datavorous/entrance-exam-dataset huggingface dataset:
Consider $I=\int_{0}^{\pi} \frac{x d x}{1+\sin x}$
What is $\int_{0}^{\pi} \frac{d x}{1+\sin x}$ equal to ?

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

The correct answer is:
2

Solution:
From eq. (iii).
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}$
$\Rightarrow \int_{0}^{\pi} \frac{\mathrm{dx}}{1+\sin x}=\frac{2}{\pi} \mathrm{I}$
$\Rightarrow \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}=\frac{2}{\pi} \times \pi=2 \quad(\therefore \mathrm{I}=\pi)$
Given, $\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin (\pi-\mathrm{x})} \mathrm{dx}$
$\left[\because \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}\right]$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin \mathrm{x}} \mathrm{dx}$
$[\because \sin (\pi-\mathrm{x})=\sin \mathrm{x}]$
Adding eqs. (i) and (ii), we get
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}$
$\begin{aligned} \Rightarrow 2 \mathrm{I}=2 \pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1+\sin \mathrm{x}} \\ &\left[\because \int_{0}^{2 \mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=2 \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}, \text { if } \mathrm{f}(2 \mathrm{a}-\mathrm{x})=\mathrm{f}(\mathrm{x})\right] \end{aligned}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{d} \mathrm{x}}{1+\left(\frac{2 \tan \frac{\mathrm{x}}{2}}{1+\tan ^{2} \frac{\mathrm{x}}{2}}\right)}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{\mathrm{x}}{2} \mathrm{dx}}{\tan ^{2} \frac{\mathrm{x}}{2}+1+2 \tan \frac{\mathrm{x}}{2}}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\left(\sec ^{2} \frac{\mathrm{x}}{2}\right) \mathrm{dx}}{\left(\tan \frac{\mathrm{x}}{2}+1\right)^{2}}$
Let $\tan \frac{x}{2}+1=t$
$\Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} \mathrm{~d} x=\mathrm{dt}$
$\Rightarrow \sec ^{2} \frac{x}{2} \mathrm{dx}=2 \mathrm{dt}$
When $\mathrm{x}=0$, then $\mathrm{t}=1$ and when $\mathrm{x}=\frac{\pi}{2}$, then $\mathrm{t}=2$
$\therefore \mathrm{I}=2 \pi \int_{1}^{2} \frac{\mathrm{dt}}{\mathrm{t}^{2}}=2 \pi\left[\frac{\mathrm{t}^{-2+1}}{-2+1}\right]_{1}^{2}=-2 \pi\left[\frac{1}{\mathrm{t}}\right]_{1}^{2}$
$=-2 \pi\left[\frac{1}{2}-1\right]$
$=-2 \pi\left(-\frac{1}{2}\right)=\pi$

Topic: Mathematics, Definite Integration, NDA, NDA 2014 (Phase 2)
Consider $I=\int_{0}^{\pi} \frac{x d x}{1+\sin x}$
What is $\int_{0}^{\pi} \frac{(\pi-x) d x}{1+\sin x}$ equal to?

Options:
A 0
B $\pi$
C $\pi 2$
D $2 \pi$

The correct answer is:
0

Solution:
Let $\mathrm{I}_{1}=\int_{0}^{\pi} \frac{(\pi-\mathrm{x}) \mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}$
$=\int_{0}^{\pi} \frac{[\pi-(\pi-\mathrm{x})] \mathrm{d} \mathrm{x}}{1+\sin (\pi-\mathrm{x})}$
$\left[\because \int_{0}^{a} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}\right]$
$\mathrm{I}_{1}=\int_{0}^{\pi} \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}=\pi \quad[\because \sin (\pi-\mathrm{x})=\sin \mathrm{x}]$
Given, $\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin (\pi-\mathrm{x})} \mathrm{dx}$
$\left[\because \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}\right]$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin \mathrm{x}} \mathrm{dx}$
$[\because \sin (\pi-\mathrm{x})=\sin \mathrm{x}]$
Adding eqs. (i) and (ii), we get
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}$
$\begin{aligned} \Rightarrow 2 \mathrm{I}=2 \pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1+\sin \mathrm{x}} \\ &\left[\because \int_{0}^{2 \mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=2 \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}, \text { if } \mathrm{f}(2 \mathrm{a}-\mathrm{x})=\mathrm{f}(\mathrm{x})\right] \end{aligned}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{d} \mathrm{x}}{1+\left(\frac{2 \tan \frac{\mathrm{x}}{2}}{1+\tan ^{2} \frac{\mathrm{x}}{2}}\right)}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{\mathrm{x}}{2} \mathrm{dx}}{\tan ^{2} \frac{\mathrm{x}}{2}+1+2 \tan \frac{\mathrm{x}}{2}}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\left(\sec ^{2} \frac{\mathrm{x}}{2}\right) \mathrm{dx}}{\left(\tan \frac{\mathrm{x}}{2}+1\right)^{2}}$
Let $\tan \frac{x}{2}+1=t$
$\Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} \mathrm{~d} x=\mathrm{dt}$
$\Rightarrow \sec ^{2} \frac{x}{2} \mathrm{dx}=2 \mathrm{dt}$
When $\mathrm{x}=0$, then $\mathrm{t}=1$ and when $\mathrm{x}=\frac{\pi}{2}$, then $\mathrm{t}=2$
$\therefore \mathrm{I}=2 \pi \int_{1}^{2} \frac{\mathrm{dt}}{\mathrm{t}^{2}}=2 \pi\left[\frac{\mathrm{t}^{-2+1}}{-2+1}\right]_{1}^{2}=-2 \pi\left[\frac{1}{\mathrm{t}}\right]_{1}^{2}$
$=-2 \pi\left[\frac{1}{2}-1\right]$
$=-2 \pi\left(-\frac{1}{2}\right)=\pi$

Topic: Mathematics, Definite Integration, NDA, NDA 2014 (Phase 2)
Consider $I=\int_{0}^{\pi} \frac{x d x}{1+\sin x}$
What is $I$ equal to ?

Options:
A $-\pi$
B 0
C $\pi$
D $2 \pi$

The correct answer is:
$\pi$

Solution:
According to the explanation, $\mathrm{I}=\pi$
Given, $\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin (\pi-\mathrm{x})} \mathrm{dx}$
$\left[\because \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}\right]$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin \mathrm{x}} \mathrm{dx}$
$[\because \sin (\pi-\mathrm{x})=\sin \mathrm{x}]$
Adding eqs. (i) and (ii), we get
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}$
$\begin{aligned} \Rightarrow 2 \mathrm{I}=2 \pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1+\sin \mathrm{x}} \\ &\left[\because \int_{0}^{2 \mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=2 \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}, \text { if } \mathrm{f}(2 \mathrm{a}-\mathrm{x})=\mathrm{f}(\mathrm{x})\right] \end{aligned}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{d} \mathrm{x}}{1+\left(\frac{2 \tan \frac{\mathrm{x}}{2}}{1+\tan ^{2} \frac{\mathrm{x}}{2}}\right)}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{\mathrm{x}}{2} \mathrm{dx}}{\tan ^{2} \frac{\mathrm{x}}{2}+1+2 \tan \frac{\mathrm{x}}{2}}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\left(\sec ^{2} \frac{\mathrm{x}}{2}\right) \mathrm{dx}}{\left(\tan \frac{\mathrm{x}}{2}+1\right)^{2}}$
Let $\tan \frac{x}{2}+1=t$
$\Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} \mathrm{~d} x=\mathrm{dt}$
$\Rightarrow \sec ^{2} \frac{x}{2} \mathrm{dx}=2 \mathrm{dt}$
When $\mathrm{x}=0$, then $\mathrm{t}=1$ and when $\mathrm{x}=\frac{\pi}{2}$, then $\mathrm{t}=2$
$\therefore \mathrm{I}=2 \pi \int_{1}^{2} \frac{\mathrm{dt}}{\mathrm{t}^{2}}=2 \pi\left[\frac{\mathrm{t}^{-2+1}}{-2+1}\right]_{1}^{2}=-2 \pi\left[\frac{1}{\mathrm{t}}\right]_{1}^{2}$
$=-2 \pi\left[\frac{1}{2}-1\right]$
$=-2 \pi\left(-\frac{1}{2}\right)=\pi$

Topic: Mathematics, Definite Integration, NDA, NDA 2014 (Phase 2)
Consider the circles $x^{2}+y^{2}+2 a x+c=0$ and $x^{2}+y^{2}+2 b y+c=0$
The two circles touch each other if

Options:
A $c=\sqrt{a^{2}+b^{2}} \quad$
B $\frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
C $\quad c=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
D $c=\frac{1}{a^{2}+b^{2}}$

The correct answer is:
$\frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

Solution:
Two circles touch each other, iff distance between two centres $=$ Sum of radius of two circles
$\sqrt{a^{2}+b^{2}}=\sqrt{a^{2}-c}+\sqrt{b^{2}-c}$
On squaring both sides, we get
$a^{2}+b^{2}=a^{2}-c+b^{2}-c+2 \sqrt{\left(a^{2}-c\right)\left(b^{2}-c\right)}$
$\Rightarrow \mathrm{c}=\sqrt{\left(\mathrm{a}^{2}-\mathrm{c}\right)\left(\mathrm{b}^{2}-\mathrm{c}\right)}$
Again, squaring both sides, we get
$$
\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~b}^{2}-\mathrm{a}^{2} \mathrm{c}-\mathrm{b}^{2} \mathrm{c}+\mathrm{c}^{2}
$$
$\Rightarrow a^{2} b^{2}=\left(a^{2}+b^{2}\right) c \Rightarrow \frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

Topic: Mathematics, Circle, NDA, NDA 2014 (Phase 2)
Consider the circles $x^{2}+y^{2}+2 a x+c=0$ and $x^{2}+y^{2}+2 b y+c=0$
What is the distance between the centres of the two circles?

Options:
A $\sqrt{a^{2}+b^{2}}$
B $a^{2}+b^{2}$
C $a+b$
D $2(a+b)$

The correct answer is:
$\sqrt{a^{2}+b^{2}}$

Solution:
Equations of circles are $x^{2}+y^{2}+2 a x+c=0$
and $x^{2}+y^{2}+2 b y+c=0$
Since, the centres of two circles are $(-a, 0)$ and $(0,-b)$
- Distance between two centres $=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$

Topic: Mathematics, Circle, NDA, NDA 2014 (Phase 2)
Consider the curve $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$.
What is $\frac{d^{2} y}{d x^{2}}$ equal to?

Options:
A $\sec ^{2} \theta$
B $-\operatorname{cosec}^{2} \theta$
C $\frac{\sec ^{3} \theta}{a \theta}$
D None of these

The correct answer is:
$\frac{\sec ^{3} \theta}{a \theta}$

Solution:
Given $\mathrm{x}=\mathrm{a}(\cos \theta+\theta \sin \theta)$
$\mathrm{y}=\mathrm{a}(\sin \theta-\theta \cos \theta)$ and we have $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \theta$
According to question $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=\sec ^{2} \theta \frac{\mathrm{d} \theta}{\mathrm{dx}}$
$=\sec ^{2} \theta\left(\frac{1}{a \theta \cos \theta}\right)$ $\quad \left[\because \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a} \theta \cos \theta\right]$
Hence, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\sec ^{3} \theta}{\mathrm{a} \theta}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2014 (Phase 2)
Consider the curve $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$.
What is $\frac{d y}{d x}$ equal to?

Options:
A $\tan \theta$
B $\cot \theta$
C $\sin 2 \theta$
D $\cos 2 \theta$

The correct answer is:
$\tan \theta$

Solution:
Given, $\mathrm{x}=\mathrm{a}(\cos \theta+\theta \sin \theta)$ and
$y=a(\sin \theta-\theta \cos \theta)$
$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(-\sin \theta+\theta \cos \theta+\sin \theta)=\mathrm{a} \theta \cos \theta$
$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(\cos \theta+\theta \sin \theta-\cos \theta)=\mathrm{a} \theta \sin \theta$
Now $\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{a} \theta \sin \theta}{\mathrm{a} \theta \cos \theta}=\tan \theta$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2014 (Phase 2)
Consider the curves $y=\sin x$ and $y=\cos x$
What is the area of the region bounded by the above two
curves and the lines $\mathrm{x}=0$ and $x=\frac{\pi}{4}$ ?

Options:
A $\sqrt{2}-1$.
B $\sqrt{2}+1$
C $\sqrt{2}$
D 2

The correct answer is:
$\sqrt{2}-1$.

Solution:

Topic: Mathematics, Area Under Curves, NDA, NDA 2015 (Phase 1)
Consider the curves $y=\sin x$ and $y=\cos x$
What is the area of the region bounded by the above two
curves and the lines $x=\frac{\pi}{4}$ and $x=\frac{\pi}{2} ?$

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

The correct answer is:
$\sqrt{2}-1$

Solution:
No Solution

Topic: Mathematics, Area Under Curves, NDA, NDA 2015 (Phase 1)
Consider the curves $\mathrm{y}=|\mathrm{x}-1|$ and $|\mathrm{x}|=2$
What is the area of the region bounded by the curves and $\mathrm{x}-$ axis?

Options:
A 3 square units
B 4 square units
C 5 square units
D 6 square units

The correct answer is:
5 square units

Solution:
Bounded region is shaded So area of bounded region has two triangles $A C B$ and $B D E$
Area of $\Delta A C B=\frac{1}{2} \times 3 \times 3=\frac{9}{2}$
Area of $\Delta B D E=\frac{1}{2} \times 1 \times 1=\frac{1}{2}$
Area of region bounded by curves and $\mathrm{x}$ -axis is
Area $=\Delta A C B+\Delta B D E=\frac{9}{2}+\frac{1}{2}=5$ square units

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
Consider the curves $\mathrm{y}=|\mathrm{x}-1|$ and $|\mathrm{x}|=2$
What is/are the point(s) of intersection of the curves?

Options:
A $(-2,3)$ only
B $(2,1)$ only
C $(-2,3)$ and $(2,1)$
D Neither $(-2,3)$ nor $(2,1)$

The correct answer is:
$(-2,3)$ and $(2,1)$

Solution:
$y=|x-1|$ and $|x|=2$
$y=\left\{\begin{array}{ll}x-1 & x \geq 1 \\ 1-x & x < 1\end{array}\right.$
and $x=2$ $x=-2$
Hence curves intersect at $(-2,3)$ and $(2,1)$.

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
Consider the diagonals of a quadrilateral formed by the vectors $3 \hat{i}+6 \hat{j}-2 \hat{k}$ and $4 \hat{i}-\hat{j}+3 \hat{k}$. The quadrilateral must be a

Options:
A Square
B Rhombus
C Rectangle
D None of these

The correct answer is:
Rhombus

Solution:
Let $\vec{d}_{1}$ and $\vec{d}_{2}$ be the two diagonals of a quadrilateral such that
$\vec{d}_{1}=3 \hat{i}+6 \hat{j}-2 \hat{k}$
and $\vec{d}_{2}=4 \hat{i}-\hat{j}+3 \hat{k}$
Now, Dot product of $\vec{d}_{1}$ and $\vec{d}_{2}$ is
$\vec{d}_{1} \cdot \vec{d}_{2}=3(4)+6(-1)-2(3)=0$
Now, $\left|\vec{d}_{1}\right|=\sqrt{3^{2}+6^{2}+2^{2}}=7$
$\left|\vec{d}_{2}\right|=\sqrt{4^{2}+1^{2}+3^{2}}=\sqrt{26}$
Since, $\left|\vec{d}_{1}\right| \neq\left|\vec{d}_{2}\right|$
Hence, given quadrilateral is a rhombus,

Topic: Mathematics, Vector Algebra, NDA, NDA 2009 (Phase 2)
Consider the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(b>a)$. Then, which one ofthe following is correct?

Options:
A Real foci donot exist
B Foci are $(\pm a e, 0)$
C Foci are $(\pm b e, 0)$
D Foci are $(0, \pm b e)$

The correct answer is:
Foci are $(0, \pm b e)$

Solution:
Given equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Since, $b>a$
Foci $=(0, \pm \mathrm{b})$

Topic: Mathematics, Ellipse, NDA, NDA 2008 (Phase 2)
Consider the equation $(x-p)(x-6)+1=0$ having integral coefficients. If the equation has integral roots, then what values can p have?

Options:
A 4 or 8
B 5 or 10
C 6 or 12
D 3 or 6

The correct answer is:
4 or 8

Solution:
Given equation is $(x-p)(x-6)+1=0$ $\Rightarrow x^{2}-6 x-p x+6 p+1=0$
$\Rightarrow x^{2}-(p+6) x+(6 p+1)=0$
Now, $b^{2}-4 a c=0$
$a=\mathrm{L}, b=-(p+6), c=6 p+1$
$\Rightarrow(p+6)^{2}-4(6 p+1)=0$
$\Rightarrow p^{2}+36+12 p-24 p-4=0$
$\Rightarrow p^{2}-12 p+32=0$
$\Rightarrow(p-4)(p-8)=0$
$\Rightarrow p=4,8$
Hence, $p$ can have 4 or 8 .

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 1)
Consider the expansion \(\left(x^{2}+\frac{1}{x}\right)^{15}\). What is the sum of the coefficients of the middle terms in the given expansion?

Options:
A $\mathrm{C}(15,9)$
B $\mathrm{C}(16,9)$
C $\mathrm{C}(16,8)$
D None of these

The correct answer is:
$\mathrm{C}(16,8)$

Solution:
Given $\left(x^{2}+\frac{1}{x}\right)^{15}$
Since, $n$ is odd
So, it has two middle terms $\mathrm{T}_{8}$ and $\mathrm{T}_{9}$.
$$
\begin{aligned}
\therefore \mathrm{T}_{8}+\mathrm{T}_{9}={ }^{15} \mathrm{C}_{7}+{ }^{15} \mathrm{C}_{8}={ }^{16} \mathrm{C}_{8} \\
&\left(\because{ }^{n} \mathrm{C}_{\mathrm{r}-1}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right)
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2015 (Phase 1)
Consider the expansion of $(1+\mathrm{x})^{2 \mathrm{n}+1}$
If the coefficients of $\mathrm{x}^{\mathrm{r}}$ and $\mathrm{x}^{\mathrm{r}+1}$ are equal in the expansion, then $\mathrm{r}$ is equal to

Options:
A $\mathrm{n}$
B $\frac{2 \mathrm{n}-1}{2}$
C $\frac{2 \mathrm{n}+1}{2}$
D $\mathrm{n}+1$

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

Solution:
$(1+\mathrm{x})^{2 \mathrm{n}+1}={ }^{(2 \mathrm{n}+1)} \mathrm{C}_{0} \mathrm{x}^{0}+{ }^{(2 \mathrm{n}+1)} \mathrm{C}_{1} \mathrm{x}^{1}$
$+\ldots+{ }^{(2 n+1)} C_{2 n+1}(x)^{2 n+1}$
Coefficient of $\mathrm{x}^{\mathrm{r}}=(2 \mathrm{n}+1) \mathrm{Cr}$
Coefficient of $\mathrm{x}^{\mathrm{r}+1}=(2 \mathrm{n}+1) \mathrm{Cr}+1$
$(2 \mathrm{n}+1) \mathrm{C}_{\mathrm{r}}$=$(2 n+1) C r+1$
$\Rightarrow \frac{(2 \mathrm{n}+1) !}{\mathrm{r} !(2 \mathrm{n}+1-\mathrm{r}) !}=\frac{(2 \mathrm{n}+1) !}{(\mathrm{r}+1) !(2 \mathrm{n}-\mathrm{r}) !}$
$\Rightarrow \frac{(2 \mathrm{n}-\mathrm{r}) !}{(2 \mathrm{n}+1-\mathrm{r})(2 \mathrm{n}-\mathrm{r}) !}=\frac{\mathrm{r} !}{(\mathrm{r}+1) \mathrm{r} !}$
$\Rightarrow(\mathrm{r}+1)=2 \mathrm{n}+1-\mathrm{r}$
$\Rightarrow \mathrm{r}=\mathrm{n}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2015 (Phase 2)
Consider the expansion of $(1+\mathrm{x})^{2 \mathrm{n}+1}$
The average of the coefficients of the two middle terms in the expansion is

Options:
A ${ }^{2 \mathrm{n}+1} \mathrm{C}_{\mathrm{n}+}$
B ${ }^{2 n+1} C_{n}$
C ${ }^{2 n+1} C_{n-1}{ }^{1+2}$
D ${ }^{2 n} C_{n+1}$

The correct answer is:
${ }^{2 n+1} C_{n}$

Solution:
Total no. of terms in the expansion is $2 \mathrm{n}+2$. The middle two terms will be $\mathrm{n}^{\text {th }},(\mathrm{n}+1)^{\text {th }}$ term. So.
Average $=\frac{(2 \mathrm{n}+1)}{\mathrm{C}_{\mathrm{n}}+{ }^{(2 \mathrm{n}+1)} \mathrm{C}_{\mathrm{n}+1}}{2}$
$=\left[\frac{(2 \mathrm{n}+1) !}{\mathrm{n} !(\mathrm{n}+1) !}+\frac{(2 \mathrm{n}+1) !}{(\mathrm{n}+1) ! \mathrm{n} !}\right] / 2$
$=\frac{(2 \mathrm{n}+1) !}{\mathrm{n} !(\mathrm{n}+1) !}={ }^{(2 \mathrm{n}+1)} \mathrm{C}_{\mathrm{n}}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2015 (Phase 2)
Consider the expansion of $(1+\mathrm{x})^{2 \mathrm{n}+1}$
The sum of the coefficients of all the terms in the expansion is

Options:
A $2^{2 n-1}$
B $4^{\mathrm{n}-1}$
C $2 \times 4^{\text {n }}$
D None of the above

The correct answer is:
$2 \times 4^{\text {n }}$

Solution:
Sum of all coefficient
$={ }^{(2 \mathrm{n}+1)} \mathrm{C}_{0}+{ }^{(2 \mathrm{n}+1)} \mathrm{C}_{1}+\ldots \ldots .+{ }^{(2 \mathrm{n}+1)} \mathrm{C}_{2 \mathrm{n}+1}$
$=(1+1)^{2 \mathrm{n}+1}=2^{(2 \mathrm{n}+1)}=2.2^{2 \mathrm{n}}=2.4^{\mathrm{n}}$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2015 (Phase 2)
Consider the expansion $\left(x^{2}+\frac{1}{x}\right)^{15}$.
Consider the following statements:
1- The term containing $\mathrm{x}^{2}$ does not exist in the given expansion.
2- The sum of the coefficients of all the terms in the given expansion is $2^{15}$
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:
1- For coefficient of $x^{2}$,
$30-3 \mathrm{r}=2 \Rightarrow \mathrm{r}=\frac{28}{3}, \mathrm{r} \notin \mathrm{N}$
So, $x^{2}$ does not exist in the expansion Hence, Statement 1 is correct.
2- Now
$$
\begin{array}{r}
\left(x^{2}+\frac{1}{4}\right)^{15}={ }^{15} C_{0}\left(x^{2}\right)^{15}+{ }^{15} C_{1}\left(x^{2}\right)^{14}\left(\frac{1}{x}\right) \\
+\ldots+{ }^{15} C_{15}\left(\frac{1}{x}\right)^{15}
\end{array}
$$
Put $\mathrm{x}=1$ both sides, we get $(1+1)^{15}={ }^{15} \mathrm{C}_{0}+{ }^{15} \mathrm{C}_{1}+\ldots .+{ }^{15} \mathrm{C}_{15}$
$\Rightarrow 2^{15}={ }^{15} \mathrm{C}_{0}+{ }^{15} \mathrm{C}_{1}+\ldots .+{ }^{15} \mathrm{C}_{15}$
Hence, Statement 2 is correct

Topic: Mathematics, Binomial Theorem, NDA, NDA 2014 (Phase 2)
Consider the expansion $\left(x^{2}+\frac{1}{x}\right)^{15}$.
Consider the following statements :
1- There are 15 terms in the given expansion.
2- The coefficient of $\mathrm{x}^{12}$ is equal to that of $\mathrm{x}^{3}$ 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- We know that, $(\mathrm{a}+\mathrm{b})^{\mathrm{n}}$ have total $(\mathrm{n}+1)$ number of terms
So, $\left(x^{2}+\frac{1}{x}\right)^{15}$ haver 16 terms.
Hence, Statement 1 is false.
2- For coefficient of $\mathrm{x}^{12}$ $30-3 \mathrm{r}=12 \Rightarrow \mathrm{r}=6 \Rightarrow{ }^{15} \mathrm{C}_{6}$
and for coefficient of $\mathrm{x}^{3}$ $30-3 r=3 \Rightarrow r=9 \Rightarrow{ }^{15} C_{9}$
${ }^{15} \mathrm{C}_{6}={ }^{15} \mathrm{C}_{9}$
Hence, statement 2 is correct.

Topic: Mathematics, Binomial Theorem, NDA, NDA 2014 (Phase 2)
Consider the expansion $\left(x^{2}+\frac{1}{x}\right)^{15}$.
What is the independent term in the given expansion?

Options:
A 2103
B 3003
C 4503
D None of these

The correct answer is:
3003

Solution:
$\left(x^{2}+\frac{1}{x}\right)^{15}$
$$
\begin{array}{rl}T_{r+1} & ={ }^{15} C_{r}\left(x^{2}\right)^{15-r}\left(\frac{1}{x}\right)^{r} \\ & =15 \\ C_{r} x & 30-2 r-r=15\end{array}
$$
$C_{r} x^{30-3 r}$
For independent term, $30-3 r=0 \Rightarrow r=10$
Put $\mathrm{r}=10$, we get
$\mathrm{T}_{10+1}={ }^{15} \mathrm{C}_{10}=\frac{15 !}{10 ! 5 !}$
$=\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{10 ! \times 1 \times 2 \times 3 \times 4 \times 5}=3003$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2014 (Phase 2)
Consider the expansion $\left(x^{2}+\frac{1}{x}\right)^{15}$.
What is the ratio of coefficient of $\mathrm{x}^{15}$ to the term independent of $\mathrm{x}$ in the given expansion?

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

The correct answer is:
1

Solution:
For coefficient of $\mathrm{x}^{15}$,
$30-3 r=15$
$\Rightarrow \mathrm{r}=5$
$\therefore$ the coefficient of $\mathrm{x}^{15} \mathrm{is}^{15} \mathrm{C}_{5}$
and coefficient of independent of $\mathrm{x}$ is
$30-3 \mathrm{r}=0$
$\Rightarrow \mathrm{r}=10$
So, coefficient of independent of $\mathrm{x}$ is ${ }^{15} \mathrm{C}_{10}$.
$\therefore$ Required ratio $=\frac{15 \mathrm{C}_{5}}{{ }^{15} \mathrm{C}_{10}}=\frac{{ }^{15} \mathrm{C}_{5}}{{ }^{15} \mathrm{C}_{5}}=1$
$\left(\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\right)$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2014 (Phase 2)
Consider the following:
1.$\quad \alpha+\beta+\alpha \beta>0$
2.$\quad \alpha^{2} \beta+\beta^{2} \alpha>0$
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:
2 only

Solution:
Sum of roots $=\alpha+\beta=-b$ Multiplication of roots $=\alpha \beta=c$ Hence $\begin{aligned} \alpha+\beta+\alpha \beta=-b+c & \alpha^{2} \beta+\beta^{2}=& \alpha \beta(\alpha+\beta) \\ &=-b c \end{aligned}$
$\because \mathrm{b}>0 \& \mathrm{c} < 0$
$\therefore-\mathrm{b}+\mathrm{c} < 0 \&-\mathrm{bc}>0$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2016 (Phase 1)
Consider the following :
$1.$ $\quad 4 \hat{\mathrm{i}} \times 3 \hat{\mathrm{i}}=\overrightarrow{0}$
$2.$ $\frac{4 \hat{\mathrm{i}}}{3 \hat{\mathrm{i}}}=\frac{4}{3}$
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:
Statement $(1): 4 \hat{\mathrm{i}} \times 3 \hat{\mathrm{i}}=0$
It is a true statement
$(\because \hat{\mathrm{i}} \times \hat{\mathrm{i}}=0)$
Statement $(2): \frac{4 \hat{\mathrm{i}}}{3 \hat{\mathrm{i}}}=\frac{4}{3}$
It is a incorrect statement
$\because \quad \frac{4 \hat{\mathrm{i}}}{3 \hat{\mathrm{i}}}=\frac{4}{3} \hat{\mathrm{i}}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2012 (Phase 2)
Consider the following:
1.$A \cup(B \cap C)=(A \cap B) \cup(A \cap C)$
2.$\quad A \cap(B \cup C)=(A \cup B) \cap(A \cup C)$
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:
Neither 1 nor 2

Solution:
According to 'Distribution law in set theorry the given
both statements are wrong. 1.A $\cup(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \cup \mathrm{B}) \cap(\mathrm{A} \cup \mathrm{C})$
2. A\cap(B\cupC) $=(A \cap B) \cup(A \cap C)$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
Consider the following :
1- $\lim _{x \rightarrow 0} \frac{1}{x}$ exists.
2- $\lim _{x \rightarrow 0} \mathrm{e}^{\frac{1}{x}}$ does not exist.
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:
2 only

Solution:
$\lim _{x \rightarrow 0} \frac{1}{x}=\frac{1}{0}=\infty$ which does not exist.
Hence, statement- 1 is incorrect. Now, $\lim _{x \rightarrow 0} \mathrm{e}^{1 / \mathrm{x}}=\mathrm{e}^{\infty}$ which also does not exist.
Hence, statement- 2 is correct.

Topic: Mathematics, Limits, NDA, NDA 2012 (Phase 2)