datavorous/entrance-exam-dataset huggingface dataset:
If $\alpha, \beta$ arethe roots of $a x^{2}+2 b x+c=0$ and $\alpha+\delta, \beta+\delta$ are the roots of $\mathrm{Ax}^{2}+2 \mathrm{Bx}+\mathrm{C}=0$, then what is $\left(\mathrm{b}^{2}-\mathrm{ac}\right) / \mathrm{B}^{2}-\mathrm{AC}$
equal to?

Options:
A $(\mathrm{b} / \mathrm{B})^{2}$
B $(a / \mathrm{A})^{2}$
C $\left(a^{2} b^{2}\right)\left(A^{2} B^{2}\right)$
D $(\mathrm{ab}) /(\mathrm{AB})$

The correct answer is:
$(a / \mathrm{A})^{2}$

Solution:
Since, $\alpha$ and $\beta$ are the roots of $a x^{2}+2 b x+c=0$
so, $k+\beta=-\frac{2 b}{a}$ and $\alpha \beta=\frac{c}{a}$
$\delta$ and $\beta+\delta$ are theroots of $A x^{2}+2 B x+C=0$
so, sum of the roots $=\alpha+\beta+2 \delta=-\frac{2 \mathrm{~B}}{\mathrm{~A}}$ and product
ofthe roots $(\alpha+\delta)(\beta+\delta)=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow-\frac{2 b}{a}+2 \delta=-\frac{2 B}{A}$
$\Rightarrow \delta=\frac{b}{a}-\frac{B}{A}$.. (i)
and $(\alpha+\delta)(\beta+\delta)=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \alpha \beta+(\alpha+\beta) \delta+\delta^{2}=\frac{\mathrm{C}}{\mathrm{A}}$...(ii)
Putting value of $\delta$ from equation (i) in equation (ii),
$\frac{\mathrm{c}}{\mathrm{a}}-\frac{2 \mathrm{~b}}{\mathrm{a}}\left(\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{B}}{\mathrm{A}}\right)+\left(\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}-\frac{2 \mathrm{~b}^{2}}{\mathrm{a}^{2}}+\frac{2 \mathrm{bB}}{\mathrm{aA}}+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}+\left(\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}-\frac{2 \mathrm{bB}}{\mathrm{aA}}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{c}}{\mathrm{a}}-\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{2}+\left(\frac{\mathrm{B}}{\mathrm{A}}\right)^{2}=\frac{\mathrm{C}}{\mathrm{A}}$
$\Rightarrow \frac{\mathrm{B}^{2}}{\mathrm{~A}^{2}}-\frac{\mathrm{C}}{\mathrm{A}}=\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{\mathrm{B}^{2}-\mathrm{AC}}{\mathrm{A}^{2}}=\frac{\mathrm{b}^{2}-\mathrm{ac}}{\mathrm{a}^{2}}$
$\Rightarrow \frac{b^{2}-a c}{B^{2}-A C}=\left(\frac{a}{A}\right)^{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 1)
If at any instant $t$, for a sphere, $r$ denotes the radius, $S$ denotes the surface area and $V$ denotes the volume, then
what is $\frac{d V}{d t}$ equal to?

Options:
A $\frac{1}{2} S \frac{d r}{d t}$
B $\frac{1}{2} r \frac{d S}{d t}$
C $r \frac{d S}{d t}$
D $\frac{1}{2} r^{2} \frac{d S}{d t}$

The correct answer is:
$\frac{1}{2} r \frac{d S}{d t}$

Solution:
Surface area of sphere $\mathrm{S}=4 \pi r^{2}$ Differentiate both sides w.r.t. $\mathrm{t}^{\prime}$
$\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=\frac{8 \pi \mathrm{rdr}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{d} \mathrm{r}}{\mathrm{dt}}=\frac{1}{8 \pi \mathrm{r}} \cdot \frac{\mathrm{d} \mathrm{s}}{\mathrm{dt}}$
and Volume $=\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}$
$\Rightarrow \quad \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi \cdot 3 \mathrm{r}^{2} \frac{\mathrm{d} \mathrm{r}}{\mathrm{dt}}=4 \pi \mathrm{r}^{2} \frac{\mathrm{d} \mathrm{r}}{\mathrm{dt}}$
$\quad=\frac{4 \pi \mathrm{r}^{2}}{8 \pi \mathrm{r}} \cdot \frac{\mathrm{d} \mathrm{S}}{\mathrm{dt}}=\frac{1}{2} \mathrm{r} \frac{\mathrm{dS}}{\mathrm{dt}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
If $\mathrm{b}_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}$ are three consecutive terms of an arithmetic progression with common difference $\mathrm{d}>0$, then what is the
value of $\mathrm{d}$ for which $\mathrm{b}_{3}^{2}=\mathrm{b}_{2} \mathrm{~b}_{3}+\mathrm{b}_{1} \mathrm{~d}+2 ?$

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

The correct answer is:
1

Solution:
$b_{1}, b_{2}, b_{3}$ are in $A P$ with common difference $d$, so $b_{2}=b_{1}+d$ and $b_{3}=b_{1}+2 d$
As given, $\mathrm{b}_{3}^{2}=\mathrm{b}_{2} \mathrm{~b}_{3}+\mathrm{b}_{1} \mathrm{~d}+2$
$\Rightarrow\left(b_{1}+2 d\right)^{2}=\left(b_{1}+d\right)\left(b_{1}+2 d\right)+b_{1} d+2$
$\Rightarrow \mathrm{b}_{1}^{2}+4 \mathrm{~d}^{2}+4 \mathrm{~b}_{1} \mathrm{~d}=\mathrm{b}_{1}^{2}+2 \mathrm{~b}_{1} \mathrm{~d}+\mathrm{b}_{1} \mathrm{~d}+2 \mathrm{~d}^{2}+\mathrm{b}_{1} \mathrm{~d}+2$
$\Rightarrow 2 \mathrm{~d}^{2}=2$
$\Rightarrow \mathrm{d}^{2}=1$
$\Rightarrow \mathrm{d}=\pm 1$
i.e. $\mathrm{d}=1$ or $-1$
Since, $\mathrm{d}>0,-1$ is discarded and $\mathrm{d}=1$

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 1)
If $\mathrm{B}=\left[\begin{array}{lll}3 & 2 & 0 \\ 2 & 4 & 0 \\ 1 & 1 & 0\end{array}\right]$, then what is adjoint of B equal to?

Options:
A $\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ -2 & -1 & 8\end{array}\right]$
B $\left[\begin{array}{ccc}0 & 0 & -2 \\ 0 & 0 & -1 \\ 0 & 0 & 8\end{array}\right]$
C $\left[\begin{array}{lll}0 & 0 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right]$
D It does not exist

The correct answer is:
$\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ -2 & -1 & 8\end{array}\right]$

Solution:
$B=\left[\begin{array}{lll}3 & 2 & 0 \\ 2 & 4 & 0 \\ 1 & 1 & 0\end{array}\right]$
C factor matrix of
$B=\left[\begin{array}{ccc}(4)(0)-(0)(1) & -[(2)(0)-(0)(1)] & (2)(1)-(4)(1) \\ -[2(0)-(0)(1)] & 3(0)-(0)(1) & -[(3)(1)-(2)(1)] \\ 2(0)-0(4) & -[3(0)-2(0)] & 3(4)-2(2)\end{array}\right]$
$=\left[\begin{array}{ccc}0 & 0 & -2 \\ 0 & 0 & -1 \\ 0 & 0 & 8\end{array}\right]$
Adjoint $B=\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ -2 & -1 & 8\end{array}\right]$

Topic: Mathematics, Matrices, NDA, NDA 2019 (Phase 1)
If $\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}$, then $a, b, c$ are in

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

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

Solution:
Suppose a, $\mathrm{b}$ and $\mathrm{c}$ are in $\mathrm{HP}$.
$b=\frac{2 a c}{a+c}$
Now, consider
$\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{\frac{2 a c}{a+c}-a}+\frac{1}{\frac{2 a c}{a+c}-c}$
$=\frac{a+c}{a(c-a)}+\frac{a+c}{c(a-c)}=\left(\frac{a+c}{c-a}\right)\left(\frac{1}{a}-\frac{1}{c}\right)$
$=\frac{a+c}{c-a} \times \frac{c-a}{c a}=\frac{a+c}{c a}=\frac{1}{a}+\frac{1}{c}$
Thus, our supposition is correct. Hence, $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are in $\mathrm{HP}$.

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 1)
If $\vec{b}$ and $\vec{c}$ are the position vectors of the points $B$ and $C$ respectively, then the position vector of the point D such
that $\overrightarrow{\mathrm{BD}}=4 \overrightarrow{\mathrm{BC}}$ is\

Options:
A $\quad 4(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}})$
B $-4(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}})$
C $\quad 4 \overrightarrow{\mathrm{c}}-3 \overrightarrow{\mathrm{b}}$
D $4 \overrightarrow{\mathrm{c}}+3 \overrightarrow{\mathrm{b}}$

The correct answer is:
$\quad 4 \overrightarrow{\mathrm{c}}-3 \overrightarrow{\mathrm{b}}$

Solution:
$\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}$
$\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}}$
$\overrightarrow{\mathrm{BD}}=4 \overrightarrow{\mathrm{BC}}$
$\overrightarrow{\mathrm{BO}}+\overrightarrow{\mathrm{OD}}=4(\overrightarrow{\mathrm{BO}}+\overrightarrow{\mathrm{OC}})$
$\overrightarrow{\mathrm{OD}}=3 \overrightarrow{\mathrm{BO}}+4 \overrightarrow{\mathrm{OC}}$
$\overrightarrow{\mathrm{OD}}=4 \overrightarrow{\mathrm{OC}}-3 \overrightarrow{\mathrm{OB}}$
$\overrightarrow{\mathrm{OD}}=\overrightarrow{4 \mathrm{c}}-3 \overrightarrow{\mathrm{b}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 2)
If $\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{c+b}=3$, then what is the value of $x$ ?

Options:
A 0
B 1
C a+b+c
D abc

The correct answer is:
a+b+c

Solution:
$\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=3$
Such an equation is possible only, if
$\frac{x-a}{b+c}=\frac{x-b}{c+a}=\frac{x-c}{a+b}=1$
$\Rightarrow x=a+b+c$

Topic: Mathematics, Functions, NDA, NDA 2007 (Phase 1)
If $\mathrm{B}$ is a non-singular matrix and $\mathrm{A}$ is a square matrix, then the value of det $\left(\mathrm{B}^{-1} \mathrm{AB}\right)$ is equal to

Options:
A $\operatorname{det}(\mathrm{B})$
B $\operatorname{det}(\mathrm{A})$
C $\operatorname{det}\left(\mathrm{B}^{-1}\right)$
D $\operatorname{det}\left(\mathrm{A}^{-1}\right)$

The correct answer is:
$\operatorname{det}(\mathrm{A})$

Solution:
$\begin{aligned} &\left|\mathrm{B}^{-1} \mathrm{AB}\right|=\left|\mathrm{B}^{-1}\right||\mathrm{A}||\mathrm{B}| \\ &=\frac{1}{|\not{B} \mid}|\mathrm{A}||\not{B}| \quad \because \quad\left|\mathrm{B}^{-1}\right|=\frac{1}{|\mathrm{~B}|} \\ &=|\mathrm{A}| \end{aligned}$

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 1)
If $\alpha, \beta, \gamma$ be angles which the vector $\overrightarrow{\mathrm{r}}=\lambda \overrightarrow{\mathrm{i}}+2 \overrightarrow{\mathrm{j}}-\overrightarrow{\mathrm{k}}$ makes
with the coordinate axes, then what is the value of $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma$ ?

Options:
A 2
B 1
C $\lambda^{2}+1$
D $1-\lambda^{2}$

The correct answer is:
2

Solution:
We know that for any line that makes $\alpha, \beta$ and $\gamma$ angle With axis $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\Rightarrow \quad 1-\sin ^{2} \alpha+1-\sin ^{2} \beta+1-\sin ^{2} \gamma=1$
$\Rightarrow \quad \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 1)
If both the roots of the equation $x^{2}-2 k x+k^{2}-4=0$ lie between $-3$ and 5 , then which one of the following is correct?

Options:
A $-2 < \mathrm{k} < 2$
B $-5 < \mathrm{k} < 3$
C $-3 < \mathrm{k} < 5$
D $-1 < \mathrm{k} < 3$

The correct answer is:
$-1 < \mathrm{k} < 3$

Solution:
$\begin{aligned} & x^{2}-2 k x+k^{2}-4=0 \\ & \Rightarrow(x-k)^{2}-2^{2}=0 \\ & \Rightarrow(x-k-2)(x-k+2)=0 \\ & \Rightarrow x=k+2, k-2 \\ & \Rightarrow k+2 < 5 \& k-2>-3 \\ & \Rightarrow k < 3 \& k>-1 \\ & \Rightarrow-1 < k < 3 \end{aligned}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 2)
If $\mathrm{c}>0$ and $4 \mathrm{a}+\mathrm{c} < 2 \mathrm{~b}$, then $\mathrm{ax}^{2}-\mathrm{bx}+\mathrm{c}=0$ has a root in
which one of the following intervals?

Options:
A $(0,2)$
B $(2,3)$
C $(3,4)$
D $(-2,0)$

The correct answer is:
$(0,2)$

Solution:
Let $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}-\mathrm{bx}+\mathrm{c}$
$\mathrm{f}(2)=4 \mathrm{a}-2 \mathrm{~b}+\mathrm{c} < 0 \quad$ (given)
$f(0)=c>0 \quad$ (given)
So, we can see that sign of $f(x)$ changes, when $x$ changes from 0 to 2, so it has a root in the interval $(0,2) .$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 2)
If $\mathrm{C}(20, \mathrm{n}+2)=\mathrm{C}(20, \mathrm{n}-2)$, then what is $\mathrm{n}$ equal to ?

Options:
A 8
B 10
C 12
D 16

The correct answer is:
10

Solution:
Given, $\mathrm{C}(20, \mathrm{n}+2)=\mathrm{C}(20, \mathrm{n}-2)$
$\Rightarrow{ }^{20} \mathrm{C}_{\mathrm{n}+2}={ }^{20} \mathrm{C}_{\mathrm{n}-2}$
$\Rightarrow 20=\mathrm{n}+2+\mathrm{n}-2 \quad\left(\because{ }^{n} c_{r}={ }^{n} c_{s} \Rightarrow n=r+s\right)$
$\Rightarrow 20=2 \mathrm{n}$
$\Rightarrow \mathrm{n}=10$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2019 (Phase 1)
If $C(28,2 r)=C(28,2 r-4)$, then what is $r$ equal to ?

Options:
A 7
B 8
C 12
D 16

The correct answer is:
8

Solution:
$\mathrm{C}(28,2 \mathrm{r})=\mathrm{C}(28,2 \mathrm{r}-4)$
${ }^{28} \mathrm{C}_{2 \mathrm{r}}={ }^{28} \mathrm{C}_{2 \mathrm{r}-4}$
$\Rightarrow 2 \mathrm{r}+2 \mathrm{r}-4=28$
$\Rightarrow 4 \mathrm{r}=32$
$\Rightarrow \mathrm{r}=8$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2013 (Phase 2)
If $\vec{c}$ is the unit vector perpendicular to both the vectors
$\overrightarrow{\text { a and }} \overrightarrow{\mathrm{b}}$, then what is another unit vector perpendicular
to both the vectors a and $\overrightarrow{\mathrm{b}}$ ?

Options:
A $\vec{c} \times \vec{a}$
B $\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$
C $-\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}$
D $\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}$

The correct answer is:
$\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}$

Solution:
Let $\vec{c}$ is the unit vector perpendicular to both the
vectors $\vec{a}$ and $\vec{b}$.
So, A unit vector which is perpendicular to both the
vectors $\vec{a}$ and $\vec{b}$ is $\frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2011 (Phase 2)
If $C(n, 12)=C(n, 8)$, then what is the value of $C(22, n)$ ?

Options:
A 131
B 231
C 256
D 292

The correct answer is:
231

Solution:
Given $C(n, 12)=C(n, 8)$ $\begin{array}{l}\Rightarrow{ }^{n} C_{12}={ }^{n} C_{8} \\ \Rightarrow \frac{n !}{(n-12) ! 12 !}=\frac{n !}{(n-8) ! 8 !} \\ \Rightarrow \frac{1}{(n-12) !(12 \times 11 \times 10 \times 9 \times 8 !)} \\ = & \frac{1}{(n-8)(n-9)(n-10)(n-11)(n-12) ! 8 !} \\ \Rightarrow \frac{1}{12 \times 11 \times 10 \times 9}=\frac{1}{(\mathrm{n}-8)(\mathrm{n}-9)(\mathrm{n}-10)(\mathrm{n}-11)} \\ \Rightarrow(n-8)(n-9)(n-10)(n-11) \\ =12 \times 11 \times 10 \times 9 \\ \Rightarrow n-8=12, n-9=11, n-10=10 \text { and } n-11=9 \\ \Rightarrow \mathrm{n}=20 \\ \Rightarrow C(22, n)={ }^{22} C_{20} \\ \quad=\frac{22 !}{2 ! 20 !}=\frac{22 \times 21}{2}=231\end{array}$

Topic: Mathematics, Permutation Combination, NDA, NDA 2009 (Phase 2)
If $C$ the velocity of light, $h$ Planck's constant and $G$ gravitational constant are taken as fundamental quantities, then the dimensional formula of mass is

Options:
A $h^{-1 / 2} G^{1 / 2} C^0$
B $h^{1 / 2} C^{1 / 2} G^{-1 / 2}$
C $h^{-1 / 2} C^{1 / 2} G^{-1 / 2}$
D $h^{-1 / 2} C^{-1 / 2} G^{-1 / 2}$

The correct answer is:
$h^{1 / 2} C^{1 / 2} G^{-1 / 2}$

Solution:
Let,
$\begin{aligned}
M & =C^a h^b G^c \\
\mathrm{ML}^0 \mathrm{~T}^0 & =\left[\mathrm{LT}^{-1}\right]^a\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{\mathrm{b}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{\mathrm{a}}
\end{aligned}$
where, $h=\frac{\text { Energy }}{\text { Frequency }}$
$\begin{aligned}
& =\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \\
C & =\frac{\text { Metre }}{\text { Second }}=\left[\mathrm{LT}^{-1}\right] \\
G & =\frac{\text { Force } \times \text { (distance) }}{\text { (mass) }} \\
& =\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]
\end{aligned}$
Comparing the coefficients $M, L, T$, of both sides we get
$\begin{array}{l}
b-c=1 \\
a+2 b+3 c=0 \\
-(a+b+2 c)=0
\end{array}$
Solve the Eqs. (ii), (iii) and (iv), we get
$a=\frac{1}{2}, b=\frac{1}{2}, c=-\frac{1}{2}$
So, $M=h^{1 / 2} C^{1 / 2} G^{-1 / 2}$

Topic: Physics, Units and Dimensions, AP EAMCET, AP EAMCET 2014
If $C$ the velocity of light, $h$ Planck's constant and $G$ gravitational constant are taken as fundamental quantities, then the dimensional formula of mass is

Options:
A $h^{-1 / 2} G^{1 / 2} C^0$
B $h^{1 / 2} C^{1 / 2} G^{-1 / 2}$
C $h^{-1 / 2} C^{1 / 2} G^{-1 / 2}$
D $h^{-1 / 2} C^{-1 / 2} G^{-1 / 2}$

The correct answer is:
$h^{1 / 2} C^{1 / 2} G^{-1 / 2}$

Solution:
Let,
$\begin{aligned}
M & =C^a h^b G^c \\
\mathrm{ML}^0 \mathrm{~T}^0 & =\left[\mathrm{LT}^{-1}\right]^a\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{\mathrm{b}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{\mathrm{a}}
\end{aligned}$
where, $h=\frac{\text { Energy }}{\text { Frequency }}$
$\begin{aligned}
& =\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \\
C & =\frac{\text { Metre }}{\text { Second }}=\left[\mathrm{LT}^{-1}\right] \\
G & =\frac{\text { Force } \times \text { (distance) }}{\text { (mass) }} \\
& =\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]
\end{aligned}$
Comparing the coefficients $M, L, T$, of both sides we get
$\begin{array}{l}
b-c=1 \\
a+2 b+3 c=0 \\
-(a+b+2 c)=0
\end{array}$
Solve the Eqs. (ii), (iii) and (iv), we get
$a=\frac{1}{2}, b=\frac{1}{2}, c=-\frac{1}{2}$
So, $M=h^{1 / 2} C^{1 / 2} G^{-1 / 2}$

Topic: Physics, Units and Dimensions, TS EAMCET, TS EAMCET 2014
If coefficient of variation is \( 60 \) and standarddeviation is \( 24 \), then Arithmetic mean is

Options:
A \( 40 \)
B \( \frac{7}{20} \)
C \( \frac{20}{7} \)
D \( \frac{1}{40} \)

The correct answer is:
\( 40 \)

Solution:
Coefficient of variation is given by
\[
C V=\frac{\sigma}{x} \times 100
\]
Here, \( \mathrm{CV}=60 \) and \( \sigma=24 \). Then,

Topic: Mathematics, Definite Integration, KCET, KCET 2017
If $\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\theta$, then what is the value of $\operatorname{cosec}^{-1}(\sqrt{5)}$ ?

Options:
A $\left(\frac{\pi}{2}\right)+\theta$
B $\left(\frac{\pi}{2}\right)-\theta$
C $\frac{\pi}{2}$
D $-\theta$

The correct answer is:
$\left(\frac{\pi}{2}\right)-\theta$

Solution:
Let, $\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\theta \Rightarrow \cos \theta=\frac{1}{\sqrt{5}}$
$\Rightarrow \quad \sec \theta=\sqrt{5} \Rightarrow \sec ^{-1}(\sqrt{5})=\theta$
$\Rightarrow \frac{\pi}{2}-\operatorname{cosec}^{-1}(\sqrt{5})=\theta \quad\left(\because \sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}\right)$
$\Rightarrow \quad \operatorname{cosec}^{-1}(\sqrt{5})=\frac{\pi}{2}-\theta$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2007 (Phase 2)
If $\cos 3 \mathrm{~A}=\frac{1}{2}$, then how many values can $\sin \mathrm{A}$ assume? $\left(0 < \mathrm{A} < 360^{\circ}\right)$

Options:
A 3
B 4
C 5
D 6

The correct answer is:
4

Solution:
Given, $\cos 3 \mathrm{~A}=\frac{1}{2}$
$\Rightarrow \cos 3 \mathrm{~A}=\cos \left(\frac{\pi}{3}\right)$ or $\left(\frac{5 \pi}{3}\right)$
Since, $\mathrm{O} < \mathrm{A} < 360^{\circ}$
A can take the values, $\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}$ and
$\frac{17 \pi}{9}$
So, $\sin$ A can assume 6 values.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2007 (Phase 2)
If $\cos A=\frac{3}{4}$, then what is the value of $\sin \left(\frac{A}{2}\right) \sin \left(\frac{3 A}{2}\right)$ ?

Options:
A $\frac{5}{8}$
B $\frac{5}{16}$
C $\frac{5}{24}$
D $\frac{7}{32}$

The correct answer is:
$\frac{5}{16}$

Solution:
$\cos A=\frac{3}{4}$
$\sin \left(\frac{A}{2}\right) \cdot \sin \left(\frac{3 A}{2}\right)=\frac{1}{2}\left[2 \sin \frac{A}{2} \sin \frac{3 A}{2}\right]$
$=\frac{1}{2}\left[\cos \left(\frac{3 A}{2}-\frac{A}{2}\right)-\cos \left(\frac{3 A}{2}+\frac{A}{2}\right)\right]$
$=\frac{1}{2}[\cos A-\cos 2 A]=\frac{1}{2}\left[\cos A-\left(2 \cos ^{2} A-1\right)\right]$
$=\frac{1}{2}\left[\cos A-2 \cos ^{2} A+1\right]$
$=\frac{1}{2}\left[\left(\frac{3}{4}\right)-2\left(\frac{3}{4}\right)^{2}+1\right]$
$=\frac{1}{2}\left[\frac{3}{4}-\frac{18}{16}+1\right]=\frac{1}{2}\left[\frac{12-18+16}{16}\right]$
$=\frac{1}{2}\left[\frac{10}{16}\right]=\frac{5}{16}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2019 (Phase 1)
If $\cos A+\cos B=m$ and $\sin A+\sin B=n$, where $\mathrm{m}, n \neq 0$,
then what is $\sin (A+B)$ equal to ?

Options:
A $\frac{m n}{m^{2}+n^{2}}$
B $\frac{2 m n}{m^{2}+n^{2}}$
C $\frac{m^{2}+n^{2}}{2 m n}$
D $\frac{m n}{m+n}$

The correct answer is:
$\frac{2 m n}{m^{2}+n^{2}}$

Solution:
Let $\cos A+\cos B=m$
and $\sin A+\sin B=n$
Consider $\sin (A+B)=\frac{\left(m^{2}+n^{2}\right) \sin (A+B)}{m^{2}+n^{2}}$
$=\frac{[2+2 \cos (A-B)] \sin (A+B)}{2+2 \cos (A-B)}$
(from i and ii)
$=\frac{2 \sin (\mathrm{A}+\mathrm{B})+2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})}{1+1+2 \cos (\mathrm{A}-\mathrm{B})}$
$=\frac{2 \sin (A+B)+\sin (A+B+A-B)+\sin (A+B-A+B)}{1+1+2 \cos (A-B)}$
$=\frac{2 \mathrm{mn}}{\mathrm{m}^{2}+\mathrm{n}^{2}}($ from (i) and (ii))
Hence, $\sin (\mathrm{A}+\mathrm{B})=\frac{2 \mathrm{mn}}{\mathrm{m}^{2}+\mathrm{n}^{2}}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 2)
If $\cos \alpha$ and $\cos \beta(0 < \alpha < \beta < \pi)$ are the roots of the quadratic equation $4 \mathrm{x}^{2}-3=0$, then what is the value of $\sec \alpha \times \sec \beta$ ?

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

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

Solution:
Product of roots $=\frac{\mathrm{c}}{\mathrm{a}}$
$\Rightarrow \cos \alpha \cdot \cos \beta=-\frac{3}{4}$
$\Rightarrow \frac{1}{\cos \alpha \cdot \cos \beta}=\sec \alpha \cdot \sec \beta=-\frac{4}{3}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 2)
If $\cos \alpha+\cos \beta+\cos \gamma=0$, where $0 < \alpha \leq \frac{\pi}{2}, 0 < \beta \leq \frac{\pi}{2}$,
$0 < \gamma \leq \frac{\pi}{2}$, then what is the value of $\sin \alpha+\sin \beta+\sin \gamma ?$

Options:
A 0
B 3
C $\frac{5 \sqrt{2}}{2}$
D $\frac{3 \sqrt{2}}{2}$

The correct answer is:
3

Solution:
$\cos \alpha+\cos \beta+\cos \gamma=0$
Given, $0 < \alpha \leq \frac{\pi}{2}, 0 < \beta \leq \frac{\pi}{2}, 0 < \gamma \leq \frac{\pi}{2}$.
(1) is satisfied when $\alpha=\frac{\pi}{2}, \beta=\frac{\pi}{2}$ and $\gamma=\frac{\pi}{2}$.
$\therefore \sin \alpha+\sin \beta+\sin \gamma=\sin \frac{\pi}{2}+\sin \frac{\pi}{2}+\sin \frac{\pi}{2} .$
$=1+1+1=3 .$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2018 (Phase 1)
If $\cos \theta < \sin \theta$ and $\theta$ lies in the first quadrant, then which one of the following is correct?

Options:
A $0 < \theta < \pi / 4$
B $\pi / 4 < \theta < \pi / 2$
C $0 < \theta \leq \pi / 3$
D $\pi / 3 < \theta < \pi / 2$

The correct answer is:
$\pi / 4 < \theta < \pi / 2$

Solution:
We know that, for $\frac{\pi}{4} < \theta < \frac{\pi}{2}$
$\cos \theta < \sin \theta$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
If $\cos x \neq-1$, then what is $\frac{\sin x}{1+\cos x}$ equal to?

Options:
A $-\cot \frac{x}{2}$
B $\cot \frac{x}{2}$
C $\tan \frac{x}{2}$
D $-\tan \frac{x}{2}$

The correct answer is:
$\tan \frac{x}{2}$

Solution:
Consider $\frac{\sin x}{1+\cos x}=\frac{2 \sin x / 2 \cos x / 2}{1+2 \cos ^{2}(x / 2)-1}$
$\left(\because \sin 2 x=2 \sin x \cos x\right.$ and $\left.\cos 2 x=2 \cos ^{2} x-1\right)$
$=\frac{2 \sin x / 2 \cos x / 2}{2 \cos ^{2} x / 2}=\frac{\sin x / 2}{\cos x / 2}=\tan x / 2$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2010 (Phase 1)
If $\cos x=\frac{1}{3}$, then what is $\sin x \cdot \cot x$. cosecx $\cdot \tan x$ equal to?

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

The correct answer is:
1

Solution:
$$
\begin{aligned} \sin x & \cdot \cot x \cdot \operatorname{cosec} x \cdot \tan x \\ &=(\sin x \cdot \operatorname{cosec} x) \cdot(\cot x \cdot \tan x) \\ &=1 \times 1=1\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 2)
If $\operatorname{cosec} \theta-\cot \theta=\frac{1}{\sqrt{3}}$ where $\theta \neq 0$, then what is the value
of $\cos \theta ?$

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

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

Solution:
Consider, $\operatorname{cosec} \theta-\cot \theta=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}=\frac{1}{\sqrt{3}} \Rightarrow \frac{1-\cos \theta}{\sin \theta}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{1-\left(1-2 \sin ^{2} \theta / 2\right)}{2 \sin \theta / 2 \cos \theta / 2}=\frac{1}{\sqrt{3}} \Rightarrow \frac{2 \sin ^{2} \theta / 2}{2 \sin \theta / 2 \cos \theta / 2}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{\sin \theta / 2}{\cos \theta / 2}=\frac{1}{\sqrt{3}} \Rightarrow \tan \frac{\theta}{2}=\tan 30^{\circ}$
$\Rightarrow \theta=60^{\circ}$
$\therefore \cos \theta=\cos 60^{\circ}=\frac{1}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)