datavorous/entrance-exam-dataset huggingface dataset:
Consider the following statements
The algebraic sum of the deviations of a set of $n$ value:
from its arithmetic mean is zere 2 . In the case of frequency distribution, mode is the value of variable which corresponds to maximum frequency. Which of the statements above given is/are correct

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

The correct answer is:
Both 1 and

Solution:
No Solution

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 2)
Consider the following statements The appropriate number of classes while constructing a frequency distribution should be chosen such that
1 . the class-frequency first increases to a peak and then declines.
2 . the class-frequency should cluster around the class mid point Which of the statements given 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 appropriate number of classes while constructing a frequency distribution should be chosen such that the class frequency should cluster around the class mid point

Topic: Mathematics, Statistics, NDA, NDA 2007 (Phase 2)
Consider the following statements There are infinitely many rational numbers between two
distinct
1.integers
2.rational numbers
3.real numbers Which of the statements above are correct

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

The correct answer is:
1,2 and 3

Solution:
There are infinitely many rational numbers between two distinct integers, so, statement 1 is correct. Same is true in case of two distinct rational numbers and real numbers. So, statement $(2)$ and $(3)$ are also correct.

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 2)
Consider the following statements with regard to correlation coefficient $r$ between random variables $x$ and $y$.
I. $\quad r=+1$ or $-1$ means there is a linear relationship between the variables.
II. $-1 \leq r \leq 1$ and $r^{2}$ is a measure of the linear relationship between the variables. Which of the statements given above is/are correct?

Options:
A Only I
B Only II
C Both I and II
D Neither Inor II

The correct answer is:
Both I and II

Solution:
Both the given statements which is related to correlation coefficient $r$ between variables $x$ and $y$ are correct.

Topic: Mathematics, Statistics, NDA, NDA 2011 (Phase 1)
Consider the following three methods of collecting data
(1) collecting data from government offices $\quad$
(2) collecting data from public libraries
(3) collecting data by telephonic interview Select the correct answer using the code given below

Options:
A All the three methods give secondary data
B 1 and 2 give secondary and 3 gives primary data
C 1 and 3 give secondary and 2 gives primary data
D 2 and 3 give secondary and 1 gives primary data

The correct answer is:
1 and 2 give secondary and 3 gives primary data

Solution:
Collecting data from government offices is secondary. collecting data from public libraries is also secondary but collecting data by telephonic interview is primary data.

Topic: Mathematics, Statistics, NDA, NDA 2008 (Phase 2)
Consider the following tissues in the stelar region of a stem showing secondary growth.
(A) Primary xylem
(B) Secondary xylem
(C) Primary phloem
(D) Secondary phloem
Arrange these in the correct sequence of their position from pith towards cortex.

Options:
A $(A),(B),(C),(D)$
B $(B),(A),(D),(C)$
C $(A),(B),(D),(C)$
D $(B),(A),(C),(D)$

The correct answer is:
$(A),(B),(D),(C)$

Solution:
The correct sequence of tissues in the stelar region of the stem showing secondary growth from pith towards cortex is

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2023 (Manipur)
Consider the following values of $\mathrm{x}$ :
1)$8$
2) $-4$
3) $\frac{1}{6}$
4) $-\frac{1}{4}$
Which of the above values of $\mathrm{x}$ is/are the solutions of the
equation $\tan ^{-1}(2 \mathrm{x})+\tan ^{-1}(3 \mathrm{x})=\frac{\pi}{4}$

Options:
A 3 only
B 2 and 3 only
C 1 and 4 only
D 4 only

The correct answer is:
3 only

Solution:
$\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x .3 x}\right)=\frac{\pi}{4}$
$\tan ^{-1} \frac{5 x}{1-6 x^{2}}=\frac{\pi}{4}$
$\frac{5 x}{1-6 x^{2}}=1$
$1-6 x^{2}=5 x$
$6 x^{2}+5 x-1=0$
$(6 x-1)(x+1)=0$
$\Rightarrow \quad x=\frac{1}{6},-1$
$\Rightarrow \quad x=\frac{1}{6}$
Here $x=-1$ rejected.

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2018 (Phase 2)
Consider the following with regard to a relation $R$ on a set of real numbers defined by $x R y$ if and only if $3 x+4 y=5$
I. $\quad 0 R 1$
II. $\quad 1 R \frac{1}{2}$
III. $\frac{2}{3} \mathrm{R} \frac{3}{4}$
Which of the above are correct?

Options:
A $\mathrm{I}$ and $\mathrm{II}$
B I and III
C II and III only
D I, II and III

The correct answer is:
II and III only

Solution:
Let on the setof real numbers, $R$ is a relation defined by $x R y$ if and only if $3 x+4 y=5$ Consider, $3 x+4 y=5$
(I) Put $x=0$ and $y=1$, we get $\mathrm{LHS}=3(0)+4(1)=4 \neq 5(=\mathrm{RHS})$
Hence 0 is not related to 1
(II) Now, Put $x=1$ and $y=\frac{1}{2}$, we get
$\mathrm{LHS}=3(1)+4 \times \frac{1}{2}=5=5(=\mathrm{RHS})$
Hence 1 is related to $\frac{1}{2}$
(III) Similarly, $\frac{2}{3}$ is related to $\frac{3}{4}$. Hence, both statements II and III are correct

Topic: Mathematics, Sets and Relations, NDA, NDA 2011 (Phase 1)
Consider the function $f(x)=\frac{27\left(x^{2 / 3}-x\right)}{4}$
How many solutions does the function $\mathrm{f}(\mathrm{x})=-1$ have?

Options:
A One
B Two
C Three
D Four

The correct answer is:
One

Solution:
Similarly for $f(x)=-1$ we will get $27 y^{3}-27 y^{2}-4=0$ and after solving it we will find that it has one real solution. $y_{0}=1.1184$
$x=\left(\mathrm{y}_{0}\right)^{3}=(1.1184)^{3}=1.4$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 1)
Consider the function $f(x)=\frac{27\left(x^{2 / 3}-x\right)}{4}$
How many solutions does the function $\mathrm{f}(\mathrm{x})=1$ have?

Options:
A One
B Two
C Three
D Four

The correct answer is:
Two

Solution:
For $f(x)=1$
$\frac{27}{4}\left(\left(x^{2}\right)^{1 / 3}-x\right)=1$
$\left(x^{1 / 3}\right)^{2}-x=\frac{4}{27}$
Put $x^{1 / 3}=y$
$x=y^{3}$
$y^{2}-y^{3}=\frac{4}{27}$
$\Rightarrow y^{3}-y^{2}+\frac{4}{27}=0$
$\Rightarrow 27 y^{3}-27 y^{2}+4=0$
This is a cubic equation.
If we put $y=-\frac{1}{3}$ then $(3 y+1)=0$ is a factor of cubic equation.
$3 y+1$ ) $27 y^{3}-27 y^{2}+4\left(9 y^{2}-12 y+4\right.$
$\frac{-27 y^{3}+9 y^{2}}{-36 y^{2}+4}$
$\frac{\mp^{-36 y^{2}} \mp^{12 y}}{12 y+4}$
$\frac{12 y+4}{x}$
$(3 y+1)\left(9 y^{2}-12 y+4\right)=0$
$(3 y+1)(3 y-2)^{2}=0$
Hence $y=-\frac{1}{3}, \frac{2}{3}$
Thus $f(x)=1$ has two solutions.

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 1)
Consider the function
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\frac{\sin 2 \mathrm{x}}{5 \mathrm{x}} \text { if } \mathrm{x} \neq 0 \\ \frac{2}{15} \text { if } \mathrm{x}=0\end{array}\right.$
Which one of the following is correct in respect of the function?

Options:
A It is not continuous at $x=0$
B It is continuous at every $\mathrm{x}$
C It is not continuous at $\mathrm{x}=\pi$
D It is continuous at $\mathrm{x}=0$

The correct answer is:
It is not continuous at $x=0$

Solution:
$\lim _{x \rightarrow 0} \frac{\sin (2 x)}{5 x}=\lim _{x \rightarrow 0} \frac{\sin (2 x)(2)}{5 \cdot(2 x)}=\frac{2}{5}$
But at $x=0, f(x)=\frac{2}{15}$
Hence, $f(x)$ is not continuous at $x=0$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2018 (Phase 2)
Consider the function
$f(x)=\left\{\begin{array}{ccc}a x-2 & \text { for } & -2 < x < -1 \\ -1 & \text { for } & -1 \leq x \leq 1 \\ a+2(x-1)^{2} & \text { for } & 1 < x < 2\end{array}\right.$
What is the value of a for which $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=-1$ $\operatorname{and} \mathrm{x}=1$ ?

Options:
A -1
B 1
C 0
D 2

The correct answer is:
-1

Solution:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{ax}-2 & -2 < \mathrm{x} < -1 \\ -1 & -1 \leq \mathrm{x} \leq 1 \\ \mathrm{a}+2(\mathrm{x}-1)^{2} & 1 < \mathrm{x} < 2\end{array}\right.$
if $f(x)$ is continuous at $x=-1$
then, $\lim _{x \rightarrow-1}(a x-2)=\lim _{x \rightarrow-1}(-1)$
$\Rightarrow a(-1)-2=-1$
$\Rightarrow a=-1$
if $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=1$
then, $\lim _{x \rightarrow 1} a+2(x-1)^{2}=\lim _{x \rightarrow 1}-1$
$\Rightarrow \mathrm{a}+2(1-1)^{2}=-1$
$\Rightarrow \mathrm{a}=-1$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2015 (Phase 2)
Consider the function $\mathrm{f}^{\prime \prime}(\mathrm{x})=\sec ^{4} \times+4$ with $\mathrm{f}(0)=0$ and $\mathrm{f}^{\prime}(0)=0$.
What is $f^{\prime}(x)$ equal to?

Options:
A $\tan x-\frac{\tan ^{3} x}{3}+4 x$
B $\tan x+\frac{\tan ^{3} x}{3}+4 x$
C $\tan x-\frac{\sec ^{3} x}{3}+4 x$
D $-\tan x-\frac{\tan ^{3} x}{3}+4 x$

The correct answer is:
$\tan x+\frac{\tan ^{3} x}{3}+4 x$

Solution:
$\mathrm{f}^{\prime}(\mathrm{x})=\int \mathrm{f}^{\prime \prime}(\mathrm{x}) \mathrm{d} \mathrm{x}+\mathrm{C}_{1}$
$=\int\left(\sec ^{4} \mathrm{x}+4\right) \mathrm{d} \mathrm{x}+\mathrm{C}_{1}$
$=\int \sec ^{2} \mathrm{x} \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}+\int 4 \mathrm{~d} \mathrm{x}+\mathrm{C}_{1}$
$=\int\left(1+\tan ^{2} \mathrm{x}\right) \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}+4 \mathrm{x}+\mathrm{C}_{1}=\mathrm{I}_{1}+4 \mathrm{x}+\mathrm{C}_{1}$
Put $\tan \mathrm{x}=\mathrm{t}$ in the integral $\mathrm{I}_{1}$, then
$\sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt}$
$\therefore \mathrm{I}_{1}=\int\left(1+\mathrm{t}^{2}\right) \mathrm{dt}=\mathrm{t}+\frac{\mathrm{t}^{3}}{3}+\mathrm{C}^{\prime}$
$=\tan \mathrm{x}+\frac{\tan ^{3} \mathrm{x}}{3}+\mathrm{C}^{\prime}$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\tan \mathrm{x}+\frac{\tan ^{3} \mathrm{x}}{3}+4 \mathrm{x}+\mathrm{C}^{\prime}$
where, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}^{\prime}$
$\because \mathrm{f}^{\prime}(0)=0 \Rightarrow \mathrm{C} 0$
Thus, $\mathrm{f}^{\prime}(\mathrm{x})=\tan \mathrm{x}+\frac{\tan ^{3} \mathrm{x}}{3}+4 \mathrm{x}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2014 (Phase 2)
Consider the function $\mathrm{f}^{\prime \prime}(\mathrm{x})=\sec ^{4} \times+4$ with $\mathrm{f}(0)=0$ and $\mathrm{f}^{\prime}(0)=0$.
What is $f(x)$ equal to?

Options:
A $\frac{2 \ln \sec x}{3}+\frac{\tan ^{2} x}{6}+2 x^{2}$
B $\frac{3 \ln \sec x}{2}+\frac{\cot ^{2} x}{6}+2 x^{2}$
C $\frac{4 \ln \sec x}{3}+\frac{\sec ^{2} x}{6}+2 x^{2}$
D $\ln \sec x+\frac{\tan ^{4} x}{12}+2 x^{2}$

The correct answer is:
$\frac{2 \ln \sec x}{3}+\frac{\tan ^{2} x}{6}+2 x^{2}$

Solution:
$\mathrm{f}(\mathrm{x})=\int \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{d} \mathrm{x}+\mathrm{C}_{2}$
$=\int\left(\tan \mathrm{x}+\frac{\tan ^{3} \mathrm{x}}{3}+4 \mathrm{x}\right) \mathrm{d} \mathrm{x}+\mathrm{C}_{2}$
$=\int \tan \mathrm{x} \mathrm{d} \mathrm{x}+\frac{1}{3} \int \tan ^{3} \mathrm{x} \mathrm{dx}+4 \int \mathrm{x} \mathrm{d} \mathrm{x}+\mathrm{C}_{2}$
$=\int \tan \mathrm{x} \mathrm{d} \mathrm{x}+\frac{1}{3} \int \tan \mathrm{x}\left(\sec ^{2} \mathrm{x}-1\right) \mathrm{d} \mathrm{x}+4 \cdot \frac{\mathrm{x}^{2}}{2}+\mathrm{C}_{2}$
$=\frac{2}{3} \int \tan \mathrm{x} \mathrm{d} \mathrm{x}+\frac{1}{3} \int \tan \mathrm{x} \cdot \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}+2 \mathrm{x}^{2}+\mathrm{C}_{2}$
$=\frac{2}{3} \ln (\sec \mathrm{x})+\frac{1}{3} \mathrm{I}_{2}+2 \mathrm{x}^{2}+\mathrm{C}_{2}$
Consider $\mathrm{I}_{2}=\int \tan \mathrm{x} \sec ^{2} \mathrm{x} \mathrm{dx}$
Put $\tan \mathrm{x}=\mathrm{t} \Rightarrow \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt}$
$\Rightarrow 1_{2}=\int \mathrm{t} \mathrm{dt}=\frac{\mathrm{t}^{2}}{2}+\mathrm{C}_{3}=\frac{\tan ^{2} \mathrm{x}}{2}+\mathrm{C}_{3}$
$\therefore \mathrm{f}(\mathrm{x})=\frac{2}{3} \ln (\sec \mathrm{x})+\frac{1}{6} \tan ^{2} \mathrm{x}+2 \mathrm{x}^{2}+\mathrm{C}_{4}$
Here, $\mathrm{C}_{4}=\mathrm{C}_{2}+\left(\frac{\mathrm{C}_{3}}{3}\right)$
$\because \mathrm{f}(0)=0$
$\Rightarrow 0=\frac{2}{3} \ln (1)+0+0+\mathrm{C}_{4}$
$\Rightarrow \mathrm{C}_{4}=0$
$\therefore \mathrm{f}(\mathrm{x})=\frac{2}{3} \ln (\sec \mathrm{x})+\frac{1}{6} \tan ^{2} \mathrm{x}+2 \mathrm{x}^{2}$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2014 (Phase 2)
Consider the function $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl}\mathrm{x}^{2}, & \mathrm{x}>2 \\ 3 \mathrm{x}-2, & \mathrm{x} \leq 2\end{array}\right.$.
Which one of the following statements is correct in respect of the above function?

Options:
A $\mathrm{f}(\mathrm{x})$ is derivable but not continuous at $\mathrm{x}=2$.
B $\mathrm{f}(\mathrm{x})$ is continuous but not derivable at $\mathrm{x}=2$.
C $\mathrm{f}(\mathrm{x})$ is neither continuous nor derivable at $\mathrm{x}=2$.
D $\mathrm{f}(\mathrm{x})$ is continuous as well as derivable at $\mathrm{x}=2$.

The correct answer is:
$\mathrm{f}(\mathrm{x})$ is continuous but not derivable at $\mathrm{x}=2$.

Solution:
First we check continuity at $\mathrm{x}=2$
L. H. $L=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} 3(2-h)-2$
$=\lim _{h \rightarrow 0} 4-3 h=4$
R. H. L. $=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}(2+h)^{2}=4$
Also, $\mathrm{f}(2)=(2)^{2}=4$
Since, $\mathrm{L} . \mathrm{H} \cdot \mathrm{L}=\mathrm{R} . \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(2)$
$\therefore \mathrm{f}(\mathrm{x})$ is continuous at $2 .$
Now, we check for differentiability $\begin{array}{ll}\text { L. H. Dat } x=2 & \text { R. H. Dat } x=2\end{array}$
$\mathrm{f}(\mathrm{x})=3 \mathrm{x}-2 \quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$
$\mathrm{f}^{\prime}(\mathrm{x})=3 \quad \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}$
$\left.\mathrm{f}^{\prime}(\mathrm{x})\right|_{\mathrm{x}=2}=\left.3 \quad \mathrm{f}^{\prime}(\mathrm{x})\right|_{\mathrm{x}=2}=4$
Since L. H. $\mathrm{D} \neq \mathrm{R}$ H. $\mathrm{D}$ $\therefore \mathrm{f}(\mathrm{x})$ is not derivable at $\mathrm{x}=2$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2013 (Phase 1)
Consider the function $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{x}^{2} \ln |\mathrm{x}| & \mathrm{x} \neq 0 \\ 0 & \mathrm{x}=0\end{array}\right.$. What is
$\begin{array}{cc}\mathrm{f}^{\prime}(0) \text { equal to? } &\end{array}$

Options:
A 0
B 1
C $-1$
D It does not exist

The correct answer is:
0

Solution:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{x}^{2} \ell \mathrm{n}|\mathrm{x}| & \mathrm{x} \neq 0 \\ 0 & \mathrm{x}=0 .\end{array}\right.$
$f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\mathrm{f}(\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}^{2} \cdot \ln |\mathrm{h}|-0}{\mathrm{~h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} . \ell \mathrm{n}|\mathrm{h}|$
$=0$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2018 (Phase 1)
Consider the function
$\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\mathrm{x}^{3} & \sin \mathrm{x} & \cos \mathrm{x} \\ 6 & -1 & 0 \\ \mathrm{p} & \mathrm{p}^{2} & \mathrm{p}^{3}\end{array}\right|$ where $\mathrm{p}$ is a constant.
What is the value of $\mathrm{f}^{\prime}(0) ?$

Options:
A $\mathrm{p}^{3}$
B $3 \mathrm{p}^{3}$
C $6 \mathrm{p}^{3}$
D $-6 \mathrm{p}^{3}$

The correct answer is:
$-6 \mathrm{p}^{3}$

Solution:
$f(x)=\left|\begin{array}{ccc}x^{3} & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right|$
$f^{\prime}(x)=\left|\begin{array}{ccc}3 x^{2} & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right|+\left|\begin{array}{ccc}x^{3} & \sin x & \cos x \\ 6 & -1 & 0 \\ 0 & 0 & 0\end{array}\right|$ $+\left|\begin{array}{ccc}x^{3} & \sin x & \cos x \\ 0 & 0 & 0 \\ p & p^{2} & p^{3}\end{array}\right|$
$\begin{aligned} f^{\prime}(x) &=\left|\begin{array}{ccc}3 x^{2} & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right| \\ f^{\prime}(0) &=\left|\begin{array}{ccc}0 & \cos 0 & -\sin 0 \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right| \\ &=-6 p^{3} \end{aligned}$

Topic: Mathematics, Determinants, NDA, NDA 2016 (Phase 1)
Consider the function
$\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\mathrm{x}^{3} & \sin \mathrm{x} & \cos \mathrm{x} \\ 6 & -1 & 0 \\ \mathrm{p} & \mathrm{p}^{2} & \mathrm{p}^{3}\end{array}\right|$ where $\mathrm{p}$ is a constant.
What is the value of p for which $\mathrm{f}^{\prime \prime}(0)=0$ ?

Options:
A $-\frac{1}{6}$ or 0
B $-1$ or 0
C $-\frac{1}{6}$ or 1
D $-1$ or 1

The correct answer is:
$-\frac{1}{6}$ or 0

Solution:
$f^{\prime \prime}(x)=\left|\begin{array}{ccc}6 x & -\sin x & -\cos x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right|$
$f^{\prime \prime}(0)=\left|\begin{array}{ccc}0 & 0 & -\cos 0 \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right|$
$\quad=-1\left(6 p^{2}+p\right)$
$f^{\prime \prime}(0)=0$
$-\left(6 p^{2}+p\right)=0$
$p(6 p+1)=0$
$p=0$ or $p=-\frac{1}{6}$

Topic: Mathematics, Determinants, NDA, NDA 2016 (Phase 1)
Consider the function
$$
f(x)=\left\{\begin{array}{cc}
\frac{\tan k x}{x}, & x < 0 \\
3 x+2 k^{2}, & x \geq 0
\end{array}\right.
$$
What is the non-zero value of $k$ for which the function is continuous at $x=0$ ?

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

The correct answer is:
$1 / 2$

Solution:
Given, $f(x)=\left\{\begin{array}{ll}\frac{\tan \mathrm{kx}}{x}, & x < 0 \\ 3 x+2 \mathrm{k}^{2}, & x \geq 0\end{array}\right.$
When function is continuous at $\mathrm{x}=0$, then
$\lim _{x-0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\therefore \lim _{x \rightarrow 0-h}\left(\frac{\tan k x}{x}\right)=\lim _{x \rightarrow 0+h}\left(3 x+2 k^{2}\right)=3(0)+2 k^{2}$
$\Rightarrow \lim _{x \rightarrow 0-h}\left[\frac{\tan k(0-h)}{(0-h)}\right]=\lim _{x \rightarrow 0+h}\left[3(0+h)+2 k^{2}\right]=2 k^{2}$
Therefore, $\lim _{h \rightarrow 0}\left(\frac{\tan \mathrm{kh}}{\mathrm{h}}\right)=2 \mathrm{k}^{2}$
$\mathrm{k}=2 \mathrm{k}^{2}$
$\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right]$
Hence, $\mathrm{k}=\frac{1}{2}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2014 (Phase 2)
Consider the functions
$\mathrm{f}(\mathrm{x})=\mathrm{xg}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})=\left[\frac{1}{\mathrm{x}}\right]$
Where $[\cdot]$ is the greatest integer function.
What is $\int_{\frac{1}{3}}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ equal to ?

Options:
A $\frac{37}{72}$
B $\frac{2}{3}$
C $\frac{17}{72}$
D $\frac{37}{144}$

The correct answer is:
$\frac{37}{72}$

Solution:
$\int_{1 / 3}^{1} f(x) d x=\int_{1 / 3}^{1 / 2} f(x) d x+\int_{1 / 2}^{1} f(x) d x$
$f(x)=x g(x)$
$g\left(\frac{1}{3}\right)=3 \quad g\left(\frac{1}{2}\right)=2 g(1)=1$
The value of $g(x)$ in value $\left(\frac{1}{2}, \frac{1}{3}\right)$ will be 2 and in range $\left(\frac{1}{2}, 1\right)$ it will be 1 form(1)
$\int_{1 / 3}^{1} f(x) d x=\int_{1 / 3}^{1 / 2} x g(x) d x+\int_{1 / 2}^{1} x g(x) d x$
$=\int_{1 / 3}^{1 / 2} x \cdot 2 d x+\int_{1 / 2}^{1} x \times 1 d x .$
$=\left[x^{2}\right]_{1 / 3}^{1 / 2}+\left[\frac{x^{2}}{2}\right]_{1 / 2}^{1}$
$=\left[\frac{1}{4}-\frac{1}{9}\right]+\frac{1}{2}\left[1-\frac{1}{4}\right]$
$=\frac{5}{36}+\frac{3}{8}=\frac{37}{72}$

Topic: Mathematics, Definite Integration, NDA, NDA 2016 (Phase 1)
Consider the functions
$\mathrm{f}(\mathrm{x})=\mathrm{xg}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})=\left[\frac{1}{\mathrm{x}}\right]$
Where $[\cdot]$ is the greatest integer function.
What is $\int_{\frac{1}{3}}^{\frac{1}{2}} \mathrm{~g}(\mathrm{x}) \mathrm{dx}$ equal to?

Options:
A $\frac{1}{6}$
B $\frac{1}{3}$
C $\frac{5}{18}$
D $\frac{5}{36}$

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

Solution:
$\int_{1 / 3}^{1 / 2} \mathrm{~g}(x) \mathrm{d} x$
$$
\mathrm{~g}(\mathrm{x})=\left\{\begin{array}{ll}2, & \text { if } \frac{1}{3} < x \leq \frac{1}{2} \\ 3, & \text { if } x=\frac{1}{3}\end{array}\right.
$$
As $\mathrm{g}(\mathrm{x})$ is a gretest integer function so value of $\mathrm{g}(\mathrm{x})$ in integral limit will be
So $\int_{1 / 3}^{1 / 2} g(x) d x=\int_{1 / 3}^{1 / 2} 2 d x$
$=2[x]_{1 / 3}^{1 / 2}=2\left[\frac{1}{2}-\frac{1}{3}\right]=\frac{1}{3}$

Topic: Mathematics, Definite Integration, NDA, NDA 2016 (Phase 1)
Consider the integral $I=\int_{0}^{\pi} \ln (\sin x) d x$
What is $\int_{0}^{\pi / 2} \ln (\cos x) d x$ equal to ?

Options:
A $\mathrm{I} / 2$
B $I$
C $2 I$
D $4 I$

The correct answer is:
$\mathrm{I} / 2$

Solution:
From eq. (ii), we have
$\mathrm{I}=2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \mathrm{x}) \mathrm{d} \mathrm{x}$
$\Rightarrow \int_{0}^{\frac{\pi}{2}} \ln (\cos x) \mathrm{d} x=\frac{1}{2} I$
\begin{array}{l}
\text { Consider } \mathrm{I}=\int_{0}^{\pi} \ln (\sin \mathrm{x}) \mathrm{d} \mathrm{x}\\
\mathrm{I}=\int_{0}^{\pi} \ln (\sin \mathrm{x}) \mathrm{d} \mathrm{x}\\
=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin x) \mathrm{d} x\\
\left[\because \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x, \text { if } f(2 a-x)=f(x)\right]\\
=2 \int_{0}^{\frac{\pi}{2}} \ln \left[\sin \left(\frac{\pi}{2}-\mathrm{x}\right)\right] \mathrm{d} \mathrm{x}\\
\left(\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)\\
=2 \int_{0}^{\frac{\pi}{2}} \ln (\cos x) \mathrm{d} x
\end{array}

Topic: Mathematics, Definite Integration, NDA, NDA 2014 (Phase 2)
Consider the integral $I=\int_{0}^{\pi} \ln (\sin x) d x$
What is $\int_{0}^{\pi / 2} \ln (\sin x) d x$ equal to ?

Options:
A $4 I$
B $2 I$
C $I$
D $I / 2$

The correct answer is:
$I / 2$

Solution:
(d) From eq. (i),
$\mathrm{I}=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin \mathrm{x}) \mathrm{d} \mathrm{x}$
$\Rightarrow \int_{0}^{\frac{\pi}{2}} \ln (\sin \mathrm{x}) \mathrm{d} \mathrm{x}=\frac{1}{2} \mathrm{I}$
Consider $\mathrm{I}=\int_{0}^{\pi} \ln (\sin \mathrm{x}) \mathrm{d} \mathrm{x}$
$\mathrm{I}=\int_{0}^{\pi} \ln (\sin \mathrm{x}) \mathrm{d} \mathrm{x}$
$=2 \int_{0}^{\frac{\pi}{2}} \ln (\sin x) \mathrm{d} x$
$\left[\because \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x\right.$, if $\left.f(2 a-x)=f(x)\right]$
$=2 \int_{0}^{\frac{\pi}{2}} \ln \left[\sin \left(\frac{\pi}{2}-\mathrm{x}\right)\right] \mathrm{d} \mathrm{x}$
$\left(\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$
$=2 \int_{0}^{\frac{\pi}{2}} \ln (\cos x) \mathrm{d} x$

Topic: Mathematics, Definite Integration, NDA, NDA 2014 (Phase 2)
Consider the integral $I_{m}=\int_{0}^{\pi \sin 2 m x} \frac{\sin x} d x$, where $m$ is a positive
integer.
Consider the following:
1 . $I_{m}-I_{m-1}$ is equal to 0 .
2 . $\quad I_{2 m}>I_{m}$
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:
1 only

Solution:
$\mathrm{I}_{2 \mathrm{~m}}>\mathrm{I}_{\mathrm{m}}$ is wrong statement
Because, $\mathrm{I}_{\mathrm{m}}=\mathrm{I}_{\mathrm{m}-1}=\ldots . .=\mathrm{I}_{\mathrm{n}}$
Thus, $\mathrm{I}_{\mathrm{m}}-\mathrm{I}_{\mathrm{m}-1}=0$ is the only correct statement.

Topic: Mathematics, Definite Integration, NDA, NDA 2015 (Phase 1)
Consider the integral $I_{m}=\int_{0}^{\pi \sin 2 m x} \frac{\sin x} d x$, where $m$ is a positive
integer.
What is $I_{1}$ equal to?

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

The correct answer is:
0

Solution:
$\begin{array}{ll} & I_{m}=\int_{0}^{\pi} \frac{\sin 2 m x}{\sin x} d x \\ \therefore \quad & I_{1}=\int_{0}^{\pi} \frac{\sin 2 x}{\sin x} d x=\int_{0}^{\pi} \frac{2 \sin x \cos x}{\sin x} d x \\ & =2 \int_{0}^{\pi} \cos x d x=2[\sin x]_{0}^{\pi}=2[\sin \pi-\sin 0] \\ & =2(0)=0\end{array}$

Topic: Mathematics, Definite Integration, NDA, NDA 2015 (Phase 1)