datavorous/entrance-exam-dataset huggingface dataset:
\(\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x\)

Options:
A \(-\frac{1}{2} \sin 2 x+C\)
B \(\frac{1}{2} \sin 2 \mathrm{x}+\mathrm{C}\)
C \(\frac{1}{2} \sin x+C\)
D \(-\frac{1}{2} \sin x+C\)

The correct answer is:
\(-\frac{1}{2} \sin 2 x+C\)

Solution:
Hints: \(\int\left(\sin ^2 x-\cos ^2 x\right) d x=-\int \cos ^{2 x} d x=-\frac{1}{2} \sin 2 x+C\)

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2011
\(\int \frac{x^3 d x}{1+x^8}=\)

Options:
A \(4 \tan ^{-1} \mathrm{x}^3+\mathrm{c}\)
B \(\frac{1}{4} \tan ^{-1} x^4+c\)
C \(x+4 \tan ^{-1} x^4+c\)
D \(x^2+\frac{1}{4} \tan ^{-1} x^4+c\)

The correct answer is:
\(\frac{1}{4} \tan ^{-1} x^4+c\)

Solution:
Hints: \(\int \frac{x^3 d y}{1+\left(x^4\right)^2}=\frac{1}{4} \tan ^{-1}\left(x^4\right)\)

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2011
\(\int_0^{2 \pi} \theta \sin ^6 \theta \cos \theta d \theta\) is equal to

Options:
A \(\frac{\pi}{16}\)
B \(\frac{3 \pi}{16}\)
C \(\frac{16 \pi}{3}\)
D 0

The correct answer is:
0

Solution:
Hint : Let \(I=\int_0^{2 \pi} \theta \sin ^6 \theta \cos \theta d \theta \ldots \ldots .(I)\)
applying King's Property,
\(\int_0^{2 \pi}(2 \pi-\theta) \sin ^6 \theta \cdot \cos \theta d \theta \ldots \ldots .(II)\)
adding (I) and (II), \(2 \mathrm{I}=2 \pi \int_0^{2 \pi} \sin ^6 \theta \cos \theta d \theta\)
\(\Rightarrow I=\pi \int_0^{2 \pi} \sin ^6 \theta \cos \theta \ldots \ldots .(III)\)
Using Queen's Property, \(\int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\), when \(f(2 a-x)=f(x)\)
applying Queen's Property on (III), I \(=2 \pi \int_0^\pi \sin ^6 \theta \cos \theta d \theta\)
again, using Queen's Property, \(\int_0^{2 a} f(x) d x=0\), when, \(f(2 a-x)=-f(x)\)
\(I=0\)

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2023
Integrating Factor (I.F.) of the defferential equation
\(\frac{d y}{d x}-\frac{3 x^2 y}{1+x^3}=\frac{\sin ^2(x)}{1+x} \text { is }\)

Options:
A \(\mathrm{e}^{1+\mathrm{x}^3}\)
B \(\log \left(1+x^3\right)\)
C \(1+x^3\)
D \(\frac{1}{1+x^3}\)

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

Solution:
\(\begin{aligned}
& \text {Hints: If } e^{\int p d x}=e^{-\int \frac{3 x^2 d x}{1+x^3}}=e^{-\log \left(1+x^3\right)}=e^{\log \left(1+x^3\right)^{-1}} \\
& =\left(1+x^3\right)^{-1}=\frac{1}{1+x^3}
\end{aligned}\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2011
\(\int_\pi^{16 \pi}|\sin x| d x=\)

Options:
A 0
B 32
C 30
D 28

The correct answer is:
30

Solution:
Hints: \(15 \int_0^\pi \sin x d x=15(-\cos x)_0^\pi=30\)

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2011
Ionic solids with Schottky defect may contain in their structure

Options:
A cation vacancies only
B cation vacancies and interstitial cations
C equal number of cation and anion vacancies
D anion vacancies and interstitial anions

The correct answer is:
equal number of cation and anion vacancies

Solution:
In Schotthy defect, equal number of cations and anions are missing creating equal number of cation and anion vacancies.

Topic: Chemistry, Solid State, WBJEE, WBJEE 2016
Ionisation potential values of noble gases decrease down the group with increase in atomic size. Xenon forms binary fluorides by the direct reaction of elements. Identify the correct statement(s) from below.

Options:
A Only the heavier noble gases form such compounds
B It happens because the noble gases have higher ionisation energies
C It happens because the compounds are formed with electronegative ligands
D Octet of electrons provide the stable arrangements

The correct answers are:
Only the heavier noble gases form such compounds, It happens because the compounds are formed with electronegative ligands

Solution:
Ionisation potential values of noble gases decrease down the group with increase in atomic size. Xenon forms binary fluorides by the direct reaction of elements.
The reason is that only the heavier noble gases form such compounds. Octet of electrons provide the stable arrangements but this statement is not very much valid in cases of higher element which can be ionised easily due to large distance between nucleus and valence electrons.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2015
$\omega$ is an imaginary cube root of unity and $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$ then one of the values of $x$ is

Options:
A 1
B 0
C -1
D 2

The correct answer is:
0

Solution:
$$
\begin{aligned}
\text { Hints: } & \stackrel{C_1^{\prime} \rightarrow C_1+C_2+C_3}{\longrightarrow}\left|\begin{array}{ccc}
x & \omega & 1 \\
x & \omega^2 & 1+x \\
x & x+\omega & \omega^2
\end{array}\right|=x\left|\begin{array}{ccc}
1 & \omega & 1 \\
1 & \omega^2 & 1+x \\
1 & x+\omega & \omega^2
\end{array}\right| \\
\quad & =x\left|\begin{array}{ccc}
1 & \omega & 1 \\
0 & \omega^2-\omega & x \\
0 & x & \omega^2-1
\end{array}\right|=x\left\{\left(\omega^2-\omega\right)\left(\omega^2-1\right)-x^2\right\}=0 \quad \Rightarrow x=0 \quad \text { One value of } x=0
\end{aligned}
$$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2010
It is difficult to cook rice in an open vessel by boiling it at high altitudes because of

Options:
A low boiling point and high pressure
B high boiling point and low pressure
C low boiling point and low pressure
D high boiling point and high pressure

The correct answer is:
low boiling point and low pressure

Solution:
At high altitude pressure is low and boiling point also low

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2009
$\mathrm{K}_{f}$ (water) $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The temperature at which ice begins to separate from a mixture of 10 mass $\%$ ethylene glycol is

Options:
A $-1.86^{\circ} \mathrm{C}$
B $-3.72^{\circ} \mathrm{C}$
C $-3.3^{\circ} \mathrm{C}$
D $-3^{\circ} \mathrm{C}$

The correct answer is:
$-3.3^{\circ} \mathrm{C}$

Solution:
Hint:
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$
$=1.86\left(\frac{10}{62} \times \frac{1000}{90}\right)$
$=3.3^{\circ} \mathrm{C}$
$\therefore$ Temparature at which ice begins to separate $=3.3^{\circ} \mathrm{C}$

Topic: Chemistry, Solutions, WBJEE, WBJEE 2020
Keto-enol taut omerism is not observed in

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COC}_6 \mathrm{H}_5$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}=\mathrm{CH}_2$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_2 \mathrm{COCH}_3$
D $\mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{COCH}_3$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{COC}_6 \mathrm{H}_5$

Solution:
as contains no $\alpha-\mathrm{H}$

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2010
ldentify the correct method for the synthesis of lhe compound shown below from the following alternatives.

Options:
A
B
C
D

The correct answers are:

Solution:
The sequence of reaction is as follows:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2015
Le the relation $\rho$ be defined on $\mathbb{R}$ by a $\rho$ b holds if and only if $a-b$ is zero or irrational, then

Options:
A $\rho$ is equivalence relation
B $\rho$ is reflexive $\&$ symmetric but is not transitive
C $\rho$ is reflexive and transitive but is not symmetric
D $\rho$ is reflexive only

The correct answer is:
$\rho$ is reflexive $\&$ symmetric but is not transitive

Solution:
Hint:
If $a-b=0$ then $b-a=0$, if $a-b$ is irrational then $b-a$ is irrational
$\therefore a \rho b \Rightarrow b \rho a \Rightarrow$ symmetric
$\forall a \in \mathbb{R}, a-a=0 a \rho a \Rightarrow$ reflexive
If $a=2, b=\sqrt{2}, c=3$, then
a $\rho$ b, b $\rho$ c but a $\rho$ c is not true $\Rightarrow$ not transitive

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2020
Let $0 < \alpha < \beta < 1$. Then $\lim {n \rightarrow \infty} \sum_{k=1}^{n} \int_{1/k+\beta}^{1 /(k+a)} \frac{d x}{1+x}$ is

Options:
A $\log _{e} \frac{\beta}{\alpha}$
B $\log _{e} \frac{1+\beta}{1+\alpha}$
C $\log _{e} \frac{1+\alpha}{1+\beta}$
D $\infty$

The correct answer is:
$\log _{e} \frac{1+\beta}{1+\alpha}$

Solution:
Hint:
$\lim _{n \rightarrow \infty} \sum_{k=1}^{n}[\log |1+x|]_{\frac{1}{k+\beta}} \frac{1}{k+a}$
$=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\log \left(1+\frac{1}{k+\alpha}\right)-\log \left(1+\frac{1}{k+\beta}\right)\right)$
$=\lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\log \left(\frac{k+\alpha+1}{k+\alpha}\right)-\log \left(\frac{k+\beta+1}{k+\beta}\right)\right)$
$=\log \left(\frac{\beta+1}{\alpha+1}\right)$

Topic: Mathematics, Limits, WBJEE, WBJEE 2020
Let $\rho_{1}$ and $\rho_{2}$ be two equivalence relations defined on a non-void set $S$. Then

Options:
A both $\rho_{1} \cap \rho_{2}$ and $\rho_{1} \cup \rho_{2}$ are equivalence relations
B $\rho_{1} \cap \rho_{2}$ is equivalence relation but $\rho_{1} \cup \rho_{2}$ is not so.
C $\rho_{1} \cup \rho_{2}$ is equivalence relation but $\rho_{1} \cap \rho_{2}$ is not so
D neither $\rho_{1} \cap \rho_{2}$ nor $\rho_{1} \cup \rho_{2}$ is equivalence relation.

The correct answer is:
$\rho_{1} \cap \rho_{2}$ is equivalence relation but $\rho_{1} \cup \rho_{2}$ is not so.

Solution:
Hint:
Union of two transitive may or may not be transitive

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2020
Let $\int \frac{x^{1 / 2}}{\sqrt{1-x^3}} d x=\frac{2}{3} g(f(x))+c$; then

Options:
A $f(x)=\sqrt{x}, g(x)=x^{3 / 2}$
B $f(x)=x^{3 / 2}, g(x)=\sin ^{-1} x$
C $f(x)=\sqrt{x}, g(x)=\sin ^{-1} x$
D $\quad f(x)=\sin ^{-1} x, g(x)=x^{3 / 2}$

The correct answer is:
$f(x)=x^{3 / 2}, g(x)=\sin ^{-1} x$

Solution:
$\int \frac{x^{1 / 2} d x}{\sqrt{1-\left(x^{3 / 2}\right)^2}}=2 / 3 \int \frac{d t}{\sqrt{1-t^2}}$
$x^{3 / 2}=t$
$3 / 2 x^{1 / 2} d x=d t$
$=2 / \sin ^{-1}(t)+c$
$=2 / 3 \sin ^{-1}\left(x^{3 / 2}\right)+c$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2022
Let $(1+x)^{10}=\sum_{r=0}^{10} c_{r} x^{r}$ and $(1+x)^{7}=\sum_{r=0}^{7} d_{r} x^{r}$
If $P=\sum_{r=0}^{5} c_{2}$ and $Q=\sum_{r=0}^{3} d_{2 r+1},$ then $\frac{P}{Q}$ is equal
to

Options:
A 4
B 8
C 16
D 32

The correct answer is:
8

Solution:
$P=\sum_{r=0}^{5} C_{2 r}$
$={ }^{10} C_{0}+{ }^{10} C_{2}+\quad+{ }^{10} C_{10}=\frac{2^{0}}{2}=2^{9}$
$Q=\sum_{r=0}^{3} d_{2 r+1}=d_{1}+d_{3}+d_{5}+d_{7}$
$={ }^{7} C_{1}+{ }^{7} C_{3}+{ }^{7} C_{5}+{ }^{7} C_{7}=\frac{2^{7}}{2}=2^{6}$
$\therefore \frac{P}{\Omega}=\frac{2^{9}}{3^{6}}=2^{3}=8$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2012
Let $\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}$
$+\ldots+a_{18} x^{18} \cdot$ Then

Options:
A $a_{0}+a_{2}+\ldots+a_{18}=a_{1}+a_{3}+\ldots+a_{17}$
B $a_{0}+a_{2}+\ldots+a_{18}$ is even
C $a_{0}+a_{2}+\ldots+a_{18}$ is divisible by 9
D $a_{0}+a_{2}+\ldots+a_{18}$ is divisible by 3 but not by 9

The correct answer is:
$a_{0}+a_{2}+\ldots+a_{18}$ is even

Solution:
$\left(1+x+x^{2}\right)^{9}=a_{0}+a_{1} x+a_{2} x^{2}+ \dots+a_{18}x^{18}$
$$
Put $x=-1,$ we get
$$
\begin{array}{l}
(1-1+1)^{9}=a_{0}-a_{1}+a_{2}+\ldots+a_{18} \\
\Rightarrow 1=a_{0}-a_{1}+a_{2}+\ldots+a_{18}
\end{array}
$$
Put $x=1,$ we get
$$
(1+1+1)^{9}=a_{0}+a_{1}+a_{2}+\ldots+a_{14}
$$
$\Rightarrow 3^{9}=a_{0}+a_{1}+a_{2}+\ldots+a_{18}$
On adding Eqs. (i) and (ii), we get
$$
\begin{array}{c}
3^{9}+1=2\left(a_{0}+a_{2}+\ldots+a_{10}\right) \\
a_{0}+a_{2}+a_{4}+\ldots+a_{18}=\frac{3^{0}+1}{2} \\
=\frac{19683+1}{2} \\
=\frac{19684}{2}
\end{array}
$$
$=9842$ which is even number.

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2017
Let $A=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & \text { cost } & \text { sint } \\ 0 & -\sin t & \cos t\end{array}\right)$
Let $\lambda_{1}, \lambda_{2}, \lambda_{3}$ be the roots of $\operatorname{det}\left(A-\lambda I_{3}\right)=0$, where $I_{3}$ denotes the identity matrix. If $\lambda_{1}+\lambda_{2}+\lambda_{3}=\sqrt{2}+1$, then the set of possible values of $t,-\pi \leq t < \pi$ is

Options:
A a void set
B $\left\{\frac{\pi}{4}\right\}$
C $\left\{-\frac{\pi}{4}, \frac{\pi}{4}\right\}$
D $\left\{-\frac{\pi}{3}, \frac{\pi}{3}\right\}$

The correct answer is:
$\left\{-\frac{\pi}{4}, \frac{\pi}{4}\right\}$

Solution:
$\left|\begin{array}{ccc}1-\lambda & 0 & 0 \\ 0 & \cos t-\lambda & \sin t \\ 0 & -\sin t & \cos t-\lambda\end{array}\right|=0$
$\Rightarrow(1-\lambda)(\cos t-\lambda)^{2}+\sin ^{2} t=0$
$\Rightarrow(1-\lambda)\left(\lambda^{2}-2 \lambda \cos t+\cos ^{2} t\right)+\sin ^{2} t=0$
$\Rightarrow \lambda^{2}-2 x \cos t+\cos ^{2} t-\lambda^{3}+2 \lambda^{2} \cos t+\lambda \cos ^{2} t+\sin ^{2} t=0$
$\Rightarrow-\lambda^{3}+\lambda^{2}(1+2 \cos t)+\lambda\left(\cos ^{2} t-2 \cos t\right)+1=0$
$\lambda_{1}+\lambda_{2}+\lambda_{3}=1+2 \operatorname{cost}=1+\sqrt{2}$
$\therefore$ cost $=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{t}=\frac{\pi}{4},-\frac{\pi}{4}$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2021
Let $A=\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)$ Then, for positive integer
$n, A^{n}$ is

Options:
A $\left(\begin{array}{lll}1 & n & n^{2} \\ 0 & n^{2} & n \\ 0 & 0 & n\end{array}\right)$
B $\left(\begin{array}{lll}1 & n & n\left(\frac{n+1}{2}\right) \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right)$
C $\left(\begin{array}{lll}1 & n^{2} & n \\ 0 & n & n^{2} \\ 0 & 0 & n^{2}\end{array}\right)$
D $\left(\begin{array}{ccc}1 & n & 2 n-1 \\ 0 & \frac{n+1}{2} & n^{2} \\ 0 & 0 & \frac{n+1}{2}\end{array}\right)$

The correct answer is:
$\left(\begin{array}{lll}1 & n & n\left(\frac{n+1}{2}\right) \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right)$

Solution:
We have, $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$\therefore \quad A^{2}=A \cdot A$
$=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1+0+0 & 1+1+0 & 1+1+1 \\ 0+0+0 & 0+1+0 & 0+1+1 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 2 & \frac{2(2+1)}{2} \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]$
Again, $A^{3}=A^{2} \cdot A$
$=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{lll}1+0+0 & 1+2+0 & 1+2+3 \\ 0+0+0 & 0+1+0 & 0+1+2 \\ 0+0+0 & 0+0+0 & 0+0+1\end{array}\right] \\=\left[\begin{array}{lll}1 & 3 & 6 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}1 & 3 & \frac{3(3+1)}{2} \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{array}\right] \\\therefore A^{n} =\left[\begin{array}{lll}1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1\end{array}\right] \end{aligned}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2017
Let $A(-1,0)$ and $B(2,0)$ be two points. $A$ point $M$ moves in the plane in such a way that $\angle M B A=2 \angle M A B .$ Then, the point $M$ moves along

Options:
A a straight line
B a parabola
C an ellipse
D a hyperbola

The correct answer is:
a hyperbola

Solution:
$$
\begin{array}{l}
\text { Let } \angle M A B=\theta \text { , then } \angle M B A=2 \theta \\
\text { tan } \theta=\frac{k}{1+h} \text { and } \tan 2 \theta=\frac{k}{2-h} \\
\text { then } \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{k}{2-h} \\
\Rightarrow \quad \frac{2(k / 1+h)}{1-(k / 1+h)^{2}}=\frac{k}{2-h} \\
\Rightarrow \quad \frac{2}{1+h} \times \frac{(1+h)^{2}}{(1+h)^{2}-k^{2}}=\frac{1}{2-h} \\
\Rightarrow \quad \frac{2(1+h)}{(1+h)^{2}-k^{2}}=\frac{1}{2-h} \\
\Rightarrow \quad 1+h^{2}+2 h-k^{2}=2(1+h)(2-h) \\
\Rightarrow \quad 1+h^{2}+2 h-k^{2}=2\left(2-h+2 h-h^{2}\right) \\
\Rightarrow \quad 1+h^{2}+2 h-k^{2}=2\left(2+h-h^{2}\right) \\
\Rightarrow \quad 1+h^{2}+2 h-k^{2}=4+2 h-2 h^{2} \\
\Rightarrow \quad 1+3 h^{2}-k^{2}=4 \\
\Rightarrow \quad 3 h^{2}-k^{2}=3
\end{array}
$$
which represents hyperbola (d).

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2017
Let $A(2,-3)$ and $B(-2,1)$ be two angular points of $\Delta A B C$. If the centroid of the triangle moves on the line $2 x+3 y=1$, then the locus of the angular point $C$ is given by

Options:
A $2 x+3 y=9$
B $2 x-3 y=9$
C $3 x+2 y=5$
D $3 x-2 y=3$

The correct answer is:
$2 x+3 y=9$

Solution:
Let the coordinates of $C$ be $(\alpha, \beta)$.
$\therefore$ Coordinates of centroid
$$
\begin{array}{l}
=\left(\frac{2-2+\alpha}{3}, \frac{-3+1+\beta}{3}\right) \\
=\left(\frac{\alpha}{3}, \frac{\beta-2}{3}\right)
\end{array}
$$
Since, centroid lie on $2 x+3 y=1$
$\therefore \quad \frac{2 \alpha}{3}+3\left(\frac{\beta-2}{3}\right)=1$
$\Rightarrow \quad \frac{2 \alpha}{3}+\frac{3 \beta-6}{3}=1$
$\Rightarrow \quad 2 \alpha+3 \beta-6=3 \Rightarrow 2 \alpha+3 \beta=9$
$\therefore$ Locus of point $C$ will be $2 x+3 y=9$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2017
Let $A=\left(\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3\end{array}\right) .$ Then, the roots of the equation $\operatorname{det}\left(A-\lambda I_{3}\right)=0$ (where $I_{3}$ is the identity matrix of order 3 ) are

Options:
A 3,0,3
B 0,3,6
C 1,0,-6
D 3,3,6

The correct answers are:
0,3,6

Solution:
We have,
$$
A=\left[\begin{array}{lll}
3 & 0 & 3 \\
0 & 3 & 0 \\
3 & 0 & 3
\end{array}\right]
$$
Again, we have
$$
\left|\begin{array}{ccc}
3-\lambda & 0 & 3 \\
0 & 3-\lambda & 0 \\
3 & 0 & 3-\lambda
\end{array}\right|=0
$$
$\Rightarrow(3-\lambda)\left[(3-\lambda)^{2}-0\right]-0+3[0-3(3-\lambda)]=0$
$\Rightarrow (3-\lambda)^{3}-9(3-\lambda)=0$
$\Rightarrow (3-\lambda)\left[(3-\lambda)^{2}-9\right]=0$
$\Rightarrow (3-\lambda)\left(9+\lambda^{2}-6 \lambda-9\right)=0$
$\Rightarrow (3-\lambda)\left(\lambda^{2}-6 \lambda\right)=0$
$\Rightarrow \quad(3-\lambda) \lambda(\lambda-6)=0$
${\lambda}=3,0,6$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2019
Let $\mathrm{A}=\left(\begin{array}{ccc}3-\mathrm{t} & 1 & 0 \\ -1 & 3-\mathrm{t} & 1 \\ 0 & -1 & 0\end{array}\right)$ and $\operatorname{det} \mathrm{A}=5$, then

Options:
A $\mathrm{t}=1$
B $\mathrm{t}=2$
C $t=-1$
D $\mathrm{t}=-2$

The correct answer is:
$\mathrm{t}=-2$

Solution:
Hint:
$|\mathrm{A}|=\left|\begin{array}{ccc}3-\mathrm{t} & 1 & 0 \\ -1 & 3-\mathrm{t} & 1 \\ 0 & -1 & 0\end{array}\right|=3-\mathrm{t}=5, \mathrm{t}=-2$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2020
Let $\mathrm{A}=\left(\begin{array}{ccc}12 & 24 & 5 \\ \mathrm{x} & 6 & 2 \\ -1 & -2 & 3\end{array}\right) .$ The value of $\mathrm{x}$ for which the matrix $\mathrm{A}$ is not invertible is

Options:
A 6
B 12
C 3
D 2

The correct answer is:
3

Solution:
Hint:
If matrix is not invertible $\Rightarrow|\mathrm{A}|=0$
$\therefore|\mathrm{A}|=\left|\begin{array}{ccc}12 & 24 & 5 \\ \mathrm{x} & 6 & 2 \\ -1 & -2 & 3\end{array}\right|=0 \Rightarrow \mathrm{x}=3$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2020
Let $\mathrm{A}=\left(\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ c & \mathrm{~d}\end{array}\right)$ be a $2 \times 2$ real matrix with $\operatorname{det} \mathrm{A}=1$. If the equation $\operatorname{det}\left(\mathrm{A}-\lambda \mathrm{I}{2}\right)=0$ has imaginary roots $\left(\mathrm{I}{2}\right.$ be the
Identity matrix of order 2), then

Options:
A $(a+d)^{2} < 4$
B $(a+d)^{2}=4$
C $(a+d)^{2}>4$
D $(a+d)^{2}=16$

The correct answer is:
$(a+d)^{2} < 4$

Solution:
Hint:
$|\mathrm{A}|=0 \therefore \mathrm{ad}-\mathrm{bc}=1$
$\left|\mathrm{~A}-\lambda \mathrm{I}_{2}\right|=0$
$\left|\begin{array}{cc}\mathrm{a}-\lambda & \mathrm{b} \\ \mathrm{c} & \mathrm{d}-\lambda\end{array}\right|=0$
$\therefore \mathrm{ad}-(\mathrm{a}+\mathrm{d}) \lambda+\lambda^{2}-\mathrm{bc}=0$
$\lambda^{2}-(\mathrm{a}+\mathrm{d}) \lambda+1=0$
$\therefore(\mathrm{a}+\mathrm{d})^{2} < 4$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2020
Let \(A\) and \(B\) are orthogonal matrices and \(\operatorname{det} A+\operatorname{det} B=0\). Then

Options:
A \(A+B\) is singular
B \(A+B\) is non-singular
C \(A+B\) is orthogonal
D \(A+B\) is skew symmetric

The correct answer is:
\(A+B\) is singular

Solution:
Hint: \(A A^{\top}=I\) and \(B B^{\top}=I\)
\(\Rightarrow \operatorname{det}\left(A A^{\top}\right)=1\) and \(\operatorname{det}\left(B B^{\top}\right)=1\)
\(\operatorname{det}(A)=-\operatorname{det}(B)\) (Given) \( \Rightarrow(\operatorname{det}(A))^2=1 \Rightarrow(\operatorname{det}(B))^2=1\)
\(\therefore \operatorname{det}(A+B)=\operatorname{det}\left(A\left(B^{\top}+A^{\top}\right) B\right)\)
\(=-\operatorname{det}\left(B^{\top}+A^{\top}\right) \quad[\because(\operatorname{det}(A))(\operatorname{det}(B))=-1]\) as \(\operatorname{det}(A)=-\operatorname{det}(B)\)
\(=-\operatorname{det}(B+A)^{\top}\)
\(=-\operatorname{det}(A+B)^{\top}\)
\(=-\operatorname{det}(A+B)\)
\(\Rightarrow 2 \operatorname{det}(A+B)=0\)
\(\Rightarrow \operatorname{det}(A+B)=0\)
\(\therefore(A+B)\) is singular

Topic: Mathematics, Determinants, WBJEE, WBJEE 2023
Let \(A\) and \(B\) are two independent events. The probability that both \(A\) and \(B\) happen is \(\frac{1}{12}\) and probability that neither A nor \(B\) happen is \(\frac{1}{2}\). Then

Options:
A \(P(A)=\frac{1}{3}, P(B)=\frac{1}{4}\)
B \(P(A)=\frac{1}{2}, P(B)=\frac{1}{6}\)
C \(P(A)=\frac{1}{6}, P(B)=\frac{1}{2}\)
D \(P(A)=\frac{2}{3}, P(B)=\frac{1}{8}\)

The correct answer is:
\(P(A)=\frac{1}{3}, P(B)=\frac{1}{4}\)

Solution:
\(\begin{array}{ll}
\text {Hint : } P(A \cap B)=\frac{1}{12} & \Rightarrow P(A) P(B)=\frac{1}{12} \cdots (1) \\
P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{2} & \Rightarrow P(A \cup B)=1-\frac{1}{2}=\frac{1}{2}
\end{array}\)
\(\begin{aligned} & \Rightarrow P(A)+P(B)-P(A) P(B)=\frac{1}{2} \\ & \Rightarrow P(A)+P(B)=\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\end{aligned}\)
\(\begin{aligned} & \text { From (1), } \quad P(A)\left(\frac{7}{12}-P(A)\right)=\frac{1}{12} \\ & \Rightarrow P(A)^2-\frac{7}{12} P(A)+\frac{1}{12}=0 \\ & \Rightarrow 12 P(A)^2-7 P(A)+1=0 \\ & \Rightarrow 12 P(A)^2-4 P(A)-3 P(A)+1=0 \\ & \Rightarrow(3 P(A)-1)-(4 P(A)-1)=0\end{aligned}\)
\(\begin{aligned} & \therefore P(A)=\frac{1}{3} \text { or } \quad P(A)=\frac{1}{4} \\ & \therefore \quad P(B)=\frac{1}{12} \times 3 \Rightarrow P(B)=\frac{1}{12} \times 4 \\ & \qquad\qquad=\frac{1}{4} \qquad \qquad \qquad=\frac{1}{3} \\ & \therefore P(A)=\frac{1}{3}, P(B)=\frac{1}{4}\end{aligned}\)

Topic: Mathematics, Probability, WBJEE, WBJEE 2023
Let $A$ and $B$ be two events such that $P(A \cap B)=\frac{1}{6}, P(A \cup B)=\frac{31}{45}$ and $P(\bar{B})=\frac{7}{10}$ then

Options:
A $A$ and $B$ are independent
B A and $B$ are mutually exclusive
C $P\left(\frac{A}{B}\right) < \frac{1}{6}$
D $P\left(\frac{B}{A}\right) < \frac{1}{6}$

The correct answer is:
$A$ and $B$ are independent

Solution:
Given, $P(A \cap B)=\frac{1}{6}$
$P(A \cup B)=\frac{31}{45}$
and $\quad P(\bar{B})=\frac{7}{10}$
$\therefore \quad \quad P(B)=1-\frac{7}{10}=\frac{3}{10}$
Now, $\quad P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow \quad \frac{31}{45}=P(A)+\frac{3}{10}-\frac{1}{6}$
$\therefore \quad P(A)=\frac{31}{45}+\frac{1}{6}-\frac{3}{10}$ $=\frac{5}{9}$
Then. $\quad P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}=\frac{1}{5}=\frac{3}{10}>\frac{1}{6}$
and $\quad P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{6}}{\frac{3}{3}}=\frac{5}{9}>\frac{1}{6}$
and, $P(A) \times P(B)=\frac{5}{9} \times \frac{3}{10}=\frac{1}{6}=P(A \cap B)$
Hence, $A$ and $B$ are independent.

Topic: Mathematics, Probability, WBJEE, WBJEE 2016
Let $A$ and $B$ be two events with $P\left(A^{C}\right)=0.3$ $P(B)=0.4 \quad$ and $\quad P\left(A \cap B^{C}\right)=0.5 . \quad$ Then
$P\left(B \mid A \cup B^{c}\right)$ is equal to

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

The correct answer is:
$\frac{1}{4}$

Solution:
$P\left(\frac{B}{A \cup B^{c}}\right)=\frac{P\left(B \cap\left(A \cup B^{e}\right)\right)}{P\left(A \cup B^{c}\right)}$
$=\frac{P(A \cap B)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}$
$=\frac{P(A)-P\left(A \cap B^{c}\right)}{P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right)}$
$=\frac{0.7-0.5}{0.7+0.6-0.5}=\frac{0.2}{0.8}=\frac{1}{4}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2012