datavorous/entrance-exam-dataset huggingface dataset:
In $\mathrm{XeF}_{2}, \mathrm{XeF}_{4}, \mathrm{XeF}_{6}$ the number of lone pairs on $\mathrm{Xe}$ are respectively

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

The correct answer is:
$3,2,1$

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2021
In Young's double slit experiment 10 th order maximum is obtained at the point of observation in the interference pattern for $\lambda=7000 \AA$. If the source is replaced by another one of wavelength $5000 \AA$ then the order of maximum at the same point will be-

Options:
A 12 th
B 14 th
C 16 th
D 18 th

The correct answer is:
14 th

Solution:
$\mathrm{n}_{1} \lambda_{1}=\mathrm{n}_{2} \lambda_{2}$

Topic: Physics, Wave Optics, BITSAT, BITSAT 2013
In Young's double slit experiment, if the slit widths are in the ratio $1: 2,$ the ratio of the intensities at minima and maxima will be

Options:
A 1: 2
B 1: 3
C 1: 4
D 1: 9

The correct answer is:
1: 9

Solution:
No Solution available

Topic: Physics, Wave Optics, BITSAT, BITSAT 2009
In Young's double slit experiment, intensity at a point is 14 of the maximum intensity. Angular position of this point is

Options:
A sin-1λd
B sin-1λ2d
C sin-1λ3d
D sin-1λ4d

The correct answer is:
sin-1λ3d

Solution:
I=Imaxcos2ϕ2

Imax4=Imaxcos2ϕ2

cosϕ2=12

ϕ2=π3

⇒ ϕ=2π3=2πλ×Δx....i

where, Δx=dsinθ

Putting the value of Δx in Eq. (i), we get

sinθ=λ3d

θ=sin-1λ3d

Topic: Physics, Wave Optics, BITSAT, BITSAT 2018
In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ1=12000Å and λ2=10000Å . At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Options:
A 8 mm
B 6 mm
C 4 mm
D 3 mm

The correct answer is:
6 mm

Solution:
Given, λ1=12000Å

and λ2=10000Å

D=2 cm

d=2 mm=2×10-3 m

We have,

λ1λ2=n2n1=1200010000=65

as X=n1λ1Dd n1=5

=5×12000×10-10×22×10-3

=5×1.2×104×10-10×103=6 mm

Topic: Physics, Wave Optics, BITSAT, BITSAT 2017
In Young's double-slit experiment, the distance between slits and screen is 2 m and distance between slits is 0.25 mm. A light of wavelength 800 nm is used to find fringes on the screen. If the screen moves with a speed of 5 m s-1, then first maxima will move with a speed of:

Options:
A 16 mm s-1
B 8 mm s-1
C 20 mm s-1
D 50 mm s-1

The correct answer is:
16 mm s-1

Solution:
Given, D=2 m, d=0.25 mm=2.5×10-4 m

Velocity of screen, dDdt=5 m s-1

Speed of first maxima =dβdt=?

In Young's double-slit experiment, fringe width β=Dλd

On differentiating with respect to t, we get

dβdt=λd·dDdt

=800×10-92.5×10-4×5

=16×10-3 m s-1=16 mm s-1

Topic: Physics, Wave Optics, BITSAT, BITSAT 2019
Increasing order of rms velocities of $\mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}$ and $\mathrm{HBr}$ is

Options:
A $\mathrm{H}_{2}>\mathrm{O}_{2}>\mathrm{N}_{2}>\mathrm{HBr}$
B $\mathrm{HBr}<\mathrm{O}_{2}<\mathrm{N}_{2}<\mathrm{H}_{2}$
C $\mathrm{H}_{2}>\mathrm{N}_{2}<\mathrm{O}_{2}>\mathrm{HBr}$
D $\mathrm{HBr}>\mathrm{N}_{2}<\mathrm{O}_{2}<\mathrm{H}_{2}$

The correct answer is:
$\mathrm{HBr}<\mathrm{O}_{2}<\mathrm{N}_{2}<\mathrm{H}_{2}$

Solution:
RMS velocity of molecules depends on mass. If mol. wt. increases, rms velocity of melocules decreases. $\mathrm{rms} \propto \frac{1}{\sqrt{\mathrm{m} \cdot \mathrm{wt}}}$

The order of increasing $\mathrm{m} . \mathrm{wt}$. is

$\mathrm{H}_{2}<\mathrm{N}_{2}<\mathrm{O}_{2}<\mathrm{HBr}$

Order of $V_{\mathrm{mx}}$ of molecules. $\mathrm{HBr}<\mathrm{O}_{2}<\mathrm{N}_{2}<\mathrm{H}_{2}$

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2013
Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories?

Options:
A An enzyme
B A hormone
C A co-enzyme
D An antibiotic

The correct answer is:
A hormone

Solution:
Insulin is a biochemically active peptide harmone secreted by pancreas.

Topic: Chemistry, Chemistry in Everyday Life, BITSAT, BITSAT 2013
\(\int 4 \cos \left(x+\frac{\pi}{6}\right) \cos 2 x \cdot \cos \left(\frac{5 \pi}{6}+x\right) d x\)

Options:
A \(-\left(x+\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c\)
B \(-\left(x+\frac{\sin 4 x}{4}-\frac{\sin 2 x}{2}\right)+c\)
C \(-\left(x-\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c\)
D \(-\left(x-\frac{\sin 4 x}{4}+\frac{\cos 2 x}{2}\right)+c\)

The correct answer is:
\(-\left(x+\frac{\sin 4 x}{4}+\frac{\sin 2 x}{2}\right)+c\)

Solution:
\(\begin{aligned}
& \int 4 \cos \left(x+\frac{\pi}{6}\right) \cos 2 x \cdot \cos \left(\frac{5 \pi}{6}+x\right) d x \\
& =2 \int\left(\cos (2 x+\pi) \cos \frac{2 \pi}{3}\right) \cos 2 x d x \\
& =2 \int\left(-\cos 2 x-\frac{1}{2}\right) \cos 2 x d x \\
& =\int\left(-2 \cos ^2 2 x-\cos 2 x\right) d x \\
& =-\int(1+\cos 4 x+\cos 2 x) d x \\
& =-x-\frac{\sin 4 x}{x}-\frac{\sin 2 x}{2}+c
\end{aligned}\)

Topic: Mathematics, Definite Integration, BITSAT, BITSAT 2010
Interparticle forces present in nylon 66 are

Options:
A van der Waal's
B hydrogen bonding
C dipole-dipole interactions
D None of the above

The correct answer is:
hydrogen bonding

Solution:
No Solution available

Topic: Chemistry, Chemical Bonding and Molecular Structure, BITSAT, BITSAT 2009
Inthetruth table for the statement $(p \wedge q) \rightarrow(q \vee \sim p)$, the last column has the truth value in the following order is

Options:
A TTFF
B FTTT
C TFTT
D $\mathrm{TTTT}$

The correct answer is:
$\mathrm{TTTT}$

Solution:
$\mathrm{T} \mathrm{T} \mathrm{T} \mathrm{T}$

Topic: Mathematics, Mathematical Reasoning, BITSAT, BITSAT 2013
Ionic hydrides reacts with water to give

Options:
A acidic solutions
B hydride ions
C basic solutions
D electrons

The correct answer is:
basic solutions

Solution:
Ionic hydrides give the basic solution when it reacts with water, e.g.,
$\mathrm{LiH}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{LiOH}+\mathrm{H}_2$

Topic: Chemistry, Hydrogen, BITSAT, BITSAT 2023 (Memory Based Paper 1)
It is given that the events \(A\) and \(B\) are such that \(P(A)=\frac{1}{4}, P(A \mid B)=\frac{1}{2} \quad\) and \(\quad P(B \mid A)=\frac{2}{3}\). Then \(P(B)\) is

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

The correct answer is:
\(\frac{1}{3}\)

Solution:
\(P(A)=1 / 4, P(A / B)=\frac{1}{2}, P(B / A)=2 / 3\)
By conditional probability,
\(\begin{aligned}
& P(A \cap B)=P(A) P(B / A)=P(B) P(A / B) \\
& \Rightarrow \frac{1}{4} \times \frac{2}{3}=P(B) \times \frac{1}{2} \Rightarrow P(B)=\frac{1}{3}
\end{aligned}\)

Topic: Mathematics, Probability, BITSAT, BITSAT 2009
$\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ is used in detecting.

Options:
A $\mathrm{Fe}^{3+} \mathrm{ion}$
B $\mathrm{Cu}^{+} \mathrm{ion}$
C $\mathrm{Cu}^{3+}$ ion
D $\mathrm{Fe}^{2+}$ ion

The correct answer is:
$\mathrm{Fe}^{3+} \mathrm{ion}$

Solution:
$\mathrm{Fe}^{3+}$ ion can be detected by $\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

$$

\begin{aligned}

4 \mathrm{Fe}^{3+}+3 \mathrm{~K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \\

& \longrightarrow \mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{~K}^{+}

\end{aligned}

$$

Topic: Chemistry, Practical Chemistry, BITSAT, BITSAT 2015
$K_{c}$ for the the reaction, $\left[\mathrm{Ag}(\mathrm{CN})_{2}\right] \rightleftharpoons \mathrm{Ag}^{+}+2 \mathrm{CN}^{-},$

the equillibrium constant at $25^{\circ} \mathrm{C}$ is $4.0 \times 10^{-19}$ then the silver ion concentration in a solution which was originally 0.1 molar in $\mathrm{KCN}$ and 0.03 molar in $\mathrm{AgNO}_{3}$ is:

Options:
A $7.5 \times 10^{18}$
B $7.5 \times 10^{-19}$
C $7.5 \times 10^{19}$
D $7.5 \times 10^{-18}$

The correct answer is:
$7.5 \times 10^{-18}$

Solution:
$\left.\begin{array}{c}2 \mathrm{KCN}+\mathrm{AgNO}=\left[\underset{0.1}{0.03}=\left[\begin{array}{c}\mathrm{Ag}(\mathrm{CN})_{2} \\ 0\end{array}\right]^{-}\right. \\ (0.1-0.06) & 0\end{array}\right] \begin{array}{c}\mathrm{KNO}_{3}+\mathrm{K}_{0}^{+} \\ (0.03)\end{array}$

$\therefore\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}=\mathrm{Ag}^{+}+\underset{\mathrm{a}}{0.03} \mathrm{a} \quad 2 \mathrm{CN}$

$(0.03-\mathrm{a}) \quad \mathrm{a} .04(\mathrm{left} \mathrm{from} \mathrm{KCN}$

$\quad 0.04 \approx 0.04$

$\therefore \quad \mathrm{K}_{\mathrm{c}}=4 \times 10^{-19}=\frac{(0.04)^{2} \times \mathrm{a}}{0.03}$

$\mathrm{a}=7.5 \times 10^{-18}$

Topic: Chemistry, Ionic Equilibrium, BITSAT, BITSAT 2014
Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance $\mathrm{r}$

between sun and planet i.e. $\mathrm{T}^{2}=\mathrm{Kr}^{3}$ here $\mathrm{K}$ is constant. If the masses of sun and planet are $\mathrm{M}$ and $\mathrm{m}$ respectively then as per Newton's law of gravitation force of attraction between them is

$\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$, here $\mathrm{G}$ is gravitational constant. The relation between $\mathrm{G}$ and $\mathrm{K}$ is described as

Options:
A $\mathrm{GMK}=4 \pi^{2}$
B $\quad \mathrm{K}=\mathrm{G}$
C $\mathrm{K}=\frac{1}{\mathrm{G}}$
D $\quad \mathrm{GK}=4 \pi^{2}$

The correct answer is:
$\mathrm{GMK}=4 \pi^{2}$

Solution:
As we know, orbital speed, $V_{\text {orb }}=\sqrt{\frac{\text { GM }}{\text { r }}}$

Time period $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}_{\text {orb }}}=\frac{2 \pi \mathrm{r}}{\sqrt{\mathrm{GM}}} \sqrt{\mathrm{r}}$

Squarring both sides,

$\mathrm{T}^{2}=\left(\frac{2 \pi \mathrm{r} \sqrt{\mathrm{r}}}{\sqrt{\mathrm{GM}}}\right)^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}$

$\Rightarrow \quad \frac{\mathrm{T}^{2}}{\mathrm{r}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}=\mathrm{K} \Rightarrow \mathrm{GMK}=4 \pi^{2}$

Topic: Physics, Gravitation, BITSAT, BITSAT 2016
Keto-enol tautomerism is observed in

Options:
A
B
C
D None

The correct answer is:

Solution:

Topic: Chemistry, General Organic Chemistry, BITSAT, BITSAT 2009
Kinetic theory of gases proves

Options:
A only Boyle's law
B only Charles' law
C only Avogadro's law
D all of these

The correct answer is:
all of these

Solution:
Kinetic theory of gases proves all the given gas laws.

Topic: Chemistry, States of Matter, BITSAT, BITSAT 2023 (Memory Based Paper 1)
Knowing that the chemistry of lanthanoids(Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect?

Options:
A The ionic size of Ln (III) decrease in general with increasing atomic number
B Ln (III) compounds are generally colourless.
C Ln (III) hydroxide are mainly basic in character.
D Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character.

The correct answer is:
Ln (III) compounds are generally colourless.

Solution:
Most of the $\mathrm{Ln}^{3+}$ compounds except $\mathrm{La}^{3+}$ and $\mathrm{Lu}^{3+}$ are coloured due to the presence of $f$ -electrons.

Topic: Chemistry, d and f Block Elements, BITSAT, BITSAT 2016
Laughing gas is

Options:
A nitrogen pentoxide
B nitrous oxide
C $\quad$ nitrogen trioxide
D nitric oxide

The correct answer is:
nitrous oxide

Solution:
Nitrous oxide (i.e, $\mathrm{N}_{2} \mathrm{O}$ ) is the laughing gas.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), BITSAT, BITSAT 2015
Lenz's law is a consequence of the law of conservation of

Options:
A charge
B mass
C energy
D momentum

The correct answer is:
energy

Solution:
No Solution available

Topic: Physics, Electromagnetic Induction, BITSAT, BITSAT 2009
Let $\alpha_{1}, \alpha_{2}$ and $\beta_{1}, \beta_{2}$ be the roots of $a x^{2}+b x+c$ $=0$ and $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0$ respectively. If the system of equations $\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}=0$ and $\beta_{1} \mathrm{y}+\beta_{2} \mathrm{z}=0$ has a non-trivial solution, then

Options:
A $\frac{b^{2}}{q^{2}}=\frac{a c}{p r}$
B $\frac{\mathrm{c}^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{ab}}{\mathrm{pq}}$
C $\frac{a^{2}}{p^{2}}=\frac{b c}{q r}$
D None of these

The correct answer is:
$\frac{b^{2}}{q^{2}}=\frac{a c}{p r}$

Solution:
Since $\alpha_{1}, \alpha_{2}$ and $\beta_{1}, \beta_{2}$ are the roots of ax $^{2}$ $+b x+c=0$ and $p x^{2}+q x+r=0$ respectively

therefore

$$

\alpha_{1}+\alpha_{2}=\frac{-b}{a}, \alpha_{1} \alpha_{2}=\frac{c}{a}

$$

and $\beta_{1}+\beta_{2}=\frac{-q}{p}, \beta_{1} \beta_{2}=\frac{r}{p}$

Since the given system of equation has a non-trivial solution

$$

\therefore\left|\begin{array}{ll}

\alpha_{1} & \alpha_{2} \\

\beta_{1} & \beta_{2}

\end{array}\right|=0 \text { i.e. } \alpha_{1} \beta_{2}-\alpha_{2} \beta_{1}=0

$$

or $\quad \frac{\alpha_{1}}{\beta_{1}}=\frac{\alpha_{2}}{\beta_{2}}=\frac{\alpha_{1}+\alpha_{2}}{\beta_{1}+\beta_{2}}=\sqrt{\frac{\alpha_{1} \alpha_{2}}{\beta_{1} \beta_{2}}}$

$\Rightarrow \frac{\mathrm{pb}}{\mathrm{qa}}=\sqrt{\frac{\mathrm{pc}}{\mathrm{ra}}} \Rightarrow \frac{\mathrm{b}^{2}}{\mathrm{q}^{2}}=\frac{\mathrm{ac}}{\mathrm{pr}}$

Topic: Mathematics, Determinants, BITSAT, BITSAT 2016
Let 2x2+3x+410=∑r=020arxr, then the value of $\frac{a_{8}}{a_{12}},$ is

Options:
A 2
B 4
C 8
D 16

The correct answer is:
4

Solution:
Given, 2x2+3x+410=∑r=020arxr

Replacing x by 2x, we get

8x2+6x+410=∑r=020ar2xr

⇒2102x2+3x+410=∑r=020ar2rx20-r

⇒ 210∑r=020arxr=∑r=020ar2rx20-r

Comparing ccefficient of x8 both sides, we get

210a8=a12212⇒a8a12=212210=4

Topic: Mathematics, Binomial Theorem, BITSAT, BITSAT 2019
Let $A=\left[\begin{array}{ll}0 & \alpha \\ 0 & 0\end{array}\right]$ and $(A+I)^{50}-50 A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$,
find $a b c+a b d+b c d+a c d$

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

The correct answer is:
0

Solution:
As $\mathrm{A}^{2}=0, \mathrm{~A}^{k}=0 \forall k \geq 2$
Thus, $\quad(\mathrm{A}+\mathrm{I})^{50}=\mathrm{I}+50 \mathrm{~A} \Rightarrow(\mathrm{A}+\mathrm{I})^{50}-50 \mathrm{~A}=\mathrm{I}$
$\therefore \quad \mathrm{a}=1, \mathrm{~b}=0, \mathrm{c}=0, \mathrm{~d}=1$
$\mathrm{abc}+\mathrm{abd}+\mathrm{bed}+\mathrm{acd}=0$

Topic: Mathematics, Matrices, BITSAT, BITSAT 2012
Let A=-1,1 and f:A→A be defined as fx=xx for all x∈A, then fx is

Options:
A many one and into function
B one-many and into function
C many-one and into function
D one-one and onto function

The correct answer is:
one-one and onto function

Solution:
fx=xx=x2,x≥0-x2,x<0

Since, -1≤x≤1, therefore -1≤fx≤1

∴ Function is one-one and onto.

Topic: Mathematics, Functions, JEE Main
Let A=-1,1 and f:A→A be defined as fx=xx for all x∈A, then fx is

Options:
A many one and into function
B one-many and into function
C many-one and into function
D one-one and onto function

The correct answer is:
one-one and onto function

Solution:
fx=xx=x2,x≥0-x2,x<0

Since, -1≤x≤1, therefore -1≤fx≤1

∴ Function is one-one and onto.

Topic: Mathematics, Functions, BITSAT, BITSAT 2017
Let $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$
If $B$ is the inverse of $A$, then the value of $\alpha$ is

Options:
A
4

B
3

C
$-4$

D
5

The correct answer is:

5

Solution:
$A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
$\therefore|A|=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
$=1(1+3)+1(2+3)+1(2-1)=4+5+1=10$
$\Rightarrow \operatorname{Adj} A=\left[\begin{array}{ccc}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{array}\right]^T=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
$\Rightarrow \quad B=A^{-1}=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\because A^{-1}=\frac{1}{|A|} \operatorname{Adj} A\right]$
$\Rightarrow 10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
Hence, $\alpha=5$

Topic: Mathematics, Matrices, BITSAT, BITSAT 2022
Let A={1,2,3,4,5} and R be a relation defined by R={(x,y):x,y∈A,x+y=5}. Then, R is

Options:
A reflexive and symmetric but not transitive
B an equivalence relation
C neither reflexive nor transitive
D neither reflexive nor symmetric but transitive

The correct answer is:
neither reflexive nor transitive

Solution:
We first write the elements of the set R.

i.e. R={(1,4),(4,1),(2,3),(3,2)}

Since, (1,1)∉R⇒R is not reflexive.

Now as,(1,4)∈R⇒(4,1)∈R and (2,3)∈R

⇒ (3,2)∈R

So, R is symmetric

Now, (1,4)∈R and (4,1)∈R

Thus, R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2017
Let $a_1, a_2, a_3 \ldots$ be a harmonic progression with $a_1=5$ and $a_{20}=25$. The least positive integer $n$ for which $a_n < 0$, is

Options:
A
22

B
23

C
24

D
25

The correct answer is:

25

Solution:
It is given that $a_1, a_2, a_3 \ldots$. are in HP. Therefore, $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots \ldots .$. are in AP. Let $d$ be the common difference of the AP.
$\therefore \quad \frac{1}{a_n}=\frac{1}{a_1}+(n-1) d$ and $\frac{1}{a_{20}}=\frac{1}{a_1}+19 d$
$\Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}+(n-1) d$ and $\frac{1}{25}=\frac{1}{5}+19 d$
$\Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}+(n-1) d$ and $d=\frac{-4}{19 \times 25}$
$\Rightarrow \quad \frac{1}{a_n}=\frac{1}{5}-\frac{4(n-1)}{19 \times 25}$
$\Rightarrow \quad \frac{1}{a_n}=\frac{95-4 n+4}{19 \times 25}$
$\Rightarrow \quad a_n=\frac{99-4 n}{19 \times 25}$
Now, $a_n < 0$
$\Rightarrow \quad \frac{99-4 n}{19 \times 25} < 0$
$\Rightarrow \quad 99-4 n < 0$
$\Rightarrow \quad 4 n>99$
$\Rightarrow \quad n>24 \frac{3}{4} \Rightarrow n \geq 25$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2022
Let $a_{1}, a_{2}, a_{3} \ldots \ldots \ldots \ldots .$ be terms on A.P. If
$\frac{a_{1}+a_{2}+\ldots \ldots \ldots a_{p}}{a_{1}+a_{2}+\ldots \ldots \ldots .+a_{q}}=\frac{p^{2}}{q^{2}}, p \neq q,$ then $\frac{a_{6}}{a_{21}}$
equals

Options:
A $\frac{41}{11}$
B $\frac{7}{2}$
C $\frac{2}{7}$
D $\frac{11}{41}$

The correct answer is:
$\frac{11}{41}$

Solution:
$\frac{\frac{p}{2}\left[2 a_{1}+(p-1) d\right]}{\frac{q}{2}\left[2 a_{1}+(q-1) d\right]}=\frac{p^{2}}{q^{2}}$
$\Rightarrow \frac{2 a_{1}+(p-1) d}{2 a_{1}+(q-1) d}=\frac{p}{q}$
$\frac{a_{1}+\left(\frac{p-1}{2}\right) d}{a_{1}+\left(\frac{q-1}{2}\right) d}=\frac{p}{q}$ For $\frac{a_{6}}{a_{21}}, \quad p=11, q=41$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2012
Let $a_1, a_2, \ldots \ldots a_{40}$ be in AP and $h_1, h_2, \ldots . h_{10}$ be in HP. If $a_1=h_1=2$ and $a_{10}=h_{10}=3$, then $a_4 h_7$ is

Options:
A
2

B
3

C
5

D
6

The correct answer is:

6

Solution:
Let $d$ be the common difference of the AP. Then,
$\begin{aligned}
& a_{10}=3 \Rightarrow a_1+9 d=3 \\
\Rightarrow & 2+9 d=3 \Rightarrow d=\frac{1}{9} \\
\therefore & a_4=a_1+3 d=2+\frac{1}{3}=\frac{7}{3}
\end{aligned}$
Let $D$ be the common difference of $\frac{1}{h_1}, \frac{1}{h_2}, \ldots .-\frac{1}{h_{10}}$.
Then, $\quad h_{10}=3$
$\begin{array}{ll}
\Rightarrow & \frac{1}{h_{10}}=\frac{1}{3} \Rightarrow \frac{1}{2}+9 D=\frac{1}{3} \\
\Rightarrow & 9 D=-\frac{1}{6} \Rightarrow D=-\frac{1}{54} \\
\therefore & \frac{1}{h_7}=\frac{1}{h_1}+6 D=\frac{1}{2}-\frac{1}{9}=\frac{7}{18} \\
\Rightarrow & h_7=\frac{18}{7} \\
\therefore & a_4 h_7=\frac{7}{3} \times \frac{18}{7}=6
\end{array}$

Topic: Mathematics, Sequences and Series, BITSAT, BITSAT 2022
Let A and B are two independent events. If the probability that both A and B occur together is 16 and the probability that neither of them occurs is 13, then the probability of occurrence of A is

Options:
A 0 or 1
B 12 or 13
C 12 or 14
D 13 or 14

The correct answer is:
12 or 13

Solution:
PA∩B=16 and PAC∩BC=13

Now, PA∪BC=PAC∩BC=13

⇒1-PA∪B=13

⇒PA∪B=23

But PA∪B=PA+PB-PA∩B

⇒PA+PB=56 ...i

∵ A and B are independent events

∴ PA∩B=PAPB

⇒PAPB=16

PA-PB2=PA+PB2-4PAPB

=2536-46=136

⇒PA-PB=±16 ...ii

On solving Eqs. (i) and (ii), we get

PA=12 or 13

Hence, option (b) is correct.

Topic: Mathematics, Probability, BITSAT, BITSAT 2018
Let a^ and b^ are two non-collinear unit vectors. If u→=a^-a^·b^b^ and v→=a^×b^, then |v→| is equal to

Options:
A u→
B u→+v→.a^
C 2v→
D v→+u→·a^+b^

The correct answer is:
u→

Solution:
We have, a^=b^=1

Given, v→=a^×b^=a^b^sinθn^=sinθn^

where, θ is the angle between a^ and b^.

∴ v→=sinθ

Now, u→=a^-a^·b^b^=a^-a^b^cosθb^

u→=a^-cosθb^

∴ u→=a^2+-cosθb^2-2a^-cosθb^cosθ

⇒ u→=1+cos2θ-2cos2θ

⇒u→=1-cos2θ=sin2θ

∴ u→=sinθ

So, v→=u→

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2017
Let A and B be two sets such that A∩X=B∩X=ϕ and A∪X=B∪X for same set X. Then,

Options:
A A=B
B A=X
C B=X
D A∪B=X

The correct answer is:
A=B

Solution:
Given, that, A∪X=B∪X

⇒A∩(A∪X)=A∩(B∪X)

⇒(A∩A)∪(A∩X)=(A∩B)∪(A∩X)

[Using distributive law]

⇒A∪ϕ=(A∩B)∪ϕ [∵A∩X=ϕ]

⇒ A=A∩B ...(i)

Again, consider A∪X=B∪X

⇒B∩(A∪X)=B∩(B∪X)

⇒(B∩A)∪(B∩X)=(B∩B)∪(B∩X)

⇒(B∩A)∪ϕ=B∪ϕ[B∩X=ϕ]

⇒A∩B=B ...(ii)

Thus from Eq. (i) and (ii), we get

A=B

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2017
Let A and B be two sets then $(\mathrm{A} \cup \mathrm{B}) \cup\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$ is equal to

Options:
A $\mathrm{A}^{\prime}$
B $\mathrm{A}$
C B'
D None of these

The correct answer is:
$\mathrm{A}^{\prime}$

Solution:
$\text { From Venn-Euler's Diagram. }$

$\therefore(\mathrm{A} \cup \mathrm{B})^{\prime} \cup\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\mathrm{A}^{\prime}$

Topic: Mathematics, Sets and Relations, JEE Main
Let A and B be two sets then $(\mathrm{A} \cup \mathrm{B}) \cup\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$ is equal to

Options:
A $\mathrm{A}^{\prime}$
B $\mathrm{A}$
C B'
D None of these

The correct answer is:
$\mathrm{A}^{\prime}$

Solution:
$\text { From Venn-Euler's Diagram. }$

$\therefore(\mathrm{A} \cup \mathrm{B})^{\prime} \cup\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\mathrm{A}^{\prime}$

Topic: Mathematics, Sets and Relations, BITSAT, BITSAT 2012
Let $a, b$ and $c$ be three vectors satisfying $a \times b=$ $(a \times c),|a|=|c|=1,|b|=4$ and $|b \times c|=\sqrt{15}$. If $b-$

$2 \mathrm{c}=\lambda \mathrm{a},$ then $\lambda$ equals

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

The correct answer is:
-4

Solution:
Let $\theta$ be the angle between $\mathbf{b}$ and $\mathbf{c}$. Given, $|\mathrm{b} \times \mathrm{c}|=\sqrt{15}$

$$

\begin{array}{l}

\Rightarrow|\mathrm{b} \| \mathrm{c}| \sin \theta=\sqrt{15} \\

\Rightarrow \sin \theta=\frac{\sqrt{15}}{|\mathrm{~b}| \mathrm{c} \mid} \\

\Rightarrow \sin \theta=\sqrt{\frac{15}{4 \times 1}}=\frac{\sqrt{15}}{4} \\

\quad \therefore \cos \theta=\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4}

\end{array}

$$

Now given, $\mathrm{b}-2 \mathrm{c}=\lambda \mathrm{a} \Rightarrow\left|\mathrm{b}-2 \mathrm{c}^{2}\right|=\mid \lambda \mathrm{a}^{2}$

$$

\Rightarrow|\mathrm{b}|^{2}+4|\mathrm{c}|-4(\mathrm{~b} \cdot \mathrm{c})=\lambda^{2}|\mathrm{a}|^{2}

$$

$$

\begin{array}{l}

\Rightarrow 16+4-4|\mathrm{~b} \| \mathrm{c}| \cos \theta=\lambda^{2} \\

\Rightarrow 20-16 \cos \theta=\lambda^{2} \\

\Rightarrow 20-4=\lambda^{2} \Rightarrow \lambda^{2}=16\left(\therefore \cos \theta=\frac{1}{4}\right) \\

\Rightarrow \lambda=\pm 4

\end{array}

$$

Topic: Mathematics, Vector Algebra, BITSAT, BITSAT 2015
Let $a, b$ be the solutions of $x^2+p x+1=0$ and $c, d$ be the solution of $x^2+q x+1=0$. If $(a-c)(b-c)$ and $(a+d)(b+d)$ are the solution of $x^2+a x+\beta=0$, then $\beta$ is equal to

Options:
A
$p+q$

B
$p-q$

C
$p^2+q^2$

D
$q^2-p^2$

The correct answer is:

$q^2-p^2$

Solution:
Since, $a+b=-p, a b=1$ .....(i)
and $c+d=-q, c d=1$
Now $(a-c)(b-c)$ and $(a+d)(b+d)$ are the roots of $x^2+a x+\beta=0$
$$
\begin{aligned}
& (a-c)(b-c)(a+d)(b+d)=\beta \\
& \Rightarrow\left(a b-a c-b c+c^2\right)\left(a b+a d+b d+d^2\right)=\beta \\
& \Rightarrow\left\{1-c(a+b)+c^2\right\}\left\{1+d(a+b)+d^2\right\}=\beta \\
& \Rightarrow \quad\left(1+p c+c^2\right)\left(1-p d+d^2\right)=\beta \\
& \Rightarrow 1-p d+d^2+p c-p^2 c d+p c d^2+c^2-p c^2 d \\
& +c^2 d^2=\beta \\
& \Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta \\
& {[\because c d=1]} \\
& \Rightarrow \quad 2+d^2+c^2-p^2=\beta \\
&
\end{aligned}
$$
$\begin{aligned} \Rightarrow \quad 2 c d+c^2+d^2-p^2 & =\beta \\ (c+d)^2-p^2 & =\beta \\ q^2-p^2 & =\beta \quad[\because(c+d)=-q]\end{aligned}$

Topic: Mathematics, Quadratic Equation, BITSAT, BITSAT 2022