datavorous/entrance-exam-dataset huggingface dataset:
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 $I_{2}+I_{3}$ equal to?

Options:
A 4
B 2
C 1
D 0

The correct answer is:
0

Solution:
$\mathrm{I}_{2}=\int_{0}^{\pi} \frac{\sin 4 \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx}=\int_{0}^{\pi} \frac{2 \sin 2 \mathrm{x} \cos 2 \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx}$
$=2 \int_{0}^{\pi} \frac{2 \sin \mathrm{x} \cos \mathrm{x} \cos 2 \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx}$
$=4 \int_{0}^{\pi} \cos \mathrm{x}\left(1-2 \sin ^{2} \mathrm{x}\right) \mathrm{dx}$
$=4 \int_{0}^{\pi} \cos x d x-8 \int_{0}^{\pi} \cos x \sin ^{2} x d x=0$
Let $\sin x=t$
$\cos x d x=d t$
$=4(\sin \pi-\sin 0)-0=0$
$I_{3}=\int_{0}^{\pi} \frac{\sin 6 x}{\sin x} d x$
$=\int_{0}^{\pi} \frac{2\left[3 \sin x-4 \sin ^{3} x\right] \cos 3 x}{\sin x} d x$
$=\int_{0}^{\pi} 2\left(3-4 \sin ^{2} x\right) \cos 3 x d x$
$=\int_{0}^{\pi} 6 \cos 3 x d x-\int_{0}^{\pi} 8 \sin ^{2} x \cos 3 x d x$
$=\frac{6 \sin 3 x}{3} \int_{0}^{\pi}-8 \cdot \frac{1}{2}\left[\int_{0}^{\pi} \sin x(\sin 4 x-\sin 2 x) d x\right]$
$=6-\frac{8}{2}\left[\int_{0}^{\pi}(\sin x \sin 4 x-\sin x \cdot \sin 2 x) d x\right]$
$=\frac{-8}{2}\left[\frac{1}{2} \int_{0}^{\pi} \cos 3 x-\cos 5 x-\cos x+\cos 3 x d x\right]$
Hence, $I_{2}+I_{3}=0+0=0$

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 $I_{m}$ equal to?

Options:
A 0
B 1
C $\mathrm{m}$
D $2 \mathrm{~m}$

The correct answer is:
0

Solution:
$\begin{aligned} & I_{m}=\int_{0}^{\pi} \frac{\sin 2 m x}{\sin x} d x \\ &=\int_{0}^{\pi} \frac{\sin 2 m(\pi-x)}{\sin (\pi-x)} d x \\ &=\int_{0}^{\pi} \frac{\sin (2 m \pi-2 m x)}{\sin x} d x \\ &=\int_{0}^{\pi-\sin 2 m x}{\sin x} d x \\ & I_{m}=-\int_{0}^{\pi} \frac{\sin 2 m x}{\sin x} d x \\ & 2 I_{m}=0 \Rightarrow I_{m}=0 \end{aligned}$

Topic: Mathematics, Definite Integration, NDA, NDA 2015 (Phase 1)
Consider the integrals
$\mathrm{A}=\int_{0}^{\pi} \frac{\sin \mathrm{x} \mathrm{dx}}{\sin \mathrm{x}+\cos \mathrm{x}}$ and $\overline{\mathrm{B}}=\int_{0}^{\pi} \frac{\sin \mathrm{x} \mathrm{d} \mathrm{x}}{\sin \mathrm{x}-\cos \mathrm{x}}$
What is the value of $\mathrm{B}$ ?

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{2}$
C $\frac{3 \pi}{4}$
D $\pi$

The correct answer is:
$\frac{\pi}{2}$

Solution:
$\begin{aligned} & \mathrm{B}=\int_{0}^{\pi} \frac{\sin \mathrm{x}}{\sin \mathrm{x}-\cos \mathrm{x}} \times \frac{(\sin \mathrm{x}+\cos \mathrm{x})}{(\sin \mathrm{x}+\cos \mathrm{x})} \mathrm{dx} \\ &=-\int_{0}^{\pi} \frac{\sin ^{2} \mathrm{x}+\sin \mathrm{x} \cos \mathrm{x}}{\cos 2 \mathrm{x}} \mathrm{dx} \\ &=\frac{-1}{2} \int_{0}^{\pi} \frac{2 \sin ^{2} \mathrm{x}}{\cos 2 \mathrm{x}} \mathrm{dx}-\frac{1}{2} \int_{0}^{\pi} \frac{2 \sin \mathrm{x} \cos \mathrm{x}}{\cos 2 \mathrm{x}} \mathrm{dx} \\ &=\frac{-1}{2} \int_{0}^{\pi} \frac{1-\cos 2 \mathrm{x}}{\cos 2 \mathrm{x}} \mathrm{dx}-\frac{1}{2} \int_{0}^{\pi} \tan 2 \mathrm{x} \mathrm{dx} \\ &=\frac{-1}{2} \int_{0}^{\pi} \sec 2 \mathrm{x} \mathrm{dx}+\frac{1}{2} \int_{0}^{\pi} \mathrm{dx}-\frac{1}{2} \int_{0}^{\pi} \tan 2 \mathrm{x} \mathrm{d} \mathrm{x} \\ &=\frac{-1}{2}\left[\frac{\log |\sec 2 \mathrm{x}+\tan 2 \mathrm{x}|}{2}\right]_{0}^{\pi}+\frac{1}{2}[\mathrm{x}]_{0}^{\pi}-\frac{1}{2}\left[\frac{\log |\sec 2 \mathrm{x}|}{2}\right]_{0}^{\pi} \\ &=\frac{-1}{4}[\log (1+0)-\log (1+0)]+\frac{1}{2}[\pi+0]-\frac{1}{4}[\log (1)-\log (1)] \\ &=0+\pi / 2-0=\pi / 2 \end{aligned}$

Topic: Mathematics, Definite Integration, NDA, NDA 2015 (Phase 2)
Consider the integrals
$\mathrm{A}=\int_{0}^{\pi} \frac{\sin \mathrm{x} \mathrm{dx}}{\sin \mathrm{x}+\cos \mathrm{x}}$ and $\overline{\mathrm{B}}=\int_{0}^{\pi} \frac{\sin \mathrm{x} \mathrm{d} \mathrm{x}}{\sin \mathrm{x}-\cos \mathrm{x}}$
Which one of the follwing is correct?

Options:
A $\quad \mathrm{A}=2 \mathrm{~B}$
B $\quad B=2 A$
C $\quad \mathrm{A}=\mathrm{B}$
D $A=3 B$

The correct answer is:
$\quad \mathrm{A}=\mathrm{B}$

Solution:
$\mathrm{A}=\int_{0}^{\pi} \frac{\sin x}{\sin x+\cos x} \mathrm{dx}$
Using property
$A=\int_{0}^{\pi} \frac{\sin (\pi-x)}{\sin (\pi-x)+\cos (\pi-x)} d x$
$A=\int_{0}^{\pi} \frac{\sin x}{\sin x-\cos x} d x=B$
$A=B$

Topic: Mathematics, Definite Integration, NDA, NDA 2015 (Phase 2)
Consider the line $x=\sqrt{3} y$ and the circle $x^{2}+y^{2}=4$
What is the area of the region in the first quadrant enclosed by the $x$ -axis, the line $x=\sqrt{3}$ and the circle?

Options:
A $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
B $\frac{\pi}{2}-\frac{\sqrt{3}}{2}$
C $\frac{\pi}{3}-\frac{1}{2}$
D None of these

The correct answer is:
$\frac{\pi}{3}-\frac{\sqrt{3}}{2}$

Solution:
$x^{2}+y^{2}=4$ and $x=\sqrt{3} y$
$P(\sqrt{3}, 1)=$ First quadrant
The point of intersection of the line and the circle in the first
quadrant is $(\sqrt{3}, 1)$. Area of $\Delta$ OPA $=$ Area of $\Delta$ OPB $+$ Area of PAB
$=\frac{1}{2} \times \mathrm{OB} \times \mathrm{PB}=\frac{1}{2} \times \sqrt{3} \times 1=\frac{\sqrt{3}}{2}$
Area of $\mathrm{PAB} \int_{\sqrt{3}}^{2} \mathrm{ydx}$
$=\int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} \mathrm{~d} x=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}$
$\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2}[\sqrt{4-3}]-2 \sin ^{-1} \frac{\sqrt{3}}{2}\right]$
$=\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}=\frac{\pi}{3}-\frac{\sqrt{3}}{2}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2015 (Phase 1)
Consider the line $x=\sqrt{3} y$ and the circle $x^{2}+y^{2}=4$
What is the area of the region in the first quadrant enclosed by the $x$ -axis, the line $x=\sqrt{3} y$ and the circle?

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{6}$
C $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
D None of these

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

Solution:
Area enclosed by $x$ -axis, the line $x=\sqrt{3}$.y, and the circle $x^{2}+y^{2}=4$ in the first quadrant $=\frac{\sqrt{3}}{2}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}$
$x^{2}+y^{2}=4$ and $x=\sqrt{3} y$
$P(\sqrt{3}, 1)=$ First quadrant
The point of intersection of the line and the circle in the first
quadrant is $(\sqrt{3}, 1)$. Area of $\Delta$ OPA $=$ Area of $\Delta$ OPB $+$ Area of PAB
$=\frac{1}{2} \times \mathrm{OB} \times \mathrm{PB}=\frac{1}{2} \times \sqrt{3} \times 1=\frac{\sqrt{3}}{2}$
Area of $\mathrm{PAB} \int_{\sqrt{3}}^{2} \mathrm{ydx}$
$=\int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} \mathrm{~d} x=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^{2}$
$\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2}[\sqrt{4-3}]-2 \sin ^{-1} \frac{\sqrt{3}}{2}\right]$
$=\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}=\frac{\pi}{3}-\frac{\sqrt{3}}{2}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2015 (Phase 1)
Consider the lines $\mathrm{y}=3 \mathrm{x}, \mathrm{y}=6 \mathrm{x}$ and $\mathrm{y}=9$
The centroid of the triangle is at which one of the following points?

Options:
A $(3,6)$
B $\left(\frac{3}{2}, 6\right)$
C $(3,3)$
D $\left(\frac{3}{2}, 9\right)$

The correct answer is:
$\left(\frac{3}{2}, 6\right)$

Solution:
Coordinates of $O, A, B$ are $(0,0)\left(\frac{3}{2}, 9\right),(3,9)$ respectively.
Centroid $C=\left[\left(\frac{0+\frac{3}{2}+3}{3}\right),\left(\frac{0+9+9}{3}\right)\right]=\left(\frac{3}{2}, 6\right)$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
Consider the lines $\mathrm{y}=3 \mathrm{x}, \mathrm{y}=6 \mathrm{x}$ and $\mathrm{y}=9$
What is the area of the triangle formed by these lines?

Options:
A $\frac{27}{4}$ square units
B $\frac{27}{2}$ square units
C $\frac{19}{4}$ square units
D $\frac{19}{2}$ square units

The correct answer is:
$\frac{27}{4}$ square units

Solution:
$O A B$ is triangle.
$\begin{aligned} \text { Area of triangle } &=\frac{1}{2} \times \text { base } \times \text { height }=\frac{1}{2} \times \mathrm{AB} \times \mathrm{OP} \\ &=\frac{1}{2} \times \frac{3}{2} \times 9=\frac{27}{4} \end{aligned}$
Area of triangle $=\frac{27}{4}$ square units

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
Consider the parabola $y=x^{2}+7 x+2$ and the straight line $y=3x-3$
What are thecoordinates of the point on the parabola which is closest to the straight line?

Options:
A $(0,2)$
B $(-2,-8)$
C $(-7,2)$
D $(1,10)$

The correct answer is:
$(-2,-8)$

Solution:
Parabola Eq: $y=x^{2}+7 x+2$ Line eq. : $y=3 x-3$ Since all the points given in the options lie on the parabola Thus we will calculate the distance from the given line to these points :
for $(0,2):$ distance $==\frac{|3(0)-(2)-3|}{\sqrt{(3)^{2}+(-1)^{2}}}=\frac{5}{\sqrt{10}}$
for $(-2,-8):$ distance $=\frac{|3(-2)-(-8)-3|}{\sqrt{10}}=\frac{1}{\sqrt{10}}$
for $(-7,2):$ distance $=\frac{|3(-7)-2-3|}{\sqrt{10}}=\frac{26}{\sqrt{10}}$
for $(1,10):$ distance $=\frac{|3(1)-10-3|}{\sqrt{10}}=\frac{10}{\sqrt{10}}$
$(-2,-8)$ is the given point.

Topic: Mathematics, Parabola, NDA, NDA 2016 (Phase 2)
Consider the parabola $y=x^{2}+7 x+2$ and the straight line $y=3 x-3$
What is the shortest distance from the above point on the parabola to the line?

Options:
A $\frac{\sqrt{10}}{2}$
B $\frac{\sqrt{10}}{5}$
C $\frac{1}{\sqrt{10}}$
D $\frac{\sqrt{5}}{4}$

The correct answer is:
$\frac{1}{\sqrt{10}}$

Solution:
No Solution

Topic: Mathematics, Parabola, NDA, NDA 2016 (Phase 2)
Consider the parabolas $S_{1}=y^{2}-4 a x=0$ and
$S_{2}=y^{2}-4 b x=0, S_{2}$ will contain $S_{1}$, if

Options:
A $a>b>0$
B $b>a>0$
C $a>0, b < 0$ but $|b|>a$
D $a < 0, b>0$ but $b>|a|$

The correct answer is:
$b>a>0$

Solution:
If a and $b>0$, then graphic representation would be as follows
$S_{2}$ will contain $S_{1}$, if latusrectum of $S_{2}>$ latusrectum of $S$ $\Rightarrow 4 b>4 a$
$\therefore b>a>0$

Topic: Mathematics, Parabola, NDA, NDA 2008 (Phase 2)
Consider the plane passing through the points $\mathrm{A}(2,2,1), \mathrm{B}(3,4,2)$ and $\mathrm{C}(7,0,6)$
What are the direction ratios of the normal to the plane?

Options:
A $ < 1,0,1>$
B $ < 0,1,0>\quad$
C $ < 1,0,-1>$
D None of these

The correct answer is:
$ < 1,0,-1>$

Solution:
Direction ratios of the nomal to the plane $\mathrm{x}-\mathrm{z}=1$ are $(1,0,-1)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 2)
Consider the plane passing through the points $\mathrm{A}(2,2,1), \mathrm{B}(3,4,2)$ and $\mathrm{C}(7,0,6)$
Which one of the following points lines on the plane?

Options:
A $(1,0,0)$
B $(1,0,1)$
C $(0,0,1)$
D None of these

The correct answer is:
$(1,0,0)$

Solution:
We know that, equation of plane passing through three non-collinear points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ and $\left(\mathrm{x}_{3}, \mathrm{y}_{3}, \mathrm{z}_{3}\right)$ is
$\left|\begin{array}{lll}\mathrm{x}-\mathrm{x}_{1} & \mathrm{y}-\mathrm{y}_{1} & \mathrm{z}-\mathrm{z}_{1} \\ \mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\ \mathrm{x}_{3}-\mathrm{x}_{1} & \mathrm{y}_{3}-\mathrm{y}_{1} & \mathrm{z}_{3}-\mathrm{z}_{1}\end{array}\right|=0$
Put the value of $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$
we get
$\left|\begin{array}{ccc}\mathrm{x}-2 & \mathrm{y}-2 & \mathrm{z}-1 \\ 1 & 2 & 1 \\ 5 & -2 & 5\end{array}\right|=0$
$\Rightarrow(x-2)(10+2)-(y-2)(5-5)+(z-1)(-2-10)=0$
$\Rightarrow 12 x-12 z=12 \Rightarrow x-z=1$
Hence the equation of plane parses through $(1,0,0)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 2)
Consider the points $(a-1, a, a+1),(a, a+1, a-1)$ and $(a+1, a-1, a)$
1 . These points always form the vertices of an equilateral triangle for any real value of a
2 . The area of the triangle formed by these points is independent of a. Which of the statement(s) 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:
Let $\mathrm{A}(\mathrm{a}-1, \mathrm{a}, \mathrm{a}+1), \mathrm{B}(\mathrm{a}, \mathrm{a}+1, \mathrm{a}-1)$ and
a) be the vertices of a triangle $\mathrm{ABC}$. $\mathrm{C}(\mathrm{a}+1, \mathrm{a}-1$
Length of $A B=\sqrt{(a-a+1)^{2}+(a+1-a)^{2}+(a-1-a-1)^{2}}$
$=\sqrt{1+1+4}=\sqrt{6}$
Length of $\begin{aligned} \mathrm{BC} &=\sqrt{(\mathrm{a}+\mathrm{l}-\mathrm{a})^{2}+(\mathrm{a}-1-\mathrm{a}-1)^{2}+(\mathrm{a}-\mathrm{a}+1)^{2}} \\ &=\sqrt{1+4+1}=\sqrt{6} \\ \text { and } \mathrm{CA}=\sqrt{(\mathrm{a}-1-\mathrm{a}-1)^{2}+(\mathrm{a}-\mathrm{a}+1)^{2}+(\mathrm{a}+1-\mathrm{a})^{2}} \end{aligned}$
$=\sqrt{4+1+1}=\sqrt{6}$
$\mathrm{AB}=\mathrm{BC}=\mathrm{CA}$
Triangle $\mathrm{ABC}$ is an equilateral triangle and these given points are vertices of an equilateral triangle for any real value of a.
Position vector of $\mathrm{A}, \overrightarrow{\mathrm{OA}}=(\mathrm{a}-1) \hat{\mathrm{i}}+\hat{\mathrm{a}}+(\mathrm{a}+1) \hat{\mathrm{k}}$
Position vector of $\mathrm{B}, \overrightarrow{\mathrm{OB}}=\mathrm{a} \hat{\mathrm{i}}+(\mathrm{a}+1) \hat{\mathrm{j}}+(\mathrm{a}-1) \hat{\hat{k}}$
Position vector of $\mathrm{C}, \overrightarrow{\mathrm{OC}}=(\mathrm{a}+1) \hat{\mathrm{i}}+(\mathrm{a}-1) \hat{\mathrm{j}}+\mathrm{a} \hat{k}$
$\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$
Now, area of a triangle $\Delta \mathrm{ABC}$
$=\frac{1}{2}|\mathrm{~A} \overline{\mathrm{B}} \times \mathrm{A} \overrightarrow{\mathrm{C}}|$
$=\frac{1}{2}\left|\begin{array}{ccc}\hat{1} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & -2 \\ 2 & -1 & -1\end{array}\right|$
$=\frac{1}{2}|-3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}|=\frac{1}{2} \sqrt{9+9+9}=\frac{\sqrt{27}}{2} \mathrm{sq}$ units.
Since, a does not appear so, area of triangle formed by these points is independent of a.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 1)
Consider the propagating sound (with velocity 330 m s-1) in a pipe of length 1.5 m with one end closed and the other open. The frequency associated with the fundamental mode is

Options:
A 11 Hz
B 55 Hz
C 110 Hz
D 165 Hz

The correct answer is:
55 Hz

Solution:
Given,
Velocity of sound =330m s-1
Length of closed pipe =15 m
In a closed pipe for the fundamental mode
λ4=l
λ=4×15=6 m
v=nλ
n=vλ
=3306
=55 Hz

Topic: Physics, Waves and Sound, NEET
Consider the set A of all matrices of order $3 \times 3$ with entries 0 or 1 only. Let $\mathrm{B}$ be the subset of A consisting of all matrices whose determinant is $1 .$ Let $\mathrm{C}$ be the subset of A consisting of all matrices whose determinant is $-1$. Then which one of the following is correct?

Options:
A $\mathrm{C}$ is empty
B $\mathrm{B}$ has as many elements as $\mathrm{C}$
C $A=B \cup C$
D $\mathrm{B}$ has thrice as many elements as $\mathrm{C}$

The correct answer is:
$\mathrm{B}$ has as many elements as $\mathrm{C}$

Solution:
determinant of $\mathrm{B}=1$
Let ' $\mathrm{B}$ ' be identity matrix
So, $B=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
If we interchange any 2 rows, determinant will be $-1$
Let, $\mathrm{C}=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right|,|\mathrm{C}|=-1$
Here, number of elements in $\mathrm{B}$ and $\mathrm{C}$ are equal.

Topic: Mathematics, Determinants, NDA, NDA 2017 (Phase 1)
Consider the spheres $x^{2}+y^{2}+z^{2}-4 y+3=0$ and $x^{2}+y^{2}+z^{2}+2 x$ $+4 z-4=0$
Consider the following statements:
1 . The two spheres intersect each other.
2 . The radius of first sphere is less than that of second sphere. 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:
$x^{2}+y^{2}+z^{2}-4 y+3=0$
$x^{2}+y^{2}-4 y+4-4+z^{2}+3=0$
$x^{2}+(y-2)^{2}+z^{2}=1$
Sphere with centre $(0,2,0)$ and radius 1 unit. $x^{2}+y^{2}+z^{2}+2 x+4 z-4=0$
$x^{2}+2 x+1-1+y^{2}+z^{2}+4 z+4-4-4=0$
$(x+1)^{2}+y^{2}+(z+2)^{2}=3^{2}$
Sphere with centre $(-1,0,-2)$ and radius 3 units.
$\mathrm{r}_{1}+\mathrm{r}_{2}=3+1=4$
$\mathrm{C}_{1} \mathrm{C}_{2} < \mathrm{r}_{1}+\mathrm{r}_{2}$
$\therefore$ Two spheres intersect each other.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 1)
Consider the spheres $x^{2}+y^{2}+z^{2}-4 y+3=0$ and $x^{2}+y^{2}+z^{2}+2 x$ $+4 z-4=0$
What is the distance between the centres of the two spheres?

Options:
A 5 units
B 4 units
C 3 units
D 2 units

The correct answer is:
3 units

Solution:
$x^{2}+y^{2}+z^{2}-4 y+3=0$
$x^{2}+y^{2}-4 y+4-4+z^{2}+3=0$
$x^{2}+(y-2)^{2}+z^{2}=1$
Sphere with centre $(0,2,0)$ and radius 1 unit. $x^{2}+y^{2}+z^{2}+2 x+4 z-4=0$
$x^{2}+2 x+1-1+y^{2}+z^{2}+4 z+4-4-4=0$
$(x+1)^{2}+y^{2}+(z+2)^{2}=3^{2}$
Sphere with centre $(-1,0,-2)$ and radius 3 units.
$\quad C_{1} C_{2}=\sqrt{(0+1)^{2}+(2-0)^{2}+(0+2)^{2}}=3$ units

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 1)
Consider the system of circles $x^2+y^2+2 f y+\lambda\left(x^2+y^2+2 g x+k\right)=0$, where $g \neq 0, f \neq 0$ and $\lambda$ a parameter. If $A$ and $B$ are the point circles of this system such that $\angle A O B=\frac{\pi}{2}$, then $g^2$ is equal to

Options:
A −k
B $\frac{k}{2}$
C 0
D $f^2$

The correct answer is:
$\frac{k}{2}$

Solution:
Given, system of family of circles is
$$
x^2+y^2+2 f y+\lambda\left(x^2+y^2+2 g x+k\right)=0
$$
$\begin{aligned} & \Rightarrow \quad(1+\lambda) x^2+(1+\lambda) y^2+2 g \lambda x+2 f y+\lambda k=0 \\ & \Rightarrow \quad x^2+y^2+2 \cdot\left(\frac{g \lambda}{1+\lambda}\right) x+2\left(\frac{f}{1+\lambda}\right) y+\frac{\lambda k}{1+\lambda}=0\end{aligned}$
For point circle, radius should be zero.
$$
\begin{array}{ll}
\therefore & \frac{g^2 \lambda^2}{(1+\lambda)^2}+\frac{f^2}{(1+\lambda)^2}=\frac{\lambda k}{1+\lambda} \\
\Rightarrow & g^2 \lambda^2+f^2=k\left(\lambda+\lambda^2\right)
\end{array}
$$
Let the quadratic Eq. (i) has roots $\lambda_1$ and $\lambda_2$, then the coordinates of point circles are
$$
c_1\left(\frac{-g \lambda_1}{1+\lambda_1}, \frac{f}{1+\lambda_1}\right) \text { and } c_2\left(\frac{-g \lambda_2}{1+\lambda_2}, \frac{f}{1+\lambda_2}\right)
$$
$\because$ Lines $O C_1$ and $O C_2$ are perpendicular to each other.
So,
$$
\frac{\frac{f}{1+\lambda_2}}{\frac{g \lambda_2}{1+\lambda_2}} \times \frac{\frac{f}{1+\lambda_1}}{\frac{g \lambda_1}{1+\lambda_1}}=-1
$$
$\because$ Product of roots of Eq. (i), we get
From Eqs. (ii) and (iii), we get
$$
\begin{array}{rlrl}
g^2\left(\frac{f^2}{g^2-k}\right) & =-f^2 \\
\Rightarrow & & g^2 & =-g^2+k \\
\Rightarrow & 2 g^2 & =k \Rightarrow g^2=\frac{k}{2}
\end{array}
$$

Topic: Mathematics, Circle, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 2)
Consider the system of equations
\(\begin{array}{cc}
x+y+z & =0 \\
\alpha x+\beta y+\gamma z & =0 \\
\alpha^{2} x+\beta^{2} y+\gamma^{2} z & =0
\end{array}\)
Then the system of equations has

Options:
A a unique solution for all values of $\alpha, \beta$ and $\gamma$.
B infinite number of solutions, if any two of $\alpha, \bar{\beta},$ yare equal.
C a unique solution, if $\alpha, \beta$ and $\gamma$ are distinct.
D more than one, but finite number of solutions depending on values of $\alpha, \beta$ and $\gamma$

The correct answers are:
infinite number of solutions, if any two of $\alpha, \bar{\beta},$ yare equal., a unique solution, if $\alpha, \beta$ and $\gamma$ are distinct.

Solution:
Given system of equations is $x+y+z=0$
$\begin{array}{l}
\alpha x+\beta y+\gamma z=0 \\
\alpha^{2} x+\beta^{2} y+\gamma^{2} z=0
\end{array}$
The coefficient matrix, $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2}\end{array}\right]$
Now, $|A|=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2}\end{array}\right|=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$
(i) The system of equations has a unique solution, if $\alpha, \beta$ and $\gamma$ are distinct
i.e. $|A| \neq 0$
(ii) The system of equations has infinite number of solutions, if any two of $\alpha, \beta$ and $\gamma$ ar equal.
i.e. $|A|=0$

Topic: Mathematics, Linear Programming, WBJEE, WBJEE 2013
Consider the triangle $A B C$ with vertices $A(-2,3), B(2,1)$ and $C(1,2)$.
What is the centroid of the tirnalge $A B C$ ?

Options:
A $\left(\frac{1}{3}, 1\right)$
B $\left(\frac{1}{3}, 2\right)$
C $\left(1, \frac{2}{3}\right)$
D $\left(\frac{1}{2}, 3\right)$

The correct answer is:
$\left(\frac{1}{3}, 2\right)$

Solution:
Centroid of the triangle
$=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}}{3}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}}{3}\right)$
$=\left(\frac{-2+2+1}{3}, \frac{3+1+2}{3}\right)=\left(\frac{1}{3}, 2\right)$
$\therefore \quad$ Option (b) is correct.

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 1)
Consider the triangle $A B C$ with vertices $A(-2,3), B(2,1)$ and $C(1,2)$.
What is the circumcentre of the triangle $\mathrm{ABC} ?$

Options:
A $\quad(-2,-2)$
B $(2,2)$
C $(-2,2)$
D $(2,-2)$

The correct answer is:
$\quad(-2,-2)$

Solution:
A circumcentre is a point at which perpendicular bisectors meet each other.
Here, ' $\mathrm{E}$ ' represents circumcentre
Mid-point of $\mathrm{BC}=\left(\frac{2+1}{2}, \frac{1+2}{2}\right)=\left(\frac{3}{2}, \frac{3}{2}\right)$
Slope of $\mathrm{BC}=\frac{2-1}{1-2}=-1$
$\therefore \quad$ Slope of $\mathrm{DE}=1$
Now, equation of $\overline{E D}$ is $\left(y-\frac{3}{2}\right)=1\left(x-\frac{3}{2}\right)$
$2 y-3=2 x-3$
$\mathrm{x}=\mathrm{y}$ ...(i)
Now, mid-point of $\mathrm{AC}=\left(\frac{-2+1}{2}, \frac{3+2}{2}\right)=\left(-\frac{1}{2}, \frac{5}{2}\right)$
Slope of $\mathrm{AC}=\frac{3-2}{-2-1}=-\frac{1}{3}$
Slope of $\mathrm{EF}=3$
Now, equation of $\overrightarrow{\mathrm{EF}}$ is $\left(\mathrm{y}-\frac{5}{2}\right)=3\left(\mathrm{x}+\frac{1}{2}\right)$
$2 y-5=6 x+3$ ...(ii)
From equations (i) and (ii), $\mathrm{x}=-2$ and $\mathrm{y}=-2$
Hence, circumcentre of $\Delta \mathrm{ABC}$ is $(\mathrm{x}, \mathrm{y})=(-2,-2)$
Option (a) is correct.

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 1)
Consider the triangle $A B C$ with vertices $A(-2,3), B(2,1)$ and $C(1,2)$.
What is the foot of the altitude from the vertex $A$ of the triangle $A B C ?$

Options:
A $(1,4)$
B $(-1,3)$
C $(-2,4)$
D $(-1,4)$

The correct answer is:
$(-1,4)$

Solution:
Slope of $\mathrm{BC}=\frac{2-1}{1-2}=-1$
Slope of $\mathrm{AD}=1$
Now, equation of $\overrightarrow{B C}$ is $y-2=-1(x-1)$
$\therefore \quad y-2=-x+1$
$\therefore \quad x+y-3=0$ ...(i)
and equation of $\mathrm{AD}$ is $(\mathrm{y}-3)=1(\mathrm{x}+2)$
$\therefore \quad \mathrm{x}-\mathrm{y}+5=0$...(ii)
From equations (i) and (ii), $\mathrm{x}=-1$ and $\mathrm{y}=4$
$\therefore$ Foot of altitude from the vertex A of the triangle $\mathrm{ABC}$ is $(-1,4)$
$\therefore$ Option (d) is correct.

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 1)
Consider the two circles $(x-1)^{2}+(y-3)^{2}=r^{2}$ and $\mathrm{x}^{2}+\mathrm{y}^{2}-8 \mathrm{x}+2 \mathrm{y}+8=0$
If the circles intersect at two distinct points, then which one of the following is correct?

Options:
A $\mathrm{r}=1$
B $\quad 1 < \mathrm{r} < 2$
C $\quad r=2$
D $2 < \mathrm{r} < 8$

The correct answer is:
$2 < \mathrm{r} < 8$

Solution:
Given equation of circles $(x-1)^{2}+(y-3)^{2}=r^{2}$
$\left(h_{1}, k_{1}\right) \equiv$ coordinates of centre $\equiv(1,3)$
$$
\begin{array}{l}
\therefore x^{2}+y^{2}-8 x+2 y+8=0 \\
\Rightarrow(x-4)^{2}+(y+1)^{2}=(3)^{2}
\end{array}
$$
$\left(h_{2}, k_{2}\right) \equiv$ coordinates of centre $\equiv(4,-1)$
Radius of circle one $=r_{1}=r$
Radius of circle two $=r_{2}=3$ ". Circle intersects at two points so distance between
circle is $d < r_{1}+r_{2}$
$5 < r+3$
$r>2$ also $\mathrm{r} \leq 5+2$. Hence, $2 < \mathrm{r} < 8$

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