datavorous/entrance-exam-dataset huggingface dataset:
If $\operatorname{cosec} \theta+\cot \theta=\mathrm{c}$, then what is $\cos \theta$ equal to $?

Options:
A $\frac{\mathrm{c}}{\mathrm{c}^{2}-1}$
B $\frac{c}{c^{2}+1}$
C $\frac{\mathrm{c}^{2}-1}{\mathrm{c}^{2}+1}$
D None of the above

The correct answer is:
$\frac{\mathrm{c}^{2}-1}{\mathrm{c}^{2}+1}$

Solution:
Let $\operatorname{cosec} \theta+\cot \theta=\mathrm{c}$
$\Rightarrow \frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=\mathrm{c} \Rightarrow \frac{1+\cos \theta}{\sin \theta}=\mathrm{c}$
$\Rightarrow \frac{1+\left(2 \cos ^{2} \frac{\theta}{2}-1\right)}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\mathrm{c} \Rightarrow \frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\mathrm{c}$
$\Rightarrow \cot \frac{\theta}{2}=\mathrm{c} \Rightarrow \cos \theta=\frac{1-\frac{1}{\mathrm{c}^{2}}}{1+\frac{1}{\mathrm{c}^{2}}}=\frac{\mathrm{c}^{2}-1}{\mathrm{c}^{2}+1}$
$\left(\because \cos \theta=\frac{1-\tan ^{2}\left(\frac{\theta}{2}\right)}{1+\tan ^{2}\left(\frac{\theta}{2}\right)}\right)$
$\Rightarrow \tan \frac{\theta}{2}=\frac{1}{\mathrm{c}}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2013 (Phase 1)
If cot $\theta=2 \cos \theta$, where $(\pi / 2) < \theta < \pi$, then what is the value of $\theta ?$

Options:
A $5 \pi 6$
B $2 \pi / 3$
C $3 \pi 4$
D $11 \pi / 12$

The correct answer is:
$5 \pi 6$

Solution:
Given $\cot \theta=2 \cos \theta, \frac{\pi}{2} < \theta < \pi$
$$
\begin{array}{l}
\Rightarrow \frac{\cos \theta}{\sin \theta}=2 \cos \theta \Rightarrow \frac{\cos \theta}{2 \cos \theta}=\sin \theta \\
\Rightarrow \frac{1}{2}=\sin \theta \Rightarrow \sin \frac{\pi}{6}=\sin \theta \Rightarrow \theta=\frac{\pi}{6}
\end{array}
$$
But $\frac{\pi}{2} < \theta < \pi \therefore \theta=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
If $\cot \theta=5 / 12$ and $\theta$ lies in the third quadrant, then what is $\begin{array}{ll}(2 \sin \theta+3 \cos \theta) \text { equal to? }

Options:
A $-4$
B $-p^{2}$ for some odd prime $p$
C $(-q / p)$ where $p$ is an odd prime and $q$ a positive integer with $(q / p)$ not an integer
D $-p$ for some odd prime $p$

The correct answer is:
$-p$ for some odd prime $p$

Solution:
Let $\cot \theta=\frac{5}{12}$
$$
\begin{array}{l}
\Rightarrow \tan \theta=\frac{12}{5}=\frac{\text { perpendicular }(\mathrm{p})}{\text { base }(\mathrm{b})} \\
\therefore \text { Hypotenuse }(\mathrm{H})=\sqrt{p^{2}+b^{2}} \\
=\sqrt{(12)^{2}+(5)^{2}}=\sqrt{144+25}=\sqrt{169}=13
\end{array}
$$
Consider $2 \sin \theta+3 \cos \theta$
$=2\left(\frac{p}{H}\right)+3\left(\frac{b}{H}\right) \quad(\mathrm{H}-\mathrm{Height})$
But $\theta$ lies in $3^{\text {rd }}$ quadrant and $\sin \theta, \cos \theta$ both are negative in $3^{\text {rd }}$ quadrant
$\begin{aligned} \therefore 2 \sin \theta+3 \cos \theta &=2\left(\frac{-p}{H}\right)+3\left(\frac{-b}{H}\right) \\ &=2\left(\frac{-12}{13}\right)+3\left(\frac{-5}{13}\right) \\ &=\frac{-24-15}{13}=\frac{-39}{13}=-3 \end{aligned}$
which is an odd prime.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 2)
If $\cot \mathrm{A}=2$ and $\cot \mathrm{B}=3$, then what is the value of $\mathrm{A}+\mathrm{B}$ ?

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

The correct answer is:
$\pi / 4$

Solution:
\cot \mathrm{A}=2 \text { and } \cot \mathrm{B}=3\\
\begin{aligned}
& \cot (A+B)=\frac{\cot A \cot B-1}{\cot A+\cot B}=\frac{6-1}{2+3}=\frac{5}{5}=1 \\
& \Rightarrow \cot (A+B)=\cot \left(\frac{\pi}{4}\right) \Rightarrow A+B=\frac{\pi}{4} \\
&=\left[\sin \left(90^{\circ}-23 \frac{1^{\circ}}{2}\right)\right]^{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\
=& \cos ^{2} 23 \frac{1^{\circ}}{2}-\sin ^{2} 23 \frac{1^{\circ}}{2} \\
=& \cos 2\left(23 \frac{1^{\circ}}{2}\right)=\cos 47^{\circ} \\
\left(\because \cos 2 \mathrm{~A}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}\right) \\

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 2)
If $\cot \alpha$ and $\cot \beta$ are the roots of the equation $x^{2}+b x+c=0$ with $b \neq 0$, then the value of $\cot (\alpha+\beta)$

Options:
A $\frac{\mathrm{c}-1}{\mathrm{~b}}$
B $\frac{1-c}{b}$
C $\frac{b}{c-1}$
D $\frac{b}{1-c}$

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

Solution:
Given equation, $\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}=0$
Roots are $\cot \alpha, \cot \beta$
Sum of roots $=\cot \alpha+\cot \beta=-b$
Product of roots $=\cot \alpha, \cot \beta=c$
$\cot (\alpha+\beta) \frac{\cot \alpha \cdot \cot \beta-1}{\cot \beta+\cot \alpha}=\frac{\mathrm{c}-1}{-\mathrm{b}}=\frac{1-\mathrm{c}}{\mathrm{b}}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2017 (Phase 1)
If cot $(x+y)=1 / \sqrt{3}, \cot (x-y)=\sqrt{3}$ then what are the smallest positive values of $x$ and $y$ respectively?

Options:
A $45^{\circ}, 30^{\circ}$
B $30^{\circ}, 45^{\circ}$
C $15^{\circ}, 60^{\circ}$
D $45^{\circ}, 15^{\circ}$

The correct answer is:
$45^{\circ}, 15^{\circ}$

Solution:
Since $\quad \cot (x+y)=\frac{1}{\sqrt{3}}=\cot 60^{\circ} \quad\left[\cot 60^{\circ}=\frac{1}{\sqrt{3}}\right]$
$\Rightarrow x+y=60^{\circ}$
and $\cot (x-y)=\sqrt{3}=\cot 30^{\circ}$
$\Rightarrow x-y=30^{\circ}$
From equations (i) and (ii), we get $x=45^{\circ}$ and $y=15^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2009 (Phase 1)
If cotA $\cot B=2$, then what is the value of $\cos (\mathrm{A}+\mathrm{B}) \sec (\mathrm{A}-\mathrm{B}) ?$

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

The correct answer is:
$\frac{1}{3}$

Solution:
Consider $\cos (\mathrm{A}+\mathrm{B}) \cdot \sec (\mathrm{A}-\mathrm{B})$
$=\frac{\cos (\mathrm{A}+\mathrm{B})}{\cos (\mathrm{A}-\mathrm{B})}=\frac{\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}}{\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}}$
Divide $\mathrm{Nr}$ and $\mathrm{Dr}$ by $\sin \mathrm{A} \sin \mathrm{B}$,
$\frac{\cot \mathrm{A} \cot \mathrm{B}-1}{\cot \mathrm{A} \cot \mathrm{B}+1}=\frac{2-1}{2+1}=\frac{1}{3}$

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2012 (Phase 1)
If $\mathrm{D}$ is determinant of order 3 and $\mathrm{D}^{\prime}$ is the determinant obtained by replacing the elements of $D$ by their cofactors, then which one of the following is correct?

Options:
A $\mathrm{D}^{\prime}=\mathrm{D}^{2}$
B $\mathrm{D}^{\prime}=\mathrm{D}^{3}$
C $\mathrm{D}^{\prime}=2 \mathrm{D}^{2}$
D $\mathrm{D}^{\prime}=3 \mathrm{D}^{3}$

The correct answer is:
$\mathrm{D}^{\prime}=\mathrm{D}^{2}$

Solution:
$\mathrm{D}^{\prime}=$ cofactor $\mathrm{D}$ $\Rightarrow\left|\mathrm{D}^{\prime}\right|=\mid$ cofactor $\mathrm{D} \mid$
$\Rightarrow\left|\mathrm{D}^{\prime}\right|=|\mathrm{D}|^{3-1}$
$\Rightarrow\left|\mathrm{D}^{\prime}\right|=|\mathrm{D}|^{2}$
So $\mathrm{D}^{\prime}=\mathrm{D}^{2}$

Topic: Mathematics, Determinants, NDA, NDA 2013 (Phase 1)
If d is the number of degrees contained in an angle, $\mathrm{m}$ is the number of minutes and s is the number of seconds, then the value of $(\mathrm{s}-\mathrm{m}) /(\mathrm{m}-\mathrm{d})$ is:

Options:
A 1
B 60
C $\frac{1}{60}$
D None of these

The correct answer is:
$\frac{1}{60}$

Solution:
$40 \%$ of total population $=0.4 \times 75 \times 10^{6}=3 \times 10^{7}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
If d is the number of degrees contained in an angle, $\mathrm{m}$ is the number of minutes and s is the number of seconds, then the value of $(\mathrm{s}-\mathrm{m}) /(\mathrm{m}-\mathrm{d})$ is:

Options:
A 1
B 60
C $\frac{1}{60}$
D None of these

The correct answer is:
$\frac{1}{60}$

Solution:
$\mathrm{d}=60 \times 60 \mathrm{~s}$
$\mathrm{~m}=60 \mathrm{~s}$
$\frac{s-m}{m-d}=\frac{s-60 s}{60 s-60 \times 60 \mathrm{~s}}=\frac{-59}{60 \times(-59)}=\frac{1}{60}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
If $\frac{d y}{d x}=1+x+y+x y$ and $y(-1)=0$, then what is $y(x)$ equal to?

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

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

Solution:
Let $\frac{d y}{d x}=1+x+y+x y$
$\Rightarrow \frac{d y}{d x}=(1+x)(1+y)$
$\Rightarrow \frac{d y}{1+y}=d x(1+x)$
$\Rightarrow \int \frac{1}{1+y} d y=\int(1+x) d x$
$\Rightarrow \log (1+y)=x+\frac{x^{2}}{2}+c$
Given At $x=-1, y=0$
$\Rightarrow \log 1=-1+\frac{1}{2}+c$
$\Rightarrow c=\frac{1}{2}$
$\therefore \quad \log (1+y)=x+\frac{x^{2}}{2}+\frac{1}{2}=\frac{(1+x)^{2}}{2}$
$\Rightarrow 1+y=e^{\frac{(1+x)^{2}}{2}}$
$\Rightarrow y=e^{\frac{(1+x)^{2}}{2}}-1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
If $\omega$ denotes the cube root of unity, then what is the real root of the equation $x^{3}-27=0$ ?

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

The correct answer is:
$3 \omega^{3}$

Solution:
The given equation is $x^{3}-27=0$ $\Rightarrow \mathrm{x}^{3}-3^{3}=0 \Rightarrow \mathrm{x}=3,3 \omega, 3 \omega^{2}$
Thus, real root of given equation is $3 \omega^{3}$, since $\omega^{3}=1$

Topic: Mathematics, Complex Number, NDA, NDA 2007 (Phase 1)
If different words are formed with all the letters of the word 'AGAIN' and are arrangedalphabeticallyamong themselves as in a dictionary, the word at the 50 th place will be

Options:
A NAAGI
B $\quad$ NAAIG
C IAAGN
D IAANG

The correct answer is:
$\quad$ NAAIG

Solution:

Topic: Mathematics, Permutation Combination, NDA, NDA 2015 (Phase 2)
If E and E0 represent the energies, t and t0 represent the times, in which of the following is dimensionally correct relation?

Options:
A E=E0e-t
B E=E0t0e-t/t0
C E=E0t0e-t2
D E=E0e-t/t0

The correct answer is:
E=E0e-t/t0

Solution:
The dimensional formula for E=E0e-t/t0 as e-t/t0 is a dimensionless quantity.

In the question, E and E0 represents the energies. E is the time-averaged energy of the oscillator which is dimensionally represented as ML2T-2 and E0 is the total energy of an undamped oscillator which is represented by E0=12mA2w02, where m is the mass, A is the amplitude, and w is the angular velocity.

The dimensional representation of the total energy of an undamped oscillator is

E0=ML2T-12=ML2T-2

Thus, the dimensional formula is E=E0e-t/t0.

Topic: Physics, Motion In One Dimension, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
If $E(\theta)=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, then $E(\alpha) E(\beta)$ is equal to

Options:
A $E(\alpha \beta)$
B $E(\alpha-\beta)$
C $E(\alpha+\beta)$
D $-E(\alpha+\beta)$

The correct answer is:
$E(\alpha+\beta)$

Solution:
$\mathrm{E}(\theta)=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
Now $\mathrm{E}(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
and $\mathrm{E}(\beta)=\left[\begin{array}{cc}\cos \beta & \sin \beta \\ -\sin \beta & \cos \beta\end{array}\right]$
Now $\mathrm{E}(\alpha) \mathrm{E}(\beta)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \beta & \sin \beta \\ -\sin \beta & \cos \beta\end{array}\right]$
$=\left[\begin{array}{lr}\cos \alpha \cdot \cos \beta & \cos \alpha \sin \beta \\ -\sin \alpha \sin \beta & +\sin \alpha \cos \beta \\ -\sin \alpha \cos \beta & -\sin \alpha \sin \beta \\ -\cos \alpha \sin \beta & +\cos \alpha \cos \beta\end{array}\right]$
$=\left[\begin{array}{cc}\cos (\alpha+\beta) & \sin (\alpha+\beta) \\ -\sin (\alpha+\beta) & \cos (\alpha+\beta)\end{array}\right]$
$=\mathrm{E}(\alpha+\beta)$
$\therefore$ Option $(\mathrm{c})$ is correct.

Topic: Mathematics, Matrices, NDA, NDA 2015 (Phase 1)
If E is the universal set and $\mathrm{A}=\mathrm{B} \cup \mathrm{C}$, then the set $\mathrm{E}-(\mathrm{E}-$ $(\mathrm{E}-(\mathrm{E}-(\mathrm{E}-\mathrm{A}))))$ is same as the set

Options:
A $\mathrm{B} \cup \mathrm{C}$
B $\mathrm{B} \cup \mathrm{C}$
C $\mathrm{B}^{\prime} \cap \mathrm{C}$
D $B \cap C$

The correct answer is:
$\mathrm{B}^{\prime} \cap \mathrm{C}$

Solution:
$\mathrm{E}$ is the universal set and $\mathrm{A}=\mathrm{B} \cup \mathrm{C}$.
Since, $\mathrm{E}$ is the universal set, $\mathrm{E}-\mathrm{A}=\mathrm{A}^{\prime}$
$\therefore \mathrm{E}-(\mathrm{E}-(\mathrm{E}-(\mathrm{E}-(\mathrm{E}-\mathrm{A}))))$
$=\mathrm{E}-\left(\mathrm{E}-\left(\mathrm{E}-\left(\mathrm{E}-\mathrm{A}^{\prime}\right)\right)\right)$
$=\mathrm{E}-(\mathrm{E}-(\mathrm{E}-\mathrm{A}))$
$=\mathrm{E}-(\mathrm{E}-\mathrm{A})^{\prime}$
$=\mathrm{E}-\mathrm{A}$
$=\mathrm{A}^{\prime}$
$=(\mathrm{B} \cup \mathrm{C})^{\prime}$
$=\mathrm{B}^{1} \cap \mathrm{C}^{\prime}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2017 (Phase 2)
If $e^{y}+x y=e$, then what is the value of $\frac{d^{2} y}{d x^{2}}$ at $x=0$?

Options:
A $e^{-1}$
B $e^{-2}$
C e
D 1

The correct answer is:
$e^{-2}$

Solution:
Given $e^{y}+x y=e$ On differentiating w.r.t. $x$, we get
$e^{y} \frac{d y}{d x}+y+x \frac{d y}{d x}=0$ ...(i)
At $x=0$ we get $e^{y}+0 . y=e \Rightarrow e^{y}=e \Rightarrow y=1$
$\therefore$ By putting $y=1$ in equation (i)
we get
$e \frac{d y}{d x}+1+0=0$
$\Rightarrow \frac{d y}{d x}=-\frac{1}{e}$
Again differentiating Eq. (i), we get
$e^{y} \frac{d^{2} y}{d x^{2}}+e^{y}\left(\frac{d y}{d x}\right)^{2}+\frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0$
$\Rightarrow \frac{d^{2} y}{d x^{2}}\left(e^{y}+x\right)+e^{y}\left(\frac{d y}{d x}\right)^{2}+\frac{2 d y}{d x}=0$
Now, At $x=0, y=1$
$\frac{d^{2} y}{d x^{2}}(e+0)+e\left(-\frac{1}{e}\right)^{2}+2\left(-\frac{1}{e}\right)=0$
$\Rightarrow e \frac{d^{2} y}{d x^{2}}-\frac{1}{e}=0$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{1}{e^{2}}=e^{-2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be___(For surface tension, Force=Surface tension×length)

Options:
A EÂ V-2Â T-1
B EÂ V-1Â T-2
C E V-2 T-2
D E-2Â V-1Â T-3

The correct answer is:
E V-2 T-2

Solution:
Dimensional formula of surface tension, S=M1L0T-2
Dimensional formula of energy, E=M1L2T-2
Dimensional formula of velocity, V=M0L1T-1
By applying dimensional analysis,
S=EaVbTc
M1L0T-2=M1L2T-2aM0L1T-1bTc
M1L0T-2=MaL2a+bT-2a-b+c
On comparing,
a=1
2a+b=0⇒b= -2
-2a-b+c=-2⇒c= -2
∴S=E1V-2T-2

Topic: Physics, Units and Dimensions, NEET, NEET 2015 (Phase 1)
If entropy of a solid is greater than zero, at $\mathrm{T}=0$, it is called

Options:
A standard entropy
B residual entropy
C formal entropy
D absolute entropy

The correct answer is:
residual entropy

Solution:
(C)

Topic: Chemistry, Thermodynamics (C), MHT CET, MHT CET 2020 (16 Oct Shift 2)
If equation of sound wave is $y=0.0015 \sin (62.8 x+316 t)$, then its wavelength will be

Options:
A 0.2 unit
B 0.3 unit
C 0.1 unit
D 2 unit.

The correct answer is:
0.1 unit

Solution:
Equation of sound wave is
$$
y=0.0015 \sin (62.4 x+316 t) \text {. }
$$
Standard equation of a sound wave is
$$
y=A \sin 2 \pi\left[\frac{x}{\lambda}+\frac{t}{T}\right]
$$
Comparing the given equation with the standard equation, we get
$$
2 \pi / \lambda=62.8 \text { or, } \lambda=\frac{2 \pi}{62.8}=0.1 \text { unit. }
$$
(where $\lambda=$ wavelength of the wave)

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2002
If equilibrium constant of a process is $3.8 \times 10^{-3}$ at $25^{\circ} \mathrm{C}$, standard free energy change of the process is
$$
\left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 0.0038=-2.42\right)
$$

Options:
A $5.7 \mathrm{~kJ} \mathrm{~mol}^1$
B $9.9 \mathrm{~kJ} \mathrm{~mol}^1$
C $13.8 \mathrm{~kJ} \mathrm{~mol}^1$
D $15.6 \mathrm{~kJ} \mathrm{~mol}^1$

The correct answer is:
$13.8 \mathrm{~kJ} \mathrm{~mol}^1$

Solution:
$$
\begin{aligned}
\Delta G^{\circ} & =-R T \ln K \\
\Delta G^{\circ} & =-2.303 R T \log K \\
\Delta G^{\circ} & =-2.303 \times 8.314 \mathrm{~mol}^{-1} \times 298 \log 3.8 \times 10^{-3} \\
\Delta G^{\circ} & =13.8 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} \\
\Delta G^{\circ} & =13.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2018 (05 May Shift 2)
If $f(9)=9$ and $f^{\prime}(9)=4$ then what is $\lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$ equal to ?

Options:
A 36
B 9
C 4
D None of these

The correct answer is:
4

Solution:
$\lim _{\mathrm{x} \rightarrow 0} \frac{\sqrt{\mathrm{f}(\mathrm{x})}-3}{\sqrt{\mathrm{x}}-3}=\lim _{\mathrm{x} \rightarrow 0} \frac{\frac{1}{2 \sqrt{\mathrm{f}(\mathrm{x})}} \mathrm{f}^{\prime}(\mathrm{x})}{\frac{1}{2 \sqrt{\mathrm{x}}} .1}$
(ByL' Hospital rule)
$=\lim _{x \rightarrow 0} \frac{\mathrm{f}^{\prime}(\mathrm{x}) \times \sqrt{\mathrm{x}}}{\sqrt{\mathrm{f}(\mathrm{x})}}=\frac{\mathrm{f}^{\prime}(9) \times \sqrt{9}}{\sqrt{\mathrm{f}(9)}}$
$=\frac{4 \times 3}{\sqrt{9}}=\frac{4 \times 3}{3}=4$

Topic: Mathematics, Limits, NDA, NDA 2014 (Phase 1)
If $f$ and $g$ are two increasing functions such that $f o g$ is defined, then which one of the following is correct?

Options:
A $f o g$ is always an increasing function
B $f o g$ is always a decreasing function
C $f o g$ is neither an increasing nor a decreasing function
D None of the above

The correct answer is:
$f o g$ is always an increasing function

Solution:
Product of two increasing function is always an increasing function. $\therefore f o g$ is always an increasing function

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 2)
If $\mathrm{f}: \mathrm{IR} \rightarrow \mathrm{IR}, \mathrm{g}: \mathrm{IR} \rightarrow \mathrm{IR}$ be two functions given by
$\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{3}+5$, then $(\mathrm{fog})^{-1}(\mathrm{x})$ is equal to

Options:
A $\left(\frac{x+7}{2}\right)^{\frac{1}{3}}$
B $\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$
C $\left(x-\frac{7}{2}\right)^{\frac{1}{3}}$
D $\left(x+\frac{7}{2}\right)^{\frac{1}{3}}$

The correct answer is:
$\left(\frac{x-7}{2}\right)^{\frac{1}{3}}$

Solution:
$\mathrm{f}(\mathrm{x})=2 \mathrm{x}-3$
$\mathrm{~g}(\mathrm{x})=\mathrm{x}^{3}+5$
$\operatorname{fog}(\mathrm{x})=\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{f}\left(\mathrm{x}^{3}+5\right)$
$=2\left(\mathrm{x}^{3}+5\right)-3$
$\operatorname{fog}(\mathrm{x})=2 \mathrm{x}^{3}+7=\mathrm{y}(\mathrm{say})$
$\Rightarrow 2 \mathrm{x}^{3}=\mathrm{y}-7$
$\Rightarrow \mathrm{x}=\left[\frac{\mathrm{y}-7}{2}\right]^{\frac{1}{3}}$
$\Rightarrow \operatorname{fog}^{-1}(\mathrm{y})=\left[\frac{\mathrm{y}-7}{2}\right]^{\frac{1}{3}}$
$\Rightarrow \mathrm{fog}^{-1}(\mathrm{x})=\left[\frac{\mathrm{x}-7}{2}\right]^{\frac{1}{3}}$

Topic: Mathematics, Functions, NDA, NDA 2015 (Phase 2)
If $f$ is defined by $f(x)=\left\{\begin{array}{cc}\frac{1-\cos a x}{x \sin x}, $(x \neq 0)$ \frac{1}{2} \quad,(x=0)\end{array}\right.$ and $f$ is continuous at $x=0$, then $a^2=$

Options:
A $1$
B $2$
C $3$
D $4$

The correct answer is:
$1$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Continuity and Differentiability, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 2)
If $F(n)$ denotes the set of all divisors of $n$ except 1, what is the least value of y satisfying $[F(20) \cap F(16)] \subseteq F(y$

Options:
A 1
B 4
C 2
D 8

The correct answer is:
4

Solution:
Given that $\mathrm{F}(\mathrm{n})=$ set of all divisors of $\mathrm{n}$ except $\mathrm{F}(20)=\{2,4,5,10,20\}$
and $\mathrm{F}(16)=\{2,4,8,16\}$
$\mathrm{F}(20) \cap \mathrm{F}(16)=\{2,4,5,10,20\} \cap\{2,4,8,16\}$
$=\{2,4\}$
Also, $\{\mathrm{F}(20) \cap \mathrm{F}(16)\} \subseteq \mathrm{F}(\mathrm{y})$
So, least value of $y=2$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
If $f: R \rightarrow R, g: R \rightarrow R$ and $g(x)=x+3$ and $(f \circ g)(x)=(x+3)^{2}$, then what is the value of $f(-3)$ ?

Options:
A -9
B 0
C 9
D 3

The correct answer is:
9

Solution:
Given $f \circ g(x)=(x+3)^{2}$ and $g(x)=x+3$ $f o g(x)=(x+3)^{2} \Rightarrow f[g(x)]=(x+3)^{2}$
$\Rightarrow \quad f[x+3]=(x+3)^{2}(\because g(x)=x+3)$
$\Rightarrow f(x)=x^{2}$
Hence, $f(-3)=(-3)^{2}=9$

Topic: Mathematics, Functions, NDA, NDA 2010 (Phase 1)
If $\mathrm{f}: \mathbb{R} \rightarrow \mathrm{S}$ defined by $\mathrm{f}(\mathrm{x})=4 \sin \mathrm{x}-3 \cos \mathrm{x}+1$ is onto,
then what is S equal to?

Options:
A $[-5,5]$
B $(-5,5)$
C $(-4,6)$
D $[-4,6]$

The correct answer is:
$[-4,6]$

Solution:
$f(x)=4 \sin x-3 \cos x+1$
To find range of this function, we have to find maximum and minimum values.
We know, for a $\sin \theta+\mathrm{b} \cos \theta+\mathrm{c}, \max$ value is
$\sqrt{a^{2}+b^{2}}+c$ and minimum value is $-\sqrt{a^{2}+b^{2}}+c$. $\therefore$ Maximum value
$\begin{array}{ll} & =\sqrt{(4)^{2}+(-3)^{2}}+1=\sqrt{16+9}+1=5+1=6 \\ & \text { Minimum value }=-\sqrt{16+9}+1=-5+1=-4 \\ \therefore \quad & \text { Range, } \mathrm{S} \in[-4,6]\end{array}$

Topic: Mathematics, Functions, NDA, NDA 2018 (Phase 1)
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}1 & \mathrm{x} \text { is a rational number } \\ 0, & \mathrm{x} \text { is an irrational number, }\end{array}\right.$ what is/are the value(s) of (f of) $(\sqrt{3)}$ ?

Options:
A 0
B 1
C Both 0 and 1
D None of these

The correct answer is:
1

Solution:
From the given direction of function
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}1, & \mathrm{x} \text { is a rational number } \\ 0, & \mathrm{x} \text { is an irrational number }\end{array}\right.$
$($ f 0 f $) \sqrt{3}=\mathrm{f}\{\mathrm{f}(\sqrt{3})\}$
$=\mathrm{f}(0) \quad(\because \sqrt{3}$ is an irrational number $)$ $=1$
$(\because 0$ is a rational number)

Topic: Mathematics, Functions, NDA, NDA 2007 (Phase 2)
If $f(x)=\frac{4 x+x^{4}}{1+4 x^{3}}$ and $g(x)=\operatorname{In}\left(\frac{1+x}{1-x}\right)$, then what is the
value of $f^{\circ} \mathrm{g}\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)$ equal to?

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

The correct answer is:
1

Solution:
$\mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}+\mathrm{x}^{4}}{1+4 \mathrm{x}^{3}}, \mathrm{~g}(\mathrm{x})=\ln \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$
$g\left(\frac{e-1}{e+1}\right)=\ell n\left(\frac{1+\left(\frac{e-1}{e+1}\right)}{1-\left(\frac{e-1}{e+1}\right)}\right)=\ell n\left(\frac{e+1+e-1}{e+1-e+1}\right)$
$=\ln \left(\frac{2 \ell}{2}\right)=\ell$ n $\mathrm{e}=1$
fo $\mathrm{g}\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)=\mathrm{f}(1)=\frac{4(1)+(1)^{4}}{1+4(1)^{3}}=\frac{4+1}{1+4}=\frac{5}{5}=1$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 2)
If $\mathrm{f}\left(\mathrm{x}_{1}\right)-\mathrm{f}\left(\mathrm{x}_{2}\right)=\mathrm{f}\left(\frac{\mathrm{x}_{1}-\mathrm{x}_{2}}{1-\mathrm{x}_{1} \mathrm{x}_{2}}\right)$ for $\mathrm{x}_{1}, \mathrm{x}_{2} \in(-1,1)$, then what
is $\mathrm{f}(\mathrm{x})$ equal to?

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

The correct answer is:
$\ln \left(\frac{1-x}{1+x}\right)$

Solution:
$f\left(x_{1}\right)-f\left(x_{2}\right)=f\left(\frac{x_{1}-x_{2}}{1-x_{1} x_{2}}\right)$
$x_{1}, x_{2} \in(-1,1)$
then $f(x)=\log \frac{(1-x)}{(1+x)}$
$\begin{aligned} f\left(x_{1}\right)=\log \frac{1-x_{1}}{1+x_{1}} & f\left(x_{2}\right)=\log \frac{1-x_{2}}{1+x_{2}} \\ f\left(x_{1}\right)-f\left(x_{2}\right) &=\log \frac{1-x_{1}}{1+x_{1}}-\log \frac{1-x_{2}}{1+x_{2}} \\=& \log \frac{\left(1-x_{1}\right)}{\left(1+x_{1}\right)} \times \frac{\left(1+x_{2}\right)}{\left(1-x_{2}\right)} \\=& \log \frac{\left(1-x_{1}+x_{2}-x_{1} x_{2}\right)}{\left(1+x_{1}-x_{2}-x_{1} x_{2}\right)} \\=& \log \frac{\left(1-x_{1} x_{2}\right)-\left(x_{1}-x_{2}\right)}{\left(1-x_{1} x_{2}\right)+\left(x_{1}-x_{2}\right)} \\ f\left(x_{1}\right)-f\left(x_{2}\right)=f\left(\frac{x_{1}-x_{2}}{1-x_{1} x_{2}}\right) \end{aligned}$

Topic: Mathematics, Functions, NDA, NDA 2016 (Phase 1)