datavorous/entrance-exam-dataset huggingface dataset:
If
\[
f(x)=\left\{\begin{array}{cl}
\frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}} ; & \text { if }-1 \leq \mathrm{x} \leq 0 \\
\frac{2 \mathrm{x}+1}{\mathrm{x}-1} ; & \text { if } 0 \leq \mathrm{x} \leq 1
\end{array}\right.
\]
is continuous at \( x=0 \), then the value of \( k \) is

Options:
A \( k=1 \)
B \( k=-1 \)
C \( k=0 \)
D \( k=2 \)

The correct answer is:
\( k=-1 \)

Solution:
Given that
\[
f(x)=\mid \begin{array}{cc}
\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & ;-1 \leq x < 0 \\
\frac{2 x+1}{x-1} & ; 0 \leq x \leq 1
\end{array}
\]
For continues at \( x=0 \), we have
\[
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)
\]
\[
\begin{array}{l}
\text { So, } \lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \\
=\lim _{x \rightarrow 0^{-}} \frac{2 k(\sqrt{1-k x}+\sqrt{1+k x})}{x(\sqrt{x})}=k \\
\text { and } \lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-1}=-1=k \\
\text { So, } k=-1
\end{array}
\]

Topic: Mathematics, Continuity and Differentiability, KCET, KCET 2018
If \(f(x)=x^2-4 x-5\), then \(f(A)\), where \(A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\) equals

Options:
A \(\mathrm{O}\)
B I
C \(-\mathrm{I}\)
D \(2 \mathrm{I}\)

The correct answer is:
\(\mathrm{O}\)

Solution:
Given \(f(x)=x^2-4 x-5\)
\(\therefore f(A)=A^2-4 A-5 I\)
Now, \(A^2=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]=\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]\)
\(\begin{aligned}
\therefore \mathrm{A}^2-4 \mathrm{~A}-5 \mathrm{I} & =\left[\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right] -4\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]-5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}\)
\(=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=\mathrm{O}\)

Topic: Mathematics, Matrices, VITEEE, VITEEE 2023
If $G(x)=\sqrt{25-x^{2}}$, then what is $\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}$ equal to?

Options:
A $-\frac{1}{2 \sqrt{6}}$
B $\frac{1}{5}$
C $-\frac{1}{\sqrt{6}}$
D $\frac{1}{\sqrt{6}}$

The correct answer is:
$-\frac{1}{2 \sqrt{6}}$

Solution:
$\mathrm{G}(\mathrm{x})=\sqrt{25-\mathrm{x}^{2}}$
Now,
$\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}$
$\lim _{x \rightarrow 1} \frac{\sqrt{25-x^{2}}-\sqrt{24}}{x-1}$
Applying L'Hospital rule,
$\lim _{x \rightarrow 1} \frac{(-2 x)}{2 \sqrt{25-x^{2}}}=\frac{-2}{2 \sqrt{25-x^{2}}}$
$=\frac{-2}{2 \sqrt{24}}=\frac{-1}{\sqrt{24}}=\frac{-1}{2 \sqrt{6}}$
Option (a) is correct.

Topic: Mathematics, Limits, NDA, NDA 2015 (Phase 1)
If $\mathrm{g}(\mathrm{x})=\frac{1}{\mathrm{f}(\mathrm{x})}$ and $\mathrm{f}(\mathrm{x})=\mathrm{x}, \mathrm{x} \neq 0$, then which one of the
following is correct $\quad[2015-I I]$

Options:
A $\quad \operatorname{f}(\mathrm{f}(\mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(\mathrm{x}))))))=\mathrm{g}(\mathrm{g}(\mathrm{f}(\mathrm{g}(\mathrm{f}(\mathrm{x})))))$
B $\quad \mathrm{f}(\mathrm{f}(\mathrm{g}(3(\mathrm{~g}(\mathrm{f}(\mathrm{x}))))))=\mathrm{g}(\mathrm{g}(\mathrm{f}(\mathrm{g}(\mathrm{f}(\mathrm{x}))))))$
C $\mathrm{f}(\mathrm{g}(\mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(\mathrm{g}(\mathrm{x}))))))=\mathrm{g}(\mathrm{g}(\mathrm{f}(\mathrm{g}(\mathrm{f}(\mathrm{x}))))))$
D $\quad \mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{x}))))))=\mathrm{f}(\mathrm{f}(\mathrm{f}(\mathrm{g}(\mathrm{f}(\mathrm{x})))))$

The correct answer is:
$\quad \mathrm{f}(\mathrm{f}(\mathrm{g}(3(\mathrm{~g}(\mathrm{f}(\mathrm{x}))))))=\mathrm{g}(\mathrm{g}(\mathrm{f}(\mathrm{g}(\mathrm{f}(\mathrm{x}))))))$

Solution:
$f(x)=x$ $\mathrm{g}(\mathrm{x})=1 / \mathrm{x}$
Putting these values in the options, only (b) is correct.

Topic: Mathematics, Functions, NDA, NDA 2015 (Phase 2)
If $\mathrm{g}(\mathrm{x})=\sin \mathrm{x}, \mathrm{x} \in \mathrm{R}$ and $\mathrm{f}(\mathrm{x})=\frac{1}{\sin \mathrm{x}}, \mathrm{x} \in\left(0, \frac{\pi}{2}\right)$ what is (gof) (x) equal to?

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

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

Solution:
Given function are :
$g(x)=\sin x$ and $f(x)=\frac{1}{\sin x}$
$($ gof $)(x)=g[f(x)]$
$=\sin \mathrm{f}(\mathrm{x})$
$=\sin \left(\frac{1}{\sin \mathrm{x}}\right)$

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 1)
If $I_{n}=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x$ then what is $I_{n}+I_{n-2}$ equal to?

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

The correct answer is:
$\frac{1}{(n-1)}$

Solution:
Let $I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x$
Consider,
$I_{n}+I_{n-2}=\int_{0}^{\pi / 4} \tan ^{n} x d x+\int_{0}^{\pi / 4} \tan ^{n-2} x d x$
$=\int_{0}^{\pi / 4} \tan ^{n-2} x\left(\tan ^{2} x+1\right) d x$
$=\int_{0}^{\pi / 4} \sec ^{2} x \tan ^{n-2} x d x$
Put $\tan x=t$
$\sec ^{2} x d x=d t$
when $x=0$ then $t=0$ and when $x=\frac{\pi}{4}, t=1$
$\therefore I_{n}+I_{n-2}=\int_{0}^{1} t^{n-2} d t$
$=\left.\frac{t^{n-2+1}}{n-2+1}\right|_{0} ^{1}=\left.\frac{t^{n-1}}{n-1}\right|_{0} ^{1}=\frac{1}{n-1}[1-0]=\frac{1}{n-1}$

Topic: Mathematics, Definite Integration, NDA, NDA 2011 (Phase 1)
If in a $\Delta A B C, \cos B=(\sin A) /(2 \sin C)$, then the triangle is

Options:
A Isosceles triangle
B Equilateral triangle
C Right angled triangle
D Scalene triangle

The correct answer is:
Isosceles triangle

Solution:
Consider $\cos B=\frac{\sin A}{2 \sin C}$
$\Rightarrow \frac{c^{2}+a^{2}-b^{2}}{2 a c}=\frac{a}{2 c}$
$\quad\left(\because \frac{a}{\sin A}=\frac{c}{\sin C} \Rightarrow \frac{\sin A}{\sin C}=\frac{a}{c}\right)$
$\Rightarrow c^{2}+a^{2}-b^{2}=\frac{2 a^{2} c}{2 c}$
$\Rightarrow c^{2}+a^{2}-b^{2}=a^{2}$
$\Rightarrow c^{2}-b^{2}=0$
$\Rightarrow c=b$
Hence, $\Delta \mathrm{ABC}$ is isosceles triangle.

Topic: Mathematics, Properties of Triangles, NDA, NDA 2009 (Phase 2)
If in a frequency distribution table with 12 classes, the width of each class is $2.5$ and the lowest class boundary is $6.1$. then what is the upper class boundary of the highest class?

Options:
A 30 .
B $27.6$
C $30.6$
D $36.1$

The correct answer is:
$36.1$

Solution:
Given : lowest class boundary $=6.1$ Class width $=2.5$, Number of elasses $=12$
$\Rightarrow$ Upper class boundary of the highest class $=6.1+(2.5 \times 12)=6.1+30=36.1$

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 2)
If in a right-angled triangle $\mathrm{ABC}$, hypotenuse $\mathrm{AC}=\mathrm{p}$, then what is $\overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B}$ equal to ?

Options:
A $\mathrm{p}^{2}$
B $2 \mathrm{p}^{2}$
C $\frac{p^{2}}{2}$
D $\mathrm{p}$

The correct answer is:
$\mathrm{p}^{2}$

Solution:
Hypotenuse, $\mathrm{AC}=\mathrm{P}$ $\mathrm{BC}$ is perpendicular to $\mathrm{AB}$.
$\therefore \overrightarrow{B C} \cdot \overrightarrow{B A}=0$
$\therefore \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B}$
$=\overrightarrow{A B} \cdot \overrightarrow{A C}+0+\overrightarrow{A C} \cdot \overrightarrow{B C}=\overrightarrow{A C}(\overrightarrow{A B}+\overrightarrow{B C})$
$=\overrightarrow{A C} \cdot \overrightarrow{A C}=\overrightarrow{A C}^{2}=\mathrm{P}^{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2019 (Phase 1)
If in general, the value of $\sin A$ is known, but the value of $\mathrm{A}$
is not known, then how many values of $\tan \left(\frac{A}{2}\right)$ can be calculated?

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
We know, $\sin A=\frac{2 \tan \frac{A}{2}}{1+\tan ^{2} \frac{A}{2}}$
If $\sin A$ is known then equation (1) becomes
quadratic equation in $\tan \left(\frac{A}{2}\right)$. This mean 2 values of
$\tan \left(\frac{A}{2}\right)$ can be calculated.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2011 (Phase 1)
If instantaneous current is given by $i=4 \cos (\omega t+\phi)$, then the rms value of current is

Options:
A $2 \sqrt{2} \mathrm{~A}$
B $4 \mathrm{~A}$
C $4 \sqrt{2} \mathrm{~A}$
D zero

The correct answer is:
$2 \sqrt{2} \mathrm{~A}$

Solution:
Instantaneous current,
$$
i=4 \cos (\omega t+\phi)
$$
Comparing with standard equation, $i=i_{0} \cos (\omega t+\phi)$, we get
$\therefore$ Peak current, $i_{0}=4 \mathrm{~A}$
$$
\text { Hence, } \begin{aligned}
i_{\mathrm{rms}} &=\frac{i_{0}}{\sqrt{2}} \\
&=\frac{4}{\sqrt{2}}=2 \sqrt{2} \mathrm{~A}
\end{aligned}
$$

Topic: Physics, Alternating Current, COMEDK, COMEDK 2015
If $\omega$ is a complex cube root of unity and $x=\omega^{2}-\omega-2$, then what is the value of $x^{2}+4 x+7 ?$

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

The correct answer is:
0

Solution:
Given $x=\omega^{2}-\omega-2$
$\Rightarrow x+2=\omega^{2}-\omega$
On squaring both sides, we get
$(x+2)^{2}=\left(\omega^{2}-\omega\right)^{2}$
$\Rightarrow x^{2}+4 x+4=\omega^{4}+\omega^{2}-2 \omega^{3}$
Add 3 on both side
$\Rightarrow x^{2}+4 x+4+3=\omega+\omega^{2}-2+3\left(\because \omega^{3}=1\right)$
$\Rightarrow x^{2}+4 x+7=1+\omega+\omega^{2}$
$\Rightarrow \quad x^{2}+4 x+7=0\left(\because 1+\omega+\omega^{2}=0\right)$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 1)
If $\omega$ is a complex cube root of unity, then what is $\omega^{10}+\omega^{-10}$ equal to ?

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

The correct answer is:
-1

Solution:
Consider $\omega^{10}+\omega^{-10}=\omega^{10}+\frac{1}{\omega^{10}}$
$=\left(\omega^{3}\right)^{3} \cdot \omega+\frac{1}{\left(\omega^{3}\right)^{3} \cdot \omega}=\omega+\frac{1}{\omega}=-1$

Topic: Mathematics, Complex Number, NDA, NDA 2010 (Phase 2)
If $\omega$ is a complex cube root of unity, then what is the value
of $1-\frac{1}{(1+\omega)}-\frac{1}{\left(1+\omega^{2}\right)} ?$

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

The correct answer is:
0

Solution:
$1-\frac{1}{(1+\omega)}-\frac{1}{\left(1+\omega^{2}\right)}=1-\frac{1}{-\omega^{2}}-\frac{1}{-\omega}$
$=\frac{\omega^{2}+1+\omega}{\omega^{2}}=\frac{0}{\omega^{2}}=0$

Topic: Mathematics, Complex Number, NDA, NDA 2008 (Phase 2)
If $\omega$ is a complex non-real cube root of unity, then $\omega$ satisfies which one of the following equations?

Options:
A $x^{2}-x+1=0$
B $x^{2}+x+1=0$
C $x^{2}+x-1=0$
D $x^{2}-x-1=0$

The correct answer is:
$x^{2}+x+1=0$

Solution:
If $\omega$ is cube root of unity then $\omega^{2}+\omega+1=0$ So, \omega satisfies the equaion $x^{2}+x+1=0$

Topic: Mathematics, Complex Number, NDA, NDA 2008 (Phase 1)
If $\alpha$ is a complex number such that $\alpha^{2}+\alpha+1=0$, then what is $\alpha^{31}$ equal to?

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

The correct answer is:
$\alpha$

Solution:
Since, $\alpha$ is a complex root therefore $\alpha^{2}+\alpha+1=0 \Rightarrow \alpha=\omega$ or $\omega^{2}$
consider $\alpha^{31}=(\omega)^{31}$
$=\left(\omega^{3}\right)^{10} \cdot \omega$
$=\omega \quad\left(\because \omega^{3}=1\right)$
$=\alpha$

Topic: Mathematics, Complex Number, NDA, NDA 2009 (Phase 2)
If $\phi$ is a null set, then which one of the following is correct?

Options:
A $\phi=0$
B $\phi=\{0\}$
C $\phi=\{\phi$
D $\phi=\{\}$

The correct answer is:
$\phi=\{\}$

Solution:
Since $\phi$ is null set therefore $\phi=$

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 2)
If $\vec{\beta}$ is perpendicular to both $\vec{\alpha}$ and $\vec{\gamma}$ where $\vec{\alpha}=\overrightarrow{\mathrm{k}}$ and
$\vec{\gamma}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$, then what is $\vec{\beta}$ equal to?

Options:
A $\quad 3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}$
B $-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}$
C $2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}$
D $-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}$

The correct answer is:
$-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}$

Solution:
Let $\vec{\beta}=a \hat{i}+b \hat{j}+c \hat{k}$
since $\vec{\beta} \cdot \vec{\alpha}=0$ and $\vec{\beta} \cdot \vec{\gamma}=0$
$\therefore \quad c=0$ and $2 a+3 b=0 \Rightarrow a=-\frac{3 b}{2}$
Hence, $\vec{\beta}=-\frac{3 b}{2} \hat{i}+b \hat{j}=0 \Rightarrow-3 \hat{i}+2 \hat{j}=0$

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 1)
If $\theta$ is the acute angle between the diagonals of a cube, then which one of the following is correct?

Options:
A $\quad \theta < 30^{\circ}$
B $\quad \theta=60^{\circ}$
C $\quad 30^{\circ} < \theta < 60^{\circ}$
D $\quad \theta>60^{\circ} \quad$

The correct answer is:
$\quad \theta>60^{\circ} \quad$

Solution:
Let there be cube of side 'a'. Co-ordinates of its vertices $\mathrm{O}, \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \mathrm{F}$ have bemarked in the figure. Diagonals
are $\mathrm{OE}, \mathrm{FC}, \mathrm{GB}$ and $\mathrm{AD}$. Direction ratios $\left(\mathrm{dr}_{3}\right.$ ) of these diagonals are : $\mathrm{OE}\langle(\mathrm{a}-0),(\mathrm{a}-0),(\mathrm{a}-0)\rangle=(\mathrm{a}, \mathrm{a}, \mathrm{a})$
$F C\langle(-a, a,-a)\rangle ; G B\langle(-a, a, a)\rangle$ and $A D\langle(a, a,-a)\rangle$
Their des are :
OE, $\left\langle\frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}, \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}, \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}\right\rangle$
$=\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$
$\mathrm{AD},\left\langle\frac{\mathrm{a}}{\sqrt{\Sigma \mathrm{a}^{2}}}, \frac{\mathrm{a}}{\sqrt{\Sigma \mathrm{a}^{2}}}, \frac{-\mathrm{a}}{\sqrt{\Sigma \mathrm{a}^{2}}}\right\rangle=\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right\rangle$
Angle, $\theta$, between $\mathrm{AD}$ and $\mathrm{OE}$ is given by
$\cos \theta=\pm \frac{\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}}{\sqrt{\left\{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}\right\}\left\{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(-\frac{1}{\sqrt{3}}\right)^{2}\right\}}}$
$=\frac{\frac{1}{3}}{1 \times 1}=\pm \frac{1}{3}$
Since the cube is in positive octant, we take $+\frac{1}{3}$.
So, $\cos \theta=\frac{1}{3} \Rightarrow \theta>60$ '
[Since value of $\cos \theta$ decreases as $\theta$ increases in 0 to $90^{\circ} \cdot \cos \theta=1$ when $\theta=0^{\circ}$ and $\cos \theta=0$ when $\theta=90^{\circ}$ ]

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 1)
If $\theta$ is the acute angle between the diagonals of a cube, then which one of the following is correct?

Options:
A $\quad \theta=30^{\circ}$
B $\quad \theta=45^{\circ}$
C $2 \cos \theta=1$
D $3 \cos \theta=1$

The correct answer is:
$3 \cos \theta=1$

Solution:
diagonals are OP and AD and Acute anngle $=\theta$
$\cos \theta=\left|\frac{\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}}{\sqrt{\mathrm{a}_{1}^{2}+\mathrm{b}_{1}^{2}+\mathrm{c}_{1}^{2}} \sqrt{\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}+\mathrm{c}_{2}^{2}}}\right|$
$=\left|\frac{\mathrm{a}(-\mathrm{a})+(\mathrm{a})(\mathrm{a})+(\mathrm{a})(\mathrm{a})}{\sqrt{\mathrm{a}^{2}+\mathrm{a}^{2}+\mathrm{a}^{2}} \sqrt{\mathrm{a}^{2}+\mathrm{a}^{2}+\mathrm{a}^{2}}}\right|$
$=\left|\frac{-\mathrm{a}^{2}+\mathrm{a}^{2}+\mathrm{a}^{2}}{\sqrt{3 \mathrm{a}^{2}} \sqrt{3 \mathrm{a}^{2}}}\right|=\left|\frac{\mathrm{a}^{2}}{3 \mathrm{a}^{2}}\right|=\frac{1}{3}$
$\Rightarrow 3 \cos \theta=1$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 2)
If $\theta$ is the angle between the vectors is $4(\hat{i}-\hat{k})$ and $\hat{i}+\hat{j}+\hat{k}$, then what is $(\sin \theta+\cos \theta)$ equal to ?

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

The correct answer is:
1

Solution:
Angle $\mathrm{b} / \mathrm{w}$ the vectors is $\cos \theta=\frac{4(1)-4(1)}{\sqrt{32} \sqrt{3}}=0$
$\Rightarrow \theta=\frac{\pi}{2}$
Hence, $\cos \theta+\sin \theta=\cos \frac{\pi}{2}+\sin \frac{\pi}{2}=0+1=1$.

Topic: Mathematics, Vector Algebra, NDA, NDA 2013 (Phase 1)
If $\omega$ is the cube root of unity, then what is one root of the equation
$\left|\begin{array}{ccc}x^{2} & -2 x & -2 \omega^{2} \\ 2 & \omega & -\omega \\ 0 & \omega & 1\end{array}\right|=0 ?$

Options:
A 1
B $-2$
C 2
D $\omega$

The correct answer is:
$-2$

Solution:
Given matrix is :
$\left|\begin{array}{ccc}x^{2} & -2 x & -2 \omega^{2} \\ 2 & \omega & -\omega \\ 0 & \omega & 1\end{array}\right|=0$
$\mathrm{By} \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}$, we get
$\Rightarrow\left|\begin{array}{ccc}\mathrm{x}^{2} & -2 \mathrm{x}-2 \omega^{2} & -2 \omega^{2} \\ 2 & 0 & -\omega \\ 0 & 1+\omega & 1\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}\mathrm{x}^{2} & -2 \mathrm{x}-2 \omega^{2} & -2 \omega^{2} \\ 2 & 0 & -\omega \\ 0 & -\omega^{2} & 1\end{array}\right|=0$
$\quad\left[\because \quad 1+\omega=-\omega^{2}\right]$
$\Rightarrow \omega^{2}\left|\begin{array}{cc}x^{2} & -2 \omega^{2} \\ 2 & -\omega\end{array}\right|+1\left|\begin{array}{cc}x^{2} & -2 x-2 \omega^{2} \\ 2 & -0\end{array}\right|=0$
$\Rightarrow \omega^{2}\left(-\omega x^{2}+4 \omega^{2}\right)-\left(-4 x-4 \omega^{2}\right)=0$
$\Rightarrow-x^{2}+4 \omega+4 x+4 \omega^{2}=0$
$\Rightarrow-x^{2}+4 \omega-4 x-4-4 \omega=0 \Rightarrow-x^{2}-4 x-4=0$
$\Rightarrow(x+2)^{2}=0 \Rightarrow x=-2$

Topic: Mathematics, Determinants, NDA, NDA 2007 (Phase 2)
If $\omega$ is the cube root of unity, then what is the conjugate of $2 \omega^{2}+3 i ?$

Options:
A $2 \omega-3 i$
B $3 \omega+2 i$
C $2 \omega+3 i$
D $3 \omega-2 i$

The correct answer is:
$2 \omega-3 i$

Solution:
Let $z=2 \omega^{2}+3 i$
Since, $\omega$ is the cube root of unity
$\therefore \omega=\frac{-1+\sqrt{3} i}{2}$ and $\omega^{2}=\frac{-1-\sqrt{3} i}{2}$
$\therefore z=2 \omega^{2}+3 i$
$\quad=2\left[\frac{-1-\sqrt{3} i}{2}\right]+3 i$
$\quad=-1-\sqrt{3} i+3 i=-1+(3-\sqrt{3}) i$
$\bar{z}=-1-(3-\sqrt{3}) i=-1+\sqrt{3} i-3 i=2\left(\frac{-1+\sqrt{3} i}{2}\right)-3 i$
$=2 \omega-3 i$

Topic: Mathematics, Complex Number, NDA, NDA 2009 (Phase 2)
If $\omega$ is the imaginary cube root of unity, then what is $\left(2-\omega+2 \omega^{2}\right)^{27}$ equal to?

Options:
A $3^{27} \omega$
B $-3^{27} \omega^{2}$
C $3^{27}$
D $-3^{27}$

The correct answer is:
$-3^{27}$

Solution:
Consider, $\left(2-\omega+2 \omega^{2}\right)^{27}=\left[2\left(1+\omega^{2}\right)-\omega\right]^{27}$
$=(-2 \omega-\omega)^{27}\left[\because 1+\omega+\omega^{2}=0\right]$
$=(-3 \omega)^{27}=-3^{27} \cdot \omega^{27}$
$=(-3)^{27} \cdot\left(\omega^{3}\right)^{9}=(-3)^{27}$

Topic: Mathematics, Complex Number, NDA, NDA 2011 (Phase 1)
If $\mu$ is the universal set and $\mathrm{P}$ is a subset of $\mu$, then what is $\mathrm{P} \cap(\mathrm{P}-\mu) \cup(\mu-\mathrm{P})\}$ equal to ?

Options:
A $\phi$
B $P$
C $\mu$
D $\mathrm{P}$

The correct answer is:
$\phi$

Solution:
Since $\mu$ is universal set and $\mathrm{P} \subseteq \mu, \mathrm{P}-\mu=\phi$ and $\mu-\mathrm{P}=\mathrm{P}$
So, $(P-\mu) \cup(\mu-P)=\phi \cup P=P^{\prime}$
Now, $\operatorname{P} \cap\{P-\mu) \cup(\mu-P)\}=\operatorname{P} \cap P=\phi$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 1)
If $\mathrm{K}=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right)$, then what is the value
of K?

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{8}$
D $\frac{1}{16}$

The correct answer is:
$\frac{1}{8}$

Solution:
$\mathrm{K}=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right)$
We know, $2 \sin A \sin B=\cos (A-B)-\cos (A+B)$
$K=\frac{1}{2} \cdot \sin \frac{\pi}{18}\left[2 \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}\right]$
$=\frac{1}{2} \cdot \sin \frac{\pi}{18}\left[\cos \frac{2 \pi}{18}-\cos \frac{2}{\frac{\not / \pi} \pi}{3^{6}}\right]$
$=\frac{1}{2} \cdot \frac{1}{2}\left[2 \sin \frac{\pi}{18} \cos \frac{2 \pi}{18}-2 \sin \frac{\pi}{18} \cos \frac{2 \pi}{3}\right]$
$=\frac{1}{4}\left[\sin \left(\frac{3 \pi}{18}\right)+\sin \left(\frac{-\pi}{18}\right)-2 \sin \frac{\pi}{18} \cos \left(\pi-\frac{\pi}{3}\right)\right]$
$=\frac{1}{4}\left[\sin \frac{\pi}{6}-\sin \frac{\pi}{18}+2 \sin \frac{\pi}{18} \cdot \cos \frac{\pi}{3}\right]$
$=\frac{1}{4}\left[\sin \frac{\pi}{6}-\sin \frac{\pi}{18}+\not 2 \sin \frac{\pi}{18} \cdot \frac{1}{\not}\right]$
$=\frac{1}{4} \sin \frac{\pi}{6}=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2017 (Phase 1)
If $l_{1}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right) \quad[2017-\mathrm{II}]$
$l_{2}=\lim _{h \rightarrow 0} \frac{e^{\sin (x+h)}-e^{\sin x}}{h}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cos \mathrm{x} \mathrm{d} \mathrm{x}$
then which one of the following is correct?

Options:
A $l_{1} \neq l_{2}$
B $\frac{\mathrm{d}}{\mathrm{dx}}\left(l_{3}\right)=l_{2}$
C $\int l_{3} \mathrm{dx}=l_{2}$
D $l_{2}=l_{3}$

The correct answer is:
$\frac{\mathrm{d}}{\mathrm{dx}}\left(l_{3}\right)=l_{2}$

Solution:
$l_{1}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right)$
$l_{2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\sin (\mathrm{x}+\mathrm{h})}-\mathrm{e}^{\sin \mathrm{x}}}{\mathrm{h}}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cdot \cos \mathrm{x} \mathrm{d} \mathrm{x}$
$l_{2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\sin (\mathrm{x}+\mathrm{h})}-\mathrm{e}^{\sin \mathrm{x}}}{\mathrm{h}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right)=l_{1}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cdot \cos \mathrm{x} \mathrm{d} \mathrm{x}$
Let $\sin \mathrm{x}=\mathrm{t} \Rightarrow \cos \mathrm{x} \mathrm{dx}=\mathrm{dt}$
$l_{3}=\int \mathrm{e}^{\mathrm{t}} \cdot \mathrm{dt}=\mathrm{e}^{\mathrm{t}}+\mathrm{c}=\mathrm{e}^{\sin \mathrm{x}}+\mathrm{c}$
$\frac{\mathrm{d}}{\mathrm{dx}}\left(l_{3}\right)=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{e}^{\sin \mathrm{x}}+\mathrm{c}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right)=l_{1}=l_{2}$

Topic: Mathematics, Limits, NDA, NDA 2017 (Phase 2)
If ' $l$ ' is the length of the open pipe, ' $r$ ' is the internal radius of the pipe and ' $V$ ' is the velocity of sound in air then fundamental frequency of open pipe is

Options:
A $\frac{\mathrm{V}}{(l+0.3 \mathrm{r})}$
B $\frac{\mathrm{V}}{(l+1.2 \mathrm{r})}$
C $\frac{\mathrm{V}}{(l+0.6 \mathrm{r})}$
D $\frac{\mathrm{V}}{2(l+1.2 \mathrm{r})}$

The correct answer is:
$\frac{\mathrm{V}}{2(l+1.2 \mathrm{r})}$

Solution:
For an open organ pipe, the length of the pipe with end correction is given as:
$$
\begin{aligned}
& \mathrm{L}=l+2 \mathrm{e}=l+2 \times 0.6 \mathrm{r} \\
& \mathrm{L}=l+1.2 \mathrm{r}
\end{aligned}
$$
$\therefore \quad$ The fundamental frequency of open pipe is:
$$
\begin{aligned}
& \mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}} \\
& \mathrm{f}=\frac{\mathrm{v}}{2(l+1.2 \mathrm{r})}
\end{aligned}
$$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (14 May Shift 1)
If $l+m+\mathrm{n}=0$, then the system of equations $-2 x+y+z=l$
$x-2 y+z=m$
$x+y-2 z=n$
has

Options:
A a trivial solution
B no solution
C a unique solution
D infinitely many solutions

The correct answer is:
infinitely many solutions

Solution:
$\begin{aligned} \text { Here, } A &=\left[\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right] \\ B &=\left[\begin{array}{c}1 \\ m \\ n\end{array}\right] \text { and } \quad X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \end{aligned}$
$\begin{aligned} \therefore|A| &=-2\left|\begin{array}{cc}-2 & 1 \\ 1 & -2\end{array}\right|-1\left|\begin{array}{cc}1 & 1 \\ 1 & -2\end{array}\right|+1\left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right| \\ &=-2(4-1)-1(-2-1)+(1+2) \\ &=-6+3+3=0 \end{aligned}$
Now, adj $A=\left[\begin{array}{lll}3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3\end{array}\right]$
$\therefore(\operatorname{adj} A) \cdot B=0$
So, the given system of equations has an infinitely many solutions.

Topic: Mathematics, Determinants, NDA, NDA 2008 (Phase 2)