datavorous/entrance-exam-dataset huggingface dataset:
Consider the following statements:
$1.$ If det $\mathrm{A}=0$, then $\operatorname{det}(\operatorname{adj} \mathrm{A})=0$
$2 .$ If A is non- singular, then $\operatorname{det}\left(\mathrm{A}^{-1}\right)=(\operatorname{det} \mathrm{A})^{-1}$

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither l nor 2

The correct answer is:
Both 1 and 2

Solution:
We know that, adj A and A has same value of determinant, if det $\mathrm{A}=0$, then $\operatorname{det}(\operatorname{adj} \mathrm{A})=0$
so, statement (1) is correct. Also If A is a matrix the determinant of $\mathrm{A}^{-1}$ equals inverse of determinant $\mathrm{A}$, so, $\operatorname{det}\left(\mathrm{A}^{-1}\right)=(\operatorname{det} \mathrm{A})^{-1}$, if $\mathrm{A}$
is non singular; Statement 2 is correct.
Thus both (1) and (2) are correct.

Topic: Mathematics, Determinants, NDA, NDA 2007 (Phase 1)
Consider the following statements:
1- If $f(x)=x^{3}$ and $g(y)=y^{3}$ then $f=g$.
2- Identity function is not always a bijection. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
1 only

Solution:
(1) $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ and $\mathrm{g}(\mathrm{y})=\mathrm{y}^{3}$.
$\Rightarrow \mathrm{f}=\mathrm{g}$ is a correct statement.
(2) Identify function is always a bijection.

Topic: Mathematics, Functions, NDA, NDA 2012 (Phase 2)
Consider the following statements:
1 . If the correlation coefficient $\mathrm{r}_{\mathrm{xy}}=0$, then the two line:
of regression are parallel to each oth
2 . If the correlation coefficien A then the two lines of regression are perpendicular to each other Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
We know, $\tan \theta=\left|\frac{1-\mathrm{r}_{\mathrm{xy}}^{22}}{\mathrm{r}_{\mathrm{xy}}}\right|\left(\frac{\sigma_{\mathrm{x}} \cdot \sigma_{\mathrm{y}}}{\sigma_{\mathrm{x}}^{2} \cdot \sigma_{\mathrm{y}}^{2}}\right)$
then $\tan \theta=\infty$ and lines are perpendicular.
If $\mathrm{r}_{\mathrm{xy}}$ is 0, then $\tan \theta=\infty$ and lines are
If $\mathrm{r}_{\mathrm{xy}}$ is 1 , then the lines are parallel.

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
Consider the following statements:
1 If $y=\ln (\sec x+\tan x)$, then $\frac{d y}{d x}=\sec x$.
2 If $y=\ln (\operatorname{cosec} x-\cot x)$, then $\frac{d y}{d x}=\operatorname{cosec} x$.
Which of the above is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
$(1) y=\ell n(\sec x+\tan x)$
$\frac{d y}{d x}=\frac{\sec x \tan x+\sec ^{2} x}{\sec x+\tan x}$
$\quad=\frac{\sec x(\sec x+\tan x)}{\sec x+\tan x}=\sec x$
$y=\ell n(\operatorname{cosec} x-\cot x)$
$\frac{d y}{d x}=\frac{\operatorname{cosec} x(\operatorname{cosec} x-\cot x)}{\operatorname{cosec} x-\cot x}=\operatorname{cosec} x$
Hence, both statements are correct.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 2)
Consider the following statements:
1- $\lim _{x \rightarrow 0} \sin \frac{1}{x}$ does not exist.
2- $\lim _{x \rightarrow 0} \sin \frac{1}{x}$ exists
Which of the above statements correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
Since $\sin \frac{1}{x}$ is an oscillatory function
$\therefore \quad \lim _{x \rightarrow 0} \sin \frac{1}{x}$ has a finite value between $-1$ and 1 .
Now, At $x=0$
L. H. $\mathrm{L}=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}(-\mathrm{h}) \sin \left(-\frac{1}{\mathrm{~h}}\right)$
$=0 \times$ a finite value between $-1$ and 1 $\quad=0$
Similarly $\mathrm{R}$. H. $\mathrm{L}=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0+\mathrm{h})=0$
Also, $\mathrm{f}(0)=0$
$\therefore \quad \lim _{\mathrm{x} \rightarrow 0} \mathrm{x} \sin \frac{1}{\mathrm{x}}$ exists.
Hence Both statements are correct.

Topic: Mathematics, Limits, NDA, NDA 2013 (Phase 1)
Consider the following statements:
1- $\lim _{x \rightarrow 0} \frac{x^{2}}{x}$ exists.
2- $\left(\frac{x^{2}}{x}\right)$ is not continuous at $x=0$
3- $\lim _{x \rightarrow 0} \frac{|x|}{x}$ does not exist.
Which of the statements given above are correct?

Options:
A 1,2 and 3
B 1 and 2 only
C 2 and 3 only
D 1 and 3 only

The correct answer is:
1 and 3 only

Solution:
1- $\lim _{x \rightarrow 0} \frac{x^{2}}{x}=\lim _{x \rightarrow 0}(x)=0$
2- $\quad \frac{x^{2}}{x}=x$,
since a polynomial is continuous everywhere, so it is continuous at $x=0$
3- LHL $=\lim _{h \rightarrow 0} \frac{|0-h|}{(0-h)}=\lim _{h \rightarrow 0} \frac{h}{-h}=-1$
RHL $=\lim _{h \rightarrow 0} \frac{|0+h|}{(0+h)}=\lim _{h \rightarrow 0} \frac{h}{h}=1$
$\therefore L H L \neq \mathrm{RHL}$
So, it does not exist.
Thus, the statements 1 and 3 are correct.

Topic: Mathematics, Limits, NDA, NDA 2008 (Phase 2)
Consider the following statements:
1 . Mean is independent of change in scale and change in
origin.
2 . Variance is independent of change in scale but not in origin. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither $1 \mathrm{n}$

The correct answer is:
Neither $1 \mathrm{n}$

Solution:
1 . Mean is dependent with change in origin. 2 . Variance is independent with change in origin.

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
Consider the following statements:
1. $n\left(\sin ^{2} 67 \frac{1^{\circ}}{2}-\sin ^{2} 22 \frac{1^{\circ}}{2}\right)>1$ for all positive integers
$n \geq 2 .$
2 If $x$ is any positive real number, then $n x>1$ for all positive integers $n \geq 2$. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only $y$
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
1 only

Solution:
Statement 1:
Given $\mathrm{n}\left(\sin ^{2} 67 \frac{1^{\circ}}{2}-\sin ^{2} 22 \frac{1^{\circ}}{2}\right)$
or $\mathrm{n}\left(\sin ^{2} \frac{135^{\circ}}{2}-\sin ^{2} \frac{45^{\circ}}{2}\right)$
$=\mathrm{n}\left(\sin \frac{135^{\circ}}{2}+\sin \frac{45^{\circ}}{2}\right)\left(\sin \frac{135^{\circ}}{2}-\sin \frac{45^{\circ}}{2}\right)$
$=n\left[2 \sin \frac{\left(\frac{135^{\circ}}{2}+\frac{45^{\circ}}{2}\right)}{2} \cdot \cos \frac{\left(\frac{135^{\circ}}{2}-\frac{45^{\circ}}{2}\right)}{2}\right]$
$\left[2 \cdot \cos \frac{\left(\frac{135^{\circ}+45^{\circ}}{2}\right)}{2} \cdot \sin \frac{\left(\frac{135^{\circ}-45^{\circ}}{2}\right)}{2}\right]$
$=\mathrm{n}\left[2 \cdot \sin \left(\frac{90^{\circ}}{2}\right) \cdot \cos \left(\frac{45^{\circ}}{2}\right)\right]$
$\left[2 \cdot \cos \left(\frac{90^{\circ}}{2}\right) \cdot \sin \left(\frac{45^{\circ}}{2}\right)\right]$
$=2 n\left(2 \sin \frac{45^{\circ}}{2} \cdot \cos \frac{45^{\circ}}{2}\right)\left(\sin 45^{\circ} \cdot \cos 45^{\circ}\right)$
$=2 \mathrm{n} \cdot \sin \left(2 \times \frac{45^{\circ}}{2}\right)\left(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\right)$
$=2 \mathrm{n} \cdot \sin 45 \times \frac{1}{2}=\mathrm{n} \cdot \frac{1}{\sqrt{2}}=\frac{\mathrm{n}}{\sqrt{2}}$
$\therefore \frac{\mathrm{n}}{\sqrt{2}}>1$ for all positive integers $\mathrm{n} \geq 2$. $\therefore$ Statement 1 is true Statement 2 $n x>1, \forall n \geq 2$
$\Rightarrow \mathrm{n}>\frac{1}{\mathrm{x}}, \forall \mathrm{n} \geq 2$
$\mathrm{x} \in(0, \infty)$, then we take $\mathrm{x}=1$
$\mathrm{n}>1$, but $\mathrm{n}$ is always greater or equal to 2 for all $\mathrm{x}$ positive real number. $\therefore$ Statement 2 is false.

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2014 (Phase 1)
Consider the following statements:
$1.$ $\quad \mathrm{P}(\overline{\mathrm{A}} \cup \mathrm{B})=\mathrm{P}(\overline{\mathrm{A}})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})$
$2.$ $\quad \mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$3.$ $\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{B}) \mathrm{P}(\mathrm{A} \mid \mathrm{B})$
Which of the above statements are correct?

Options:
A 1 and 2 only
B 1 and 3 only
C 2 and 3 only
D 1,2 and 3

The correct answer is:
1 and 3 only

Solution:
The statements (1) and (3) are true.

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 1)
Consider the following statements
1 . Pie diagrams are suitable for categorical data
2 . The arc length of a sector of a pie diagram proportional to the value of the component representec by the sector
Which of the statements given above is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
No Solution

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 2)
Consider the following statements:
1. $\sin |x|+\cos |x|$ is always positive.
2. $\sin \left(x^{2}\right)+\cos \left(x^{2}\right)$ is always positive.
Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
Statement 1 : $\mathrm{f}_{1}(\mathrm{x})=\sin |\mathrm{x}|+\cos |\mathrm{x}|$, the value of $|\sin \mathrm{x}|$
and $|\cos x|$ depends on its angles. $\sin |x|+\cos |x|$ is not always positive. Statement $2: \mathrm{f}_{2}(\mathrm{x})=\sin \left(\mathrm{x}^{2}\right)+\cos \left(\mathrm{x}^{2}\right)$, the value of $\mathrm{x}^{2}$
between any value which lies in the interval $\left(\pi, \frac{3 \pi}{2}\right)$, then value of $\mathrm{f}_{2}(\mathrm{x})=\sin \left(\mathrm{x}^{2}\right)+\cos \left(\mathrm{x}^{2}\right)$ is always
negative.

Topic: Mathematics, Trigonometric Ratios & Identities, NDA, NDA 2014 (Phase 1)
Consider the following statements:
1) $\tan ^{-1} 1+\tan ^{-1}(0.5)=\pi / 2$
2) $\sin ^{-1}(1 / 3)+\cos ^{-1}(1 / 3)=\pi / 2$
Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
1) L.H.S. $\begin{aligned} \tan ^{-1}(1)+\tan ^{-1}\left(\frac{1}{2}\right) \\=\tan ^{-1}(1)+\cot ^{-1}\left(\frac{1}{\frac{1}{2}}\right) &=\tan ^{-1}+\cot ^{-1} 2 \neq \frac{\pi}{2} \\ \text { So, L.H.S. } \neq \mathrm{RHS} \text { . } \\ \text { 2) } \sin ^{-1} \frac{1}{3}+\cos ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2} \\\left\{\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}\right\} \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2014 (Phase 1)
Consider the following statements:
1 . The algebraic sum of deviations of a set of values from their arithmetic mean is always zero.
2 . Arithmetic mean $>$ Median $>$ Mode for a symmetric distribution. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
1 only

Solution:
Only the statement 1 is correct.

Topic: Mathematics, Statistics, NDA, NDA 2019 (Phase 1)
Consider the following statements:
1- The derivative where the function attains maxima or minima be zero.
2- If a function is differentiable at a point, then it must be continuous at that point. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
$1.$ Not necessarily true
$2.$ True $(\because$ Differentiability $\Rightarrow$ Continuity)
#

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2013 (Phase 1)
Consider the following statements:
1) The distance between the lines
$\mathrm{y}=\mathrm{mx}+\mathrm{c}_{1}$ and $\mathrm{y}=\mathrm{mx}+\mathrm{c}_{2}$ is $\frac{\left|\mathrm{c}_{1}-\mathrm{c}_{2}\right|}{\sqrt{1-\mathrm{m}^{2}}}$
2) The distance between the lines $\mathrm{ax}+\mathrm{by}+\mathrm{c}_{1}$ and $\mathrm{ax}+\mathrm{by}+\mathrm{c}_{2}=0$ is $\frac{\left|\mathrm{c}_{1}-\mathrm{c}_{2}\right|}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}$
3) The distance between the lines $\mathrm{x}=\mathrm{c}$ and $\mathrm{x}=\mathrm{c}_{2}$ is $\left|\mathrm{c}_{1}-\mathrm{c}_{2}\right|$. Which of the above statements are correct?

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
2 and 3 only

Solution:
Using distance between two parallel lines formula.

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 2)
Consider the following statements:
1- The function $f(x)=\sqrt[3]{x}$ is continuous at all $x$ except at $x=0$.
2- The function $f(x)=[x]$ is continuous at $x=2.99$ where [.]is the bracket function. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
$\begin{aligned} & \text { LHLf }(2.99-0)=\lim _{h \rightarrow 0}(2.99-h) \\ & \lim _{h \rightarrow 0}(2.99-h)=\lim _{h \rightarrow 0} 2=2 \\ & \text { RHL } f(2.99-0)=\lim _{h \rightarrow 0} f(2.99+h) \\ &=\lim _{h \rightarrow 0}(2.99+h)=\lim _{h \rightarrow 0} 2=2 \\ & \text { LHL }=\text { RHL } \end{aligned}$
$\therefore \mathrm{f}(\mathrm{x})$ is continous at $\mathrm{x}=2.99$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 1)
Consider the following statements:
1- The function $\mathrm{f}(\mathrm{x})=$ greatest integer $\leq \mathrm{x}, \mathrm{x} \in \mathrm{R}$ is a continuous function.
2- All trigonometric functions are continuous on $\mathrm{R}$. Which of the statements given above is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
Here, greatest integer function [x] is discontinuous at its integral value of $\mathrm{x}, \cot \mathrm{x}$ and $\operatorname{cosec} \mathrm{x}$ are discontinuous at $0, \pi, 2 \pi$ etc. and $\tan \mathrm{x}$ and sec $\mathrm{x}$ are discontinuous at
$\mathrm{x}=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}$ etc. Therefore the greatest integer function and all trigonometric functions are not continuous for $\mathrm{x} \in \mathrm{R}$ Therefore, neither (1) nor (2) are true.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2006 (Phase 2)
Consider the following statements:
1- The function $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$.
2- The function $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=\frac{\pi}{2}$. Which of the above statements is/are correct?
A function $\mathrm{f}(\mathrm{x})$ is defined as follows: $\quad[2016-\mathrm{I}]$ $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{lll}\mathrm{x}+\pi & \text { for } & \mathrm{x} \in[-\pi, 0) \\ \pi \cos \mathrm{x} & \text { for } & \mathrm{x} \in\left[0, \frac{\pi}{2}\right] \\ \left(\mathrm{x}-\frac{\pi}{2}\right)^{2} & \text { for } & \mathrm{x} \in\left(\frac{\pi}{2}, \pi\right]\end{array}\right.$

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
Given
$f(x)=\left\{\begin{array}{cc}(x+\pi) & \text { for } x \in[-\pi, 0) \\ \pi \cos x & \text { for } x \in\left[0, \frac{\pi}{2}\right] \\ \left(x-\frac{\pi}{2}\right)^{2} & \text { for } x \in\left(\frac{\pi}{2}, \pi\right]\end{array}\right.$
For continuity, $f(a)=$ L.H.L. $=$ R.H.L.
At $x=0$
$f(0)=\pi \cos 0=\pi$
L.H.L. $=\lim _{x \rightarrow 0^{-}} f(x-h)$
$=\lim _{h \rightarrow 0} f(-h)$
$=\lim _{h \rightarrow 0}(-h+\pi)=\pi$
R.H.L. $=\lim _{x \rightarrow 0^{+}} f(x+h)$
$=\lim _{h \rightarrow 0} f(0+h)$
$=\lim _{h \rightarrow 0} \pi \cos h$
$\quad=\pi \cos 0=\pi$
$f(0)=$ L.H.L. $=$ R.H.L.
Hence function is continuous at $x=0$.
Statement (1) is correct.
At $x=\frac{\pi}{2}$
$\begin{aligned} \text { L.H.L. } &=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x-h) \\ &=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \end{aligned}$
$$
\begin{array}{l}
=\lim _{h \rightarrow 0} \pi \cos \left(\frac{\pi}{2}-h\right) \\
=\pi \cos \frac{\pi}{2}=0 \\
\mathrm{RH} . \mathrm{L} .=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x+h) \\
=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right) \\
=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}+h-\frac{\pi}{2}\right)^{2}=0 \\
f\left(\frac{\pi}{2}\right)=\pi \cos \frac{\pi}{2}=0 \\
\text { L.H.L. }=\mathrm{R} . \mathrm{H.L.}=f\left(\frac{\pi}{2}\right)
\end{array}
$$
Hence function is continuous at $x=\frac{\pi}{2}$
Statement $(2)$ is correct.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2016 (Phase 1)
Consider the following statements:
1- The function $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$.
2- The function $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=\frac{\pi}{2}$. Which of the above statements is/are correct?
A function $\mathrm{f}(\mathrm{x})$ is defined as follows: $\quad[2016-\mathrm{I}]$ $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{lll}\mathrm{x}+\pi & \text { for } & \mathrm{x} \in[-\pi, 0) \\ \pi \cos \mathrm{x} & \text { for } & \mathrm{x} \in\left[0, \frac{\pi}{2}\right] \\ \left(\mathrm{x}-\frac{\pi}{2}\right)^{2} & \text { for } & \mathrm{x} \in\left(\frac{\pi}{2}, \pi\right]\end{array}\right.$

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
For differentiability,
L.H.D. = R.H.D. Thus at $x=0$
$\begin{aligned} \text { L.H.D. } &=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{f(-h)-f(0)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{(-h+\pi)-(\pi \cos 0)}{-h}=1 \\ \text { L.H.D. } &=1 \\ \text { R.H.D. } &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ &=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\pi \cos h-\pi \cos 0}{h} \\ &=\lim _{h \rightarrow 0} \frac{\pi[\cos h-1]}{h} \\ &=\lim _{h \rightarrow 0} \frac{\pi\left[1-\frac{h^{2}}{2 !}+\frac{h^{4}}{4 !} \ldots . .-1\right]}{h}
\\=\lim _{h \rightarrow 0} \frac{\pi\left[1-\frac{1}{2} h^{2}+\frac{1}{24} h^{4} \ldots \ldots-1\right]}{h} \end{aligned}$
$$
=\lim _{h \rightarrow 0} \frac{\pi\left[-\frac{1}{2} h^{2}+\frac{1}{24} h^{4} \ldots \ldots\right]}{h}=0
$$
L.H.D. $\neq$ R.H.D.
So at $x=0$ function is not differentiable.
Statement (1) is not correct.
$$
\begin{array}{l}
\text { At } x=\frac{\pi}{2} \\
f\left(\frac{\pi}{2}\right)=\pi \cos \frac{\pi}{2}=0 \\
\text { R.H.D. }=\lim _{h \rightarrow 0} \frac{f\left(\frac{\pi}{2}+h\right)-f\left(\frac{\pi}{2}\right)}{-h} \\
\text { L.H.D. }=\lim _{h \rightarrow 0} \frac{f\left(\frac{\pi}{2}-h\right)-f\left(\frac{\pi}{2}\right)}{-h}
\end{array}
$$
$$
\begin{array}{l}
=\lim _{h \rightarrow 0} \frac{\pi \cos \left(\frac{\pi}{2}-h\right)-0}{-h} \\
=\lim _{h \rightarrow 0} \pi\left(\frac{\sin h}{-h}\right) \\
=-\pi \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)=-\pi(1)=-\pi \\
\text { L.H.D. } \neq \text { RH.D. }
\end{array}
$$
Hence function is not differentiable at $x=\frac{\pi}{2}$. Statement $(2)$ is not correct.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2016 (Phase 1)
Consider the following statements:
1- The function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+2 \cos \mathrm{x}$ is increasing in the interval $(0, \pi)$
2- The function $\mathrm{f}(\mathrm{x})=\ln \left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)$ is decreasing in
the interval $(-\infty, \infty)$ Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+2 \cos \mathrm{x}$
$\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-2 \sin \mathrm{x}$
$=2[\mathrm{x}-\sin \mathrm{x}]$
Between $(0, \pi) ;(\mathrm{x}-\sin \mathrm{x})$ is always $+\mathrm{ve}$, so $\mathrm{f}^{\prime}(\mathrm{x})$ is always $+\mathrm{ve}$.
Hence it is increasing.
$f(x)=\ln \left(\sqrt{1+x^{2}}-x\right)$
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}} \times\left[\frac{2 \mathrm{x}}{2 \sqrt{\left(1+\mathrm{x}^{2}\right)}}-1\right]$
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}} \times\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}-1\right)$
As $\mathrm{f}^{\prime}(\mathrm{x})$ is $-\mathrm{ve}$ always, so this function is decreasing alwavs.

Topic: Mathematics, Application of Derivatives, NDA, NDA 2015 (Phase 2)
Consider the following statements:
1- The function $f(x)=|x|$ is not differentiable at $x=1$.
2- The function $f(x)=e^{x}$ is not differentiable at $x=0$. Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
Statement $1: \mathrm{f}(\mathrm{x})=\mid \mathrm{x}$
From the graph, the curve has sharp turn at $\mathrm{x}=0$. Therefore, the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is not differentiable only $\mathrm{x}=0$, it is differentiable at $\mathrm{x}=1$ Statement $2: \mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$
$\mathrm{Rf}^{\prime}(0)=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{(0+\mathrm{h})}-\mathrm{e}^{\circ}}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\mathrm{h}}-1}{\mathrm{~h}}$
Use L'Hospital rule
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\mathrm{h}}-0}{1}=\mathrm{e}^{\mathrm{o}}=1$
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{1-\mathrm{e}^{-\mathrm{h}}}{\mathrm{h}}$
Use L' Hospital rule
$=\lim _{h \rightarrow 0} \frac{\mathrm{e}^{-\mathrm{h}}}{1}=\mathrm{e}^{-0}=1$
Therefore $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$ is differentiable at $\mathrm{x}=1$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 1)
Consider the following statements:
1- The function $f(x)=[x]$, where $[.]$ is the greatest integer function defined on $R$, is continuous at all points except at $x=0 . \quad[2014-I I]$
2- The function $f(x)=\sin |x|$ is continuous for all $x \in R$.
Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
The greatest integer function is continuous at all statement points except integer. Hence, statement 1is incorrect. Statement $2:$ Let $h(a)=\sin x$ and $g(x)=|x|$
$\operatorname{hog}(x)=\sin \mid x$
$\Rightarrow \mathrm{f}(\mathrm{x})=\operatorname{hog}(\mathrm{x})=\sin |\mathrm{x}|$
Therefore, $\mathrm{g}(\mathrm{x})$ is continuous, $\forall \mathrm{x} \in \mathrm{R}$ and $\mathrm{h}(\mathrm{x})$ is continous $\forall \mathrm{x} \in \mathrm{R}$
When both are continuous then $\operatorname{hog}(\mathrm{x})$ is also continuous. Thus, statement 2 is correct.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 2)
Consider the following statements :
1- The function is discontinuous at $x=3$.
2- The function is not differentiable at $x=0$. What of the above statements is/are correct ?
Consider the function $f(x)=\left\{\begin{array}{ll}x^{2}-5 & x \leq 3 \\ \sqrt{x+13} & x>3\end{array} \quad[2014-I I]\right.$

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Neither 1 nor 2

Solution:
$1.$ $\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=f(3)$
$\therefore \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=4$
Therefore $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=4$
$2.$ Given $f(x)=x^{2}-5 \forall x \leq 3$
$\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}$
$\mathrm{f}^{\prime}(0)=0$
So, $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=0$ Therefore, neither statement 1 nor 2 is correct.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 2)