datavorous/entrance-exam-dataset huggingface dataset:
The point ([P+1],[P]) (where, [x] is the greatest integer function) lying inside the region bounded by the circle x2+y2-2x-15=0 and x2+y2-2x-7=0, then

Options:
A P∈[-1,2)-{0,1}
B P∈[-1,0)∪(0,1)∪(1,2]
C P∈(-1,2)
D None of these

The correct answer is:
None of these

Solution:
[P+1]=[P]+1, Let [P]=n, then n is integer

∴ ([P+1],[P])=(n+1,n) lie inside the region of circles

S1=x2+y2-2x-15=0,C1=(1,0),r1=4

and S2=x2+y2-2x-7=0,C2=(1,0),r2=22

Both circles are concentric.

∴ (n+1)2+n2-2(n+1)-7>0

and (n+1)2+n2-2(n+1)-15<0⇒4<n2<8

Which is not possible for any integer.

Topic: Mathematics, Area Under Curves, BITSAT, BITSAT 2019
The points $(0,7,10),(-1,6,6)$ and $(-4,9,6)$ form

Options:
A a right angled isosceles triangle
B a scalene triangle
C a right angled triangle
D an equilateral triangle

The correct answer is:
a right angled isosceles triangle

Solution:
Let $\mathrm{P}(0,7,10), \mathrm{Q}(-1,6,6)$ and $\mathrm{R}(-4,9,6)$ be the vertices of a triangle Here, $\mathrm{PQ}=\sqrt{1+1+16}=3 \sqrt{2}$

$\begin{array}{l}

\mathrm{QR}=\sqrt{9+9+0}=3 \sqrt{2} \\

\mathrm{PR}=\sqrt{16+4+16}=6 \\

\text { Now, } \mathrm{PQ}^{2}+\mathrm{QR}^{2}=(3 \sqrt{2})^{2}+(3 \sqrt{2})^{2}=36=(\mathrm{PR})^{2}

\end{array}$

Therefore, $\triangle \mathrm{PQR}$ is a right angled triangle at $\mathrm{Q}$. Also, $\mathrm{OQ}=\mathrm{QR}$. Hence, $\triangle \mathrm{PQR}$ is a right angled isosceles triangle.

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2021
The points at which the tangent passes through the origin for the curve $y=4 x^{3}-2 x^{5}$ are

Options:
A $(0,0),(2,1)$ and $(-1,-2)$
B $(0,0),(2,1)$ and $(-2,-1)$
C $(2,0),(2,1)$ and $(-3,1)$
D $(0,0),(1,2)$ and $(-1,-2)$

The correct answer is:
$(0,0),(1,2)$ and $(-1,-2)$

Solution:
The equation of the given curve is

$\begin{array}{l}

y=4 x^{3}-2 x^{5} \\

\frac{\mathrm{dy}}{\mathrm{dx}}=12 x^{2}-10 x^{4}

\end{array}$

Therefore, the slope of the tangent at point $(x, y)$ is $12 x^{2}-10 x^{4}$. The equation of the tangent at $(x, y)$ is given by

$\mathrm{Y}-y=\left(12 x^{2}-10 x^{4}\right)(\mathrm{X}-x) \ldots \text { (i) }$

When, the tangent passes through the origin $(0,0)$, then $\mathrm{X}=\mathrm{Y}=0$ Therefore, eq. (i) reduce to

$\begin{array}{l}

-y=\left(12 x^{2}-10 x^{4}\right)(-x) \\

\Rightarrow y=12 x^{3}-10 x^{5}

\end{array}$

Also, we have $y=4 x^{3}-2 x^{5}$

$\begin{array}{l}

\therefore 12 x^{3}-10 x^{5} \\

=4 x^{3}-2 x^{5} \\

\Rightarrow 8 x^{5}-8 x^{3}=0 \\

\Rightarrow x^{5}-x^{3}=0 \\

\Rightarrow x^{3}\left(x^{2}-1\right)=0 \\

\Rightarrow x=0, \pm 1

\end{array}$

When, $x=0$

$y=4(0)^{3}-2(0)^{5}=0$

When, $x=1$,

$y=4(1)^{3}-2(1)^{5}=2$

When, $x=-1$,

$y=4(-1)^{3}-2(-1)^{5}=-2$

Hence, the require points are $(0,0),(1,2)$ and $(-1,-2)$.

Topic: Mathematics, Application of Derivatives, BITSAT, BITSAT 2021
The position of a projectile launched from the origin at $t=0$ is given by $\vec{r}=(40 \hat{i}+50 \hat{j}) \mathrm{m}$ at $t=$ $2 \mathrm{~s}$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is $\left(\right.$ take $\left.g=10 \mathrm{~ms}^{-2}\right)$\

Options:
A $\tan ^{-1} \frac{2}{3}$
B $\tan ^{-1} \frac{3}{2}$
C $\tan ^{-1} \frac{7}{4}$
D $\tan ^{-1} \frac{4}{5}$

The correct answer is:
$\tan ^{-1} \frac{7}{4}$

Solution:
From question, Horizontal velocity (initial),

$$

u_{x}=\frac{40}{2}=20 \mathrm{~m} / \mathrm{s}

$$

Vertical velocity (initial), $50=u_{y} t+\frac{1}{2} g t^{2}$

$$

\begin{array}{l}

\Rightarrow u_{y} \times 2+\frac{1}{2}(-10) \times 4 \\

\text { or, } \quad 50=2 u_{y}-20

\end{array}

$$

or, $u_{y}=\frac{70}{2}=35 \mathrm{~m} / \mathrm{s}$

$$

\begin{array}{l}

\therefore \quad \tan \theta=\frac{u_{y}}{u_{x}}=\frac{35}{20}=\frac{7}{4} \\

\Rightarrow \quad \text { Angle } \theta=\tan ^{-1} \frac{7}{4}

\end{array}

$$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2015
The position $\mathrm{x}$ of a particle varies with time $(\mathrm{t})$ as $\mathrm{x}=\mathrm{A} \mathrm{t}^{2}-\mathrm{B} \mathrm{t}^{3}$. The acceleration at time $\mathrm{t}$ of the particle will be equal to zero. What is the value of $t ?$

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

The correct answer is:
$\frac{A}{3 B}$

Solution:
Given that $x=A t^{2}-B t^{3}$
$$
\therefore \text { velocity }=\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{~A} \mathrm{t}-3 \mathrm{~B} \mathrm{t}^{2}
$$
and acceleration
$$
=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)=2 \mathrm{~A}-6 \mathrm{Bt}
$$
For acceleration to be zero $2 \mathrm{~A}-6 \mathrm{Bt}=0$.
$$
\therefore \quad \mathrm{t}=\frac{2 \mathrm{~A}}{6 \mathrm{~B}}=\frac{\mathrm{A}}{3 \mathrm{~B}}
$$

Topic: Physics, Motion In One Dimension, BITSAT, BITSAT 2010
The potential energy for a force field $\overrightarrow{\mathrm{F}}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is $-$

Options:
A $-\frac{1}{\sqrt{2}}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$
B $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
C $\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)$
D $\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{\sqrt{3}}{2} \hat{\mathrm{j}}\right)$

The correct answer is:
$\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$

Solution:
$\quad \mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$
$$
\begin{array}{l}
\mathrm{F}_{\mathrm{y}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y}) \\
\left.\mathrm{F}_{\mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\
\left.\mathrm{~F}_{\mathrm{y}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\
\therefore \mathrm{F}=\frac{1}{\sqrt{2}}[\hat{\mathrm{i}}+\hat{\mathrm{j}}]
\end{array}
$$

Topic: Physics, Work Power Energy, BITSAT, BITSAT 2010
The potential energy of a satellite of mass $\mathrm{m}$ and revolving at a height $\mathrm{R}_{\mathrm{e}}$ above the surface of earth where $\mathrm{R}_{\mathrm{e}}=$ radius of earth, is

Options:
A $-\mathrm{m} \mathrm{g} \mathrm{R}_{\mathrm{e}}$
B $\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{2}$
C $\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{3}$
D $\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{4}$

The correct answer is:
$\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{2}$

Solution:
At a height h above the surface of earth the gravitational potential energy of the particle of mass m is $\mathrm{U}_{\mathrm{h}}=-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}}$

Where $M_{e} \& R_{e}$ are the mass $\&$ radius of earth respectively. In this question, since $\mathrm{h}=\mathrm{R}_{\mathrm{e}}$

So $U_{h-R_{e}}=-\frac{G M_{e} m}{2 R_{e}}=\frac{-m g R_{e}}{2}$

Topic: Physics, Gravitation, BITSAT, BITSAT 2020
The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

Options:
A $20 \Omega$
B $15 \Omega$
C $10 \Omega$
D $30 \Omega$

The correct answer is:
$10 \Omega$

Solution:
$P=\frac{V^{2}}{R_{\mathrm{eq}}}$
$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R} R_{\mathrm{eq}}=\left(\frac{5 R}{5+R}\right) P=30 \mathrm{~W}$
Substituting the values in equation (i)
$$
30=\frac{(10)^{2}}{\left(\frac{5 R}{5+R}\right)} \Rightarrow R=10 \Omega
$$

Topic: Physics, Current Electricity, JEE Main
The power dissipated in the circuit shown in the figure is 30 Watts. The value of $R$ is

Options:
A $20 \Omega$
B $15 \Omega$
C $10 \Omega$
D $30 \Omega$

The correct answer is:
$10 \Omega$

Solution:
$P=\frac{V^{2}}{R_{\mathrm{eq}}}$
$\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R} R_{\mathrm{eq}}=\left(\frac{5 R}{5+R}\right) P=30 \mathrm{~W}$
Substituting the values in equation (i)
$$
30=\frac{(10)^{2}}{\left(\frac{5 R}{5+R}\right)} \Rightarrow R=10 \Omega
$$

Topic: Physics, Current Electricity, BITSAT, BITSAT 2010
The prefix $10^{18}$ is

Options:
A giga
B kilo
C exa
D nano

The correct answer is:
exa

Solution:
$\mathrm{Exa}=10^{18}$

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2015
The preparation of ethyl acetoacetate involves:

Options:
A Wittig reaction
B Cannizzaro's reaction
C Reformatsky reaction
D Claisen condensation.

The correct answer is:
Claisen condensation.

Solution:
In Claisen condensation intermolecular condensation of esters containing $\alpha$ -hydrogen atom in presence of strong base produce $\beta$ -keto ester.
$\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H.CH}_{2} \cdot \mathrm{CO} . \mathrm{OC}_{2} \mathrm{H}_{5}$
Ethyl acetate

Topic: Chemistry, Aldehydes and Ketones, BITSAT, BITSAT 2012
The pressure versus temperature graph of an ideal gas is shown in the figure below. If the density of gas at point A is ρ0, then the density of the gas at point B will be

Options:
A 2ρ0
B 23ρ0
C 32ρ0
D 3ρ0

The correct answer is:
23ρ0

Solution:
Given, density of gas at point A, ρA=ρ0

General equation for an ideal gas

pV=nRT

⇒pV=mMRT [m→mass, M→ molecular mass]

⇒p=mV·RM·T

⇒p=ρRMT ...i [where density ρ=mV]

Applying Equation. (i) at point A and B, respectively.

p=ρARMT0 ...ii

2p=ρBRM3T0 ...iii

From Equations. (ii) and (iii), we get

12=ρAρB·13

⇒ρB=23ρA=23ρ0

Topic: Physics, Thermodynamics, BITSAT, BITSAT 2019
The probability density plots of $1 s$ and $2 s$ orbitals are given in figure.
The density of dots in a region represents the probability density of finding electrons in the region.
On the basis of above diagram which of the following statements is incorrect?

Options:
A $1 s$ and $2 s$ orbitals are spherical in shape.
B The probability of finding the electron is maximum near the nucleus.
C The probability of finding the electron at a given distance is equal in all directions.
D The probability density of electrons for $2 s$ orbital decreases uniformly as distance from the nucleus increases.

The correct answer is:
The probability density of electrons for $2 s$ orbital decreases uniformly as distance from the nucleus increases.

Solution:
The probability density of electrons in $2 s$ orbital first increases then decreases and after that it increases again as distance increases from nucleus.

Topic: Chemistry, Structure of Atom, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The probability of getting sum more than 7 when a pair of dice are thrown is:

Options:
A $\frac{7}{36}$
B $\frac{5}{12}$
C $\frac{7}{12}$
D None of these

The correct answer is:
$\frac{5}{12}$

Solution:
Here $n(S)=6^2=36$
Let $\mathrm{E}$ be the event "getting sum more than 7" i.e. sum of pair of dice $=8,9,10,11,12$
i.e. $E=\left\{\begin{array}{lllll}(2,6) & (3,5) & (4,4) & (5,3) & (6,2) \\ (3,6) & (4,5) & (5,4) & (6,3) \\ (4,6) & (5,5) & (6,4) & \\ (5,6) & (6,5) & (6,6) & \end{array}\right\}$
$\therefore \quad n(E)=15 . \therefore$ Req. prob $=\frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The probability of India winning a test match against West Indies is $1 / 2$. Assuming independence from match to match, the probability that in a 5 match series India's second win occurs at the third test, is -

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

The correct answer is:
$1 / 4$

Solution:
The sample space is [LWW, WLW]

$$

\therefore P(\mathrm{LWW})+\mathrm{P}(\mathrm{WLW})

$$

= Probability that in 5 match series, it is India's second win

$$

\begin{array}{l}

=P(\mathrm{~L}) \mathrm{P}(\mathrm{W}) \mathrm{P}(\mathrm{W})+\mathrm{P}(\mathrm{W}) \mathrm{P}(\mathrm{L}) \mathrm{P}(\mathrm{W}) \\

=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}

\end{array}

$$

Topic: Mathematics, Probability, BITSAT, BITSAT 2013
The probability of simultaneous occurrence of atleast one of two events $A$ and $B$ is $p .$ If the probability that exactly one of $A, B$ occurs is $q$ then $P\left(A^{\prime}\right)+P\left(B^{\prime}\right)$ is equal to

Options:
A $2-2 p+q$
B $2+2 p-q$
C $3-3 p+q$
D $2-p+q$

The correct answer is:
$2-2 p+q$

Solution:
Since, $P($ exactly one of $A, B$ occurs $)=q$.
$\therefore \quad P(A \cup B)-P(A \cap B)=q$
$\Rightarrow \quad p-P(A \cap B)=q \Rightarrow P(A \cap B)=p-q$
$\Rightarrow \quad 1-P\left(A^{\prime} \cup B^{\prime}\right)=p-q \Rightarrow P\left(A^{\prime} \cup B^{\prime}\right)=1-p+q$
$\Rightarrow \quad P\left(A^{\prime}\right)+P\left(B^{\prime}\right)-P\left(A^{\prime} \cap B^{\prime}\right)=1-p+q$
$\Rightarrow \quad P\left(A^{\prime}\right)+P\left(B^{\prime}\right)=(1-p+q)+[1-P(A \cup B)]$
$=(1-p+q)+(1-p)=2-2 p+q$

Topic: Mathematics, Probability, BITSAT, BITSAT 2012
The probability that a card drawn from a pack of 52 cards will be a diamond or king is:

Options:
A $\frac{1}{52}$
B $\frac{2}{13}$
C $\frac{4}{13}$
D $\frac{1}{13}$

The correct answer is:
$\frac{4}{13}$

Solution:
Total no. of cards $=52$
13 cards are diamonds and 4 cards are king.
There is only one card which is a king of diamond.
$\therefore \mathrm{P}($ card is diamond $)=\frac{13}{52}, \mathrm{P}$ (card is king)
$=\frac{4}{52}$
$\mathrm{P}($ card is king of diamond $)=\frac{1}{52}$
$\therefore \quad P$ (card is diamond or king)
$=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}$

Topic: Mathematics, Probability, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The probability that certain electronic component fails when first used is 0.10 . If it does not fail immediately, the probability that it lasts for one year is $0.99 .$ The probability that a new component will last for one year is

Options:
A 0.99
B 0.871
C 0.891
D 0.762

The correct answer is:
0.891

Solution:
Probability that the electronic component fails when first used is $P(F)=0.10$. Therefore, $P\left(F^{\prime}\right)=1-P(F)=0.90$

Let $E$ be the event that a new component will last for one year.

$\text { Then, } P(E)=P(F) P\left(\frac{E}{F}\right)+P\left(F^{\prime}\right) P\left(\frac{E}{F^{\prime}}\right)$

[total probability theorem] $=0.10 \times 0+0.90 \times 0.99=0.891$

Topic: Mathematics, Probability, BITSAT, BITSAT 2020
The product of atomic weight and specific heat of any element is a constant, approximately $6.4 .$ This is known as:

Options:
A Dalton's law
B Avogadro's law
C Newton's law
D Dulong Pettit law

The correct answer is:
Dulong Pettit law

Solution:
According to Dulong and Pettit's law
Atomic weight $\times$ Specific heat $=6.4$ (approx) This law is applicable only to solid elements but it fails to explain very high specific heat of diamond.

Topic: Chemistry, Some Basic Concepts of Chemistry, BITSAT, BITSAT 2011
The product of $n$ positive numbers is unity, then their sum is:

Options:
A a positive integer
B divisible by $n$
C equal to $n+\frac{1}{n}$
D never less than $n$

The correct answer is:
never less than $n$

Solution:
Since, product of $n$ positive number is unity. $\Rightarrow x_{1} x_{2} x_{3} \ldots \ldots x_{n}=1$

Using A.M. $\geq$ GM

$$

\begin{array}{l}

\Rightarrow \frac{x_{1}+x_{2}+\ldots . .+x_{n}}{n} \geq\left(x_{1} x_{2} \ldots \ldots x_{n}\right)^{\frac{1}{n}} \\

\Rightarrow x_{1}+x_{n}+\ldots \ldots+x_{n} \geq n(1)^{\frac{1}{n}}\left[\text { From eq }^{\mathrm{n}}(\mathrm{i})\right]

\end{array}

$$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2014
The quadratic equation whose roots are the $\mathrm{x}$ and yintercepts of the line passing through (1,1) and making a triangle of area $\mathrm{A}$ with the co-ordinate axes is

Options:
A $x^{2}+A x+2 A=0$
B $x^{2}-2 A x+2 A=0$
C $x^{2}-A x+2 A=0$
D None of these

The correct answer is:
$x^{2}-2 A x+2 A=0$

Solution:
Equation of the line making intercepts a and

$b$ on the axes is $\frac{x}{a}+\frac{y}{b}=1$ Since, it passes through (1,1) $\Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=1$

Also the area of the triangle formed by the line and the axes is $\mathrm{A}$.

$\therefore \frac{1}{2} \mathrm{ab}=\mathrm{A} \Rightarrow \mathrm{ab}=2 \mathrm{~A}$

From eqs. (i) and (ii), we get, $a+b=2 A$ Hence, a and b are the roots of the eq. $x^{2}-(a+b) x+a b=0 \Rightarrow x^{2}-2 A x+2 A=0$

Topic: Mathematics, Straight Lines, BITSAT, BITSAT 2013
The radiation emitted by a perfectly black body is proportional to

Options:
A temperature on ideal gas scale
B fourth root of temperature on ideal gas scale
C fourth power of temperature on ideal gas scale
D square of temperature on ideal gas scale

The correct answer is:
fourth power of temperature on ideal gas scale

Solution:
No Solution available

Topic: Physics, Thermal Properties of Matter, BITSAT, BITSAT 2011
The random variable \(\mathrm{X}\) has the following probability distribution
\(\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{x} & 0 & 1 & 2 & 3 & 4 \\
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & \mathrm{k} & 3 \mathrm{k} & 5 \mathrm{k} & 2 \mathrm{k} & \mathrm{k} \\
\hline
\end{array}\)
Then the value of \(P(X \geq 2)\) is

Options:
A \(\frac{1}{3}\)
B \(\frac{2}{3}\)
C \(\frac{3}{4}\)
D \(\frac{1}{4}\)

The correct answer is:
\(\frac{2}{3}\)

Solution:
Since, \(\Sigma \mathrm{P}_{\mathrm{i}}(\mathrm{X}=\mathrm{x})=1\)
\(\begin{aligned}
& \therefore \mathrm{k}+3 \mathrm{k}+5 \mathrm{k}+2 \mathrm{k}+\mathrm{k}=1 \therefore 12 \mathrm{k}=1 \therefore \mathrm{k}=\frac{1}{12} \\
& \text { Now, } \mathrm{P}(\mathrm{X} \geq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \\
& =5 \mathrm{k}+2 \mathrm{k}+\mathrm{k}=8 \mathrm{k}=8\left(\frac{1}{12}\right)=\frac{2}{3}
\end{aligned}\)

Topic: Mathematics, Properties of Triangles, BITSAT, BITSAT 2009
The range of the function $f(x)=\sqrt{3 x^2-4 x+5}$ is

Options:
A $\left(-\infty, \sqrt{\frac{11}{3}}\right]$
B $\left(-\infty, \sqrt{\frac{11}{5}}\right)$
C $\left[\sqrt{\frac{11}{3}}, \infty\right)$
D $\left(\sqrt{\frac{11}{5}}, \infty\right)$.

The correct answer is:
$\left[\sqrt{\frac{11}{3}}, \infty\right)$

Solution:
$f(x)$ is defined if $3 x^2-4 x+5 \geq 0$
$\Rightarrow 3\left[x^2-\frac{4}{3} x+\frac{5}{3}\right] \geq 0 \Rightarrow 3\left[\left(x-\frac{2}{3}\right)^2+\frac{11}{9}\right] \geq 0$
Which is true for all real $x \therefore$ Domain of $f(x)=$ $(-\infty, \infty)$
Let $y=\sqrt{3 x^2-4 x+5}$
$\Rightarrow y^2=3 x^2-4 x+5$ i.e. $3 x^2-4 x+\left(5-y^2\right)=0$
For $x$ to be real , $16-12\left(5-y^2\right) \geq 0 \Rightarrow y \geq \sqrt{\frac{11}{3}}$
$\therefore \quad$ Range of $y=\left[\sqrt{\frac{11}{3}}, \infty\right)$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The range of the function \(f(x)=\frac{x^2-x+1}{x^2+x+1}\) where \(\mathrm{x} \in \mathrm{R}\), is

Options:
A \((-\infty, 3]\)
B \((-\infty, \infty)\)
C \([3, \infty)\)
D \(\left[\frac{1}{3}, 3\right]\)

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

Solution:
\(\begin{aligned}
& \text {Let } y=\frac{x^2-x+1}{x^2+x+1} \\
& \Rightarrow \mathrm{x}^2(\mathrm{y}-1)+\mathrm{x}(\mathrm{y}+1)+(\mathrm{y}-1)=0 \\
& \Rightarrow x=\frac{-(y+1) \pm \sqrt{(y+1)^2-4(y-1)^2}}{2(y-1)} \\
& =\frac{-(y+1) \pm \sqrt{-3 y^2+10 y-3}}{2(y-1)} \text { is real iff } \\
& \mathrm{y}-1 \neq 0 \Rightarrow \mathrm{y} \neq 1 \\
& \text { If } y=1 \text { then original equation gives } \mathrm{x}=0 \text {, } \\
& \text { so taking } \mathrm{y}=1 \\
& \text { Also } 3 y^2-10 y+3 \leq 0 \\
& \Rightarrow(3 y-1)(y-3) \leq 0 \\
& \Rightarrow \mathrm{y} \in\left[\frac{1}{3}, 3\right] \quad \therefore \text { Range is }\left[\frac{1}{3}, 3\right] \\
&
\end{aligned}\)

Topic: Mathematics, Functions, BITSAT, BITSAT 2009
The range of the particle when launched at an angle of $15^{\circ}$ with the horizontal is $1.5 \mathrm{~km}$. What is the range of the projectile when launched at an angle of $45^{\circ}$ to the horizontal.

Options:
A $1.5 \mathrm{~km}$
B $3.0 \mathrm{~km}$
C $6.0 \mathrm{~km}$
D $0.75 \mathrm{~km}$

The correct answer is:
$3.0 \mathrm{~km}$

Solution:
$1.5=\frac{\mathrm{u}^{2} \sin 30}{\mathrm{~g}} ; \mathrm{R}=\frac{\mathrm{u}^{2} \sin 90}{\mathrm{~g}}=3 \mathrm{~km}$

Topic: Physics, Motion In Two Dimensions, BITSAT, BITSAT 2012
The rate of a reaction triples when temperature changes from 200C to 500C. The energy of activation for the reaction is R=8.314JK-1 mol-1

Options:
A 181.327 J mol-1
B 428.141 J mol-1
C 32.4321kJmol-1
D 28.8118kJmol-1

The correct answer is:
28.8118kJmol-1

Solution:
Arrhenius equation is given by,
log10K2K1=Ea2.303×RT2-T1T1T2

Given,

K2K1=3;R=-8.314 JK-1mol-1

T1=20+273=293 K
and T2=50+273=323 K

Substituting the given values in Arrhenius equation,

log103=Ea8.314×2.303323-293323×293

Ea=2.303×8.314×323×293×0.47730

=28811.8 J mol1

=28.8118kJmol-1

Topic: Chemistry, Chemical Kinetics, BITSAT, BITSAT 2018
The ratio in which the YZ-plane divide the line segment formed by joining the points $(-2,4,7)$ and $(3,-5,8)$ is $2: \mathrm{m}$. The value of $\mathrm{m}$ is

Options:
A 2
B 3
C 4
D 1

The correct answer is:
3

Solution:
Let the points be $A(-2,4,7)$ and $B(3,-5,8)$ on $Y Z$-plane, $x$-coordinate $=0$.
Let the ratio be $K: 1$.
The coordinates of $C$ are $\left(\frac{3 K-2}{K+1}, \frac{-5 K+4}{K+1}, \frac{8 K+7}{K+1}\right)$
Clearly $\frac{3 K-2}{K+1}=0 \Rightarrow 3 K=2 \Rightarrow K=\frac{2}{3}$
Hence required ratio is $2: 3$.

Topic: Mathematics, Three Dimensional Geometry, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The ratio $\frac{K_{p}}{K_{c}}$ for the reaction

$\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)$ is:

Options:
A $\frac{1}{\sqrt{R T}}$
B $(R T)^{1 / 2}$
C $\mathrm{RT}$
D 1

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

Solution:
$K_{P}=K_{C}(R T)^{\Delta n_{\text {g }}}$

For the reaction

$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(g) \leftrightharpoons \mathrm{CO}_{2}(\mathrm{~g})$

$\Delta n_{g}=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}$

$\therefore \quad \mathrm{K}_{\mathrm{P}}=\frac{\mathrm{K}_{\mathrm{C}}}{\sqrt{\mathrm{RT}}} ; \quad \frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{C}}}=\frac{1}{\sqrt{\mathrm{RT}}}$

Topic: Chemistry, Chemical Equilibrium, BITSAT, BITSAT 2020
The ratio of angular momentum L to the atomic dipole moment μi for hydrogen like atoms and ions is

Options:
A always constant and is equal to the ratio of mass to the charge of electron
B always constant and is equal to twice the ratio of mass to the charge of electron to the principal quantum number n
C proportional to the principal quantum number n
D proportional to 1n2

The correct answer is:
always constant and is equal to the ratio of mass to the charge of electron

Solution:
The electron moving with a speed v in the circular Bohr orbit of radius r constitutes a current of magnitude.

i=eT. where T is the orbital period of the electron.

T=2πr/v

i=ev2πr

By definition of atomic dipole moment μi, its magnitude is given by

μi=iA=eV2πr·πr2=evr2

∴ Angular momentum, L=mvr
∴ Lμi=2mvrevr=2me= constant

Topic: Physics, Magnetic Effects of Current, BITSAT, BITSAT 2018
The ratio of average electric energy density and total average energy density of electromagnetic wave is :

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

The correct answer is:
$\frac{1}{2}$

Solution:
We have $\frac{\mathrm{U}_{\mathrm{E}}}{\mathrm{U}_{\mathrm{T}}}=\frac{\mathrm{U}_{\mathrm{E}}}{\mathrm{U}_{\mathrm{E}}+\mathrm{U}_{\mathrm{B}}}=\frac{\mathrm{U}_{\mathrm{E}}}{2 \mathrm{U}_{\mathrm{E}}}=\frac{1}{2}$
$\left[\because \mathrm{U}_{\mathrm{E}}=\mathrm{U}_{\mathrm{B}}=\frac{1}{2} \in_0 \mathrm{E}_0^2=\frac{\mathrm{B}_0^2}{2 \mu_0}\right]$

Topic: Physics, Electromagnetic Waves, BITSAT, BITSAT 2023 (Memory Based Paper 1)
The ratio of intensities at two points $\mathrm{P}$ and $\mathrm{Q}$ on the screen in a Young's double slit experiment where phase difference between two wave of same amplitude are $\frac{\pi}{3}$ and $\frac{\pi}{2}$, respectively are

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

The correct answer is:
$3: 2$

Solution:
Intensity at a point in Young's double slit experiment is given by
$\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi$
Here $\mathrm{I}_1=\mathrm{I}_2=\mathrm{I}_0$ (say)
At $P$
$\therefore \mathrm{I}_{\mathrm{p}}=\mathrm{I}_0+\mathrm{I}_0+2 \mathrm{I}_0 \cos \frac{\pi}{3}=2 \mathrm{I}_0+2 \mathrm{I}_0 \times \frac{1}{2}=3 \mathrm{I}_0$
At $Q$
$\mathrm{I}_{\mathrm{Q}}=\mathrm{I}_0+\mathrm{I}_0+2 \mathrm{I}_0 \cos 90^{\circ}=2 \mathrm{I}_0$
$\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{Q}}}=\frac{3}{2}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2023 (Memory Based Paper 1)