datavorous/entrance-exam-dataset huggingface dataset:
The threshold wavelength of the tungsten is 2300 $Ã… .$ If ultraviolet light of wavelength $1800 Ã…$ is incident on it, then the maximum kinetic energy of photoelectrons would be about -

Options:
A $1.49 \mathrm{eV}$
B $2.2 \mathrm{eV}$
C $3.0 \mathrm{eV}$
D $5.0 \mathrm{eV}$

The correct answer is:
$1.49 \mathrm{eV}$

Solution:
$\mathrm{K}_{\max }=\mathrm{hv}-\mathrm{hv}_{0}=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)$

$1.24 \times 10^{-6}\left(\frac{10^{8}}{18}-\frac{10^{8}}{23}\right)=1.49 \mathrm{eV}$

Topic: Physics, Dual Nature of Matter, BITSAT, BITSAT 2014
The time period of a bob performing simple harmonic motion in water is 2 s. If density of bob is 43×103 kg m-3, then time period of bob performing simple harmonic motion in air will be

Options:
A 3 s
B 4 s
C 2 s
D 1 s

The correct answer is:
1 s

Solution:
Given, density of bob, ρ=43×103 kg m-3
Density of water, σ=103 kg m-3

If g' be gravitational acceleration in water then,

g'=g 1-σρ=g 1-10343×103=g4

As, Tair =2πIg,

Similarly, Twater =2πIg' Twater =2 s g'=g4

2=2π Ig/4=2π Ig ·2

2=2.Tair

⇒ Tair =1 s

Topic: Physics, Oscillations, BITSAT, BITSAT 2019
The torque of a force $5 \hat{i}+3 \hat{j}-7 \hat{k}$ about the origin is $\tau$. If the force acts on a particle whose position vector is $2 \hat{i}+2 \hat{j}+\hat{k}$, then the value of $\tau$ will be :

Options:
A $11 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$
B $-11 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}-16 \hat{\mathrm{k}}$
C $-17 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$
D $17 \hat{i}+9 \hat{j}+16 \hat{k}$

The correct answer is:
$-17 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$

Solution:
Given that torque, $\vec{\tau}=5 \hat{i}+3 \hat{j}-7 \hat{k}$
Position vector, $\overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}++2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\begin{aligned} & \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7\end{array}\right| \\ & =\hat{\mathrm{i}}(-14-3)-\hat{\mathrm{j}}(-14-5)+\hat{\mathrm{k}}(6-10) \\ & =-17 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\end{aligned}$

Topic: Physics, Rotational Motion, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The total energy of an electron in the first excited state of hydrogen is about -3.4 eV. Its kinetic energy in this state is

Options:
A -3.4 eV
B -6.8 eV
C 6.8 eV
D 3.4 eV

The correct answer is:
3.4 eV

Solution:
We have, Kinetic energy of electron,

K=Ze28πε0·r

also, Potential energy of electron,

U=-14πε0·Ze2r

∴ Total energy

E=K+U

=Ze28πε0·r-Ze24πε0·r

E=-Ze28πε0r or K=-E=--3.4

=3.4 eV

Topic: Physics, Atomic Physics, BITSAT, BITSAT 2017
The total number of 4 -digit numbers in which the digits are in descending order, is

Options:
A ${ }^{n} C_{4} \times 4 !$
B ${ }^{10} \mathrm{C}_{4}$
C $\frac{10 !}{4 !}$
D None of these

The correct answer is:
${ }^{10} \mathrm{C}_{4}$

Solution:
Total number of arrangements of 10 digits $0,1,2, \ldots, 9$ bytaking 4 at atime $={ }^{10} \mathrm{C}_{4} \times 4 !$

We observe that in every arrangement of 4 selected digits there is just one arrangement in which the digits are in descending order.

$\therefore$ Required number of 4 -digit numbers.

$$

=\frac{{ }^{10} \mathrm{C}_{4} \times 4 !}{4 !}={ }^{10} \mathrm{C}_{4}

$$

Topic: Mathematics, Permutation Combination, BITSAT, BITSAT 2015
The total number of $\mathrm{P}-\mathrm{OH}$ bonds for pyrophosphoric acid

Options:
A
4

B
5

C
6

D
8

The correct answer is:

4

Solution:
The structure of pyrophosphoric acid is

There are $4 \mathrm{P}-\mathrm{OH}$ bonds, $1 \mathrm{P}-\mathrm{O}-\mathrm{P}$ bond and two $\mathrm{P}=\mathrm{O}$ bonds.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2022
The transfer RNA anticodon for the messenger RNA codon G-C-A is

Options:
A C-G-U
B C-C-U
C U-C-C
D G-U-C

The correct answer is:
C-G-U

Solution:
The transfer RNA anticodon for messanger RNA codon G-C-A is

C-G-U, because in anticodon of RNA,

G replaced by C
C replaced by G
A replaced by U

Here, C(cytosine) forms triple bond with G (guanine) and A (adenine) forms double bond with U (uracil).

Topic: Chemistry, Biomolecules, BITSAT, BITSAT 2017
The transverse displacement $y(x, t)$ of a wave on a string is given by
$y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b}\right.}$ xt $)$
This represents a

Options:
A wave moving in $-x$ direction, speed $\sqrt{\frac{b}{a}}$
B standing wave of frequency $\sqrt{b}$
C standing wave of frequency $\frac{1}{\sqrt{b}}$
D wave moving in $+x$ direction, speed $\sqrt{\frac{a}{b}}$

The correct answer is:
wave moving in $-x$ direction, speed $\sqrt{\frac{b}{a}}$

Solution:
$y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}=e^{-(\sqrt{a x}+\sqrt{b t})^{2}}$
It is a function of type $y=f(\omega t+k x)$ $\therefore y(x, t)$ represents wave travelling along $-x$ direction.
$\text { Speed of wave }=\frac{\omega}{k}=\frac{\sqrt{b}}{\sqrt{a}}=\sqrt{\frac{b}{a}}$

Topic: Physics, Waves and Sound, BITSAT, BITSAT 2012
The truth table given below correspond to the logic gate
\begin{tabular}{cc|c}
$\mathrm{A}$ & $\mathrm{B}$ & $\mathrm{X}$ \\
\hline 0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{tabular}

Options:
A $\mathrm{OR}$
B NOR
C AND
D NAND

The correct answer is:
NOR

Solution:
The given gate is a NOR gate. The output is high, when all inputs are low.
Truth Table:
\begin{tabular}{|c|c|c|}
\hline $\mathbf{A}$ & $\mathbf{B}$ & $\mathbf{Y}$ \\
\hline 0 & 0 & 1 \\
\hline 1 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 1 & 0 \\
\hline \multicolumn{2}{|c|} { Logic Symbols } \\
\hline
\end{tabular}
Boolean expression: $\overline{\mathrm{A}+\mathrm{B}}=\mathrm{Y}$

Topic: Physics, Semiconductors, BITSAT, BITSAT 2009
The two ends of a rod of length $L$ and a uniform crosssectional area $A$ are kept at two temperatures $T_{1}$ and $\mathrm{T}_{2}\left(\mathrm{~T}_{1}>\mathrm{T}_{2}\right)$. The rate of heat transfer, $\frac{\mathrm{dQ}}{\mathrm{dt}}$ through the rod in a steady state is given by:

Options:
A $\frac{k\left(T_{1}-T_{2}\right)}{L A}$
B $k L A\left(T_{1}-T_{2}\right)$
C $\frac{k A\left(T_{1}-T_{2}\right)}{L}$
D $\frac{k L\left(T_{1}-T_{2}\right)}{A}$

The correct answer is:
$\frac{k A\left(T_{1}-T_{2}\right)}{L}$

Solution:
$\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{L}}$ $\left[\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)\right.$ is the temperature difference $]$

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2021
The two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ are drawn from a common point and $\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}},$ then angle
between $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is -
(1) $90^{\circ}$ if $\mathrm{C}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}$
(2) greater than $90^{\circ}$ if $\mathrm{C}^{2}<\mathrm{A}^{2}+\mathrm{B}^{2}$
(3) greater than $90^{\circ}$ if $\mathrm{C}^{2}>\mathrm{A}^{2}+\mathrm{B}^{2}$
(4) less than $90^{\circ}$ if $\mathrm{C}^{2}>\mathrm{A}^{2}+\mathrm{B}^{2}$
Correct options are -

Options:
A 1,2
B 1,2,3,4
C 2,3,4
D 1,2,4

The correct answer is:
1,2,4

Solution:
No Solution available

Topic: Physics, Mathematics in Physics, BITSAT, BITSAT 2011
The unit of rate constant for reactions of second order is

Options:
A $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$
B $\mathrm{L}^{-1} \mathrm{~mol} \mathrm{~s}^{-1}$
C $\mathrm{L} \mathrm{mol} \mathrm{s}^{-1}$
D $\mathrm{s}^{-1}$

The correct answer is:
$\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$

Solution:
Unit of rate constant for second order reaction is $\mathrm{L} \mathrm{mol}^{-1} \mathrm{sec}^{-1}$

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2015
The unit vector perpendicular to the vectors $6 \hat{i}+2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-6 \hat{j}-2 \hat{k}$ is $-$

Options:
A $\frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7}$
B $\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$
C $\frac{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$
D $\frac{2 \hat{i}+3 \hat{j}+6 \hat{k}}{7}$

The correct answer is:
$\frac{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$

Solution:
Unit vector perpendicular to both the given vectors is, $\frac{(6 \hat{i}+2 \hat{j}+3 \hat{k}) \times(3 \hat{i}-6 \hat{j}-2 \hat{k})}{|(6 \hat{i}+2 \hat{j}+3 \hat{k}) \times(3 \hat{i}-6 \hat{j}-2 \hat{k})|}=\frac{2 \hat{i}+3 \hat{j}-6 \hat{k}}{7}$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2012
The unsaturated hydrocarbon which on ozonolysis gives one mole each of formaldehyde, acetaldehyde and methyl glyoxal $\left(\mathrm{CH}_{3} \mathrm{COCHO}\right)$ is

Options:
A $\quad \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_{3}\right)-\mathrm{CH}_{3}$
B $\quad \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}$
C $\quad \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{C}\left(\mathrm{CH}_{3}\right)=\mathrm{CH}-\mathrm{CH}_{3}$
D $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{3}$

The correct answer is:
$\quad \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{C}\left(\mathrm{CH}_{3}\right)=\mathrm{CH}-\mathrm{CH}_{3}$

Solution:

Topic: Chemistry, Hydrocarbons, BITSAT, BITSAT 2010
The upper end of a wire of diameter $12 \mathrm{~mm}$ and length $1 \mathrm{~m}$ is clamped and its other end is twisted through an angle of $30^{\circ} .$ The angle of shear is

Options:
A $18^{\circ}$
B $0.18^{\circ}$
C $36^{\circ}$
D $0.36^{\circ}$

The correct answer is:
$0.18^{\circ}$

Solution:
$r \theta=\ell \phi \Rightarrow \phi=\frac{r \theta}{\ell}=\frac{6 \mathrm{~mm} \times 30^{\circ}}{1 \mathrm{~m}}=0.18^{\circ}$

Topic: Physics, Mechanical Properties of Solids, BITSAT, BITSAT 2010
The upper half of an inclined plane with inclination $\phi$ is perfectlysmooth while the lower halfis rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

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

The correct answer is:
$2 \tan \phi$

Solution:
Acceleration of block while sliding down upper half $=\mathrm{g} \sin \phi$

retardation of block while sliding down lower half $=-(\mathrm{g} \sin \phi-\mu \mathrm{g} \cos \phi)$

For the block to come to rest at the bottom, acceleration in I half = retardation in II half.

$\mathrm{g} \sin \phi=-(\mathrm{g} \sin \phi-\mu \mathrm{g} \cos \phi)$

$\Rightarrow \mu=2 \tan \phi$

Alternative method: According to work-energy theorem, $\mathrm{W}=\Delta \mathrm{K}=0$

(Since initial and final speeds are zero)

$\therefore$ Work done by friction $+$ Work done by gravity $=0$ i.e., $-(\mu \mathrm{mg} \cos \phi) \frac{\ell}{2}+\mathrm{mg} \ell \sin \phi=0$

or $\frac{\mu}{2} \cos \phi=\sin \phi$ or $\mu=2 \tan \phi$

Topic: Physics, Laws of Motion, BITSAT, BITSAT 2020
The valency shell of element A contains 3 electrons while the valency shell of element $\mathrm{B}$ contains 6 electrons. If A combines with $\mathrm{B}$, the probable formula of the compound formed will be

Options:
A $\mathrm{AB}_{2}$
B $A_{2} B$
C $\mathrm{A}_{2} \mathrm{~B}_{3}$
D $\mathrm{A}_{3} \mathrm{~B}_{2}$

The correct answer is:
$\mathrm{A}_{2} \mathrm{~B}_{3}$

Solution:
The element A is $\mathrm{ns}^{2} p^{1}$ and $\mathrm{B}$ is $\mathrm{ns}^{2} p^{4}$. They can form compound of the type $\mathrm{A}_{2} \mathrm{~B}_{3}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, BITSAT, BITSAT 2014
The value of ∫01cot-11-x+x2dx is

Options:
A loge2
B π2-loge2
C π2+loge2
D -loge2

The correct answer is:
π2-loge2

Solution:
cot-11-x+x2=tan-111-x+x2

⇒cot-11-x+x2=tan-111-x1-x

⇒cot-11-x+x2=tan-1x+1-x1-x1-x

⇒cot-11-x+x2=tan-1x+tan-11-x

∴ ∫01cot-11-x+x2dx=∫01tan-1xdx+∫01tan-11-xdx

⇒∫01cot-11-x+x2dx=∫01tan-1xdx+∫01tan-1xdx ∵ ∫0afxdx=∫0afa-xdx

⇒∫01cot-11-x+x2dx=2∫01tan-1xdx

On evaluating by integration by parts, we have

⇒∫01cot-11-x+x2dx=2tan-1x·x01-∫01x1+x2dx

⇒∫01cot-11-x+x2dx=2π4-12ln1+x201

⇒∫01cot-11-x+x2dx=2π4-12loge2=π2-loge2

Hence, ∫01cot-11-x+x2dx=π2-loge2

So, option (b) is correct.

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2018
The value of 21!+2+42!+2+4+63!+…. is

Options:
A e
B 2e
C 3e
D None of these

The correct answer is:
3e

Solution:
Here,

Tn=n(n+1)n!=n-1+2(n-1)!=1(n-2)!+2(n-1)!

∴ S=∑n=1∞Tn=∑n=1∞1(n-2)!+2∑n=1∞1(n-1)!

=e+2e=3e

Topic: Mathematics, Basic of Mathematics, BITSAT, BITSAT 2017
The value of $\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+\ldots$ upto n terms is

Options:
A $n-\frac{4^{n}}{3}-\frac{1}{3}$
B $\mathrm{n}+\frac{4^{-\mathrm{n}}}{3}-\frac{1}{3}$
C $n+\frac{4^{n}}{3}-\frac{1}{3}$
D $\mathrm{n}-\frac{4^{-\mathrm{n}}}{3}+\frac{1}{3}$

The correct answer is:
$\mathrm{n}+\frac{4^{-\mathrm{n}}}{3}-\frac{1}{3}$

Solution:
Consider $\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+\ldots$ upto $\mathrm{n}$ terms

$=\frac{2^{2}-1}{2^{2}}+\frac{2^{4}-1}{2^{4}}+\frac{2^{6}-1}{2^{6}}$ upto n term

$=\left(1-\frac{1}{2^{2}}\right)+\left(1-\frac{1}{2^{4}}\right)+\left(1-\frac{1}{2^{6}}\right)+\ldots$ upton

terms

$=(1+1+1+\ldots .$ upto $\mathrm{n}$ terms $)$

$-\left(\frac{1}{2^{2}}+\frac{1}{2^{4}}+\frac{1}{2^{6}}+\ldots\right.$ upto n terms $)$

$=\mathrm{n}-\frac{1}{2^{2}}\left[\frac{1-\left(\frac{1}{2^{2}}\right)^{\mathrm{n}}}{1 \frac{1}{2^{2}}}\right]=\mathrm{n}-\frac{1}{3}\left(1-4^{-\mathrm{n}}\right)$

$=\mathrm{n}+\frac{4^{-\mathrm{n}}}{3}-\frac{1}{3}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2015
The value of \(a\) in order that \(f(x)=\sin x-\cos x-a x+b\) decreases for all real values is given by

Options:
A \(a \geq \sqrt{2}\)
B a \( < \sqrt{2}\)
C \(a \geq 1\)
D \(a < 1\)

The correct answer is:
\(a \geq \sqrt{2}\)

Solution:
We have; \(f(x)=\sin x-\cos x-a x+b\)
\(\Rightarrow f^{\prime}(x)=\cos x+\sin x-a\)
\(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x}) < 0 ~\forall~ \mathrm{x} \in \mathrm{R}\)
\(\Rightarrow(\cos x+\sin x) < a ~\forall~ x \in R\)
As the max. value of \((\cos x+\sin x)\) is \(\sqrt{2}\)
The above is possible when \(a \geq \sqrt{2}\)

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2009
The value of a, so that limx→01x2eax-ex-x=32, is

Options:
A 1
B 0
C 2
D 4

The correct answer is:
2

Solution:
We have,

limx→01x2eax-ex-x=32

⇒limx→0aeax-ex-12x=32 (using L'Hospital rule to the left hand side)

As x→0, the denominator is zero and as the value of the limit is finite so, numerator must be zero for x→0.

Hence, a-1-1=0⇒a=2.

Topic: Mathematics, Limits, BITSAT, BITSAT 2018
The value of \(\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)\); \(\frac{1}{2} \leq x \leq 1\) is

Options:
A \(-\frac{\pi}{3}\)
B \(\frac{\pi}{3}\)
C \(\frac{3}{\pi}\)
D \(-\frac{3}{\pi}\)

The correct answer is:
\(\frac{\pi}{3}\)

Solution:
Let \(\cos ^{-1} x=y\)
\(\Rightarrow x=\cos y\), so that \(\frac{1}{2} \leq x \leq 1\) or \(\quad 0 \leq y \leq \frac{\pi}{3}\)
and \(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}=\frac{1}{2} \cos y+\frac{\sqrt{3}}{2} \sin y\)
\(=\cos \frac{\pi}{3} \cos y+\sin \frac{\pi}{3} \sin y=\cos \left(\frac{\pi}{3}-y\right)\)
\(\Rightarrow \cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)=\frac{\pi}{3}-y\)
\(\therefore\) the given expression is equal to
\(y+\frac{\pi}{3}-y, \text { i.e., } \frac{\pi}{3}\)

Topic: Mathematics, Matrices, BITSAT, BITSAT 2010
The value of

$\cos \left[\frac{1}{2} \cos ^{-1}\left(\cos \left(\sin ^{-1} \frac{\sqrt{63}}{8}\right)\right)\right]$ is $-$

Options:
A $3 / 16$
B $3 / 8$
C $3 / 4$
D $3 / 2$

The correct answer is:
$3 / 4$

Solution:
The given trigonometric ratio

$$

\begin{array}{l}

=\cos \left[\frac{1}{2}\left(\cos \left(\cos ^{-1} \frac{1}{8}\right)\right)\right] \\

=\cos \left(\frac{1}{2} \cos ^{-1} \frac{1}{8}\right) \\

=\sqrt{\frac{1+\cos \left(\cos ^{-1} \frac{1}{8}\right)}{2}}=\frac{3}{4}

\end{array}

$$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2013
The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$ is

Options:
A $-\frac{2 \sqrt{6}}{5}$
B $-2 \sqrt{6}$
C $-\frac{\sqrt{6}}{5}$
D
None of these

The correct answer is:
$-\frac{2 \sqrt{6}}{5}$

Solution:
$\begin{aligned} & \cos \left[2 \cos ^{-1} x+\sin ^{-1} x\right] \\ & =\cos \left[\cos ^{-1} x+\cos ^{-1} x+\sin ^{-1} x\right] \\ & =\cos \left[\cos ^{-1} x+\pi / 2\right]=-\sin \left[\cos ^{-1} x\right] \\ & =-\sin \left[\sin ^{-1} \sqrt{1-x^2}\right]=-\sqrt{1-x^2} \\ & =-\sqrt{1-\left(\frac{1}{5}\right)^2}=-\sqrt{\frac{24}{25}}=-\frac{2 \sqrt{6}}{5} \\ & \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2023 (Memory Based Paper 2)
The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$ is

Options:
A $-\frac{2 \sqrt{6}}{5}$
B $-2 \sqrt{6}$
C $-\frac{\sqrt{6}}{5}$
D None of these

The correct answer is:
$-\frac{2 \sqrt{6}}{5}$

Solution:
$\cos \left[2 \cos ^{-1} x+\sin ^{-1} x\right]$

$=\cos \left[\cos ^{-1} x+\cos ^{-1} x+\sin ^{-1} x\right]$

$=\cos \left[\cos ^{-1} x+\frac{\pi}{2}\right]=-\sin \left[\cos ^{-1} x\right]$

$=-\sin \left[\sin ^{-1} \sqrt{1-x^{2}}\right]=-\sqrt{1-x^{2}}$

$=-\sqrt{1-\left(\frac{1}{5}\right)^{2}}=-\sqrt{\frac{24}{25}}=-\frac{2 \sqrt{6}}{5}$

Topic: Mathematics, Inverse Trigonometric Functions, BITSAT, BITSAT 2021
The value of $\int \cos (\log x) \mathrm{d} \mathrm{x}$ is :

Options:
A $\frac{1}{2}[\sin (\log x)+\cos (\log x)]+C$
B $\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C$
C $\frac{x}{2}[\sin (\log x)-\cos (\log x)]+C$
D $\frac{1}{2}[\sin (\log x)-\cos (\log x)]+C$

The correct answer is:
$\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C$

Solution:
Let

\(\mathrm{I}=\int \cos (\log \mathrm{x}) \cdot 1 \mathrm{~d} \mathrm{x} \ldots . . \text { (i) }\)

Use integral by parts,

\(\begin{aligned} &I=\cos (\log x) \cdot x-\int[-\sin (\log x)] \cdot \frac{1}{x} \cdot x d x \\ &=x \cdot \cos (\log x)+\int \sin (\log x) \cdot 1 d x \\ &=x \cdot \cos (\log x)+\left[\sin (\log x) \cdot x-\int \cos (\log x) \cdot \frac{1}{x} \cdot x d x\right]+C \\ &=x \cdot(\log x)+\left[x \sin (\log x)-\int \cos (\log x) d x\right]+C \\ &=x\{\sin (\log x)+\cos (\log x)\}-I+C[f r o m \text { Eq. (i) }] \\ &\Rightarrow I=\frac{x}{2}\{\sin (\log x)+\cos (\log x)\}+C \end{aligned}\)

Topic: Mathematics, Indefinite Integration, BITSAT, BITSAT 2021
The value of current I as shown in the given circuit diagram is

Options:
A 2 A
B 1.5 A
C 0.9 A
D 0.2 A

The correct answer is:
0.9 A

Solution:
The given circuit can be redrawn as

Since, RABRBC=RADRDC=12, hence, it is a balanced Wheatstone bridge, therefore current flowing in arm BD=0

∴ Req=10+20∥6+12=30∥18

=30×1830+18=11.25 Ω

∴ I=VReq=1011.25=0.89 A=0.9 A

Topic: Physics, Current Electricity, JEE Main
The value of current I as shown in the given circuit diagram is

Options:
A 2 A
B 1.5 A
C 0.9 A
D 0.2 A

The correct answer is:
0.9 A

Solution:
The given circuit can be redrawn as

Since, RABRBC=RADRDC=12, hence, it is a balanced Wheatstone bridge, therefore current flowing in arm BD=0

∴ Req=10+20∥6+12=30∥18

=30×1830+18=11.25 Ω

∴ I=VReq=1011.25=0.89 A=0.9 A

Topic: Physics, Current Electricity, BITSAT, BITSAT 2019
The value of f(0), so that the function fx=2x-sin-1x2x+tan-1x is continuous at each point in its domain, is

Options:
A 13
B -13
C 23
D -23

The correct answer is:
13

Solution:
Since, x,sin-1x,tan-1x are continuous functions, so the
function f is clearly continuous at each point of its domain except possibly at x=0.

So, for f to be continuous at x=0

f0=limx→0fx=limx→02-sin-1xx2+tan-1xx=13

∵limx→0sin-1xx=1,limx→0tan-1xx=1

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2017
The value of $\lambda$, for which the lines $3 x-4 y=13,8 x-11 y=33$ and $2 x-3 y+\lambda=0$

are concurrent is

Options:
A -1
B -7
C $\frac{1}{7}$
D 9

The correct answer is:
-7

Solution:
For concurrency of 3 lines the determinant of coefficients of equations should be 0 .

i.e. $\left|\begin{array}{ccc}3 & -4 & -13 \\ 8 & -11 & -33 \\ 2 & -3 & \lambda\end{array}\right|=0$

$\Rightarrow 3(-11 \lambda-99)+4(8 \lambda+66)-13(-24+22)=0$

$\Rightarrow-33 \lambda-297+32 \lambda+264+312-286=0$

$\Rightarrow-\lambda-583+576=0 \Rightarrow \lambda=-7$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2014
The value of λ for which the loci argz=π6 and |z-23i|=λ on the argand plane touch each other is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
3

Solution:
z=x+iy

argz=tan-1yx

argz=π6

∴ tan-1yx=π6

yx=tanπ6=13

⇒ x=3y

Also given,

|z-23i|=λ

|x+iy-23i|=λ

x2+y-232=λ2

This is the equation of circle whose centre x-3y=0 is (0,23) and r=λ. Now, the line touches the circle

∴ λ=0-3231+3

λ=62=3

∴ λ=3

Topic: Mathematics, Complex Number, BITSAT, BITSAT 2019
The value of \(\left|\begin{array}{ccc}1 & 2 & 3 \\ -4 & 3 & 6 \\ 2 & -7 & 9\end{array}\right|\) is

Options:
A 213
B -231
C 231
D 39

The correct answer is:
231

Solution:
\(\begin{aligned} & \left|\begin{array}{ccc}1 & 2 & 3 \\ -4 & 3 & 6 \\ 2 & -7 & 9\end{array}\right|=1\left|\begin{array}{cc}3 & 6 \\ -7 & 9\end{array}\right|-2\left|\begin{array}{cc}-4 & 6 \\ 2 & 9\end{array}\right|+3 \\ & \left|\begin{array}{cc}-4 & 3 \\ 2 & -7\end{array}\right| \\ & \end{aligned}\)
\(\begin{aligned}
& =1(3 \times 9-6(-7))-2(-4 \times 9-2 \times 6)+3 \\
& ((-4)(-7)-3 \times 2) \\
& =(27+42)-2(-36-12)+3(28-6)=231
\end{aligned}\)

Topic: Mathematics, Continuity and Differentiability, BITSAT, BITSAT 2010
The value of $\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e+\frac{1}{2} e x}{x^2}$ is

Options:
A
$\frac{11}{24} e$

B
$-\frac{11}{24} e$

C
$\frac{e}{24}$

D
None of these

The correct answer is:

$\frac{11}{24} e$

Solution:
Let $y=(1+x)^{1 / x}$
$\Rightarrow \log y=\frac{1}{x} \log (1+x)=\frac{1}{x}\left[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots\right]$
$=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots$
$\begin{aligned} y & =e^{\left(1-\frac{x}{2}+\frac{x^2}{3}+\ldots .\right)}=e e^{\left(\frac{-x}{2}+\frac{x^2}{3} \ldots . .\right)} \\ & =e\left[1+\left(-\frac{x}{2}+\frac{x^2}{3}-\ldots\right)+\frac{1}{2 !}\left(-\frac{x}{2}+\frac{x^2}{3}-\ldots\right)^2+\ldots\right]\end{aligned}$
$y-e+\frac{1}{2} e x=e x^2\left[\frac{1}{3}+0(x)+\frac{1}{2}\left(-\frac{1}{2}+0(x)\right)^2+\ldots.\right]$
$[\because 0(x)$ in terms containing $x]$
$\lim \frac{y-e+\frac{1}{2} e x}{x^2}=e\left[\frac{1}{3}+\frac{1}{8}\right]=\frac{11}{24} e$

Topic: Mathematics, Limits, BITSAT, BITSAT 2022
The value of $\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{x^4}$ is

Options:
A
$\frac{1}{6}$

B
$\frac{1}{8}$

C
$\frac{1}{10}$

D
$\frac{1}{12}$

The correct answer is:

$\frac{1}{8}$

Solution:
$1-\cos (1-\cos x)=2 \sin ^2\left(\frac{1-\cos x}{2}\right)$
$=2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)$
$\therefore \lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{x^4}=\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)}{x^4}$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\sin ^2 \frac{x}{2}\right)}{\left(\sin ^2 \frac{x}{2}\right)^2} \times \frac{\sin ^4 \frac{x}{2}}{\left(\frac{x}{2}\right)^4 \times 16}$
$=\frac{1}{8}$

Topic: Mathematics, Limits, BITSAT, BITSAT 2022
The value of limx→3π44sin2xcosx-cosx+sinxsinx+cosx is

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

The correct answer is:
-1

Solution:
L=limx→3π44sin2xcosx-cosx+sinxsinx+cosx

L=limx→3π42sinxsin2x+cos2x+2sinxcosx-1-cosx+sinxsinx+cosx

L=limx→3π42sinx(sinx+cosx)2-1-cosx+sinxsinx+cosx

L=limx→3π42sinx(sinx+cosx)2-2sinx-cosx+sinxsinx+cosx

L=limx→3π4sinx+cosx2sinxsinx+cosx-1sinx+cosx

L=limx→3π42sinxsinx+cosx-1=-1

Topic: Mathematics, Limits, BITSAT, BITSAT 2019
The value of \(\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)^3}{\sin \frac{x^2}{4} \log (1+3 x)}\), is

Options:
A \(\frac{4}{3}(\ln 4)^2\)
B \(\frac{4}{3}(\ln 4)^3\)
C \(\frac{3}{2}(\ln 4)^2\)
D \(\frac{3}{2}(\ln 4)^3\)

The correct answer is:
\(\frac{4}{3}(\ln 4)^3\)

Solution:
\(\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\left(4^x-1\right)^3}{\sin \frac{x^2}{4} \log (1+3 x)} \\
& =\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)^3}{x^3} \cdot \frac{(x / 2)^2}{\sin x^2 / 4} \cdot \frac{3 x}{\log (1+3 x)} \cdot \frac{4}{3} \\
& =\frac{4}{3}\left(\log _e 4\right)^3 \cdot 1 \cdot \log _e(e)=\frac{4}{3}\left(\log _e 4\right)^3 .
\end{aligned}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2009