datavorous/entrance-exam-dataset huggingface dataset:
If the sum of odd numbered terms and the sum of even numbered terms in the expansion of $(x+a)^{n}$ are $A$ and $B$ respectively, then the value of $\left(x^{2}-\right.$ $\left.a^{2}\right)^{n}$ is

Options:
A $\mathrm{A}^{2}-\mathrm{B}^{2}$
B $\mathrm{A}^{2}+\mathrm{B}^{2}$
C $4 \mathrm{AB}$
D None

The correct answer is:
$\mathrm{A}^{2}-\mathrm{B}^{2}$

Solution:
$$

\begin{array}{r}

(x+a)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} a+{ }^{n} C_{2} x^{n-2} a^{2} \\

+{ }^{n} C_{3} x^{n-3} a^{3}+{ }^{n} C_{4} x^{n-4} a^{4}+\ldots . \\

=\left({ }^{n} C_{0} x^{n}+{ }^{n} C_{2} x^{n-2} a^{2}+{ }^{n} C_{4} x^{n-4} a^{4}+\ldots .\right)+ \\

+\left({ }^{n} C_{1} x^{n-1} a+{ }^{n} C_{3} x^{n-3} a^{3}+{ }^{n} C_{5} x^{n-5} a^{5}\right)+\ldots .

\end{array}

$$

$=A+B$ ....(i)

Similarly, $(x-a)^{n}=A-B$ ....(ii)

Multiplying eqns. (1) and (2), we get

$$

\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)^{\mathrm{n}}=\mathrm{A}^{2}-\mathrm{B}^{2}

$$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2016
If the sum of squares of distances of a point from the planes x+y+z=0, x-z=0 and x-2y+z=0 is p2 then locus of the point is

Options:
A x2+z2=p2
B x2+2xy+y2+z2=p2
C x+y+z=p2
D x2+y2+z2=p2

The correct answer is:
x2+y2+z2=p2

Solution:
Let distances of a point Px,y,z from the planes x+y+z=0, x-z=0 and x-2y+z=0 are x+y+z3, x-z2 and x-2y+z6 respectively, then the sum of the squares of distances, is as

⇒x+y+z32+x-z22+x-2y+z62=p2

⇒2x+y+z2+3x-z2+x-2y+z2=6p2

⇒ 2x2+2y2+2z2+4xy+4yz+4zx+3x2+3z2-6xz+x2+4y2+z2-4xy-4yz+2xz=6p2

⇒6x2+6y+6z2=6p2

⇒x2+y2+z2=p2

Hence, option (d) is correct.

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2018
If the sum of the coefficients in the expansion of $(x+y)^n$ is 1024 , then the value of the greatest coefficient in the expansion is

Options:
A
$356$

B
$252$

C
$210$

D
$120$

The correct answer is:

$252$

Solution:
Given, sum of the coefficient $=1024$
$\begin{array}{ll}\text { i.e. } & 2^n=1024=2^{10} \\ \Rightarrow & n=10\end{array}$
Since, $n$ is even, so greatest coefficient
$={ }^n C_{n / 2}={ }^{10} C_5=252$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2022
If the sum of the first \(2 n\) terms of \(2,5,8, \ldots \ldots\) is equal to the sum of the first \(n\) terms of 57,59 , \(61 \ldots \ldots\). , then \(\mathrm{n}\) is equal to

Options:
A 10
B 12
C 11
D 13

The correct answer is:
11

Solution:
Given,
\(\frac{2 n}{2}\{2.2+(2 n-1) 3\}=\frac{n}{2}\{2.57+(n-1) 2\}\)
or \(2(6 n+1)=112+2 n\) or \(10 n=110, ~\therefore n=11\)

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2011
If the tangent at $P(1,1)$ on $y^{2}=x(2-x)^{2}$ meets the curve again at $Q$, then $Q$ is

Options:
A $(2,2)$
B $(-1,-2)$
C $\left(\frac{9}{4}, \frac{3}{8}\right)$
D None of these

The correct answer is:
$\left(\frac{9}{4}, \frac{3}{8}\right)$

Solution:
$\begin{array}{l}

y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x \ldots \text { (i) } \\

\Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}=3 x^{2}-8 x+4 \\

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 x^{2}-8 x+4}{2 y} \\

\Rightarrow\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{P}}=\frac{3-8+4}{2}=-\frac{1}{2}

\end{array}$

$y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x \ldots$ (i) $\Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}=3 x^{2}-8 x+4$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 x^{2}-8 x+4}{2 y}$ $\Rightarrow\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{P}}=\frac{3-8+4}{2}=-\frac{1}{2}$ $\therefore$ Equation of tangent at $\mathrm{P}$ is: $y-1=-\frac{1}{2}(x-1)$ $\Rightarrow x+2 y-3=0$ $\Rightarrow x+2 y-3=0$

Using $y=\frac{3-x}{2}$ in (i), we get: $\left(\frac{3-x}{2}\right)^{2}$

$\begin{array}{l}

=x^{3}-4 x^{2}+4 x \\

\Rightarrow 4 x^{3}-17 x^{2}+22 x-9=0 \text {...(ii) }

\end{array}$

which has two roots 1,1 (Because of (ii) being tangent at $(1,1)$ ).

Sum of 3 roots $=\frac{17}{4}$

$\begin{array}{l}

\therefore 3 \mathrm{rd} \text { root }=\frac{17}{4}-2=\frac{9}{4} \\

\text { Then, } y=\frac{3-\frac{9}{4}}{2}=\frac{3}{8} \\

\therefore \mathrm{Q} \text { is }\left(\frac{9}{4}, \frac{3}{8}\right)

\end{array}$

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2021
If the the earth is at one-fourth of its present distance from the sun, the duration of year will be

Options:
A half the present year
B one-eight the present year
C one-fourth the present year
D one-sixth the present year

The correct answer is:
one-eight the present year

Solution:
No Solution

Topic: Physics, Gravitation, BITSAT, BITSAT 2020
If the third term in the expansion of $\left[\mathrm{x}+\mathrm{x}^{\log _{10} \mathrm{x}}\right]^{5}$

is $10^{6},$ then $\mathrm{x}$ may be

Options:
A 1
B $\sqrt{10}$
C 10
D $10^{-2 / 5}$

The correct answer is:
10

Solution:
Put $\log _{10} x=y,$ the given expression becomes $\left(x+x^{y}\right)^{5}$

$\mathrm{T}_{3}={ }^{5} \mathrm{C}_{2} \cdot \mathrm{x}^{3}\left(\mathrm{x}^{\mathrm{y}}\right)^{2}=10 \mathrm{x}^{3+2 \mathrm{y}}=10^{6}$ (given)

$\Rightarrow(3+2 \mathrm{y}) \log _{10} \mathrm{x}=5 \log _{10} 10=5$

$\Rightarrow(3+2 \mathrm{y}) \mathrm{y}=5$

$\Rightarrow \mathrm{y}=1,-\frac{5}{2}$

$\Rightarrow \log _{10} \mathrm{x}=1$ or $\log _{10} \mathrm{x}=-\frac{5}{2}$

$\therefore \mathrm{x}=10$ or $\mathrm{x}=(10)^{-5 / 2}$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2016
If the time period is doubled, then the angular momentum of the body will (provided the moment of inertia of the body is constant)

Options:
A remain constant
B quadruple
C become half
D double

The correct answer is:
become half

Solution:
As we know.

L=Iω

⇒L=I×2πT

L∝1T

Given, T2=2T

Hence L1L2=2TT

⇒ L2=L12

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2018
If the total binding energies of ${ }_{1}^{2} \mathrm{H},{ }_{2}^{4} \mathrm{He},$ ${ }_{26}^{56} \mathrm{Fe} \&{ }_{92}^{235} \mathrm{U}$ nuclei are 2.22,28.3,492 and 1786 MeV respectively, identify the most stable nucleus of the following.

Options:
A ${ }_{26}^{56} \mathrm{Fe}$
B ${ }_{1}^{2} \mathrm{H}$
C ${ }_{92}^{235} \mathrm{U}$
D ${ }_{2}^{4} \mathrm{He}$

The correct answer is:
${ }_{26}^{56} \mathrm{Fe}$

Solution:
$\begin{aligned} \mathrm{B.E}_{\mathrm{H}} &=\frac{2.22}{2}=1.11 \\ \mathrm{B.E}_{\mathrm{He}} &=\frac{28.3}{4}=7.08 \\ \mathrm{B.E}_{\mathrm{Fe}} &=\frac{492}{56}=8.78=\text { maximum } \\ \mathrm{B.E}_{\mathrm{U}} &=\frac{1786}{235}=7.6 \end{aligned}$
$\frac{56}{26} \mathrm{Fe}$ is most stable as it has maximum binding energy per nucleon.

Topic: Physics, Nuclear Physics, BITSAT, BITSAT 2012
If the unit of force and length be each increased by four times, then the unit of energy is increased by

Options:
A 16 times
B 8 times
C 2 times
D 4 times

The correct answer is:
16 times

Solution:
Since unit of energy $=$ (unit of force). (unit of length) so if we increase unit of length and force, each by four times, then unit of energy will increase by sixteen times.

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2012
If the value of mode and mean is 60 and 66 respectively, then the value of median is

Options:
A 70
B 64
C 90
D 50

The correct answer is:
64

Solution:
$\quad$ Mode $=3$ Median -2 Mean

$\therefore \quad$ Median $=\frac{1}{3}($ mode +2 mean $)=\frac{1}{3}(60+2 \times 66)=64$

Topic: Mathematics, Statistics, BITSAT, BITSAT 2013
If the variance of the observations x1,x2,……,xn is σ2, then the variance of αx1,αx2,……,αxn,α≠0 is

Options:
A σ2
B ασ2
C α2σ2
D σ2α2

The correct answer is:
α2σ2

Solution:
Variance =1nΣ(x-x¯)2=σ2

New variance =1nΣ(αx-αx¯)2

=α21nΣ(x-x¯)2=α2σ2

Topic: Mathematics, Statistics, BITSAT, BITSAT 2017
If three vertices of a regular hexagon are chosen at random, then the chance that they form an equilateral triangle is:

Options:
A $\frac{1}{3}$
B $\frac{1}{5}$
C $\frac{1}{10}$
D $\frac{1}{2}$

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

Solution:
Three vertices can be selected in ${ }^{6} \mathrm{C}_{3}$ ways.

The only equilateral triangles possible are $\mathrm{A}_{1} \mathrm{~A}_{3} \mathrm{~A}_{5}$ and $\mathrm{A}_{2} \mathrm{~A}_{4} \mathrm{~A}_{6}$

$\mathrm{p}=\frac{2}{{ }^{6} \mathrm{C}_{3}}=\frac{2}{20}=\frac{1}{10}$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2016
If two distinct chords drawn from the point (p,q) on the circle x2+y2=px+qy (where pq≠0) are bisected by the X-axis, then

Options:
A p2=q2
B p2=8q2
C p2<8q2
D p2>8q2

The correct answer is:
p2>8q2

Solution:
Let (t, m) be the other end of the chord drawn from the point (p, q) on the circle

x2+y2=px+qy

Their mid-point is t+p2,m+q2

since, mid-point lies on X-axis i.e. y=0

m+q=0 ...(i)

Also, (t, m) lies on the circle.

∴ t2+m2-pt-qm=0 ...(ii)

From Eqs. (i) and (ii), we get t2-pt+2q2=0

Which is quadratic in t such that, Discriminant >0

⇒ p2-8q2>0⇒p2>8q2

Topic: Mathematics, Circle, BITSAT, BITSAT 2017
If $\mathrm{V}$ is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:

Options:
A $\frac{3 V}{2}$
B
V
C $\frac{4}{3} \mathrm{~V}$
D $\frac{\mathrm{V}}{2}$

The correct answer is:
$\frac{3 V}{2}$

Solution:
Potential due to sphere
$\mathrm{V}=\frac{\mathrm{GM}}{2 \mathrm{R}^3}\left(3 \mathrm{R}^2-\mathrm{r}^2\right)$
At surface $r=R \Rightarrow V=\left(\frac{G M}{R}\right)$
At centre, $r=0$
$\therefore \mathrm{V}_0=\frac{3 \mathrm{GM}}{2 \mathrm{R}}=\left(\frac{3 \mathrm{~V}}{2}\right)$

Topic: Physics, Gravitation, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If \((\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=676\) and \(|\vec{b}|=2\) then \(|\vec{a}|\) is equal to

Options:
A 13
B 26
C 39
D None of these

The correct answer is:
13

Solution:
\(\begin{aligned} &(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=676 \\ &(|\vec{a}| \cdot|\vec{b}| \sin \theta \hat{n})^2+(|\vec{a}| \cdot|\vec{b}| \cos \theta)^2=676 \\ & \Rightarrow a^2 b^2 \sin ^2 \theta+a^2 b^2 \cos ^2 \theta=676\left[(\hat{n})^2=1\right] \\ & a^2 b^2\left(\sin ^2 \theta+\cos ^2 \theta\right)=676 \Rightarrow a^2=\frac{676}{b^2}=\frac{676}{4} \\ &|\vec{a}|=\sqrt{\frac{676}{4}} \Rightarrow|\vec{a}|=\frac{26}{2} \Rightarrow|\vec{a}|=13\end{aligned}\)

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2010
If vectors $2 \mathrm{i}-\mathrm{j}+\mathrm{k}, \mathrm{i}+2 \mathrm{j}-3 \mathrm{k}$ and $3 \mathrm{i}+\mathrm{aj}+5 \mathrm{k}$ are

coplanar, then the value of a is

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

The correct answer is:
-4

Solution:
If given vectors are coplanar, then there exists two scalar quantities $\mathrm{x}$ and $\mathrm{y}$ such that $2 \hat{i}-\hat{j}+\hat{k}=x(\hat{i}+2 \hat{j}-3 \hat{k})+y(3 \hat{i}+a \hat{j}+5 \hat{k})$

Comparing coefficient of $\hat{i}, \hat{j}$ and $\hat{k}$ on both sides of (1) we get $x+3 y=2,2 x+a y=-1,-3 x+5 y=1 \ldots(2)$

Solving first and third equations, we get $\mathrm{x}=1 / 2, \mathrm{y}=1 / 2$

Since the vectors are coplanar, therefore these values of $x$ and $y$ will satisfy the equation $2 x+a y=-1$

$\therefore 2(1 / 2)+a(1 / 2)=-1 \Rightarrow a=-4$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2013
If vertex of a parabola is $(2,-1)$ and the equation of its directrix is $4 x-3 y=21$, then the length of its latus rectum is

Options:
A
2
B
8
C
12
D
16

The correct answer is:
8

Solution:
We are given that directrix of the parabola is $4 x-3 y=21$ and vertex of the parabola is $(2,-1)$
Now $a=\frac{|8+3-21|}{5}=\frac{10}{5}=2$
$\therefore \quad$ latus rectum of the parabola $=4 a=8$

Topic: Mathematics, Parabola, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If $|w|=2$, then the set of points $z=w-\frac{1}{w}$ is contained in or equal to the set of points $z$ satisfying

Options:
A
$\operatorname{Im}(z)=0$

B
$|\operatorname{Im}(z)| \leq 1$

C
$|\operatorname{Re}(z)| \leq 2$

D
$|z| \leq 3$

The correct answer is:

$|z| \leq 3$

Solution:
We have, $|w|=2 \Rightarrow w=2(\cos \theta+i \sin \theta)$
$\therefore \quad z=w-\frac{1}{w}$
$=2(\cos \theta+i \sin \theta)-\frac{1}{2}(\cos \theta-i \sin \theta)$
$\Rightarrow \quad z=\frac{3}{2} \cos \theta+\frac{5}{2} i \sin \theta$
$\Rightarrow \quad|z|=\sqrt{\frac{9}{4}+\frac{25}{4}}=\frac{\sqrt{17}}{2} < 3$
Hence, option (d) is correct.

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2022
If $x \in\left(0, \frac{\pi}{2}\right)$ then the value of $\cos ^{-1}\left(\frac{7}{2}(1+\cos 2 x)+\sqrt{\left(\sin ^2 x-48 \cos ^2 x\right) \sin x}\right)$ is equal to

Options:
A $x-\cos ^{-1}(7 \cos x)$
B $x+\sin ^{-1}(7 \cos x)$
C $x+\cos ^{-1}(6 \cos x)$
D $x+\cos ^{-1}(7 \cos x)$

The correct answer is:
$x-\cos ^{-1}(7 \cos x)$

Solution:
$y=\cos ^{-1}\left(\frac{7}{2}(1+\cos 2 x)+\sqrt{\left(\sin ^2 x-48 \cos ^2 x\right)} \sin x\right)$
$=\cos ^{-1}\left((7 \cos x)(\cos x)+\sqrt{1-49 \cos ^2 x} \sqrt{1-\cos ^2 x}\right)$
$=\cos ^{-1}(\cos x)-\cos ^{-1}(7 \cos x) \quad[\because \cos x < 7 \cos x]$
$=x-\cos ^{-1}(7 \cos x)$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2022
If $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$, then $\frac{d y}{d x}$ is equal to :

Options:
A $-\frac{y}{x}$
B $\frac{y}{x}$
C $-\frac{x}{y}$
D $\frac{x}{y}$

The correct answer is:
$-\frac{x}{y}$

Solution:
Let $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 t}{1+t^{2}}$

Put $t=\tan \theta$, we get

$x=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ and $y=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

$\Rightarrow x=\cos 2 \theta$ and $y=\sin 2 \theta$

$\therefore \frac{\mathrm{dx}}{\mathrm{d} \theta}=-2 \sin 2 \theta$ and $\frac{\mathrm{dy}}{\mathrm{d} \theta}=2 \cos 2 \theta$

Now, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{dx}}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2021
If $x=\omega-\omega^{2}-2,$ then the value of $x^{4}+3 x^{3}+2 x^{2}-11 x-6$ is

Options:
A 1
B -1
C 2
D None of these

The correct answer is:
1

Solution:
We have, $x=\omega-\omega^{2}-2$ or $x+2=\omega-\omega^{2}$
Squaring, $x^{2}+4 x+4=\omega^{2}+\omega^{4}-2 \omega^{3}$
$=\omega^{2}+\omega^{3} \omega \cdot-2 \omega^{3}=\omega^{2}+\omega-2 \quad\left[\omega^{3}=1\right]$
$=-1-2=-3 \Rightarrow$
$x^{2}+4 x+7=0$
Dividing $x^{4}+3 x^{3}+2 x^{2}-11 x-6$ by $x^{2}+4 x+7$
we get $x^{4}+3 x^{3}+2 x^{2}-11 x-6=\left(x^{2}+4 x+7\right)\left(x^{2}-x-1\right)+1$
$=(0)\left(\mathrm{x}^{2}-\mathrm{x}-1\right)+1=0+1=1$

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2012
If x2=sinθcosθ0-cosθsinθ1sinθcosθ2 then the value of 4x2+xsin3π2+5 is

Options:
A 13-2
B 13+2
C 2-13
D Both (A) and (B)

The correct answer is:
Both (A) and (B)

Solution:
Given, x2=sinθcosθ0-cosθsinθ1sinθcosθ2

On expanding sinθcosθ0-cosθsinθ1sinθcosθ2 along C3, we get

x2=0-1sinθcosθsinθcosθ+2sinθcosθ-cosθsinθ=-1sinθcosθ-sinθcosθ+2sin2θ+cos2θ

=-1×0+2×1

⇒x2=2⇒x=±2

If x=2, then

4x2+xsin3π2+5=4×22-2+5

=8-2+5=13-2

If x=-2, then

4x2+xsin3π2+5=4×-22+2+5

=8+2+5=13+2

Topic: Mathematics, Determinants, BITSAT, BITSAT 2018
If $\int \frac{3 x+1}{(x-3)(x-5)} d x=\int \frac{-5}{(x-3)} d x+\int \frac{B}{(x-5)} d x$ then the value of $B$ is

Options:
A 3
B 4
C 6
D 8

The correct answer is:
8

Solution:
We have,

$\begin{array}{l}

\frac{3 x+1}{(x-3)(x-5)}=\frac{-5}{x-3}+\frac{\mathrm{B}}{x-5} \\

3 x+1=-5(x-5)+\mathrm{B}(x-3) \\

\text { Put } x=5 \\

3(5)+1=\mathrm{B}(5-3) \\

16=2 \mathrm{~B} \text { or } \mathrm{B}=8

\end{array}$

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2021
If ∫(x)5dx(x)7+x6=λlogx9x9+1+C, then a+λ equal to

Options:
A 2
B >2
C <2
D >3

The correct answer is:
>2

Solution:
Let I=∫dxx2+x7

=∫dxx71(x)5+1

Put 1+1x5=t

⇒ -52·1(x)7dx=dt

∴ I=-25∫dtt=-25logt+C

=-25log1+1x5+C

=-25logx5/2+1x5/2+C

⇒ λ=-25 and a=52

∴ a+λ=2.1>2

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2017
If $x^{18}$ occurs in the rth term in the expansion of $\left(x^4+\frac{1}{x^3}\right)^{15}$, then what is the value of $r$ ?

Options:
A
3
B
5
C
7
D
9

The correct answer is:
7

Solution:
In the expansion of $\left(x^4+\frac{1}{x^3}\right)^{15}$, let $T_r$ is the $r^{\text {th }}$ term
$\begin{aligned}
T_r & ={ }^{15} C_{r-1}\left(x^4\right)^{15-r+1}\left(\frac{1}{x^3}\right)^{r-1} \\
& ={ }^{15} C_{r-1} x^{64-4 r-3 r+3}=15_{C_{r-1}} x^{67-7 r}
\end{aligned}$
$\mathrm{x}^{18}$ occurs in this term
$\Rightarrow 18=67-7 r \Rightarrow 7 r=49 \Rightarrow r=7 .$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If $x=a \sin \theta$ and $y=b \cos \theta,$ then $\frac{d^{2} y}{d x^{2}}$ is

Options:
A $\frac{a}{b^{2}} \sec ^{2} \theta$
B $\frac{-b}{a} \sec ^{2} \theta$
C $\frac{-b}{a^{2}} \sec ^{3} \theta$
D $\frac{b}{a^{2}} \sec ^{3} \theta$

The correct answer is:
$\frac{-b}{a} \sec ^{2} \theta$

Solution:
(b) Given $x=a \sin \theta$ and $y=b \cos \theta$
$\Rightarrow \frac{d x}{d \theta}=a \cos \theta$ and $\frac{d y}{d \theta}=-b \sin \theta$
$\therefore \quad \frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}=-\frac{b}{a} \tan \theta \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-b}{a} \sec ^{2} \theta$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2012
If $\phi(\mathrm{x})$ is a differential function, then the solution of the differential equation $\mathrm{dy}+\left\{\mathrm{y} \phi^{\prime}(\mathrm{x})-\phi(\mathrm{x})\right.$

$\left.\phi^{\prime}(x)\right\} d x=0,$ is

Options:
A $y=\{\phi(x)-1\}+C e^{-\phi(x)}$
B $y \phi(x)=\{\phi(x)\}^{2}+C$
C $y e^{\varphi(x)}=\phi(x) e^{\phi(x)}+C$
D $y-\phi(x)=\phi(x) e^{-\phi(x)}$

The correct answer is:
$y=\{\phi(x)-1\}+C e^{-\phi(x)}$

Solution:
Given differential equation is

$\mathrm{dy}+\left\{y \phi^{\prime}(x)-\phi(x) \phi^{\prime}(x)\right\} d x=0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\phi^{\prime}(\mathrm{x}) \mathrm{y}=\phi(\mathrm{x}) \phi^{\prime}(\mathrm{x})$

which is a linear differential equation with $\quad P=\phi^{\prime}(x), Q=\phi(x) \cdot \phi^{\prime}(x)$ and

$I \cdot F=e^{\int \phi^{\prime}(x) d x}=e^{\phi(x)}$

$\therefore$ Solution is $y \cdot e^{\phi(x)}=\int \phi(x) \cdot \phi^{\prime}(x) e^{\phi(x)} d x+C$

$\Rightarrow y \cdot e^{\phi(x)}=\int \phi(x) \cdot e^{\phi(x)} \phi^{\prime}(x) d x+C$

$\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-\int \phi^{\prime}(x) e^{\phi(x)} d x+C$

$\Rightarrow y \cdot e^{\phi(x)}=\phi(x) e^{\phi(x)}-e^{\phi(x)}+C$

$\Rightarrow y=\lceil\phi(x)-1]+C e^{-\phi(x)}$

Topic: Mathematics, Differential Equations, BITSAT, BITSAT 2015
If $x$ is positive then the sum to infinity of the
series $\frac{1}{1+3 x}-\frac{1-3 x}{(1+3 x)^{2}}+\frac{(1-3 x)^{2}}{(1+3 x)^{3}}-\frac{(1-3 x)^{3}}{(1+3 x)^{4}}$ \ldots $\infty$ is

Options:
A $1 / 2$
B $\frac{1}{6 x}$
C $\frac{1}{6 x(1+3 x)}$
D $\frac{1}{2(1+3 x)}$

The correct answer is:
$1 / 2$

Solution:
The series is a G.P. with common ratio
$=\left(\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right)$ and $|\mathrm{r}|=\left|\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right|$ is less than 1 since $\mathrm{x}$ is
positive $\mathrm{S}_{\infty}=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\frac{1}{1+3 \mathrm{x}}}{1-\left\{-\left(\frac{1-3 \mathrm{x}}{1+3 \mathrm{x}}\right)\right\}}=\frac{1}{2}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2012
If x=log53+log75+log97, then

Options:
A x>32
B x>123
C x>323
D x>23

The correct answer is:
x>323

Solution:
We have, x=log53+log75+log97

Using AM≥GM for three positive real numbers a+b+c3≥abc13. Equality holds when a=b=c.

∴ log53+log75+log973>log53·log75·log971/3⇒ x>3log3log5×log5log7×log7log91/3

⇒ x>3121/3⇒x>323

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2019
If x satisfies $|3 x-2|+|3 x-4|+|3 x-6| \geq 12$, then

Options:
A $0 \leq \mathrm{x} \geq \frac{8}{3}$
B $x \geq \frac{8}{3}$ or $\frac{-4}{3}$
C $x \leq 0$ or $x \geq \frac{8}{3}$
D $x \geq 2$ only

The correct answer is:
$x \leq 0$ or $x \geq \frac{8}{3}$

Solution:
Dividing $\mathrm{R}$ at $\frac{2}{3}, \frac{4}{3}$ and $2,$ analyse 4 cases. When $\mathrm{x} \leq \frac{2}{3},$ the inequality becomes $2-3 x+4-3 x+6-3 x \geq 12$

implying $-9 x \geq 0 \Rightarrow x \leq 0$

when $\mathrm{x} \geq 2$ the ineqality becomes $3 x-2+3 x-4+3 x-6 \geq 12$

Implying $9 \mathrm{x} \geq 24 \Rightarrow \mathrm{x} \geq 8 / 3$

The inequality in invalid in the other two sections.

$\therefore$ either $x \leq 0$ or $x \geq 8 / 3$

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2013
If $x \sin ^{3} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$ and

$x \sin \theta=y \cos \theta,$ then $x^{2}+y^{2}=$

Options:
A 1
B 2
C 0
D None of these

The correct answer is:
1

Solution:
$x \sin ^{3} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$

and $x \sin \theta=y \cos \theta$

Equation (i) may be written as $x \sin \theta \cdot \sin ^{2} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$

$\Rightarrow y \cos \theta \sin ^{2} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$

$\Rightarrow y \cos \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\sin \theta \cos \theta$

$\Rightarrow y \cos \theta=\sin \theta \cos \theta \therefore y=\sin \theta \quad \ldots$

Putting the value of y from

(iii) in (ii), we get $\mathrm{x} \sin \theta=\sin \theta \cdot \cos \theta \Rightarrow \mathrm{x}=\cos \theta \quad \ldots$ (iv)

Squaring (iii) and (iv) and adding, we get $\mathrm{x}^{2}+\mathrm{y}^{2}=\cos ^{2} \theta+\sin ^{2} \theta=1$

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2013
If x, v and a denote the displacement, the velocity and the acceleration of a particle executing SHM of time period T. Then, which of the following does not change with time?

Options:
A aTx
B aT+2πv
C aTv
D a2T2+4π2v2

The correct answer is:
aTx

Solution:
We know, instantaneous displacement,

x=r sinωt

∵ Instantaneous velocity, v=dxdt=rω cosωt

So. aTx=-ω2x×Tx=-ω2T=-4π2T2×T

=-4π2T= constant

and,

⇒ aTV=-ω2rsinωt×Tωrcosωt=-ωTtanωT

=-2πT×T tanωt=not constant

Similarly, aT+2πv and a2T2+4π2v2 is also not constant,
i.e. both are function of t

Topic: Physics, Oscillations, BITSAT, BITSAT 2018
If $x, y$ and $z$ are real numbers, the $x^{2}+4 y^{2}+9 z^{2}-6 y z-3 z x-2 x y$ is always

Options:
A positive
B non-positive
C zero
D non-negative

The correct answer is:
non-negative

Solution:
$x^{2}+4 y^{2}+9 z^{2}-6 y z-3 z x-2 x y$

$=x^{2}+(2 y)^{2}+(3 z)^{2}-(2 y)(3 z)-(3 z)(x)-$ $x(2 y) \geq 0 \forall x, y, z$

$\left[\because \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0\right]$

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2020
If $\rho=\left\{(x, y) \mid x^{2}+y^{2}=1 ; x, y \in \mathbb{R}\right\}$. Then, $\rho$ is

Options:
A reflexive
B symmetric
C transitive
D anti-symmetric

The correct answer is:
symmetric

Solution:
Obviously, the relation is not reflexive and transitive, but it is symmetric, because $x^{2}+x^{2}=2 x^{2} \neq 1$

$$

\begin{array}{l}

\text { and } x^{2}+y^{2}=1, y^{2}+z^{2}=1 \\

\Rightarrow x^{2}+z^{2}=1 \\

\text { But } x^{2}+y^{2}=1 \Rightarrow y^{2}+x^{2}=1

\end{array}

$$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2015
If $x y+y^{2}=\tan x+y,$ then find $\frac{d y}{d x}$ is

Options:
A $\frac{\sec ^{2} x}{x+2 y}$
B $\frac{\sec ^{2} x-y}{(x+2 y-1)}$
C $(x+2 y-1) \sec ^{2} x$
D $\sec ^{2} x \cdot y$

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

Solution:
The given relation is $x y+y^{2}=\tan x+y$. Differentiating both sides with respect to $x$, we get

$\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}$

or $\left[y \cdot 1+x \cdot \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}$

or $(x+2 y-1) \frac{d y}{d x}=\sec ^{2} x-y$

$\therefore \frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}$

Topic: Mathematics, Differentiation, BITSAT, BITSAT 2020
If x,y,z∈R, x+y+z=5, x2+y2+z2=9, then length of interval in which x lies is

Options:
A 8/3
B 4/3
C 2/3
D 1/3

The correct answer is:
4/3

Solution:
We have,

x+y+z=5

x2+y2+z2=9

⇒ x2+y2+(5-x-y)2=9

⇒ y2+x2-5x-5y+xy+8=0

⇒ y2+y(x-5)+x2-5x+8=0

⇒ Since y∈R

(x-5)2-4x2-5x+8≥0

⇒ x2-10x+25-4x2+20x-32≥0

⇒ 3x2-10x+7≤0

⇒ 3x2-7x-3x+7≤0

⇒ (3x-7)(x-1)≤0

⇒ x∈1,73
Length of interval =73-1=43

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2019