datavorous/entrance-exam-dataset huggingface dataset:
For an ideal gas with initial pressure and volume $p_{i}$ and $V_{i}$ respectively, a reversible isothermal expansion happens, when its volume becomes $V_{0}$. Then, it is compressed to its original volume $V_{i}$ by a reversible adiabatic process. If the final pressure is $p_{f}$ then which of the following statement(s) is/are true?

Options:
A $p_{f}=p_{i}$
B $p_{f}>p_{i}$
C $p_{f} < p_{i}$
D $\frac{\rho_{f}}{V_{0}}=\frac{P_{i}}{V_{i}}$

The correct answer is:
$p_{f}>p_{i}$

Solution:
In $p-V$ diagram, the slope of an adiabatic curve at any point in steeper than that of isothermal curve at that point.

Thus, $p_{f}>p_{i}$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2018
For any complex number $z$, the minimum value of $|z|+|z-1|$ is

Options:
A 0
B 1
C 2
D -1

The correct answer is:
1

Solution:
$$
\begin{gathered}
\text { Hints }: 1=|z-(z-1)| \\
\qquad 1 \leq|z|+|z-1|
\end{gathered}
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2009
For any non-zero complex number $z$, the minimum value of $|z|+|z-1|$ is

Options:
A 1
B $\frac{1}{2}$
C 0
D $\frac{3}{2}$

The correct answer is:
1

Solution:
$|z|+|z-1| a$ is complex number
$$
\begin{array}{c}
|z|+|-(1-z)| \quad|a|=|-a| \\
|z|+|1-z| \quad\{\because|a|=|-a|\} \\
|z|+|1-z| \geq|z+1-z| \\
|z|+|1-z| \geq|1| \\
|z|+|1-z| \geq 1 \\
|z|+|-(z-1)| \geq 1 \\
|z|+|z-1| \geq 1
\end{array}
$$
Minimum value of $|z|+|z-1|=1$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2019
For any two real numbers $a$ and $b$, we define $a R b$ if and only if $\sin ^{2} a+\cos ^{2} b=1$. The relation $R$ is

Options:
A reflexive but not symmetric
B symmetric but not transitive
C transitive but not reflexive
D an equivalence relation

The correct answer is:
an equivalence relation

Solution:
Let the given relation defined as
$R=\left\{(a, b) \mid \sin ^{2} a+\cos ^{2} b=1\right\}$
For reflexive, $\sin ^{2} a+\cos ^{2} a=1$
$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1, \forall \theta \in R\right)$
$\Rightarrow { }_{a} R_{a} \Rightarrow(a, a) \in R$
$\Rightarrow R$ is reflexive.
For symmetric, $\sin ^{2} a+\cos ^{2} b=1$
$\Rightarrow 1-\cos ^{2} a+1-\sin ^{2} b=1$
$\Rightarrow \sin ^{2} b+\cos ^{2} a=1$
$\Rightarrow$ bRa
Hence, $R$ is symmetric.
For transitive
Let $a R b, b R c$ $\Rightarrow \sin ^{2} a+\cos ^{2} b=1$
and $\sin ^{2} b+\cos ^{2} c=1$
On adding Eqs. (0) and (ii), we get $\sin ^{2} a+\left(\sin ^{2} b+\cos ^{2} b\right)+\cos ^{2} c=2$
$\Rightarrow \sin ^{2} a+\cos ^{2} c+1=2$
$\Rightarrow \sin ^{2} a+\cos ^{2} c=1$
Hence, $A$ is transitive also. Therefore, relation $A$ is an equivalence relation.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2013
For any two real numbers $\theta$ and $\phi$ we define $\theta$. R $\phi$ if and only if $\sec ^{2} \theta-\tan ^{2} \phi=1$. The relation $R$ is

Options:
A reflexive but not transitive
B symmetric but not reflexive
C both reflexive and symmetric but not transitive
D an equivalence relation

The correct answer is:
an equivalence relation

Solution:
Given relation is defined as $\theta R \phi$ such that $\sec ^{2} \theta-\tan ^{2} \phi=1$
For Reflexive
When $\theta R \theta$
$\sec ^{2} \theta-\tan ^{2} \theta=1$
$\Rightarrow 1=1,$ which is true.
Thus, it is reflexive.
For Symmetric When $\theta R \phi$
$\begin{array}{l}
\sec ^{2} \theta-\tan ^{2} \phi=1 \\
\Rightarrow\left(1+\tan ^{2} \theta\right)-\left(\sec ^{2} \phi-1\right)=1 \\
\Rightarrow 2+\tan ^{2} \theta-\sec ^{2} \phi=1 \\
\Rightarrow \sec ^{2} \phi-\tan ^{2} \theta=1
\end{array}$
$\Rightarrow \phi F \theta$
Thus, it is symmetric.
For Transitive
When $\theta R \phi$ and $\phi$ Fiy, then
$\sec ^{2} \theta-\tan ^{2} \phi=1$
and $\sec ^{2} \phi-\tan ^{2} \psi=1$
Now, $\theta$ R $\psi$
Then. $\sec ^{2} \theta-\tan ^{2} \psi=1$
$\Rightarrow \sec ^{2} \theta-\tan ^{2} \psi+1=1+1$
$\Rightarrow \sec ^{2} \theta-\tan ^{2} \psi+\sec ^{2} \phi-\tan ^{2} \phi=1+1$
$\Rightarrow \theta R \phi$ and $\phi R \psi$
Thus, it is transitive. Hence, it is an equivalence relation.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2014
For any two sets $\mathrm{A}$ and $\mathrm{B}, \mathrm{A}-(\mathrm{A}-\mathrm{B})$ equals

Options:
A $\mathrm{B}$
B $\mathrm{A}-\mathrm{B}$
C $\mathrm{A} \cap \mathrm{B}$
D $\mathrm{A}^{\mathrm{C}} \cap \mathrm{B}^{\mathrm{C}}$

The correct answer is:
$\mathrm{A} \cap \mathrm{B}$

Solution:
Hints: $A-(A-B)=A-\left(A \cap B^c\right)=A \cap\left(A \cap B^c\right)^c=A \cap\left(A^c \cup B\right)=\left(A \cap A^c\right) \cup(A \cap B)=A \cap B$

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2009
For any vector $\mathbf{x},$ where $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$ have their usual meanings the value of $(\mathbf{x} \times \hat{\mathbf{i}})^{2}+(\mathbf{x} \times \mathbf{j})^{2}+(\mathbf{x} \times \mathbf{k})^{2}$ where $\mathbf{i}, \mathbf{j}, \mathbf{k}$ have their usual meanings, is equal to

Options:
A $|\mathbf{x}|^{2}$
B $2|\mathbf{x}|^{2}$
C $3|\mathbf{x}|^{2}$
D $4|\mathbf{x}|^{2}$

The correct answer is:
$2|\mathbf{x}|^{2}$

Solution:
Let $\mathbf{x}=\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k}$
Then, $\mathbf{x} \times \mathbf{i}=-\beta \hat{\mathbf{k}}+\gamma \hat{\mathbf{j}}$
$\begin{aligned} \mathbf{x} \times \hat{\mathbf{j}} &=\hat{\mathbf{k}}-\gamma \hat{\mathbf{i}} \\ \mathbf{x} \times \mathbf{k} &=-\alpha \hat{\mathbf{j}}+\beta \hat{\mathbf{i}} \end{aligned}$
$\begin{aligned} \text {Now, }(\mathbf{x} \times \mathbf{i})^{2} &=(\mathbf{x} \times \mathbf{i}) \cdot(\mathbf{x} \times \hat{\mathbf{i}}) \\ &=(-\beta \hat{\mathbf{k}}+\gamma \hat{\mathbf{j}}) \cdot(-\beta \hat{\mathbf{k}}+\gamma \hat{\mathbf{j}}) \\ &=\beta^{2}+\gamma^{2} \end{aligned}$
Similarly, $(\mathbf{x} \times \mathbf{j})^{2}=\alpha^{2}+\gamma^{2}$
and $(\mathbf{x} \times \mathbf{k})^{2}=\alpha^{2}+\beta^{2}$
$\therefore(\mathbf{x} \times \mathbf{i})^{2}+(\mathbf{x} \times \mathbf{j})^{2}+(\mathbf{x} \times \mathbf{\mathbf { k }})^{2}$
$=\beta^{2}+\gamma^{2}+\alpha^{2}+\gamma^{2}+\alpha^{2}+\beta^{2}$
$=2\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)=2|\mathbf{x}|^{2}$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2017
For $\mathrm{BCl}_{3}, \mathrm{AlCl}_{3}$ and $\mathrm{GaCl}_{3}$ the increasing order of ionic character is

Options:
A $\mathrm{BCl}_{3} < \mathrm{AlCl}_{3} < \mathrm{GaCl}_{3}$
B $\mathrm{GaCl}_{3} < \mathrm{AlCl}_{3} < \mathrm{BCl}_{3}$
C $\mathrm{BCl}_{3} < \mathrm{GaCl}_{3} < \mathrm{AlCl}_{3}$
D $\mathrm{ACCl}_{3} < \mathrm{BCl}_{3} < \mathrm{GaCl}_{3}$

The correct answer is:
$\mathrm{BCl}_{3} < \mathrm{GaCl}_{3} < \mathrm{AlCl}_{3}$

Solution:
Since, ionic character is inversely proportional to polarising power of cation therefore correct order is
$$
\mathrm{AICl}_{3}>\mathrm{GaCl}_{3}>\mathrm{BCl}_{3}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2013
For different values of $\alpha$, the locus of the point of intersection of the two straight lines $\sqrt{3} x-y-4 \sqrt{3} \alpha=0$ and $\sqrt{3} \alpha x+\alpha y-4 \sqrt{3}=0$ is

Options:
A a hyperbola with eccentricity 2
B an ellipse with eccentricity $\sqrt{\frac{2}{3}}$
C a hyperbola with eccentricity $\sqrt{\frac{19}{16}}$
D an ellipse with eccentricity $\frac{3}{4}$

The correct answer is:
a hyperbola with eccentricity 2

Solution:
Hints : $\sqrt{3} x-y=4 \sqrt{3} \alpha$
(1) $\sqrt{3} x+y=\frac{4 \sqrt{3}}{\alpha}$
(1) $x(2) \Rightarrow 3 x^2-y^2=48 \Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1$
$$
\mathrm{e}=\sqrt{\frac{48+16}{16}}=2
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2010
For each $n \in N, 2^{3 n}-1$ is divisible by
where $\mathrm{N}$ is a set of natural numbers

Options:
A 7
B 8
C 6
D 16

The correct answer is:
7

Solution:
$$
\begin{aligned}
& \text { Hints : } 2^{3 \mathrm{n}}=(8)^{\mathrm{n}}=(1+7)^{\mathrm{n}}=1+{ }^{\mathrm{n}} \mathrm{C}_1 7+{ }^{\mathrm{n}} \mathrm{C}_2 7^2 \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} 7^{\mathrm{n}} \\
& 2^{3 \mathrm{n}}-1=7\left[{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2 7+\ldots\right]
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2009
If \(1, \log _9\left(3^{1-x}+2\right), \log _3\left(4.3^{\mathrm{x}}-1\right)\) are in A.P, then \(x\) equals

Options:
A \(\log _3 4\)
B \(1-\log _3 4\)
C \(1-\log _4 3\)
D \(\log _4 3\)

The correct answer is:
\(1-\log _3 4\)

Solution:
\(\begin{aligned}
& \text {Hint: } 2 \log _9\left(3^{1-x}+2\right)=\log _3\left(4.3^x-1\right)+1 \\
& \Rightarrow \log _3\left(3^{1-x}+2\right)=\log _3 3\left(4.3^x-1\right) \\
& \Rightarrow 3^{1-x}+2=3\left(4.3^x-1\right) \\
& \Rightarrow \frac{3}{3^x}+2=4.3^{x+1}-3 \\
& \Rightarrow \frac{3}{3^x}+2=12 \cdot 3^x-3 \\
& \Rightarrow 12 \cdot\left(3^x\right)^2-5\left(3^x\right)-3=0 \\
& \Rightarrow\left(4\left(3^x\right)-3\right)\left(3\left(3^x\right)+1\right)=0 \\
& \because 3^x > 0 \quad \therefore 4\left(3^x\right)-3=0 \\
& \therefore 3\left(3^x\right)+1 \neq 0 \quad \Rightarrow 3^x=\frac{3}{4}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow x=\log _3 \frac{3}{4}=\log _3 3-\log _3 4 \\ & \Rightarrow \mathrm{x}=1-\log _3 4\end{aligned}\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2023
If $\left(1-x+x^2\right)^n=a_0+a_1 x+\ldots . .+a_{2 n} x^{2 n}$, then the value of $a_0+a_2+a_4+\ldots \ldots .+a_{2 n}$ is

Options:
A $3^{\mathrm{n}}+\frac{1}{2}$
B $3^{\mathrm{n}}-\frac{1}{2}$
C $\frac{3^{\mathrm{n}}-1}{2}$
D $\frac{3^{\mathrm{n}}+1}{2}$

The correct answer is:
$\frac{3^{\mathrm{n}}+1}{2}$

Solution:
Hints: $x=1$
$$
\begin{gathered}
1=a_0+a_1+a_2+a_3+\ldots \ldots \ldots+a_{2 n} \\
x=-1,3^{\mathrm{n}}=a_0-a_1+a_2-a_3+\ldots \ldots \ldots \ldots+a_{2 n} \\
1+3^{\mathrm{n}}=2\left[a_0+a_2+a_4+\ldots \ldots \ldots \ldots+a_{2 \mathrm{n}}\right] \\
\Rightarrow a_0+a_2+a_4+\ldots \ldots \ldots \ldots+a_{2 \mathrm{n}}=\frac{1+3^{\mathrm{n}}}{2}
\end{gathered}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2010
If $\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ then the value of $\sqrt{4 \cos ^{4} \theta+\sin ^{2} 2 \theta}+4 \cot \theta \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$ is

Options:
A $-2 \cot \theta$
B $2 \cot \theta$
C $2 \cos \theta$
D $2 \sin \theta$

The correct answer is:
$2 \cot \theta$

Solution:
$\sqrt{4 \cos ^{4} \theta+\sin ^{2} 2 \theta}+4 \cot \theta \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
$=\sqrt{4 \cos ^{4} \theta+(2 \sin \theta \cos \theta)^{2}} +2 \cot \theta\left[2 \cos ^{2}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right]$
$=\sqrt{4 \cos ^{4} \theta+4 \sin ^{2} \theta \cos ^{2} \theta} +2 \cot \theta\left[1+\cos \left(\frac{\pi}{2}-\theta\right)\right]$
$=\sqrt{4 \cos ^{2} \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}+2 \cot \theta(1+\sin \theta)$
$=|2 \cos \theta|+2 \cot \theta+2 \cos \theta$
$=-2 \cos \theta+2 \cot \theta+2 \cos \theta$
$\quad\left[\right.$ for $\left.\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right),|\cos \theta|=-\cos \theta\right]$
$=2 \cot \theta$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2015
If $\int 2^{2^{x}} \cdot 2^{x} d x=A \cdot 2^{2^{x}}+C,$ then $A$ is equal to

Options:
A $\frac{1}{\log 2}$
B $\log 2$
C $(\log 2)^{2}$
D $\frac{1}{(\log 2)^{2}}$

The correct answer is:
$\frac{1}{(\log 2)^{2}}$

Solution:
$$
\begin{aligned}
&\text { (d) Let } I=\int 2^{2^{x}} \cdot 2^{x} d x\\
&\begin{array}{l}
\text { Put } \quad t=2^{x} \\
\Rightarrow \quad d t=2^{x} \log 2 d x \\
\therefore \quad I=\int \frac{2^{t}}{\log 2} d t \\
=\frac{2^{t}}{(\log 2)^{2}}+C=\frac{2^{2^{x}}}{(\log 2)^{2}}+C \\
\therefore \quad A=\frac{1}{(\log 2)^{2}}
\end{array}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2019
If $\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x+i y)$ where $x$ and $y$
are real, then the ordered pair $(x, y)$ is

Options:
A (-3,0)
B (0,3)
C (0,-3)
D $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

The correct answer is:
$\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

Solution:
Let $z=\frac{3}{2}+i \frac{\sqrt{3}}{2}$
$\therefore$
$r=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{\frac{12}{4}}=\sqrt{3}$
$\left.\begin{array}{rl}\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right. \\ \frac{3}{2}\end{array}\right)=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$
$\frac{3}{2}+\frac{i \sqrt{3}}{2}=\sqrt{3} e^{\frac{i \pi}{6}}$
$\therefore$
$\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=\left(\sqrt{3} e^{\frac{i \pi}{6}}\right)^{50}$
$=(\sqrt{3})^{50}\left(e^{\frac{i \pi}{6}}\right)^{50}=3^{25} e^{\frac{50 \pi}{6}}$
$\begin{aligned} \Rightarrow\left(\frac{3}{2}+\frac{i \sqrt{3}}{2}\right)^{50} &=3^{25} e^{\frac{i 25 \pi}{3}} \\ &=3^{25}\left(\cos \frac{25 \pi}{3}+i \sin \frac{25 \pi}{3}\right) \\ &=3^{25}(\cos 1500+i \sin 1500) \end{aligned}$
$=3^{25}[\cos (360 \times 4+60)+i \sin (360$
$\times 4+60)]$
$=3^{25}(\cos 60+i \sin 60)$
$\Rightarrow\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)$
According to question, $\left(\frac{3}{2}+i \frac{\sqrt{3}}{2}\right)^{50}=3^{25}(x+i y)$
$3^{25}\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)=3^{25}(x+6 y)$
which is true only when $x=\frac{1}{2}, y=\frac{\sqrt{3}}{2}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2012
If $(2+i)$ and $(\sqrt{5}-2 i)$ are the roots of the equation $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=0$ where $a, b, c$ and $d$ are real constants, then product of all the roots of the equation is

Options:
A 40
B $9 \sqrt{5}$
C 45
D 35

The correct answer is:
45

Solution:
If one root of a quadratic equation is of the form
$a+i b,$ then other root will be $a-i b$ So, all the roots are $2 \pm 1, \sqrt{5} \pm 2 i$
$\therefore$ Product of all the roots $=(2+i)(2-i)(\sqrt{5}+2 i)(\sqrt{5}-2 i)$
$=(4+1)(5+4)=5 \times 9=45$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2015
If $2 \log (x+1)-\log \left(x^{2}-1\right)=\log 2$, then $x=$

Options:
A only 3
B $-1$ and 3
C only -1
D 1 and 3

The correct answer is:
only 3

Solution:
Hint:
$\log \left\{\frac{(x+1)^{2}}{x^{2}-1}\right\}=\log 2 \Rightarrow(x+1)^{2}=2\left(x^{2}-1\right) \Rightarrow x^{2}-2 x-3=0 \Rightarrow(x-3)(x+1)=0$
$x=3 ; x \neq-1$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2020
$$
\text { If }(2 \leq r \leq n), \text { then }{ }^{n} C_{r}+2 \cdot{ }^{n} C_{r+1}+{ }^{n} C_{r+2} \text { is }
$$
equal to

Options:
A $2 \cdot{ }^{n} C_{r+7}$
B $"^{-1} C_{r+1}$
C ${ }^{n+2} C_{r+2}$
D $^{n+1} C_{r}$

The correct answer is:
${ }^{n+2} C_{r+2}$

Solution:
${ }^{n} C_{r}+2 \cdot{ }^{n} C_{r+1}+{ }^{n} C_{r+2}$
$={ }^{n} C_{r}+{ }^{n} C_{r+1}+{ }^{n} C_{r+1}+{ }^{n} C_{r+2}$
$={ }^{n+1} C_{r+1}+{ }^{n+1} C_{r+2}$
$\quad\left(\because{ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}\right)$
$={ }^{n+2} C_{r+2}$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2018
If $\frac{\cos A}{3}=\frac{\cos B}{4}=\frac{1}{5},-\frac{\pi}{2} < A < 0,-\frac{\pi}{2} < B < 0$ then value of $2 \sin A+4 \sin B$ is

Options:
A 4
B -2
C -4
D 0

The correct answer is:
-4

Solution:
Hints : $\cos A=\frac{3}{5} \quad \sin A=-\frac{4}{5}$
$$
\begin{aligned}
& \cos B=\frac{4}{5} \quad \sin B=-\frac{3}{5} \\
& =2\left(-\frac{4}{5}\right)+4\left(-\frac{3}{5}\right)=-\frac{20}{5}=-4
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2010
If \(4 a^2+9 b^2-c^2+12 a b=0\), then the family of straight lines \(a x+b y+c=0\) is concurrent at

Options:
A \((2,3)\) or \((-2,-3)\)
B \((-2,3)\) or \((2,3)\)
C \((3,2)\) or \((-3,2)\)
D \((-3,2)\) or \((2,3)\)

The correct answer is:
\((2,3)\) or \((-2,-3)\)

Solution:
Hint : \(\because 2 a+3 b-c=0\) or \(2 a+3 b+c=0\)
\(\begin{aligned}
& \Rightarrow c= \pm(2 a+3 b) \\
& \because \quad a x+b y+c=0 \\
& \Rightarrow \quad a x+b y \pm(2 a+3 b)=0 \\
& \Rightarrow \quad a(x \pm 2)+b(y \pm 3)=0 \text { (family of lines) } \\
& \Rightarrow \quad(-2,-3) \quad \text { or }(2,3)
\end{aligned}\)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2023
If $\frac{1}{{ }^{5} C_{r}}+\frac{1}{{ }^{6} C_{r}}=\frac{1}{{ }^{4} C_{r}},$ then the value of $r$ is

Options:
A 4
B 2
C 5
D 3

The correct answer is:
2

Solution:
Given, $\frac{1}{5 C_{r}} +\frac{1}{6 C_{r}} =\frac{1}{4 C_{r}}$
$\Rightarrow \quad \frac{r !(5-r) !}{5 !}+\frac{r !(6-r) !}{6 !}=\frac{r !(4-r) !}{4 !}$
$\Rightarrow \quad \frac{(5-r) !}{5}+\frac{(6-r) !}{30}=(4-r) !$
$\Rightarrow \quad \frac{(5-r)}{5}+\frac{(6-r)(5-r)}{30}=1$
$\Rightarrow \quad 30-6 r+30-11r+r^{2}=30$
$\Rightarrow \quad r^{2}-17 r+30=0$
$\Rightarrow \quad r^{2}-15 r-2 r+30=0$
$\Rightarrow \quad r(r-15)-2(r-15)=0$
$\Rightarrow \quad(r-2)(r-15)=0$
$\therefore \quad r=15,2$
Hence, the value of $r$ is 2 .

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2016
If $5 \cos 2 \theta+2 \cos ^2 \frac{\theta}{2}+1=0$, when $(0 < \theta < \pi)$, then the values of $\theta$ are :

Options:
A $\frac{\pi}{3} \pm \pi$
B $\frac{\pi}{3}, \cos ^{-1}\left(\frac{3}{5}\right)$
C $\cos ^{-1}\left(\frac{3}{5}\right) \pm \pi$
D $\frac{\pi}{3}, \pi-\cos ^{-1}\left(\frac{3}{5}\right)$

The correct answer is:
$\frac{\pi}{3}, \pi-\cos ^{-1}\left(\frac{3}{5}\right)$

Solution:
Hints : $5 \cos 2 \theta+1+\cos \theta+1=0$
$$
\begin{aligned}
& 5\left(2 \cos ^2 \theta-1\right)+\cos \theta+2=0 \\
& 10 \cos ^2 \theta+\cos \theta-3=0 \\
& (5 \cos \theta+3)(2 \cos \theta-1)=0 \\
& \cos \theta=\frac{1}{2} \\
& \theta=\frac{\pi}{3}
\end{aligned}
$$
$$
\mid \begin{aligned}
& \cos \theta=-\frac{3}{5} \\
& \theta=\cos ^{-1}\left(-\frac{3}{5}\right) \\
& =\pi-\cos ^{-1}\left(\frac{3}{5}\right)
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2009
If 64,27,36 are the Pth $Q$ th and $R$ th terms of a GP, then $P+2 Q$ is equal to

Options:
A $R$
B $2 R$
C $3 R$
D $4 R$

The correct answer is:
$3 R$

Solution:
Let a be the first term and \(r\) be the common ration of a GP. \(\therefore P^{\text {th }}, Q^{\text {th }}\) and \(R^{\text {th }}\) terms of a GP are respectively ar \({ }^{\mathrm{P}-1}\), ar \(\mathrm{Q}^{-1}\) and \({a r}^{\mathrm{R}-1}\).
According to question, \(a r^{P-1}=64 \ldots . .\) (i)
\(a r^{Q-1}=27 \ldots . .\) (ii)
\(a r^{R-1}=36 \ldots . .\) (iii)
Dividing Eq. (i) by Eq. (ii), we get
\(\mathrm{r}^{\mathrm{P}-\mathrm{Q}}=\left(\frac{4}{3}\right)^{3} \ldots \ldots\) (iv)
Dividing Eq. (ii) by Eq. (iii), we get
\(\begin{array}{l}
\mathrm{r}^{\mathrm{Q}-\mathrm{R}}=\frac{3}{4} \\
\Rightarrow \mathrm{r}^{3 \mathrm{Q}-3 \mathrm{R}}=\left(\frac{3}{4}\right)^{3}
\end{array}\)
Multiplying Eq. (iv) and Eq. (v), we get
\(\begin{array}{l}
\mathrm{r}^{\mathrm{P}-\mathrm{Q}} \times \mathrm{r}^{3 \mathrm{Q}-3 \mathrm{R}}=1 \\
\Rightarrow \mathrm{r}^{\mathrm{P}-\mathrm{Q}+3 \mathrm{Q}-3 \mathrm{R}}=1 \\
\Rightarrow \mathrm{r}^{\mathrm{P}+2 \mathrm{Q}-3 \mathrm{R}}=\mathrm{r}^{0} \\
\Rightarrow \mathrm{P}+2 \mathrm{Q}-3 \mathrm{R}=0 \\
\Rightarrow \mathrm{P}+2 \mathrm{Q}=3 \mathrm{R}
\end{array}\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2012
If $a>0, b>0$ then the maximum area of the parallelogram whose three vertices are $O(0,0), A(a \cos \theta, b \sin \theta)$ and $\mathrm{B}(\mathrm{a} \cos \theta,-\mathrm{b} \sin \theta)$ is

Options:
A ab when $\theta=\pi / 4$
B 3 ab when $\theta=\pi / 4$
C ab when $\theta=\pi / 2$
D $2 \mathrm{ab}$

The correct answer is:
ab when $\theta=\pi / 4$

Solution:
Area $\mathrm{OABC}=2 \times$ area $\triangle \mathrm{OAB}$
$\begin{array}{l}
=2 \times\left(\frac{1}{2} \times \mathrm{OM} \times \mathrm{AB}\right) \\
=2 \times \frac{1}{2}|\mathrm{a} \cos \theta \times 2 \mathrm{~b} \sin \theta|=2 \times \mathrm{ab}|\sin \theta \cos \theta| \\
=\mathrm{ab}|\sin 2 \theta|
\end{array}$
$\therefore$ Maximum is ab when $\theta=\pi / 4$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2021
If $A=\left(\begin{array}{ll}1 & 1 \\ 0 & i\end{array}\right)$ and $A^{2018}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$, then $(a+d)$ equals

Options:
A $1+\mathrm{i}$
B 0
C 2
D 2018

The correct answer is:
0

Solution:
$A^2=\left(\begin{array}{cc}1 & 1+i \\ 0 & -1\end{array}\right), A^3=\left(\begin{array}{cc}1 & i \\ 0 & -i\end{array}\right), A^4=I$
$\begin{aligned}
&\therefore A^{2018}=A^{2016} \times A^2=A^2=\left(\begin{array}{ll}
1 & 1+i \\
0 & -1
\end{array}\right) \\
&\therefore a+d=1+(-1)=0
\end{aligned}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2022
If $\mathrm{A}=\left[\begin{array}{rr}1 & 2 \\ -4 & -1\end{array}\right]$ then $\mathrm{A}^{-1}$ is

Options:
A $\frac{1}{7}\left[\begin{array}{rr}-1 & -2 \\ 4 & 1\end{array}\right]$
B $\frac{1}{7}\left[\begin{array}{rr}1 & 2 \\ -4 & -1\end{array}\right]$
C $\frac{1}{7}\left[\begin{array}{rr}-1 & -2 \\ 4 & -1\end{array}\right]$
D Does not exist

The correct answer is:
$\frac{1}{7}\left[\begin{array}{rr}-1 & -2 \\ 4 & 1\end{array}\right]$

Solution:
Hints: $|\mathrm{A}|=-1+8=7$
$$
\begin{aligned}
& \operatorname{adj}(\mathrm{A})=\left[\begin{array}{rr}
+(-1) & -(2) \\
-(-4) & +(1)
\end{array}\right]=\left[\begin{array}{rr}
-1 & -2 \\
4 & 1
\end{array}\right] \\
& \mathrm{A}^{-1}=\frac{1}{7}\left[\begin{array}{rr}
-1 & -2 \\
4 & 1
\end{array}\right] \quad \operatorname{Both}(\mathrm{A} \text { and } C)
\end{aligned}
$$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2010
If $\mathrm{a}=2 \sqrt{2}, \mathrm{~b}=6, \mathrm{~A}=45^{\circ}$, then

Options:
A no triangle is possible
B one triangle is possible
C two triangle are possible
D either no triangle or two triangles are possible

The correct answer is:
no triangle is possible

Solution:
Hints: $\mathrm{a}=2 \sqrt{2} ; \mathrm{b}=6 ; \mathrm{A}=45^0$
$$
\begin{aligned}
& \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}} \Rightarrow \sin \mathrm{B}=\frac{\mathrm{b}}{\mathrm{a}} \sin \mathrm{A} \\
& \Rightarrow \sin \mathrm{B}=\frac{6}{2 \sqrt{2}} \sin 45^{\circ}=\frac{3}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{3}{2} \Rightarrow \text { No triangle is possible since } \sin \mathrm{B}>1
\end{aligned}
$$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2009
If $\left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \\ a^{2}+a b & b^{2} & c a \\ a b & b^{2}+b c & c^{2}\end{array}\right|=k a^{2} b^{2} c^{2}$, then $k=$

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

The correct answer is:
4

Solution:
Hint:
$\left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \\ a^{2}+a b & b^{2} & c a \\ a b & b^{2}+b c & c^{2}\end{array}\right|=(a b c)\left|\begin{array}{ccc}a & c & a+c \\ a+b & b & a \\ b & b+c & c\end{array}\right|$
opening through $\mathrm{R}-1=4 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{C}^{2}$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2020
If $A^2-A+I=0$, then the inverse of the matrix $A$ is

Options:
A $\mathrm{A}-\mathrm{I}$
B $\quad \mathrm{I}-\mathrm{A}$
C $\mathrm{A}+\mathrm{I}$
D $\mathrm{A}$

The correct answer is:
$\quad \mathrm{I}-\mathrm{A}$

Solution:
Hints : $\mathrm{A}^2-\mathrm{A}+\mathrm{I}=0 \Rightarrow \mathrm{A}^2=\mathrm{A}-\mathrm{I} \Rightarrow \mathrm{A}^2 \cdot \mathrm{A}^{-1}=\mathrm{A} \cdot \mathrm{A}^{-1}-\mathrm{A}^{-1} \Rightarrow \mathrm{A}=\mathrm{I}-\mathrm{A}^{-1} \Rightarrow \mathrm{A}^{-1}=\mathrm{I}-\mathrm{A}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2009
If $A=\left(\begin{array}{ccc}5 & 5 x & x \\ 0 & x & 5 x \\ 0 & 0 & 5\end{array}\right)$ and $\left|A^{2}\right|=25,$ then $|x|$ is equal to

Options:
A $\frac{1}{5}$
B 5
C $5^{2}$
D 1

The correct answer is:
$\frac{1}{5}$

Solution:
Given
$$
\begin{aligned}
A &=\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right] \\
\therefore \quad A^{2}=A \cdot A
&=\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right]\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right] \\
&\left[\begin{array}{ccc}
25 & 25 x+5 x^{2} & 5 x+25 x^{2}+5 x \\
0 & x^{2} & 5 x^{2}+25 \\
0 & 0 & 25
\end{array}\right] \\
&=25\left(25 x^{2}-0\right) \\
&=25\left(25 x^{2}\right)
\end{aligned}
$$
since, given that $|A|^{2}=25$
Now, $25\left(25 x^{2}\right)=25$
$\begin{array}{lr}\Rightarrow & 25 x^{2}=1 \\ \Rightarrow & x^{2}=\frac{1}{25} \\ \Rightarrow & x=\pm \frac{1}{5}\end{array}$
$\Rightarrow \quad |x|=\frac{1}{5}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2019