{E}\left(\mathrm{X}^2\right)$
$$
\begin{aligned}
&=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2 \mathrm{P}\left(\mathrm{X}_{\mathrm{i}}\right)=0+\frac{3}{8}+\frac{3}{2}+\frac{9}{8}=\frac{24}{8}=3 \\
&\text { and }[\mathrm{E}(\mathrm{X})]^2 \\
&=\left[\sum_{\mathrm{i}=0}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^2 \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)\right]^2=\left[0+\frac{3}{8}+\frac{3}{4}+\frac{3}{8}\right]^2=\left[\frac{12}{8}\right]^2=\frac{9}{4} \\
&\left.\therefore \operatorname{Var}(\mathrm{X})=3-\frac{9}{4}=\frac{3}{4} \quad \text { [using Eq. (i) }\right]
\end{aligned}
$$
and S.D of $X=\sqrt{\operatorname{Var}(X)}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Probability
If $x=k(\theta+\sin \theta)$ and $y=k(1+\cos \theta)$, then what is the derivative of $y$ with respect to $x$ at $\theta=\pi / 2 ?$

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

The correct answer is:
$-1$

Solution:
Let $x=k(\theta+\sin \theta)$ and $y=k(1+\cos \theta)$
Differentiate both the functions w.r.t. ' $\theta$ '
$\Rightarrow \quad \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{k}(1+\cos \theta)$
and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=-\mathrm{k} \sin \theta$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d} \mathrm{y} / \mathrm{d} \theta}{\mathrm{d} \mathrm{x} / \mathrm{d} \theta}$
$=\frac{-\mathrm{k} \sin \theta}{\mathrm{k}(1+\cos \theta)}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}=-\tan \frac{\theta}{2}$
$\left(\because \sin 2 \theta=2 \sin \theta \cos \theta\right.$ and $\left.\cos 2 \theta=2 \cos ^{2} \theta-1\right)$
$\Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}=-\tan \frac{\pi}{4}=-1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
If $x^{\ln \left(\frac{y}{z}\right)} \cdot y^{\ln (x z)^{2}} \cdot z^{\ln \left(\frac{x}{y}\right)}=y^{4 / n y}$ for any $x>1, y>1$ and $z>1$,
then which one of the following is correct?

Options:
A $\ell \mathrm{n} \mathrm{y}$ is the $\mathrm{GM}$ of $\ell \mathrm{n} \mathrm{x}, \ell \mathrm{n} \mathrm{x}, \ell \mathrm{n} \mathrm{x}$ and $\ell \mathrm{n} \mathrm{z}$
B $\ell \mathrm{n} \mathrm{y}$ is the $\mathrm{AM}$ of $\ell \mathrm{n} \mathrm{x}, \ell \mathrm{n} \mathrm{x}, \ln \mathrm{x}$ and $\ell \mathrm{n} \mathrm{z}$
C $\ell \mathrm{n} \mathrm{y}$ is the $\mathrm{HM}$ of $\ell \mathrm{n} \mathrm{x}, \ell \mathrm{n} \mathrm{x}$ and $\ell \mathrm{n} \mathrm{z}$
D $\ell \mathrm{n} \mathrm{y}$ is the $\mathrm{AM}$ of $\ell \mathrm{n}, \operatorname{In} \mathrm{x}, \ell \mathrm{n} \mathrm{z}$ and $\ln \mathrm{z}$

The correct answer is:
$\ell \mathrm{n} \mathrm{y}$ is the $\mathrm{AM}$ of $\ell \mathrm{n} \mathrm{x}, \ell \mathrm{n} \mathrm{x}, \ln \mathrm{x}$ and $\ell \mathrm{n} \mathrm{z}$

Solution:
$x^{\ln \left(\frac{y}{z}\right)} \cdot y^{\ln (x z)^{2}} \cdot z^{\ln \left(\frac{x}{y}\right)}=y^{4 \ln y}$
$\Rightarrow \ln \left[x^{ \left.\ln \left(\frac{y}{z}\right)\right]}+\ln \left[y^{\ln (x z)^{2}}\right]+\ln \left[z^{ \left.\ln \left(\frac{x}{y}\right)\right]=\ln \left[y^{4 \ln y}\right]}\right.\right.$
$\Rightarrow\left[\ln \left(\frac{y}{z}\right) \ln x\right]+[2 \ln (x z) \ln y]+\left[\ln \left(\frac{x}{y}\right) \ln z\right]=4[\ln y]^{2}$
$\Rightarrow \ln x[\ln y-\ln z]+2 \ln y[\ln x+\ln z]$
$\quad+\ln z[\ln x-\ln y]=4[\ln y]^{2}$
$\Rightarrow 3 \ln x+\ln z=4 \ln y$
$\Rightarrow \frac{\ln x+\ln x+\ln x+\ln z}{4}=\ln y$
$\therefore \quad$ lny is the AM of $\ln x, \ln x, \ln x, \ln x \& \ln z .$

Topic: Mathematics, Sequences and Series, NDA, NDA 2016 (Phase 2)
If $x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6$ then $x$ is equal to

Options:
A $2,-3$
B 2 only
C 1
D 3

The correct answer is:
1

Solution:
$\begin{aligned} & x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6 \\ & \Rightarrow x-x \log _{10} 5=\log _{10} 6-\log _{10}\left(1+2^{x}\right) \\ & \Rightarrow x\left(1-\log _{10} 5\right)=\log _{10} 6-\log _{10}\left(1+2^{x}\right) \\ & \Rightarrow x\left(\log _{10} 10-\log _{10} 5\right)=\log _{10}\left(\frac{6}{1+2^{x}}\right) \\ & \Rightarrow x\left(\log _{10}\left(\frac{10}{5}\right)\right)=\log _{10}\left(\frac{6}{1+2^{x}}\right) \\ \Rightarrow & x \log _{10} 2=\log _{10}\left(\frac{6}{1+2^{x}}\right) \end{aligned}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 2)
If $x+\log _{15}\left(1+3^{x}\right)=x \log _{15} 5+\log _{15} 12$, where $x$ is an integer, then what is $x$ equal to?

Options:
A -3
B 2
C 1
D 3

The correct answer is:
1

Solution:
$x+\log _{15}\left(1+3^{x}\right)=x \log _{15} 5+\log _{15} 12$
$\Rightarrow \quad x \cdot \log _{15} 15+\log _{15}\left(1+3^{x}\right)=x \log _{15} 5+\log _{15} 12$
$\left(\because \log _{15} 15=1\right)$
$\Rightarrow \log _{15} 15^{x}+\log _{15}\left(1+3^{x}\right)=\log _{15} 5^{x}+\log _{15} 12$
$\Rightarrow \log _{15} 15^{x}\left(1+3^{x}\right)=\log _{15}\left(5^{x} \times 12\right)$
$(\because \log \mathrm{a}+\log \mathrm{b}=\log \mathrm{ab})$
$\Rightarrow \quad 15^{x}\left(1+3^{x}\right)=5^{x} \times 12$
$3^{x}\left(1+3^{x}\right)=12$
$3^{x}+3^{2 x}=12$
$\mathrm{x}=1$ satisfies the above equation.

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
If $x=\log t$ and $y=t^{2}-1$, then what is $\frac{d^{2} y}{d x^{2}}$ at $t=1$ equal to?

Options:
A 2
B 3
C -4
D 4

The correct answer is:
4

Solution:
Let $x=\log t$ and $y=t^{2}-1$
$x=\log t$
$\Rightarrow 2 x=2 \log t$
$\Rightarrow 2 x=\log t^{2}$
$\Rightarrow 2 x=\log (y+1) \Rightarrow e^{2 x}=y+1$
On differentiating w.r.t. $x$, twice, we get
$e^{2 x} 2=\frac{d y}{d x} \Rightarrow 4 e^{2 x}=\frac{d^{2} y}{d x^{2}}$
At $t=1, x=0$
$\frac{d^{2} y}{d x^{2}}=4 e^{2(0)}=4 \quad\left(\because e^{0}=1\right)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
If $x^{m}+y^{m}=1$ such that $\frac{d y}{d x}=-\frac{x}{y}$, then what should be the value of $\mathrm{m} ?$

Options:
A 0
B 1
C 2
D None of the above

The correct answer is:
2

Solution:
Let $x^{\mathrm{m}}+\mathrm{y}^{\mathrm{m}}=1$
Differentiate both the sides w.r.t $^{\prime} x^{\prime}$
$\mathrm{m} \cdot \mathrm{x}^{\mathrm{m}-1}+\mathrm{m} \cdot \mathrm{y}^{\mathrm{m}-1} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=0$
$\Rightarrow \quad \mathrm{m} \mathrm{x}^{\mathrm{m}-1}=-\mathrm{m} \mathrm{y}^{\mathrm{m}-1} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow \frac{\mathrm{x}^{\mathrm{m}-1}}{\mathrm{y}^{\mathrm{m}-1}}=-\frac{\mathrm{dy}}{\mathrm{dx}}$
$\Rightarrow \left(\frac{x}{y}\right)^{m-1}=\frac{x}{y} \quad\left(\because \frac{d y}{d x}=-\frac{x}{y}\right)$
$\Rightarrow \quad m-1=1 \Rightarrow m=2$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 2)
If $\mathrm{x}=\mathrm{r} \sin \theta \cos \phi, \mathrm{y}=\mathrm{r} \sin \theta \sin \phi$ and $\mathrm{z}=\mathrm{r} \cos \theta$, then
$\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}$ is independent of which of the following?

Options:
A ronly
B $\mathrm{r}, \phi$
C $\theta, \phi$
D $\theta$

The correct answer is:
$\theta, \phi$

Solution:
As given, $x=r \sin \theta \cos \phi, y=r \sin \theta \sin \phi$.
and $z=r \cos \theta$
Now, $x^{2}+y^{2}+z^{2}=r^{2} \sin ^{2} \theta \cos ^{2} \phi+r^{2} \sin ^{2} \phi \sin ^{2} \theta$
$+r^{2} \cos ^{2} \theta$
$=r^{2} \sin ^{2} \theta\left(\sin ^{2} \phi+\cos ^{2} \phi\right)+r^{2} \cos ^{2} \theta$
$=r^{2} \sin ^{2} \theta+r^{2} \cos ^{2} \theta$
$=r^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$
$=r^{2}$
Thus, $x^{2}+y^{2}+z^{2}$ is independent of $\theta$ and $\phi$.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 1)
If $\mathrm{x}=\sin 70^{\circ} \sin 50^{\circ}$ and $\mathrm{y}=\cos 60^{\circ} \cos 80^{\circ}$, then what is $\mathrm{xy}$
equal to?

Options:
A $1 / 16$
B $1 / 8$
C $1 / 4$
D $1 / 2$

The correct answer is:
$1 / 16$

Solution:
$x=\sin 70^{\circ} \cdot \sin 50^{\circ}$ and $y=\cos 60^{\circ} \cdot \cos 80^{\circ}$
$\begin{aligned} \Rightarrow x y=& \cos 60^{\circ} \cdot \sin 70^{\circ} \cdot \sin 50^{\circ} \cdot \cos 80^{\circ} \\ x y=\frac{1}{2} \cdot & \sin (90-20) \cdot \sin (90-40) \cdot \cos 80 \\ \Rightarrow x y=& \frac{1}{2} \cdot \cos 20 \cdot \cos 40 \cdot \cos 80 \\(\because \sin (90-x)=\cos x) \\ \Rightarrow x y=& \frac{1}{2} \cdot \cos 20^{\circ} \cdot \cos (60-20)^{\circ} \cdot \cos (60+20)^{\circ} \\ \Rightarrow x y=& \frac{1}{2}\left[\frac{1}{4} \cos 3\left(20^{\circ}\right)\right]=\frac{1}{2} \times \frac{1}{4} \times \cos 60^{\circ}=\frac{1}{16} . \\ &\left[\because \cos \theta \cdot \cos (60-\theta) \cdot \cos (60+\theta)=\frac{1}{4} \cos 3 \theta\right] \\

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2016 (Phase 2)
If $x=\sin \theta+\cos \theta$ and $y=\sin \theta \cdot \cos \theta$, then what is the value
of $x^{4}-4 x^{2} y-2 x^{2}+4 y^{2}+4 y+1 ?$

Options:
A 0
B 1
C 2
D None of these

The correct answer is:
0

Solution:
Let $x=\sin \theta+\cos \theta$ and $y=\sin \theta \cdot \cos \theta$
Now, $x^{4}-4 x^{2} y-2 x^{2}+4 y^{2}+4 y+1$
$=(\sin \theta+\cos \theta)^{4}-4(\sin \theta+\cos \theta)^{2} y-$
$\quad 2(\sin \theta+\cos \theta)^{2}+4 y^{2}+4 y+1$
$=\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right)^{2}-$
$\quad 4\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right) y$
$\quad-2\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right)+4 y^{2}+4 y+1$
$=(1+2 y)^{2}-4(1+2 y) y-2(1+2 y)+4 y^{2}+4 y+1$
$=1+4 y^{2}+4 y-4 y-8 y^{2}-2-4 y+4 y^{2}+4 y+1=0$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 1)
If $x=\sin t-t \cos t$ and $y=t \sin t+\cos t$, then what is $\frac{d y}{d x}$ at point $\mathrm{t}=\frac{\pi}{2} ?$

Options:
A 0
B $\frac{\pi}{2}$
C $-\frac{\pi}{2}$
D $1$

The correct answer is:
0

Solution:
As given :
$\mathrm{x}=\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t}$ and $\mathrm{y}=\mathrm{t} \sin \mathrm{t}+\cos \mathrm{t}$
On differentiating w.r.t. t, we get
$\frac{\mathrm{dx}}{\mathrm{dt}}=\cos \mathrm{t}-\{\cos \mathrm{t}+\mathrm{t}(-\sin \mathrm{t})\}$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\cos \mathrm{t}-\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}=\mathrm{t} \sin \mathrm{t}$
and, $\frac{d y}{d t}=t \cos t+\sin t-\sin t=t \cos t$
Hence, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{t \cos t}{t \sin t}=\cot t$
$\Rightarrow\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{2}}=\cot \frac{\pi}{2}=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 1)
If $x=t^{2}, y=t^{3}$, then what is $\frac{d^{2} y}{d x^{2}}$ equal to?

Options:
A 1
B $\frac{3}{2 t}$
C $\frac{3}{4 t}$
D $\frac{3}{2}$

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

Solution:
Let $x=t^{2}$ and $y=t^{3}$
$\Rightarrow \frac{d x}{d t}=2 t$ and $\frac{d y}{d t}=3 t^{2}$
$\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3 t^{2}}{2 t}=\frac{3}{2} t$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{3}{2} \cdot \frac{d t}{d x}=\frac{3}{2} \cdot \frac{1}{2 t}\left(\because \frac{d x}{d t}=2 t\right)$
$=\frac{3}{4 t}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
If $\mathrm{X}=\left\{\mathrm{x}: \mathrm{x}>0, \mathrm{x}^{2} < 0\right\}$, and $\mathrm{Y}=$ flower, Churchill, moon, Kargil), then which one of the following is a correct statement?

Options:
A $\mathrm{X}$ is well defined but $\mathrm{Y}$ is not a well defined set
B $\mathrm{Y}$ is well defined but $\mathrm{X}$ is not a well defined set
C Both $\mathrm{X}$ and $\mathrm{Y}$ are well defined sets
D Neither X nor Y isa well defined set

The correct answer is:
Both $\mathrm{X}$ and $\mathrm{Y}$ are well defined sets

Solution:
$X=\left\{x: x>0, x^{2} < 0\right\}$
We know that the square of each number greater thar zero is always greater than zero. So, X contains member and so, $\mathrm{X}$ is null set but a well defined set Also, $\mathrm{Y}=\{$ flower, Churchill, Moon, Kargil\} is wel
defined. So, Y is also a well defined set

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
If $\mathrm{x}, \mathrm{x}-\mathrm{y}$ and $\mathrm{x}+\mathrm{y}$ are the angles of a triangle (not an equilateral triangle) such that $\tan (\mathrm{x}-\mathrm{y}), \tan \mathrm{x}$ and $\tan (\mathrm{x}+\mathrm{y})$
are in GP, then what is $x$ equal to?

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{3}$
C $\frac{\pi}{6}$
D $\frac{\pi}{2}$

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

Solution:
$\mathrm{x}, \mathrm{x}-\mathrm{y}, \mathrm{x}+\mathrm{y}$ are angles of a triangle. $\tan (\mathrm{x}-\mathrm{y}), \tan \mathrm{x}, \tan (\mathrm{x}+\mathrm{y})$ are in G.P.
Now, $\mathrm{x}+\mathrm{x}-\mathrm{y}+\mathrm{x}+\mathrm{y}=\pi$ (Sum of angles in triangle $\left.=180^{\circ}=\pi\right)$
$\Rightarrow 3 x=\pi$
$\Rightarrow x=\frac{\pi}{3}$

Topic: Mathematics, Properties of Triangles, NDA, NDA 2018 (Phase 1)
If $\left[\begin{array}{ccc}x & -3 i & 1 \\ y & 1 & i \\ 0 & 2 i & -i\end{array}\right]=6+11$ i, then what are the values of $\mathrm{x}$ and
y respectively ?

Options:
A $-3,4$
B 3,4
C $3,-4$
D $-3,-4$

The correct answer is:
$-3,4$

Solution:
$\begin{array}{l}{\left[\begin{array}{lcc}x & -3 i & 1 \\ y & 1 & i \\ 0 & 2 i & -i\end{array}\right]=6+11 i} \\ & \Rightarrow \mathrm{x}[-\mathrm{i}-(2 \mathrm{i})(\mathrm{i})]-\mathrm{y}[(-3 \mathrm{i})(-\mathrm{i})-(2 \mathrm{i})(1)]+0=6+11 \mathrm{i} \\ & \Rightarrow \mathrm{x}(-\mathrm{i}+2)-\mathrm{y}(-3-2 \mathrm{i})=6+11 \mathrm{i} \\ & \Rightarrow-\mathrm{x} \mathrm{i}+2 \mathrm{x}+3 \mathrm{y}+2 \mathrm{yi}=6+11 \mathrm{i} \\ & \Rightarrow(2 \mathrm{x}+3 \mathrm{y})+(-\mathrm{x}+2 \mathrm{y}) \mathrm{i}=6+11 \mathrm{i} \\ & \therefore 2 \mathrm{x}+3 \mathrm{y}=6 \text { and }-\mathrm{x}+2 \mathrm{y}=11\end{array}$
Solving the equations,
$2 \not x+3 y=6$
$\frac{-2 \not x+4 y=22}{7 y=28} \Rightarrow y=4$
$2 x+3 y=6 \Rightarrow 2 x+12=6$
$\Rightarrow 2 x=-6 \Rightarrow x=-3$
$\therefore \mathrm{x}=-3, \mathrm{y}=4$

Topic: Mathematics, Complex Number, NDA, NDA 2019 (Phase 1)
If $\sqrt{x}+\sqrt{y}=2$, then what is $\frac{d y}{d x}$ at $y=1$ and $\mathrm{x}=1$ equal to?

Options:
A 5
B 2
C 4
D $-1$

The correct answer is:
$-1$

Solution:
Let $\sqrt{x}+\sqrt{y}=2$
Differentiate w.r.t.x, we get
$\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0$ ...(1)
Put $\mathrm{y}=1, \mathrm{x}=1$ in equation (1)
$\frac{1}{2}+\frac{1}{2} \frac{d y}{d x}=0$
$\Rightarrow \quad \frac{d y}{d x}=-1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
If $x+y \leq 4$, then the how many non-zero positive integer ordered pair $(x, y) ?$

Options:
A 4
B 5
C 6
D 8

The correct answer is:
6

Solution:
Since $x$ and $y$ are non-zero positive integers therefore $x=1,2,3, \ldots \ldots \ldots \ldots$ and $y=1,2,3, \ldots$
Now, given $x+y \leq 4$ $(x, y)$ can be $(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)$
Required number of ordered pairs $=6$

Topic: Mathematics, Basic of Mathematics, NDA, NDA 2011 (Phase 1)
If $\mathrm{x}+\mathrm{y}=12$, what is the maximum value of $\mathrm{xy}$ ?

Options:
A 25
B 36
C 49
D 64

The correct answer is:
36

Solution:
Given $x+y=12$ $\mathrm{y}=12-\mathrm{x}$
so, $x y=x(12-x)=12 x-x^{2}$
Let $f(x)=12 x-x^{2}$
$\mathrm{f}^{\prime}(\mathrm{x})=12-2 \mathrm{x}$
To get maximum or minimum value $\mathrm{f}^{\prime}(\mathrm{x})=0$ and $\mathrm{f}^{\prime \prime}(\mathrm{x}) < 0$ it is maximum
$\mathrm{f}^{\prime \prime}(\mathrm{x})=-2 < 0$
so, $\mathrm{f}^{\prime}(\mathrm{x})=0$ will give maximum value. so, $12-2 x=0 \Rightarrow x=6$ and $x+y=12 \Rightarrow y=6$
Hence, $\mathrm{y}=6$ and $\mathrm{f}(\mathrm{x})=12 \mathrm{x}-\mathrm{x}^{2}=12 \times 6-36=36$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2007 (Phase 1)
If $x=y \cos \left(\frac{2 \pi}{3}\right)=z \cos \left(\frac{4 \pi}{3}\right)$, then what is
$x y+y z+z x$ equal to?

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

The correct answer is:
0

Solution:
Let $x=y \cos \left(\frac{2 \pi}{3}\right)=z \cos \left(\frac{4 \pi}{3}\right)$
$\Rightarrow x=y \cos \left(\pi-\frac{\pi}{3}\right)=-y \cos \frac{\pi}{3}=\frac{-y}{2}$
and $x=z \cos \left(\pi+\frac{\pi}{3}\right)=-z \cos \frac{\pi}{3}=\frac{-z}{2}$
from (1) and (2)
$\frac{-y}{2}=\frac{-z}{2} \Rightarrow y=z$
$\begin{aligned} \text { Thus, } x y+y z+z x &=z x+z^{2}+x z=2 x z+z^{2} \\ &=-y \cdot(y)+y^{2}=-y^{2}+y^{2}=0

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 2)