datavorous/entrance-exam-dataset huggingface dataset:
If $f(x)=\log _{e}\left(\frac{1+x}{1-x}\right), g(x)=\frac{3 x+x^{3}}{1+3 x^{2}}$ and g of $(\mathrm{t})=\mathrm{g}(\mathrm{f}(\mathrm{t}))$,
then what is $g \circ f\left(\frac{e-1}{e+1}\right)$ equal to?

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

The correct answer is:
1

Solution:
No Solution

Topic: Mathematics, Functions, NDA, NDA 2015 (Phase 1)
If $\mathrm{f}(\mathrm{x})=\log _{\mathrm{e}}\left[\log _{\mathrm{e}} \mathrm{x}\right]$, then what is $\mathrm{f}^{\prime}(\mathrm{e})$ equal to?

Options:
A $\mathrm{e}^{-1}$
B e
C 1
D 0

The correct answer is:
$\mathrm{e}^{-1}$

Solution:
The given funcion is: $\mathrm{f}(\mathrm{x})=\log _{\mathrm{e}}\left[\log _{\mathrm{e}} \mathrm{x}\right]$
Differentiating w.r.t. $\mathrm{x}$, we get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\log _{\mathrm{e}} \mathrm{x}} \cdot \frac{1}{\mathrm{x}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{\log _{\mathrm{e}} \mathrm{e}} \cdot \frac{1}{\mathrm{e}}=\frac{1}{\mathrm{c}}=\mathrm{e}^{-1}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 1)
If $f(x)=\log |x|, x \neq 0$, then what is $f^{\prime}(x)$ equal to?

Options:
A $\frac{1}{|x|}$
B $\frac{1}{x}$
C $\frac{-1}{x}$
D None of these

The correct answer is:
$\frac{1}{|x|}$

Solution:
$f(x)=\log |x|$
$f^{\prime}(x)=\frac{1}{|x|}$

Topic: Mathematics, Differentiation, NDA, NDA 2008 (Phase 2)
If $f(x)=\left\{\begin{array}{ll}m x+1 & x \leq \frac{\pi}{2} \\ \sin x+n & x>\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then
which one of the following is correct?

Options:
A $\mathrm{m}=1, \mathrm{n}=0$
B $\mathrm{m}=\frac{\mathrm{n} \pi}{2}+1$
C $\mathrm{n}=\mathrm{m}\left(\frac{\pi}{2}\right)$
D $\mathrm{m}=\mathrm{n}=\frac{\pi}{2}$

The correct answer is:
$\mathrm{m}=\mathrm{n}=\frac{\pi}{2}$

Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}$ and $\mathrm{g}(\mathrm{x})=\mathrm{cx}+\mathrm{d}$
$\mathrm{f}[\mathrm{g}(\mathrm{x}))]=\mathrm{a}(\mathrm{cx}+\mathrm{d})+\mathrm{b}$
$=\mathrm{acx}+\mathrm{ad}+\mathrm{b}$
and $\mathrm{g}[\mathrm{f}(\mathrm{x})]=\mathrm{c}(\mathrm{ax}+\mathrm{b})+\mathrm{d}$
$=\mathrm{acx}+\mathrm{dc}+\mathrm{d}$
Now from $\mathrm{f}[(\mathrm{gx})]=\mathrm{g}[\mathrm{f}(\mathrm{d})]$
$\Rightarrow \mathrm{ad}+\mathrm{b}=\mathrm{bc}+\mathrm{d}$
$\Rightarrow \mathrm{f}(\mathrm{d})=\mathrm{g}(\mathrm{b})$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 2)
If $f(x)=\left\{\begin{array}{ll}m x+1 & x \leq \frac{\pi}{2} \\ \sin x+n & x>\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then
which one of the following is correct?

Options:
A $\mathrm{m}=1, \mathrm{n}=0$
B $\mathrm{m}=\frac{\mathrm{n} \pi}{2}+1$
C $\mathrm{n}=\mathrm{m}\left(\frac{\pi}{2}\right)$
D $\mathrm{m}=\mathrm{n}=\frac{\pi}{2}$

The correct answer is:
$\mathrm{n}=\mathrm{m}\left(\frac{\pi}{2}\right)$

Solution:
Given function is
$f(x)=\left\{\begin{array}{ll}m x+1, & x \leq \frac{\pi}{2} \\ \sin x+n, & x>\frac{\pi}{2}\end{array}\right.$
As given this function is continuous at $x=\frac{\pi}{2}$.
So, limit of function when $\mathrm{x} \rightarrow \frac{\pi}{2}=\mathrm{f}\left(\frac{\pi}{2}\right)$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}+}(\sin x+n)=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \lim _{h \rightarrow 0}\left(\sin \left(\frac{\pi}{2}+h\right)+n\right)=\frac{m \pi}{2}+1$
$\Rightarrow \quad \sin \frac{\pi}{2}+n=\frac{m \pi}{2}+1$
$\Rightarrow 1+\mathrm{n}=\frac{\mathrm{m} \pi}{2}+1$
$\Rightarrow \mathrm{n}=\frac{\mathrm{m} \pi}{2}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2006 (Phase 2)
If $f(x)=\sin ^{2} x^{2}$, then what is $f^{\prime}(x)$ equal to?

Options:
A $4 x \sin \left(x^{2}\right) \cos \left(x^{2}\right)$
B $2 \sin \left(x^{2}\right) \cos \left(x^{2}\right)$
C $4 \sin \left(x^{2}\right) \sin ^{2} x$
D $2 \mathrm{x} \cos ^{2}\left(\mathrm{x}^{2}\right)$

The correct answer is:
$4 x \sin \left(x^{2}\right) \cos \left(x^{2}\right)$

Solution:
Given $f(x)=\sin ^{2} x^{2}$
$\therefore f^{\prime}(x)=2 \sin x^{2} \cos x^{2} \cdot 2 x$
$=4 x \sin x^{2} \cos x^{2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
If $f(\mathrm{x})=\tan \mathrm{x}+\mathrm{e}^{-2 \mathrm{x}}-7 \mathrm{x}^{3}$, then what is the value of $f^{\prime}(0)$?

Options:
A $-2$
B $-1$
C 0
D 3

The correct answer is:
$-1$

Solution:
Given $f(x)=\tan x+e^{-2 x}-7 x^{3}$
On differentiating w.r.t. $\mathrm{x}$, we get
$f^{\prime}(x)=\sec ^{2} x-2 e^{-2 x}-21 x^{2}$
Put $x=0$
$\Rightarrow f^{\prime}(0)=\sec ^{2} 0-2 e^{0}-21 \times 0=1-2=-1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
If $f(x)=(x+1)^{\text {cotx }}$ is continuous at $x=0$, then what is $f(0)$ equal to?

Options:
A 1
B e
C $\frac{1}{\mathrm{e}}$
D $\mathrm{e}^{2}$

The correct answer is:
e

Solution:
For a function to be continuous at a point the limit should exist and should be equal to the value of the function at that point.
Here point is $\mathrm{x}=0$
and $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(x+1)^{\cot x}$
$=\lim _{x \rightarrow 0}(1+x)^{\cot x}=\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x} \cdot x \cot x}$
$=\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x} \lim _{x \rightarrow 0} \frac{x}{\tan x}}=e^{1}=e$
Since limiting value of $f(x)=e$, when $x \rightarrow 0, f(0)$ should also be equal to e.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2007 (Phase 1)
If $f(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}-1}$, then what is $\frac{f(\mathrm{a})}{f(\mathrm{a}+1)}$ equal to?

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

The correct answer is:
$f\left(\mathrm{a}^{2}\right)$

Solution:
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\mathrm{x}-1}$
$\mathrm{f}(\mathrm{a})=\frac{\mathrm{a}}{\mathrm{a}-1}$
$\mathrm{f}(\mathrm{a}+1)=\frac{\mathrm{a}+1}{\mathrm{a}+1-1}=\frac{\mathrm{a}+1}{\mathrm{a}}$
$\therefore \quad \frac{\mathrm{f}(\mathrm{a})}{\mathrm{f}(\mathrm{a}+1)}=\frac{\frac{\mathrm{a}}{\mathrm{a}-1}}{\frac{\mathrm{a}+1}{\mathrm{a}}}=\frac{\mathrm{a}^{2}}{\mathrm{a}^{2}-1}$
$\mathrm{f}\left(\mathrm{a}^{2}\right)=\frac{\mathrm{a}^{2}}{\mathrm{a}^{2}-1}$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 1)
If $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}-9}{\mathrm{x}^{2}-2 \mathrm{x}-3}, \mathrm{x} \neq 3$ is continuous at $\mathrm{x}=3$, then
which one of the following is correct?

Options:
A $\quad f(3)=0$
B $\mathrm{f}(3)=1.5$
C $\quad f(3)=3$
D $f(3)=-1.5$

The correct answer is:
$\mathrm{f}(3)=1.5$

Solution:
$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}-9}{\mathrm{x}^{2}-2 \mathrm{x}-3}$
Given, the function is continuous at $\mathrm{x}=3$.
$f(3)=\lim _{x \rightarrow 3} \frac{x^{2}-9}{x^{2}-2 x-3}$
$=\lim _{x \rightarrow 3} \frac{(x+3)(x-3)}{(x-3)(x+1)}$
$=\frac{3+3}{3+1}=\frac{6}{4}=\frac{3}{2}=1.5$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2018 (Phase 1)
If $f(x)=x^{2}-6 x+8$ and there exists a point $c$ in the interval $[2,4]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$, then what is the value of $\mathrm{c}$?

Options:
A 2.5
B 2.8
C 3
D $3.5$

The correct answer is:
3

Solution:
Let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-6 \mathrm{x}+8$
$\begin{aligned} \mathrm{f}^{\prime}(\mathrm{x}) &=2 \mathrm{x}-6 \\ \mathrm{f}^{\prime}(\mathrm{c}) &=2 \mathrm{c}-6 \\ \mathrm{f}^{\prime}(\mathrm{c}) &=0 \\ \Rightarrow 2 \mathrm{c}-6 &=0 \Rightarrow \mathrm{c}=3 \end{aligned}$

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

Options:
A $\frac{4(x+2)}{x-2}$
B $\frac{x+2}{4(x-2)}$
C $\frac{x+2}{x-2}$
D $\frac{2(1+x)}{1-x}$

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

Solution:
$f(x) \mathrm{y}=4 \frac{x-2}{x+2}, x \neq-2$
$\frac{y}{1}=\frac{x-2}{x+2} \Rightarrow \frac{y+1}{y-1}=\frac{x-2+x+2}{x-2-x-2}$
$\Rightarrow \frac{y+1}{y-1}=\frac{2 x}{-4}$
$\Rightarrow \frac{y+1}{y-1}=\frac{-x}{2} \Rightarrow x=-2\left(\frac{y+1}{y-1}\right)$
Now, $\mathrm{y}=\frac{-2(x+1)}{x-1}=\frac{2(x+1)}{1-x}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2019 (Phase 1)
If $\mathrm{f}(\mathrm{x})=\mathrm{x}$ and $\mathrm{g}(\mathrm{x})=|\mathrm{x}|$, then what is $(\mathrm{f}+\mathrm{g})(\mathrm{x})$
equal to ?

Options:
A 0 for all $\mathrm{x} \in \mathrm{R}$
B $2 \mathrm{x}$ for all $\mathrm{x} \in \mathrm{R}$
C $\left\{\begin{array}{l}2 \mathrm{x}, \text { for } \mathrm{x} \geq 0 \\ 0, \text { for } \mathrm{x} < 0\end{array}\right.$
D $\left\{\begin{array}{l}0, \text { for } \mathrm{x} \geq 0 \\ 2 \mathrm{x}, \text { for } \mathrm{x} < 0\end{array}\right.$

The correct answer is:
$\left\{\begin{array}{l}2 \mathrm{x}, \text { for } \mathrm{x} \geq 0 \\ 0, \text { for } \mathrm{x} < 0\end{array}\right.$

Solution:
Given functions are : $\mathrm{f}(\mathrm{x})=\mathrm{x}$ and $\mathrm{g}(\mathrm{x})=|\mathrm{x}|$
$\therefore \quad(\mathrm{f}+\mathrm{g})(\mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\mathrm{x}+|\mathrm{x}|$
According to defintion of modulus function,
$(f+g)(x)=\left\{\begin{array}{ll}x+x, & x \geq 0 \\ x-x, & x < 0\end{array}\right.$
$=\left\{\begin{array}{cl}2 \mathrm{x}, & \mathrm{x} \geq 0 \\ 0, & \mathrm{x} < 0\end{array}\right.$

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 1)
If $f(x)=x \ell \operatorname{n} x$, then $f(x)$ attains minimum value at which one of the following points?

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

The correct answer is:
$x=e^{-1}$

Solution:
Let $f(x)=x \ln x$
$f^{\prime}(x)=\frac{x}{x}+\ln x=1+\ln x$
Put $f(x)=0 \Rightarrow 1+\ln x=0$
$\Rightarrow \ln x=-1 \Rightarrow x=\mathrm{e}^{-1}$
Now, $f^{\prime \prime}(x)=\frac{1}{x}$
$\left.f^{\prime \prime}(x)\right|_{x=e^{-1}}=\frac{1}{e^{-1}}=e>0$
Hence, $f(x)$ attains minimum value at $x=e^{-1}$.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 2)
If $f(x)=\left(x-x_{0}\right) \phi(x)$ and $\phi(x)$ is continuous at $x=x_{0}$, then what is $\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)$ equal to ?

Options:
A $\phi^{\prime}\left(\mathrm{x}_{0}\right)$
B $\phi\left(\mathrm{x}_{0}\right)$
C $\mathrm{x}_{0} \phi\left(\mathrm{x}_{0}\right)$
D $2 \phi\left(\mathrm{x}_{0}\right)$

The correct answer is:
$\phi\left(\mathrm{x}_{0}\right)$

Solution:
$\quad$ Given that $\mathrm{f}(\mathrm{x})=\left(\mathrm{x}-\mathrm{x}_{0}\right) \phi(\mathrm{x})$
Differentiating w. r. t. $\mathrm{x}$ $\mathrm{f}^{\prime}(\mathrm{x})=\left(\mathrm{x}-\mathrm{x}_{0}\right) \phi^{\prime}(\mathrm{x})+\phi(\mathrm{x})(1)$
Putting $\mathrm{x}=\mathrm{x}_{0}$
$\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)=\phi\left(\mathrm{x}_{0}\right)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 1)
If $\mathrm{f}(\mathrm{x})=|\mathrm{x}|+|\mathrm{x}-1|$, then which one of the following is correct?

Options:
A $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ and $\mathrm{x}=1$
B $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ but not at $\mathrm{x}=1$
C $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=1$ but not at $\mathrm{x}=0$
D $\mathrm{f}(\mathrm{x})$ is neither continuous at $\mathrm{x}=0$ nor at $\mathrm{x}=1$

The correct answer is:
$\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ and $\mathrm{x}=1$

Solution:
Given function is continuous at $\mathrm{x}=0$ and 1 .

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2018 (Phase 1)
If $f(x)=x(\sqrt{x}-\sqrt{x+1})$, then $f(x)$ is

Options:
A continuous but not differentiable at $x=0$
B differentiable at $\mathrm{x}=0$
C not continuous at $\mathrm{x}=0$
D None of the above

The correct answer is:
differentiable at $\mathrm{x}=0$

Solution:
$\begin{aligned} \mathrm{f}(\mathrm{x}) &=\mathrm{x}(\sqrt{\mathrm{x}}-\sqrt{\mathrm{x}+1}) \\ \mathrm{f}(\mathrm{x}) &=\frac{\mathrm{x}(\sqrt{\mathrm{x}}-\sqrt{\mathrm{x}+1})(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+1})}{(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+1})} \\ &=\frac{\mathrm{x}(\mathrm{x}-\mathrm{x}-1)}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+1}}=\frac{-\mathrm{x}}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+1}} \end{aligned}$
Hence, $\mathrm{f}(\mathrm{x})$ is continuous as well as differentiable at $\mathrm{x}=0$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 2)
If $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\ldots . \infty},}}}$ then what is $\mathrm{f}^{\prime}(\mathrm{x})$ equal
to?

Options:
A $\frac{1}{1-2 f(x)}$
B $\frac{1}{2 \mathrm{f}(\mathrm{x})-1}$
C $\frac{1}{1+2 \mathrm{f}(\mathrm{x})} \quad$
D $\frac{1}{2+\mathrm{f}(\mathrm{x})} \quad$

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

Solution:
Given function is:
$f(x)=\sqrt{x+\sqrt{x+\sqrt{x \sqrt{\infty}}}}$
$\Rightarrow \quad f(x)=\sqrt{x+f(x)} \Rightarrow(f(x))^{2}=x+f(x)$
On differentiating both sides wrt. $\mathrm{x}$, we get $2 \mathrm{f}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})=1+\mathrm{f}^{\prime}(\mathrm{x})$
$\mathrm{f}^{\prime}(\mathrm{x})\{2 \mathrm{f}(\mathrm{x})-1\}=1$
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \mathrm{f}(\mathrm{x})-1}$

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 2)
If $f[x y]=f[x] f[y]$, then $f[t]$ may be of the form:
where $\mathrm{k}, \mathrm{c}$ are constants

Options:
A $\mathrm{t}+\mathrm{k}$
B $\mathrm{ct}+\mathrm{k}$
C $\mathrm{t}^{\mathrm{k}}+\mathrm{c}$
D $\mathrm{t}^{\mathrm{k}}$

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

Solution:
By observing the options Let $f(t)=t^{k}$
Suppose $\mathrm{t}=\mathrm{xy}$
$\therefore \quad \mathrm{f}(\mathrm{xy})=(\mathrm{xy})^{\mathrm{k}}=\mathrm{x}^{\mathrm{k}} \cdot \mathrm{y}^{\mathrm{k}}=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y})$
Hence, $\mathrm{f}(\mathrm{t})=\mathrm{t}^{\mathrm{k}}$ where $\mathrm{k}$ ' is a constant.

Topic: Mathematics, Functions, NDA, NDA 2012 (Phase 1)
If for a distribution $\Sigma(x-5)=3, \Sigma(x-5)^2=43$ and the total number of items is 18 , find the mean and standard deviation.

Options:

None

Solution:
$$
\begin{aligned}
&\text { We have, } \Sigma(x-5)=\Sigma x-\Sigma 5=\Sigma x-5(18)=\Sigma x-90\\
&\therefore \quad \Sigma \mathbf{x}-90=3[\text { Given })]\\
&\Rightarrow \quad \Sigma \mathrm{x}=93\\
&\therefore \quad \overline{\mathrm{x}}=\frac{\Sigma \mathrm{x}}{18}=\frac{93}{18}=5.167\\
&\text { Also } 43=\Sigma(x-5)^2=\Sigma\left(x^2-10 x+25\right)\\
&\Rightarrow \quad 43=\Sigma x^2-10 \Sigma x+25(18)\\
&\Rightarrow \quad 43=\Sigma x^2-10(93)+450\\
&\Rightarrow \quad \Sigma x^2=523
\end{aligned}
$$
$$
\begin{aligned}
\therefore \text { S.D. } &=\sqrt{\frac{523}{18}-(5.167)^2} \\
&=\sqrt{2.357}=1.535=1.53
\end{aligned}
$$

Topic: Mathematics, Statistics
If, for positive real numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$, the numbers $\mathrm{x}+\mathrm{y}, 2 \mathrm{y}$ and $\mathrm{y}+\mathrm{z}$ are in harmonic progression, then which one of the following is correct?

Options:
A $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in geometric progression
B $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in arithmetic progression
C $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in harmonic progression
D None of the above

The correct answer is:
$\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in geometric progression

Solution:
As given:
$\mathrm{x}+\mathrm{y}, 2 \mathrm{y}$ and $\mathrm{y}+\mathrm{z}$ are in harmonic progression. $2 y=\frac{(x+y)(y+z)}{x+y+y+z}$
$\Rightarrow \quad \frac{1}{2 y}=\frac{1}{x+y}+\frac{1}{y+z}$
$\Rightarrow \quad y(x+2 y+z)=\left(x y+x z+y^{2}+y z\right)$
$\Rightarrow x y+2 y^{2}+y z=x y+x z+y^{2}+y z$
$\Rightarrow \quad y^{2}=x z$
$\Rightarrow \mathrm{x}, \mathrm{y}, \mathrm{z}$ are in geometric progession.

Topic: Mathematics, Sequences and Series, NDA, NDA 2007 (Phase 2)
If for the given reaction taking place at 25Â°C in a cell is,
1 2 Cu s + 1 2 Cl 2 g â 1 2 Cu 2+ + Cl â
having standard e.m.f. of 1.02 volts . Calculate standard free energy change.

Options:
A -98430 J
B 98430 J
C 96500 J
D -96500 J

The correct answer is:
-98430 J

Solution:
The relation between free energy change and standard e.m.f. is given by ΔG=-nFE°
ΔG=-1×96500×1.02 ;
ΔG=-98430 .

Topic: Chemistry, Electrochemistry, NEET
If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.

Options:
A [F][A][T]
B [F][A]T2
C [F][A]T-1
D [F]A-1[T]

The correct answer is:
[F][A]T2

Solution:
Let consider,

[Energy]=[F]a[A]b[T]c

Now put the dimensional formula of the given quantities,

ML2T-2=MLT-2aLT-2b[T]c

ML2T-2=MaLa+bT-2a-2b+c

On comparing powers of [M]

a=1

On comparing powers of [L]

a+b=2

⇒1+b=2

⇒b=1

On comparing powers of [T]

-2 a-2 b+c=-2

⇒-2-2+c=-2

⇒c=2

[ Energy ]=[F][A]T2

Topic: Physics, Units and Dimensions, NEET, NEET 2021
If force F, length L and time T be considered fundamental units, then the units of mass will be

Options:
A [F L T-2]
B [F L-2T-1]
C [F L-1T2]
D [F2L T-2]

The correct answer is:
[F L-1T2]

Solution:
Let M∝FaLbTc
Writing dimensions on both sides and using the principle of homogeneity of dimensions we have,
M1L0T0=kM L T-2aLbTc
On comparing the power on both sides we get a=1,a+b=0 and -2a+c=0
On solving we have b=-1,c=2,a=1
∴ units of mass is F L-1T2

Topic: Physics, Units and Dimensions, NEET
If force F, velocity V and time T are taken as fundamental units, the dimensions of mass are

Options:
A FVT-1
B FVT-2
C FV-1T-1
D FV-1T

The correct answer is:
FV-1T

Solution:
The force is equal to the mass and the acceleration of the body.

F=ma=mvt

It can be dimensionally represented as

F=M×LT2=MLTT
F=MVT

So, the dimension of the mass is
M=FTV-1

Topic: Physics, Units and Dimensions, NEET, NEET 2014
If four dice are thrown together, then what is the probability that the sum of the numbers appearing on them is 25 ?

Options:
A 0
B $1 / 2$
C 1
D $1 / 1296$

The correct answer is:
0

Solution:
Maximum sum of numbers appearing on four dice together $=24$ $\therefore$ Required probability $=0$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 1)
If \(\frac{3}{(x-1)\left(x^2+x+1\right)}=\frac{1}{x-1}\)
\(\begin{aligned} & -\frac{x+2}{x^2+x+1}=f_1(x)-f_2(x) \text { and } \frac{x+1}{(x-1)^2\left(x^2+x+1\right)}=A f_1(x)+\left(B+\frac{D}{x-1}\right) \\ & f_2(x)+\frac{C}{(x-1)^2}, A+B+C+D= \end{aligned}\)

Options:
A 1
B \(\frac{-1}{3}\)
C 0
D \(\frac{1}{3}\)

The correct answer is:
0

Solution:
It is given that
\(\begin{aligned}
& \frac{3}{(x-1)\left(x^2+x+1\right)}=\frac{1}{x-1}-\frac{x+2}{x^2+x+1} \\
& =f_1(x)-f_2(x) \\
& \text { and } \frac{x+1}{(x-1)^2\left(x^2+x+1\right)}=A f_1(x)+\left(B+\frac{D}{x-1}\right) \\
& f_2(x)+\frac{C}{(x-1)^2}
\end{aligned}\)
From above informations, it is necessary, that
\(\begin{aligned}
f_1(x) & =\frac{1}{x-1} \\
\text { and } \quad f_2(x) & =\frac{x+2}{x^2+x+1}
\end{aligned}\)
So,
\(\begin{array}{r}
\frac{x+1}{(x-1)^2\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{(B x-B+D)(x+2)}{\left(x^2+x+1\right)(x-1)} \\
+\frac{C}{(x-1)^2}
\end{array}\)
\(\begin{aligned}
\Rightarrow x+1 & =A(x-1)\left(x^2+x+1\right)+B(x-1) \\
& (x+2)(x-1)+D(x+2)(x-1)+C\left(x^2+x+1\right)
\end{aligned}\)
On comparing the coefficient, we get coefficient of \(x^3=0 \Rightarrow A+B=0\) coefficient of \(x^2=0 \Rightarrow A-A+2 B-2 B+D+C=0\)
\(\begin{array}{llll}
\Rightarrow & D+C =0 \\
\therefore & A+B+C+D & =0
\end{array}\)
Hence, option (3) is correct.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
If from the point of intersection of two ogives, a perpendicular is drawn on the $x$ -axis, what does the $x-$ coordinate give?

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

The correct answer is:
Median

Solution:
The x-coordinate of the point of intersection of two ogives, gives median.

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 2)
If fundamental frequency is 50 and next successive frequencies are 150 and 250 then it is

Options:
A a pipe closed at both end
B a pipe closed at one end
C an open pipe
D a stretched pipe.

The correct answer is:
a pipe closed at one end

Solution:
For a pipe closed at one end, the frequencies are odd multiples of the fundamental frequencies.

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2001
If f(x) = sin(cosx) , then f'(x) is equal to ?

Options:
A $\cos (\cos x)$
B $\sin (-\sin x)$
C $(\sin x) \cos (\cos x)$
D $(-\sin x) \cos (\cos x)$

The correct answer is:
$(-\sin x) \cos (\cos x)$

Solution:
$\mathrm{f}(\mathrm{x})=\sin (\cos \mathrm{x})$
$\mathrm{f}^{\prime}(\mathrm{x})=\cos (\cos \mathrm{x}) \cdot(-\sin \mathrm{x})$
$=-\sin \mathrm{x} \cdot \cos (\cos \mathrm{x})$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2019 (Phase 1)
If \( f(x)=\left\{\begin{aligned} K x^{3} ; & \text { if } x \leq 2 \\ 3 ; & \text { if } x>2 \end{aligned}\right. \) is continuous at \( x=2 \), then the value of \( K \) is

Options:
A \( 13 \)
B \( 04 \)
C 3/4
D \( 4 / 3 \)

The correct answer is:
3/4

Solution:
Given that \( f(x)=\left\{\begin{array}{cc}K \chi^{2} & x \leq 2 \\ 3 & x>2\end{array}\right. \)
Since, \( f(x) \) is continuous at \( x=2 \). Then, L.H.L = R.H.L
So, \( \lim _{x \rightarrow 2} K x^{2}=\lim _{x \rightarrow 2} 3 \) \( \Rightarrow K \cdot 2^{2}=3 \Rightarrow K=\frac{3}{4} \)

Topic: Mathematics, Sets and Relations, KCET, KCET 2017
If \(f(x)=\left|\begin{array}{ccc}x-3 & 2 x^2-18 & 3 x^3-81 \\ x-5 & 2 x^2-50 & 4 x^3-500 \\ 1 & 2 & 3\end{array}\right|\), then \(f(\mathrm{l}) f(3)+f(3) f(5)+f(5) f(\mathrm{l})\) is equal to

Options:
A \(f(1)\)
B \(f(3)\)
C \(f(1)+f(3)\)
D \(f(1)+f(5)\)

The correct answer is:
\(f(3)\)

Solution:
\(f(x)=(x-3)(x-5)\left|\begin{array}{ccc}1 & 2(x+3) & 3\left(x^2+9+3 x\right) \\ 1 & 2(x+5) & 4\left(x^2+25+5 x\right) \\ 1 & 2 & 3\end{array}\right|\)
\(\Rightarrow \quad f(3)=f(5)=0\)
So, \(f(1) f(3)+f(3) \cdot f(5)+f(5) f(1)=0=f(3)\)

Topic: Mathematics, Determinants, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 1)
If \( f(x)=\left\{\begin{array}{cc}\frac{\log _{e} x}{x-1} & ; x \neq 1 \\ k & ; x=1\end{array}\right. \) is continuous at \( x=1 \), then the value of \( k \) is

Options:
A e
B \( 11 \)
C \( -1 \)
D \( 00 \)

The correct answer is:
\( 11 \)

Solution:
Given that, \( f(x)=\left\{\begin{array}{cc}\frac{\log _{e} x}{x-1} & x \neq 1 \\ k & x=1\end{array}\right. \)
For continuous at \( x=1 \), we have
\[
\begin{array}{l}
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\
k=\lim _{x \rightarrow 1} \frac{\log _{e} x}{x-1}=\lim _{x \rightarrow 1} \frac{1 / x}{1}=1 \\
\text { So, } k=1
\end{array}
\]

Topic: Mathematics, Continuity and Differentiability, KCET, KCET 2018
If \( f(x)=\left\{\begin{array}{cc}\frac{\sin 3 x}{e^{2 x}-1} & x \neq 0 \\ k-2 & x=0\end{array}\right. \) is continuous at \( x=0 \), then \( k= \)

Options:
A \( \frac{9}{5} \)
B \( \frac{2}{3} \)
C \( \frac{3}{2} \)
D none

The correct answer is:
none

Solution:
Given Options are not matching
\( f(x)=\left\{\begin{array}{c}\frac{\sin 3 x}{e^{2 x}-1} x \neq 0 \\ k-2 \quad x=0\end{array}\right. \) Since \( f \) is continuous at \( x=0 \) \( \Rightarrow \operatorname{lt}_{x \rightarrow 0} f(x)=f(0) \) \( \lim _{x \rightarrow 0} \frac{\sin 3 x}{2 x-1}=k-2 \) \( \Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{x}}{\frac{e^{2 x}-1}{x}}=k-2 \) \( \Rightarrow \frac{\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}}{\text { lt } \frac{e^{2 x}-1}{x}}=k-2 \) \( \Rightarrow \frac{3}{2}=k-2 \Rightarrow k=\frac{3}{2}+2=\frac{7}{2} \) \( \therefore k=\frac{7}{2} \)
\( \Rightarrow \operatorname{lt}_{x \rightarrow 0} f(x)=f(0) \) \( \lim _{x \rightarrow 0} \frac{\sin 3 x}{2 x-1}=k-2 \)
\( \Rightarrow \lim _{x \rightarrow 0} \frac{\frac{\sin 3 x}{x}}{\frac{e^{2 x}-1}{x}}=k-2 \)
\( \rightarrow \frac{\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}}{\operatorname{lt}_{x \rightarrow 0} \frac{e^{2 x}-1}{x}}=k-2 \)
\( \Rightarrow \frac{3}{2}=k-2 \Rightarrow k=\frac{3}{2}+2=\frac{7}{2} \)
\( \therefore \mathrm{k}=\frac{7}{2} \)

Topic: Mathematics, Continuity and Differentiability, KCET, KCET 2019