datavorous/entrance-exam-dataset huggingface dataset:
If $\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}=3,$ then $\cos \theta_{1}+\cos \theta_{2}$$+\cos \theta^{3}=$

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Since, $\sin \theta_{1}+\sin \theta_{2}+\sin \theta_{3}=3$

$\begin{array}{ll}\therefore & \sin \theta_{1}=\sin \theta_{2}=\sin \theta_{3}=1 \Rightarrow \theta_{1}=\theta_{2}=\theta_{3}= \\ & \frac{\pi}{2} \\ \therefore & \cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}=0\end{array}$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2016
If sin-1x+sin-1y+sin-1z=3π2 andf(1)=2,f(p+q)=f(p)·f(q),∀p,q∈R, then xf(1)+yf(2)+zf(3)-(x+y+z)xf(1)+yf(2)+zf(3) is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
∵ -π2≤sin-1x≤π2,-π2≤sin-1y≤π2

and -π2≤sin-1z≤π2

Given that,

sin-1x+sin-1y+sin-1z=3π2

Which is possible only when

sin-1x=sin-1y=sin-1z=π2

⇒ x=y=z=1

Put p=q=1

Then, f(2)=f(1)f(1)=2×2=4

and put p=1,q=2

then, f(3)=f(1) f(2)

=2·22=8

∴ xf(1)+yf(2)+zf(3)-x+y+zxf(1)+yf(2)+zf(3)

=1+1+1-31+1+1=3-1=2

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2017
If sin-1x+tan-1x=π/2, then 2x2+1=

Options:
A 5
B 5-12
C 2
D 5+12

The correct answer is:
5

Solution:
We have,

sin-1x+tan-1x=π2⇒0<x<1

∴ tan-1x=π2-sin-1x=cos-1x

⇒ tan-1x=tan-11-x2x

⇒ x2=1-x2

⇒ x4+x2-1=0⇒x2=-1±52

as x2>0

∴ x2=5-12

⇒2x2+1=5

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2019
If $\sin ^{2} \theta+\sin ^{2} \phi=1 / 2, \cos ^{2} \theta+\cos ^{2} \phi=3 / 2,$ then $\cos ^{2}(\theta-\phi)$ is equal to

Options:
A $3 / 8$
B $5 / 8$
C $3 / 4$
D $5 / 4$

The correct answer is:
$5 / 8$

Solution:
Using cosine formula
$2 \sin (\theta+\phi) \cos (\theta-\phi)=1 / 2$ ...(i)
$2 \cos (\theta+\phi) \cos (\theta-\phi)=3 / 2$ ...(ii)
Squaring (1) and (2) and then adding $\cos ^{2}(\theta-\phi)=\frac{1}{4}+\frac{9}{4}=\frac{5}{2} \Rightarrow \cos ^{2}(\theta-\phi)=\frac{5}{8}$

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2012
If \(\sin \theta=-\frac{1}{2}\) and \(\tan \theta=1 / \sqrt{3}\), then \(\theta=\)

Options:
A \(2 \mathrm{n} \pi+\pi / 6\)
B \(2 \mathrm{n} \pi+11 \pi / 6\)
C \(2 \mathrm{n} \pi+7 \pi / 6\)
D \(2 \mathrm{n} \pi+\pi / 4\)

The correct answer is:
\(2 \mathrm{n} \pi+7 \pi / 6\)

Solution:
We shall first consider values of \(\theta\) between 0
\(\begin{aligned}
& \text { and } 2 \pi \sin \theta=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right) \\
& \text { or } \sin (2 \pi-\pi / 6) \\
& \therefore \theta=7 \pi / 6 ; 11 \pi / 6 \\
& \tan \theta=1 / \sqrt{3}=\tan (\pi / 6) \text { or } \tan (\pi+\pi / 6) \\
& \therefore \theta=\pi / 6,7 \pi / 6
\end{aligned}\)
The value of \(\theta\) which satisfies both the equations is \(7 \pi / 6\)
Hence the general value of \(\theta\) is \(2 n \pi+7 \pi / 6\) where \(\mathrm{n} \in \mathrm{I}\)

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2011
If sound travels in air with the speed of 340 m s-1, then the number of tones present in an open organ pipe of length 2 m for a maximum frequency of 1200 Hz, are

Options:
A 17
B 11
C 9
D 14

The correct answer is:
14

Solution:
Given, speed of sound, v=340 m s-1
Length of open organ pipe, l=2 m

Frequency, f=1200 Hz
For open organ pipe,

Fundamental frequency, f0=v2l=3402×2=85

∴ Number of tones present in the open organ pipe
=ff0=120085=14.11≃14

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2019
If system of equation ax+y+z=a,x+by+z=b and x+y+cz=c is inconsistent, then which of the following is correct?

Options:
A abc-a-b-c+2=0
B abc-a-b-c+3=0, a=1
C abc-a-b-c+3=0
D abc-a-b-c+2=0, a≠1,b≠1,c≠1

The correct answer is:
abc-a-b-c+2=0, a≠1,b≠1,c≠1

Solution:
We have,

a x+y+z=a ...(i)

x+by+z=b ...(ii)

x+y+cz=c ...(iii)

are inconsistent.

∴a111b111c=0 and a11bb1c1c≠0

⇒ a111b111c=0

⇒ a(bc-1)-1(c-1)+1(1-b)=0

abc-a-c+1+1-b=0

abc-a-b-c+2=0

and a11bb1c1c≠0

It is possible only when a≠1,b≠1 and c≠1

Topic: Mathematics, Determinants, BITSAT, BITSAT 2019
If $\mathrm{T}_{0}, \mathrm{~T}_{1}, \mathrm{~T}_{2} \ldots \ldots \mathrm{T}_{\mathrm{n}}$ represent the terms in the

expansion of $(x+a)^{n}$, then $\left(\mathrm{T}_{0}-\mathrm{T}_{2}+\mathrm{T}_{4}-\ldots \ldots .\right)^{2}+$

$\left(\mathrm{T}_{1}-\mathrm{T}_{3}+\mathrm{T}_{5}-\ldots . .\right)^{2}=$

Options:
A $\left(x^{2}+a^{2}\right)$
B $\left(x^{2}+a^{2}\right)^{n}$
C $\left(x^{2}+a^{2}\right)^{1 / n}$
D $\left(x^{2}+a^{2}\right)^{-1 / n}$

The correct answer is:
$\left(x^{2}+a^{2}\right)^{n}$

Solution:
From the given condition, replacing $a$ by a and $-a i$ respectively, we get

$(x+a i)^{n}=\left(T_{0}-T_{2}+T_{4}-\ldots \ldots\right)+i\left(T_{1}-T_{3}+T_{5}-\ldots .\right)$ ...(i)

and

$(x-a i)^{n}=\left(T_{0}-T_{2}+T_{4}-\ldots \ldots\right)-i\left(T_{1}-T_{3}+T_{5}-\ldots\right)$...(ii)

Multiplying (ii) and

(i) we get required result i.e.,

$$

\left(x^{2}+a^{2}\right)^{n}=\left(T_{0}-T_{2}+T_{4}-\ldots .\right)^{2}+\left(T_{1}-T_{3}+T_{5}-\ldots\right)^{2}

$$

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2014
If $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{ML}^{3}}{3 \mathrm{Yq}}}$ then find the dimensions of q. Where $\mathrm{T}$ is the time period of bar of mass $\mathrm{M}$ length $\mathrm{L}$ and Young modulus $\mathrm{Y}$.

Options:
A $[\mathrm{L}]$
B $\left[\mathrm{L}^{2}\right]$
C $\left[\mathrm{L}^{4}\right]$
D $\left[\mathrm{L}^{3}\right]$

The correct answer is:
$\left[\mathrm{L}^{4}\right]$

Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{ML}^{3}}{3 \mathrm{Yq}}},$ writing dimensions of both
the sides, we get $[\mathrm{T}]=\left[\frac{\mathrm{ML}^{3}}{\mathrm{ML}^{-1} \mathrm{~T}^{-2} \mathrm{q}}\right]^{1 / 2}$
or $\quad q=\left[L^{4}\right]$

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2011
If $\tan (\cot x)=\cot (\tan x)$, then $\sin 2 x$ is equal to:

Options:
A $\frac{2}{(2 n+1) \pi}$
B $\frac{4}{(2 n+1) \pi}$
C $\frac{2}{n(n+1) \pi}$
D $\frac{4}{n(n+1) \pi}$

The correct answer is:
$\frac{4}{(2 n+1) \pi}$

Solution:
Given, $\tan (\cot x)=\cot (\tan x)=\tan \left(\frac{\pi}{2}-\tan x\right)$

$\Rightarrow \cot x=n \pi+\frac{\pi}{2}-\tan x$

$\Rightarrow \cot x+\tan x=n \pi+\frac{\pi}{2}$

$\Rightarrow \frac{1}{\sin x \cos x}=n \pi+\frac{\pi}{2} \Rightarrow \frac{1}{\sin 2 x}=\frac{n \pi}{2}+\frac{\pi}{4}$

$\Rightarrow \sin 2 x=\frac{1}{\frac{n \pi}{2}+\frac{\pi}{4}}=\frac{4}{(2 n+1) \pi}$

Topic: Mathematics, Trigonometric Equations, BITSAT, BITSAT 2016
If, \(\tan \mathrm{A}=\frac{1}{2}\) and \(\tan B=\frac{1}{3}\), then find the value of \(\mathrm{A}+\mathrm{B}\)

Options:
A \(\pi\)
B \(\frac{\pi}{2}\)
C \(\frac{\pi}{4}\)
D \(\frac{-\pi}{4}\)

The correct answer is:
\(\frac{\pi}{4}\)

Solution:
\(\begin{aligned}
& \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}=1 \\
& \therefore A+B=45^{\circ}=\frac{\pi}{4}
\end{aligned}\)

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2011
If temperature of a black body increases from 300 K to 900 K, then the rate of energy radiation increases by

Options:
A 81
B 3
C 9
D 2

The correct answer is:
81

Solution:
According to Stefan's law ρ=Et=σAeT4

So, we can write as

E2E1=T2T14

E2E1=9003004⇒E2E1=34

E2=81E1 ⇒E2E1=81

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2017
If the $(2 \mathrm{p})^{\text {th }}$ term of a $\mathrm{H}$. $\mathrm{P}$. is $\mathrm{q}$ and the $(2 \mathrm{q})^{\text {th }}$ term is $\mathrm{p},$ then the $2(\mathrm{p}+\mathrm{q})^{\mathrm{th}}$ term is-

Options:
A $\frac{p q}{2(p+q)}$
B $\frac{2 \mathrm{pq}}{\mathrm{p}+\mathrm{q}}$
C $\frac{\mathrm{pq}}{\mathrm{p}+\mathrm{q}}$
D $\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}$

The correct answer is:
$\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}$

Solution:
If a is the first term and d is the common difference of the associated A.P. $\frac{1}{\mathrm{q}}=\frac{1}{\mathrm{a}}+(2 \mathrm{p}-1) \mathrm{d}, \frac{1}{\mathrm{p}}=\frac{1}{\mathrm{a}}+(2 \mathrm{q}-1) \mathrm{d}$

$\Rightarrow \mathrm{d}=\frac{1}{2 \mathrm{pq}}$

If h is the $2(\mathrm{p}+\mathrm{q})^{\mathrm{th}}$ term $\frac{1}{\mathrm{~h}}=\frac{1}{\mathrm{a}}+(2 \mathrm{p}+2 \mathrm{q}-1) \mathrm{d}$

$=\frac{1}{\mathrm{q}}+\frac{1}{\mathrm{p}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2014
If the $B-H$ curves of two samples of $X$ and $Y$ of iron are as shown below, then which one of the following statement is correct?

Options:
A
Both $X$ and $Y$ are suitable for making electromagnets.

B
Both $X$ and $Y$ are suitable for making permanent magnet.

C
$X$ is suitable for making permanent magnet and $Y$ for making electromagnet.

D
$X$ is suitable for making electromagnet and $Y$ is suitable for permanent magnet.

The correct answer is:

$X$ is suitable for making permanent magnet and $Y$ for making electromagnet.

Solution:
For permanent magnet we prefer a material with high retentivity and high coercivity. For electromagnet, we prefer high saturated magnetism, low coercivity and least possible area of hystersis loop. Therefore, $X$ is suitable for making permanent magnet and $Y$ for making electromagnet.

Topic: Physics, Magnetic Properties of Matter, JEE Main
If the $B-H$ curves of two samples of $X$ and $Y$ of iron are as shown below, then which one of the following statement is correct?

Options:
A
Both $X$ and $Y$ are suitable for making electromagnets.

B
Both $X$ and $Y$ are suitable for making permanent magnet.

C
$X$ is suitable for making permanent magnet and $Y$ for making electromagnet.

D
$X$ is suitable for making electromagnet and $Y$ is suitable for permanent magnet.

The correct answer is:

$X$ is suitable for making permanent magnet and $Y$ for making electromagnet.

Solution:
For permanent magnet we prefer a material with high retentivity and high coercivity. For electromagnet, we prefer high saturated magnetism, low coercivity and least possible area of hystersis loop. Therefore, $X$ is suitable for making permanent magnet and $Y$ for making electromagnet.

Topic: Physics, Magnetic Properties of Matter, BITSAT, BITSAT 2022
If the coefficient of x3 and x4 in the expansion of 1+ax+bx21-2x18 in powers of x are both zero, then a, b is equal to

Options:
A 16,2513
B 14,2513
C 14,2723
D 16,2723

The correct answer is:
16,2723

Solution:
We have,

1+ax+bx21-2x18

1+ax+bx21-C1182x+C2182x2-C3182x3+C4182x4…

Coefficient of x3 is -C31823+a·C21822-b18C12

and coefficient of x4 is -C41824-C31823a+C21822b

Coefficient of x3 and x4 are zero.

∴ -C31823+C21822a-C1182b=0

⇒ 4×17×163×2-17a+b=0 ...i

and 80-323a+b=0 ...ii

Solving Eqs. (i) and (ii), we get

a=16,b=2723

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2019
If the coordinates at one end of a diameter of the circle $x^{2}+y^{2}-8 x-4 y+c=0$ are (-3,2) , then the coordinates at the other end are

Options:
A (5,3)
B (6,2)
C (1,-8)
D (11,2)

The correct answer is:
(11,2)

Solution:
The centre of the given circle is $\mathrm{C} \equiv(4,2)$

Let $A \equiv(-3,2)$

If $(\alpha, \beta)$ are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre,

$\therefore \quad \frac{\alpha-3}{2}=4$ and $\frac{\beta+2}{2}=2 \Rightarrow \alpha=11, \beta=2$

Thus, the coordinates of the other end of diameter are (11,2)

Topic: Mathematics, Circle, BITSAT, BITSAT 2020
If the co-ordinates of the point in which the line joining the points $(3,5,-7)$ and $(-2,1,8)$ is intersected by the plane $y z$ is $\left[\mathrm{a}, \frac{13}{\mathrm{~b}}, \mathrm{c}\right]$, then $(a+b-c)=$

Options:
A
2
B
3
C
4
D
5

The correct answer is:
3

Solution:
Ratio $=-\left(\frac{3}{-2}\right)=\frac{3}{2}$
$\therefore \quad$ Required co-ordinates of the points are
$\begin{aligned}
& {\left[\frac{6-6}{5}, \frac{10+3}{5}, \frac{-14+24}{5}\right]=\left(0, \frac{13}{5}, 2\right) .} \\
& \therefore a+b-c=0+5-2=3
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2023 (Memory Based Paper 2)
If the coordinates of the points $A$ and $B$ be $(3,3)$ and $(7,6)$, then the length of the portion of the line $\mathrm{AB}$ intercepted between the axes is

Options:
A $\frac{5}{4}$
B $\frac{\sqrt{10}}{4}$
C $\frac{\sqrt{13}}{3}$
D None of these

The correct answer is:
$\frac{5}{4}$

Solution:
Equation of line $A B$ is $y-3=\frac{6-3}{7-3}(x-3)$
$\Rightarrow 3 x-4 y+3=0 \Rightarrow \frac{x}{-1}+\frac{y}{3 / 4}=1$
Hence required length is $\sqrt{(-1)^2+\left(\frac{3}{4}\right)^2}=\frac{5}{4}$.

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If the curve y=ax and y=bx intersect at angle α, then tanα is equal to

Options:
A a-b1+ab
B loga-logb1+logalogb
C a+b1-ab
D loga+logb1-logalogb

The correct answer is:
loga-logb1+logalogb

Solution:
Clearly, the point of intersection of curves is (0,1).

Now, slope of tangent of first curve

m1=dydx=axloga

⇒dydx(0,1)=m1=loga

Slope of tangent of second curve,

m2=dydx=bxlogb

⇒ m2=dydx(0,1)=logb

∴ tanα=m1-m21+m1m2=loga-logb1+logalogb

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2017
If the distance between the first maxima and fifth minima of a double slit pattern is $7 \mathrm{~mm}$ and the slits are separated by $0.15 \mathrm{~mm}$ with the screen 50 $\mathrm{cm} .$ from the slits, then the wavelength of the light used is:

Options:
A $200 \mathrm{nm}$
B $100 \mathrm{nm}$
C $800 \mathrm{nm}$
D $600 \mathrm{nm}$

The correct answer is:
$600 \mathrm{nm}$

Solution:
There are three and a half fringes from first maxima to fifth minima as shown.
$\Rightarrow \beta=\frac{7 \mathrm{~mm}}{3.5}=2 \mathrm{~mm} \Rightarrow \lambda=\frac{\beta \mathrm{D}}{\mathrm{d}}=600 \mathrm{nm}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2011
If the eccentricity and length of latus rectum of a hyperbola are $\frac{\sqrt{13}}{3}$ and $\frac{10}{3}$ units respectively, then what is the length of the transverse axis?

Options:
A $\frac{7}{2}$ unit
B 12 unit
C $\frac{15}{2}$ unit
D $\frac{15}{4}$ unit

The correct answer is:
$\frac{15}{4}$ unit

Solution:
Length of latus rectum of a hyperbola is $\frac{2 b^2}{a}$ where $a$ is the half of the distance between two vertex of the hyperbola.
Latus rectum $=\frac{2 b^2}{a}=\frac{10}{3}$ or, $b^2=\frac{5 a}{3}$ ...(i)
In case of hyperbola, $\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)$ ...(ii)
Putting value of $b^2$ from equation (i) and $e=$
$\frac{\sqrt{13}}{3}$ in equation (ii), $\frac{5 \mathrm{a}}{3}=\mathrm{a}^2\left(\frac{13}{9}-1\right)$ or,
$\frac{5 a}{3}=\frac{4 a^2}{9}$
$\Rightarrow 4 a^2-15 a=0$ or, $a(4-15 a)=0$
$a \neq 0$, hence, $a=\frac{15}{4}$

Topic: Mathematics, Hyperbola, BITSAT, BITSAT 2023 (Memory Based Paper 1)
If the elevation in boiling point of a solution of $10 \mathrm{~g}$ of solute $(\mathrm{mol} . \mathrm{wt}=100)$ in $100 \mathrm{~g}$ of water is $\Delta T b$, the ebullioscopic constant of water is

Options:
A $\frac{\Delta T_{b}}{10}$
B $\Delta T_{b}$
C $10 \Delta T_{b}$
D $100 \Delta T_{b}$

The correct answer is:
$\Delta T_{b}$

Solution:
$\mathrm{m}=\frac{1000 \times \mathrm{k}_{b} \times \mathrm{w}}{W \times \Delta T_{b}}$
or $\quad \begin{aligned} k_{\mathrm{b}} &=\frac{\mathrm{m} \times W \times \Delta T_{b}}{1000 \times \mathrm{w}}=\frac{100 \times 100 \times \Delta T_{b}}{1000 \times 10} \\ &=\Delta T_{b} \end{aligned}$

Topic: Chemistry, Solutions, BITSAT, BITSAT 2012
If the energy of a photon is $10 \mathrm{eV}$, then its momentum is

Options:
A $5.33 \times 10^{-23} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
B $5.33 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
C $5.33 \times 10^{-29} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
D $5.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$5.33 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

Solution:
Momentum of a photon $\propto \frac{\mathrm{E}}{\mathrm{c}}$
$$
=5.33 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1}
$$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2011
If the enthalpy of vaporization of water is 186.5 $\mathrm{kJmol}^{-1}$, the entropy if its vaporization will be:

Options:
A $0.5 \mathrm{kJK}^{-1} \mathrm{~mol}^{-1}$
B $1.0 \mathrm{kJK}^{-1} \mathrm{~mol}^{-1}$
C $1.5 \mathrm{kJK}^{-1} \mathrm{~mol}^{-1}$
D $2.0 \mathrm{kJK}^{-1} \mathrm{~mol}^{-1}$

The correct answer is:
$0.5 \mathrm{kJK}^{-1} \mathrm{~mol}^{-1}$

Solution:
Given enthalpy of vaporization,

$$

\Delta \mathrm{H}=186.5 \mathrm{~kJ} \mathrm{~mol}^{-1}

$$

Boiling point of water

$$

=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K}

$$

Entropy change,

$$

\begin{array}{r}

\Delta \mathrm{S}=\frac{\Delta \mathrm{H}}{\mathrm{T}}=\frac{186.5 \mathrm{~kJ} \mathrm{~mol}^{-1}}{373 \mathrm{~K}} \\

=0.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}

\end{array}

$$

Topic: Chemistry, Thermodynamics (C), BITSAT, BITSAT 2014
If the equations of the pairs of opposite sides of a parallelogram are \(x^2-5 x+6=0\) and \(y^2-6 y+5=0\), then equations of its diagonals are

Options:
A \(x+4 y=13, y=4 x-7\)
B \(4 x+y=13,4 y=x-7\)
C \(4 x+y=13, y=4 x-7\)
D \(y-4 x=13, y+4 x=7\)

The correct answer is:
\(4 x+y=13, y=4 x-7\)

Solution:
Equations of the sides of the parallelogram are
\((x-3)(x-2)=0\) and \((y-5)(y-1)=0\)
i.e. \(x=3, x=2 ; y=5, y=1\)
Hence its vertices are : \(\mathrm{A}(2,1) ; \mathrm{B}(3,1)\); \(\mathrm{C}(3,5) ; \mathrm{D}(2,5)\)
Equation of the diagonal \(\mathrm{AC}\) is
\(y-1=\frac{4}{1}(x-2) \Rightarrow y=4 x-7\)
Equation of the diagonal \(\mathrm{BD}\) is
\(y-1=\frac{4}{-1}(x-3) \Rightarrow 4 x+y=13\)

Topic: Mathematics, Parabola, BITSAT, BITSAT 2010
If the equilibrium constant of the reaction $2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}$ is $0.25,$ then the equilibrium constant for the reaction $\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}$ would be

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
When the reaction is reversed,
$$
\mathrm{K}^{\prime}=\frac{1}{\mathrm{~K}}=\frac{1}{0.25}=4
$$

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2009
If the expression \(x^2-11 x+a\) and \(x^2-14 x+2 a\) must have a common factor and \(\mathrm{a} \neq 0\), then, the common factor is

Options:
A \((x-3)\)
B \((x-6)\)
C \((x-8)\)
D None of these

The correct answer is:
\((x-8)\)

Solution:
Here Let \(x-\alpha\) is the common factor then \(\mathrm{x}=\alpha\) is root of the corresponding equation \(\therefore \alpha^2-11 \alpha+a=0\)
\(\alpha^2-14 \alpha+2 \mathrm{a}=0\)
Subtracting \(3 \alpha-a=0 \Rightarrow \alpha=a / 3\)
Hence \(\frac{a^2}{9}-11 \frac{a}{3}+a=0, a=0\) or \(a=24\) since \(\mathrm{a} \neq 0, \mathrm{a}=24\)
\(\therefore\) the common factor of \(\left\{\begin{array}{l}x^2-11 x+24 \\ x^2-14 x+48\end{array}\right.\) is clearly \(x -8\)

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2009
If the force is given by $\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}$ with $\mathrm{t}$ as time. The dimensions of a and b are

Options:
A $\left[\mathrm{MLT}^{-4}\right]$ and $\left[\mathrm{MLT}^{-2}\right]$
B $\left[\mathrm{MLT}^{-3}\right]$ and $\left[\mathrm{MLT}^{-4}\right]$
C $\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]$ and $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$
D $\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]$ and $\left[\mathrm{ML}^{3} \mathrm{~T}^{-4}\right]$

The correct answer is:
$\left[\mathrm{MLT}^{-3}\right]$ and $\left[\mathrm{MLT}^{-4}\right]$

Solution:
Dimension of at = Dimension of F
$[a t]=[F] \Rightarrow[a]=\left[\frac{F}{t}\right]$
$[b]=\left[\frac{M L T^{-2}}{T}\right] \Rightarrow[a]=\left[M L T^{-3}\right]$
Dimension of $\mathrm{bt}^{2}=$ Dimension of $\mathrm{F}$
$\left[\mathrm{bt}^{2}\right]=[\mathrm{F}] \Rightarrow[\mathrm{b}]=\left[\frac{\mathrm{F}}{\mathrm{t}^{2}}\right]$
$[\mathrm{b}]=\left[\frac{\mathrm{ML} \mathrm{T}^{-4}}{\mathrm{~T}^{2}}\right] \Rightarrow[\mathrm{b}]=\left[\mathrm{MLT}^{-4}\right]$

Topic: Physics, Units and Dimensions, BITSAT, BITSAT 2012
If the forward voltage in a semiconductor diode is changed from $0.5 \mathrm{~V}$ to $0.7 \mathrm{~V}$, then the forward current changes by $1.0 \mathrm{~mA}$. The forward resistance of diode junction will be

Options:
A $100 \Omega$
B $120 \Omega$
C $200 \Omega$
D $240 \Omega$

The correct answer is:
$200 \Omega$

Solution:
Forward resistance

$=\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}=\frac{0.7-0.5}{1.0 \times 10^{-3}}=200 \Omega$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2020
If the function $f(x)=\begin{cases}A x-B, & x \leq 1 \\ 3 x & , 1is continuous at $x=2$ and $4,$ then the values of a and $b$ are

Options:
A 3,5
B 3,-5
C 0,3
D 0,5

The correct answer is:
3,-5

Solution:
Since $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=2$
$\therefore \mathrm{f}(2)=\lim _{\mathrm{x} \rightarrow 2^{+}} \mathrm{f}(\mathrm{x}) \Rightarrow 1=\lim _{\mathrm{x} \rightarrow 2^{+}}(\mathrm{ax}+\mathrm{b})$
$\therefore 1=2 \mathrm{a}+\mathrm{b}$
Again $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=4$
$\therefore \mathrm{f}(4)=\lim _{x \rightarrow 4^{-}} \mathrm{f}(\mathrm{x}) \Rightarrow 7=\lim _{\mathrm{x} \rightarrow 4^{-}}(\mathrm{ax}+\mathrm{b})$
$\therefore 7=4 \mathrm{a}+\mathrm{b}$
Solving (1) and $(2),$ we get $a=3, b=-5$

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2012
If the function \(f:(-\infty, \infty) \rightarrow B\) defined by \(f(x)=-x^2+6 x-8\) is bijective, then \(B=\)

Options:
A \([1, \infty)\)
B \((-\infty, 1]\)
C \((-\infty, \infty)\)
D None of these

The correct answer is:
\((-\infty, 1]\)

Solution:
Since the function \(f\) is bijective, therefore \(f\) is onto. Therefore range of \(f=B\).
Let \(y=-x^2+6 x-8\)
\(\Rightarrow \mathrm{x}^2-6 \mathrm{x}+(8+\mathrm{y})=0\)
\(\Rightarrow x=\frac{6 \pm \sqrt{36-4(8+y)}}{2}=\frac{6 \pm \sqrt{4-4 y}}{2}\)
For \(x\) to be real, \(4-4 y \geq 0 \Rightarrow y \leq 1\)
\(\therefore \mathrm{B}=\) range of \(\mathrm{F}=(-\infty, 1]\)

Topic: Mathematics, Trigonometric Ratios & Identities, BITSAT, BITSAT 2009
If the harmonic mean between $a$ and $b$ be $H$, then the value of $\frac{1}{\mathrm{H}-a}+\frac{1}{\mathrm{H}-b}$ is

Options:
A $a+b$
B $a b$
C $\frac{1}{a}+\frac{1}{b}$
D $\frac{1}{a}-\frac{1}{b}$

The correct answer is:
$\frac{1}{a}+\frac{1}{b}$

Solution:
Putting $\mathrm{H}=\frac{2 a b}{a+b}$, we have

$\begin{array}{l}

\frac{1}{\mathrm{H}-a}+\frac{1}{\mathrm{H}-b} \\

=\frac{1}{\left(\frac{2 a b}{a+b}-a\right)}+\frac{1}{\left(\frac{2 a b}{a+b}-b\right)}=\frac{a+b}{a b-a^{2}}+\frac{a+b}{a b-b^{2}}

\end{array}$

$=\left(\frac{a+b}{b-a}\right)\left(\frac{1}{a}-\frac{1}{b}\right)=\left(\frac{a+b}{b-a}\right)\left(\frac{b-a}{a b}\right)=\frac{a+b}{a b}=\frac{1}{a}+\frac{1}{b}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2021
If the length of the major axis of the ellipse x2a2+y2b2=1 is three times the length of minor axis, its eccentricity is

Options:
A 13
B 13
C 23
D 223

The correct answer is:
223

Solution:
We have

x2a2+y2b2=1

Given,

Length of the major axis=3Length of the minor axis

⇒2a=32b

⇒a=3b

Now,

e=1-b2a2

⇒e=1-b29b2

⇒ e=1-19=83=223

Topic: Mathematics, Ellipse, BITSAT, BITSAT 2019
If the line $2 \mathrm{x}-3 \mathrm{y}=\mathrm{k}$ touches the parabola $\mathrm{y}^{2}=6 \mathrm{x},$ then find the value of $\mathrm{k}$

Options:
A $-15 / 4$
B $-27 / 4$
C $-1 / 4$
D $-3 / 4$

The correct answer is:
$-27 / 4$

Solution:
$$

\text { } \begin{aligned}

\text { Given } x=\frac{3 y+k}{2} & \ldots \ldots . . \\

\text { and } y^{2}=6 x & \\

\Rightarrow y^{2}=6\left(\frac{3 y+k}{2}\right) \\

\Rightarrow y^{2}=3(3 y+k) \Rightarrow y^{2}-9 y-3 k=0 & \ldots \ldots . .

\end{aligned}

$$

If line (1) touches parabola (2) then roots of quadratic equation ( 3 ) is equal $\therefore(-9)^{2}=4 \times 1 \times(-3 k) \Rightarrow k=-27 / 4$

Topic: Mathematics, Parabola, BITSAT, BITSAT 2013