datavorous/entrance-exam-dataset huggingface dataset:
Let $p, q$ and $r$ be the sides opposite to the angles $P, Q$ and $R,$ respectively in a $\Delta P Q R .$ If $r^{2} \sin P \sin Q=p q,$ then the triangle is

Options:
A equilateral
B acute angled but not equilateral
C obtuse angled
D right angled

The correct answer is:
right angled

Solution:
We know that in $\Delta A B C$ $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$
$\therefore \quad r^{2} \sin P \sin Q=p q$
$\Rightarrow r^{2} \cdot \frac{p}{2 R_{1}} \cdot \frac{q}{2 R_{1}}=p q,$ where $R_{1}$ is circumras
of $\Delta P Q R$. $\begin{array}{ll}\Rightarrow & r^{2}=4 R_{1}^{2} \\ \Rightarrow & r=2 R_{1} \\ \Rightarrow & 2 R_{1} \sin R=2 R_{1}\end{array}$
$\Rightarrow \quad R_{1}=90^{\circ}$
$\therefore \Delta P Q R$ is right angled.

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2012
Let $p, q$ be real numbers. If $\alpha$ is the root of $x^{2}+3 p^{2} x+5 q^{2}=0, \quad \beta \quad$ is $\quad$ a $\quad$ root $\quad$ of $x^{2}+9 p^{2} x+15 q^{2}=0$ and $0 < \alpha < \beta,$ then the equation $x^{2}+6 p^{2} x+10 q^{2}=0$ has a root $\gamma$ that always satisfies

Options:
A $\gamma=\frac{\alpha}{4}+\beta$
B $\beta < \gamma$
C $\gamma=\frac{\alpha}{2}+\beta$
D $\alpha < \gamma < \beta$

The correct answer is:
$\alpha < \gamma < \beta$

Solution:
Since, $\alpha$ is a root of
$\therefore$
$$
\begin{array}{l}
x^{2}+3 p^{2} x+5 q^{2}=0 \\
\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0
\end{array}
$$
and $\beta$ is a root of
$\therefore$
$$
\begin{array}{l}
x^{2}+9 p^{2} x+15 q^{2}=0 \\
\beta^{2}+9 p^{2} \beta+15 q^{2}=0
\end{array}
$$
Let $f(x)=x^{2}+6 p^{2} x+10 q^{2}$
$\begin{aligned} \text { Then, } f(\alpha) &=\alpha^{2}+6 p^{2} \alpha+10 q^{2} \\ &=\left(\alpha^{2}+3 p^{2} \alpha+5 q^{2}\right)+3 p^{2} \alpha+5 q^{2} \end{aligned}$
$=0+3 p^{2} \alpha+5 q^{2}$
$\Rightarrow \quad f(\alpha)>0$
and $\quad f(\beta)=\beta^{2}+6 p^{2} \beta+10 q^{2}$
$=\left(\beta^{2}+9 p^{2} \beta+15 q^{2}\right)-\left(3 p^{2} \beta+5 q^{2}\right)$
$=0-\left(3 p^{2} \beta+5 q^{2}\right) \quad$ I from Eq. (ii)]
$\Rightarrow \quad f(\beta) < 0$
Thus, $f(x)$ is a polynomial such that $f(\alpha)>0$ and $f(\beta) < 0$
Therefore, there exists $\gamma$ satisfying $\alpha < \gamma < \beta$ such that $f(\gamma)=0$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2014
Let $p(x)$ be a polynomial with real co-efficients, $p(0)=1$ and $p^{\prime}(x) > 0$ for all $x \in \mathbb{R}$. Then

Options:
A $p(x)$ has at least two real roots
B $p(x)$ has only one positive real root
C $\mathrm{p}(\mathrm{x})$ may have negative real root
D $p(x)$ has infinitely many real roots

The correct answers are:
$\mathrm{p}(\mathrm{x})$ may have negative real root

Solution:
$P(x)=0$ has exactly one negative real root.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2022
Let \(P(n)=3^{2 n+1}+2^{n+2}\) where \(n \in \mathbb{N}\). Then

Options:
A \(\mathrm{P}(\mathrm{n})\) is not divisible by any prime integer.
B there exists prime integer which divides \(P(n)\).
C \(P(n)\) is divisible by 5 for all \(n \in \mathbb{N}\).
D \(\mathrm{P}(\mathrm{n})\) is divisible by 3 for all \(\mathrm{n} \in \mathbb{N}\).

The correct answer is:
there exists prime integer which divides \(P(n)\).

Solution:
Hint : \(P(1)=3^3+2^3\) which is divisible by \((3+2)=5\) which is prime
\(\therefore\) There exists prime integer which divides \(\mathrm{P}(\mathrm{n})\)

Topic: Mathematics, Mathematical Induction, WBJEE, WBJEE 2023
Let $Q=\left(\begin{array}{cc}\cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)$ and$x=\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)$ then $Q^{3} x$ is equal to

Options:
A $\left(\begin{array}{l}0 \\ 1\end{array}\right) \quad$
B $\left(\begin{array}{c}-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)$
C $\left(\begin{array}{c}-1 \\ 0\end{array}\right)$
D $\left(\begin{array}{c}-\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{array}\right)$

The correct answer is:
$\left(\begin{array}{c}-1 \\ 0\end{array}\right)$

Solution:
Given,
$Q=\left[\begin{array}{cc}\cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right]$ and $x=\left[\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right]$
$\begin{aligned} \text { Let } Q(\theta) &=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right] \\ Q^{3}(\theta) &=Q(3 \theta) \end{aligned}$
$\therefore Q^{3}\left(\frac{\pi}{4}\right)=\left[\begin{array}{ll}\cos \frac{3 \pi}{4} & -\sin \frac{3 \pi}{4} \\ \sin \frac{3 \pi}{4} & \cos \frac{3 \pi}{4}\end{array}\right]=\left[\begin{array}{cc}-1 / \sqrt{2} & -1 / \sqrt{2} \\ 1 / \sqrt{2} & -1 / \sqrt{2}\end{array}\right]$
Now, $Q^{3}\left(\frac{\pi}{4}\right) x=\left[\begin{array}{cc}\frac{-1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right]=\left[\begin{array}{c}-\frac{1}{2}-\frac{1}{2} \\ \frac{1}{2}-\frac{1}{2}\end{array}\right]$
$\therefore \quad Q^{3}\left(\frac{\pi}{4}\right) x=\left[\begin{array}{c}-1 \\ 0\end{array}\right]$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2016
Let $R$ and $S$ be two equivalence relations on a non-void set $A$. Then

Options:
A $\mathrm{R} \cup \mathrm{S}$ is equivalence relation
B $\mathrm{R} \cap \mathrm{S}$ is equivalence relation
C $\mathrm{R} \cap \mathrm{S}$ is not equivalence relation
D $\mathrm{R} \cup \mathrm{S}$ is not equivalence relation

The correct answers are:
$\mathrm{R} \cap \mathrm{S}$ is equivalence relation

Solution:
(B) $\mathrm{R} \cap \mathrm{S}$ is equivalence relation

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2022
Let $R$ be a relation defined on the set $Z$ of all integers and $x R y,$ when $x+2 y$ is divisible by 3 , then

Options:
A $R$ is not transitive
B $A$ is symmetric only
C $R$ is an equivalence relation
D $A$ is not an equivalence relation

The correct answer is:
$R$ is an equivalence relation

Solution:
Reflexivity For reflexive $(x, x) \in R .$
$x+2 x=3 x$
which is divisible by 3.
$\Rightarrow$ xRx $\in R$ Hence, xRy is reflexive.
Symmetric Let $x+2 y=3 \lambda$
$\Rightarrow \quad x=3 \lambda-2 y$
$\begin{aligned}
\text {Now, } & \begin{aligned}
y+2 x &=y+2(3 \lambda-2 y) \\
&=y+6 \lambda-4 y \\
&=6 \lambda-3 y
\end{aligned}
\end{aligned}$
$\Rightarrow \quad y+2 x=3(2 \lambda-y)$
which is divisible by 3.
$\therefore$ $(x, y) \in R \Rightarrow(y, x) \in R$
Hence, xRy is symmetric.
$\begin{array}{ll}\text {Transifive } & x+2 y=3 \lambda \\ \text { and } & y+2 z=3 \mu\end{array}$
$x+3 y+2 z=3 x+3 \mu$
$\Rightarrow \quad x+2 z=3(\lambda+\mu-y)$
which is divisible by $3 .$
$\because (x, y) \in$ R and $(y, z) \in R$
$\Rightarrow \quad(x, z) \in R$
Hence, xRy is transitive
$\because$ R is reflexive, symmetric and transitive.
$\therefore$ R is an equivalence relation.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2016
Let $R$ be the real line. Let the relations $S$ and $T$ on $R$ be defined by $S=\{(x, y): y=x+1,0 < x < 2\}, T=\{(x, y):(x-y)$ is an integer $\}$. Then

Options:
A both $S$ and $T$ are equivalence relations on $R$
B $\mathrm{T}$ is an equivalence on $\mathrm{R}$ but $\mathrm{S}$ is not
C neither $S$ nor $T$ is an equivalence relation on $R$
D $\mathrm{S}$ is an equivalence relation on $\mathrm{R}$ but $\mathrm{T}$ is not

The correct answer is:
$\mathrm{T}$ is an equivalence on $\mathrm{R}$ but $\mathrm{S}$ is not

Solution:
$\mathrm{T}$ is an equivalence but $\mathrm{S}$ is not

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2021
Let $R$ be the set of all real numbers and	$f:[-1,1] \rightarrow R$ be defined $f(x)=\left\{\begin{array}{ll}x \sin \frac{1}{x}, & x \neq 0 \text { . Then } \\ 0, & x=0\end{array}\right.$

Options:
A $f$ satisfies the conditions of Rolle's theorem on [-1,1]
B $f$ satisfies the conditions of Lagrange's mean value theorem on [-1,1]
C $f$ satisfies the conditions of Rolle's theorem on [0,1]
D $f$ satisfies the conditions of Lagrange's mean value theorem on [0,1]

The correct answer is:
$f$ satisfies the conditions of Lagrange's mean value theorem on [0,1]

Solution:
Given, $\quad f(x)=\left\{\begin{array}{cl}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.$
Continuity at $x=0$. $\mathrm{LHL}=\lim _{x \rightarrow 0^{-}} x \sin \frac{1}{x}=0$
$\mathrm{RHL}=\lim _{x \rightarrow 0^{*}} x \sin \frac{1}{x}=0$
and
$\mathrm{LHL}=\mathrm{RHL}=f(0)$
Hence, $f(x)$ is continuous for all values of $x$. Differentiability at $x=0$
$$
\begin{aligned}
\mathrm{LHD} &=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}}{h} \\
&=-\lim _{h \rightarrow 0} \sin \frac{1}{h} \\
&=-\sin \left(\frac{1}{0}\right)=\text { some definite value }
\end{aligned}
$$
Thus, $f(x)$ is not differentiable at $x=0 .$ Now, $f(0)=0$ and $f(1)=1 \sin 1$
$\therefore f(0) \neq f(1)$. Rolle's theorem is not satisfy.
Now.
$$
\begin{aligned}
f^{\prime}(c) &=\frac{f(1)-f(0)}{b-0} \\
&=\frac{1 \sin (1)-0}{1-0}=\sin 1=\sin 1
\end{aligned}
$$
which lies between 0 to 1 .
Hence, there exist a real number $c \in(0,1)$ Hence,
Lagrange's mean value theorem is satisfy on $(0,1)$.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2014
Let $R$ be the set of real numbers and the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=x^{2}+2 x-3$ and $g(x)=x+1 .$ Then, the value of $x$ for which $f(g(x))=g(f(x))$ is

Options:
A -1
B 0
C 1
D 2

The correct answer is:
-1

Solution:
According to question,
$$
f(g(x))=g(f(x))
$$
$\Rightarrow \quad f(x+1)=g\left(x^{2}+2 x-3\right)$
$\Rightarrow \quad(x+1)^{2}+2(x+1)-3=x^{2}+2 x-3+1$
$\Rightarrow \quad x^{2}+1+2 x+2 x+2-3=x^{2}+2 x-2$
$\Rightarrow \quad x^{2}+4 x=x^{2}+2 x-2$
$\Rightarrow \quad x^{2}+4 x-x^{2}-2 x+2=0$
$\Rightarrow \quad 2 x+2=0$
$\Rightarrow \quad 2 x=-2$
$$
\Rightarrow \quad x=-1
$$

Topic: Mathematics, Functions, WBJEE, WBJEE 2012
Let $R$ be the set of real numbers and the mapping $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=5-x^2$ and $g(x)=3 x-4$, then the value of $(f \circ g)(-1)$ is

Options:
A -44
B -54
C -32
D -64

The correct answer is:
-44

Solution:
Hints: $\mathrm{f}(\mathrm{g}(-1))=\mathrm{f}(-3-4)=\mathrm{f}(-7)=5-49=-44$

Topic: Mathematics, Functions, WBJEE, WBJEE 2010
Let $\mathrm{r}, \mathrm{v}, \mathrm{E}$ be the radius of orbit, speed of electron and total energy of electron respectively in a $\mathrm{H}$-atom. Which of the following quantities according to Bohr theory, is proportional to the quantum number $\mathrm{n}$ ?

Options:
A vr
B $\mathrm{rE}$
C $\frac{r}{E}$
D $\frac{r}{v}$

The correct answer is:
vr

Solution:
$\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \quad \therefore \mathrm{n}=\frac{\mathrm{m} 2 \pi}{\mathrm{h}} \cdot \mathrm{vr} \quad \therefore \mathrm{n} \propto \mathrm{vr}$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2021
Let $S=\frac{2}{1}{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}$ $+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n} .$ Then, $S$ equals

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

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

Solution:
We know that $(1+x)^{n}={ }^{n} C_{0}+x{ }^{-} C_{1}+x^{2}{ }^{n} C_{2}+\ldots+x^{n}{ }^{n} C_{n}$
On integrating both sides from 0 to 2 , we get
$\left[\frac{(1+x)^{n+1}}{n+1}\right]_{0}^{2}$
$=\left[x^{n} C_{0}+\frac{x^{2}}{2}^{n} C_{1}+\frac{x^{3}}{3}^{n} C_{2}+\ldots+\frac{x^{n+1}}{n+1}^{n} C_{n}\right]_{0}^{2}$
$\Rightarrow \frac{(3)^{n+1}}{n+1}-\frac{1}{n+1}=2{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}$
$\quad+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}-0$
$\Rightarrow \frac{2}{1}{ }^{n} C_{0}+\frac{2^{2}}{2}{ }^{n} C_{1}+\frac{2^{3}}{3}{ }^{n} C_{2}+\ldots+\frac{2^{n+1}}{n+1}{ }^{n} C_{n}$
$\quad=\frac{3^{n+1}-1}{n+1}$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2014
Let $S=\{(a, b, c) \in N \times N \times N: a+b+c=21$, $a \leq b \leq c\}$ and $T=\{(a, b, c) \in N \times N \times N: a, b, c$ are in $A P \}$ where $N$ is the set of all natural numbers. Then, the number of elements in the set $\mathrm{S} \cap T$

Options:
A 6
B 7
C 13
D 14

The correct answer is:
7

Solution:
We have, $a+b+c=21$ and $\frac{a+c}{2}=b$ $\Rightarrow \quad a+c+\frac{a+c}{2}=21$
$\begin{array}{ll}\Rightarrow & a+c=14 \Rightarrow \frac{a+c}{2}=7 \\ \Rightarrow & b=7\end{array}$
So, a can take values from 1 to 6 . C can have values from 8 to 13
or a=b=c=7
So, there are 7 such triplets.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2015
Let \(S\) be the sample space of the random experiment of throwing simultaneously two unbiased dice and \(E_k=\{a, b) \in\) \(S: a b=k\}\). If \(p_k=P\left(E_k\right)\), then the correct among the following is :

Options:
A \(p_1 < p_{10} < p_4\)
B \(\mathrm{p}_2 < \mathrm{p}_8 < \mathrm{p}_{14}\)
C \(\mathrm{p}_4 < \mathrm{p}_8 < \mathrm{p}_{17}\)
D \(\mathrm{p}_2 < \mathrm{p}_{16} < \mathrm{p}_5\)

The correct answer is:
\(p_1 < p_{10} < p_4\)

Solution:
Hint: \(E_1=\{(1,1)\}, \quad E_4=\{(2,2),(1,4),(4,1)\} \quad E_{10}=\{(2,5),(5,2)\}\)
\(p_1=\frac{1}{36}\)
\(p_4=\frac{3}{36}=\frac{1}{12}\),
\(p_{10}=\frac{2}{36}=\frac{1}{18}\)

Topic: Mathematics, Probability, WBJEE, WBJEE 2023
Let $S$ be the set of points, whose abscissae and ordinates are natural numbers. Let $P \in S$, such that the sum of the distance of $P$ from (8,0) and (0,12) is minimum among all elements in $S .$ Then, the number of such points $P$ in $S$ is

Options:
A 1
B 3
C 5
D 11

The correct answer is:
3

Solution:
Sum of distances of point $P$ from point (8,0) and (0,12) will be minimum, if points are collinear.
Equation of line at point (8,0) and (0,12)
$$
\begin{aligned}
\frac{x}{8}+\frac{y}{12} &=1 \\
\Rightarrow \quad \frac{y}{12} &=1-\frac{x}{8} \\
\Rightarrow \quad y &=12-\frac{x}{8} \times 12 \\
\Rightarrow \quad y &=12-\frac{3}{2} x
\end{aligned}
$$
$\therefore$ Points $(x, y)=(2,9),(4,6)$ and (6,3) . Hence, the number of such points $P$ in S is 3.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2016
Let $S$ denote the sum of the infinite series $1+\frac{8}{2 !}+\frac{21}{3 !}+\frac{40}{4 !}+\frac{65}{5 !}+\ldots .$ Then

Options:
A $S < 8$
B $S>12$
C $8 < S < 12$
D $S=8$

The correct answer is:
$8 < S < 12$

Solution:
Let $S=1+\frac{8}{2 !}+\frac{21}{3 !}+\frac{40}{4 !}+\frac{65}{5 !}+\ldots$
Again, let $S_{1}=1+8+21+40+65+\ldots+T_{n}$
and $\quad S_{1}=+1+8+21+40+\ldots+T_{n}$
$$
\begin{aligned} 0=1+7+13+19+25+\ldots-T_{n} \\ T_{n} &=1+7+13+19+25+\ldots+n \text { terms } \\ &=\frac{n}{2}[2(1)+(n-1) 6] \\ &=n[1+3(n-1)]=n(3 n-2) \end{aligned}
$$
$S=\sum \frac{3 n-2}{(n-1) !}$
$=\sum \frac{3 n-3+1}{(n-1) !}$
$=\sum \frac{3}{(n-2) !}+\sum \frac{1}{(n-1) !}$
$\quad=3 e+e \quad\left[\because e=1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right]$
$=4 e$
We know $\quad 2 < e < 3$
$\therefore \quad 8 < 4 e < 12$
$\Rightarrow \quad 8 < S < 12$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2014
Let $S_{k}$ be the sum of an infinite GP series whose first term is $k$ and common ratio is $\frac{k}{k+1}(k>0)$ Then, the value of $\sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_{k}}$ is equal to

Options:
A $\log _{e} 4$
B $\log _{e} 2-1$
C $1-\log _{e} 2$
D $1-\log _{e} 4$

The correct answer is:
$1-\log _{e} 4$

Solution:
\(S_{1}=1+\frac{1}{2}+\frac{1}{2^{2}}+\infty=\frac{1}{1-\frac{1}{2}}=2\)
\(S_{2}=2+2 \cdot \frac{2}{3}+2\left(\frac{2}{3}\right)^{2}+\ldots \infty=\frac{2}{1-\frac{2}{3}}=6\)
\(S_{3}=3+3\left(\frac{3}{4}\right)+3\left(\frac{3}{4}\right)^{2}+\infty=\frac{3}{1-\frac{3}{4}}=12\)
\(S_{4}=4+4\left(\frac{4}{5}\right)+4\left(\frac{4}{5}\right)^{2}+\infty=\frac{4}{1-\frac{4}{5}}=20\)
\(\therefore \sum_{k=1}^{\infty} \frac{(-1)^{k}}{S_{k}}=-\frac{1}{S_{1}}+\frac{1}{S_{2}}-\frac{1}{S_{3}}+\frac{1}{S_{4}}-\infty\)
\(=-\frac{1}{2}+\frac{1}{6}-\frac{1}{12}+\frac{1}{20}-\infty\)
\(=-\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}-\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}-\infty\)
\(=-\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)-\ldots \infty\)
\(=-1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\infty\right)\)
\(=-1-2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots \infty\right)+2 \infty\)
\(=-1-2 \log _{e} 2+2=1-\log _{e} 4\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2012
Let $S_{n}=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots \ldots .$ to $n^{\text {th }}$ term. Then $\lim _{n \rightarrow \infty} S_{n}$ is

Options:
A $\frac{\pi}{3}$
B $\frac{\pi}{4}$
C $\frac{\pi}{6}$
D $\frac{\pi}{8}$

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

Solution:
$\begin{aligned} & t_{n}=\cot ^{-1} 2 n^{2} \\=& \tan ^{-1} \frac{1}{2 n^{2}}=\tan ^{-1} \frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)} \\=& \tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1) \\ \therefore & S_{n}=\tan ^{-1}(2 n+1)-\tan ^{-1} 1 \\ \therefore & \operatorname{Lim}_{n \rightarrow \infty} S_{n}=\pi / 2-\pi / 4=\pi / 4 \end{aligned}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2021
Let $\mathrm{S}, \mathrm{T}, \mathrm{U}$ be three non-void sets and $\mathrm{f}: \mathrm{S} \rightarrow \mathrm{T}, \mathrm{g}: \mathrm{T} \rightarrow \mathrm{U}$ and composed mapping $\mathrm{gof}: \mathrm{S} \rightarrow U$ be defined. Let $\mathrm{gof}$ be injective mapping. Then

Options:
A $f, g$ both are injective.
B neither $f$ nor $g$ is injective.
C $\mathrm{f}$ is obviously injective.
D $\mathrm{g}$ is obviously injective.

The correct answer is:
$\mathrm{f}$ is obviously injective.

Solution:
Let, $x_1, x_2 \in S$ and $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow g \circ f\left(x_1\right)=g\left(f\left(x_1\right)\right)=g\left(f\left(x_2\right)\right)=g$ of $\left(x_2\right)$
$\Rightarrow \mathrm{x}_1=\mathrm{x}_2$ (as gof is injective)
$\Rightarrow \mathrm{f}(\mathrm{x})$ is obviously injective.

Topic: Mathematics, Functions, WBJEE, WBJEE 2022
Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ be so that $g o f: s \rightarrow U$ is surjective. Then,

Options:
A $g$ and $f$ are both surjective
B $g$ is surjective, $f$ may not be so
C $f$ is surjective, $g$ may not be so
D $f$ and $g$ both may not be surjective

The correct answer is:
$g$ is surjective, $f$ may not be so

Solution:
We have,
gof : $S \rightarrow U$ is an onto function.
Let $Z$ be any arbitrary element such that $Z \in U$. Now, $g o f: S \rightarrow U$ is onto $\Rightarrow g_{0} f(x)=Z,$ for $x \in S$
$\Rightarrow g(f(x))=Z$
$\Rightarrow g(y)=Z,$ where $y=f(x) \in T$ for all $Z \in U$
$\therefore$ For all $Z \in U,$ there exists $y=f(x) \in T$ such that $g(y)=z$
$\therefore g: T \rightarrow U$ is an onto function.

Topic: Mathematics, Functions, WBJEE, WBJEE 2019
Let $\Delta=\left|\begin{array}{ccc}\sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0\end{array}\right|$. Then

Options:
A $\Delta$ is independent of $\theta$
B $\Delta$ is independent of $\varphi$
C $\Delta$ is a constant
D $\left(\frac{d \Delta}{d \theta}\right)_{\theta-\frac{\pi}{2}}=0$

The correct answers are:
$\Delta$ is independent of $\varphi$, $\left(\frac{d \Delta}{d \theta}\right)_{\theta-\frac{\pi}{2}}=0$

Solution:
$\Delta=\left|\begin{array}{ccc}\sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0\end{array}\right|$
$\begin{aligned}
\mathrm{C}_1 \longrightarrow \mathrm{C}_1-\left(\frac{\cos \phi}{\sin \phi}\right) \mathrm{C}_2 \Rightarrow \Delta=\left|\begin{array}{ccc}
0 & \sin \theta \sin \phi & \cos \theta \\
0 & \cos \theta \sin \phi & -\sin \theta \\
-\frac{\sin \theta}{\sin \phi} & \sin \theta \cos \phi & \theta
\end{array}\right| \\
&=\frac{\sin \theta}{\sin \phi} \cdot \sin \phi\left(\cos ^2 \theta+\sin ^2 \theta\right)= \sin \theta \rightarrow \text { Independent of } \phi \\
\frac{\mathrm{d} \Delta}{\mathrm{d} \theta}=\left.\cos \theta \Rightarrow \frac{\mathrm{d} \Delta}{\mathrm{d} \theta}\right|_{\theta-\frac{\pi}{2}}=0
\end{aligned}$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2022
Let $t_{n}$ denotes the nth term of the infinite series $\frac{1}{1 !}+\frac{10}{2 !}+\frac{21}{3 !}+\frac{34}{4 !}+\frac{49}{5 !}+\ldots .$ Then $\lim _{n \rightarrow \infty} t_{n}$ is

Options:
A $e$
B $0$
C $e^{2}$
D 1

The correct answer is:
$0$

Solution:
Let $\quad S=1+10+21+34+49+\ldots+t_{n}^{\prime}$
and $S=1+10+21+34+\ldots+t_{n}^{\prime}$
$0=1+9+11+13+15+\ldots-t_{n}^{\prime}$
$$
\begin{aligned}
\Rightarrow t_{n}^{\prime} &=1+[9+11+13+15+\ldots(n-1) \text { term }] \\
&=1+\left[\frac{n-1}{2}\{2 \times 9+(n-2) 2\}\right] \\
&=1+(n-1)[9+n-2] \\
&=1+(n-1)(n+7) \\
\therefore \quad t_{n} &=\frac{t^{\prime}}{n !}=\frac{1+(n-1)(n+7)}{n !} \\
&=\frac{1+(n-1)(n+7)}{n !} \\
&=\frac{1+n^{2}+6 n-7}{n !} \\
&=\frac{n}{(n-1) !}+\frac{6}{(n-1) !}-\frac{n^{2}+6 n-6}{n !} \\
&=\frac{n-1+1}{(n-1) !}+\frac{6}{(n-1) !}-\frac{1}{n !} \\
&=\frac{1}{(n-2) !}+\frac{7}{(n-1) !}-\frac{1}{n !} \\
&=\frac{1}{n \rightarrow \infty} t_{n}=\lim _{n \rightarrow \infty}\left[\frac{1}{(n-2) !}+\frac{7}{(n-1) !}-\frac{1}{n !}\right]=0
\end{aligned}
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2014
Let $T \& U$ be the set of all orthogonal matrices of order 3 over $R \&$ the set of all non-singular matrices of order 3 over $\mathrm{R}$ respectively
Let $A=\{-1,0,1\}$, then

Options:
A there exists bijective mapping between $\mathrm{A}$ and $\mathrm{T}, \mathrm{U}$.
B there does not exist bijective mapping between $\mathrm{A}$ and $\mathrm{T}, \mathrm{U}$.
C there exists bijective mapping between $\mathrm{A}$ and $\mathrm{T}$ but not between $\mathrm{A} \& \mathrm{U}$.
D there exists bijective mapping between $\mathrm{A}$ and $\mathrm{U}$ but not between $\mathrm{A} \& \mathrm{~T}$.

The correct answer is:
there does not exist bijective mapping between $\mathrm{A}$ and $\mathrm{T}, \mathrm{U}$.

Solution:
As $\mathrm{n}(\mathrm{A}) \neq \mathrm{n}(\mathrm{T})$ and $\mathrm{n}(\mathrm{A}) \neq \mathrm{n}(\mathrm{u})$

Topic: Mathematics, Functions, WBJEE, WBJEE 2021
Let \(\tan \alpha=\frac{a}{a+1}\) and \(\tan \beta=\frac{1}{2 a+1}\) then \(\alpha+\beta\) is

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

The correct answer is:
\(\frac{\pi}{4}\)

Solution:
\(\begin{aligned}
& \text { Hints: } \tan \alpha=\frac{a}{a+1}, \tan \beta=\frac{1}{2 a+1} \\
& \tan (\alpha+\beta)=\frac{\frac{a}{a+1}+\frac{1}{2 a+1}}{1-\frac{a}{(a+1)(2 a+1)}}=\frac{\frac{a(2 a+1)+a+1}{(a+1)(2 a+1)}}{\frac{(a+1)(2 a+1)-a}{(a+1)(2 a+1)}}=\frac{2 a^2+2 a+1}{2 a^2+2 a+1}=1 \\
& \alpha+\beta=\frac{\pi}{4}
\end{aligned}\)

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2011
Let the coefficients of powers of $x$ in the $2 n d$, 3rd and 4 th terms in the expansion of $(1+x)^{n}$ where $n$ is a positive integer, be in arithmetic progression. Then, the sum of the coefficients of odd powers of $x$ in the expansion is

Options:
A 32
B 64
C 128
D 256

The correct answer is:
64

Solution:
According to question, ${ }^{n} C_{1},{ }^{n} C_{2}$ and ${ }^{n} C_{3}$ are in $\mathrm{AP}$
$\Rightarrow \quad \frac{2 n(n-1)}{2 !}=n+\frac{n(n-1)(n-2)}{3 !}$
$\Rightarrow \quad n^{2}-9 n+14=0$
$\Rightarrow \quad(n-7)(n-2)=0$
$\Rightarrow \quad n=7$ since $\underline{n \neq 2}$
¿The sum of the coefficients of odd powers of $x$ in the expansion of $(1+x)^{n}$ is $\frac{2^{n}}{2}=\frac{2^{\gamma}}{2}=2^{6}=64$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2012
Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be reciprocal to that of the ellipse $x^{2}+9 y^{2}=9,$ then the ratio $a^{2}: b^{2}$ equals

Options:
A 8: 1
B 1: 8
C 9: 1
D 1: 9

The correct answer is:
8: 1

Solution:
Given equation of ellipses is $x^{2}+9 y^{2}=9$
$\Rightarrow \quad \frac{x^{2}}{9}+\frac{y^{2}}{1}=1$
Here, $a=3, b=1$
$c=\sqrt{(3)^{2}-(1)^{2}}=\sqrt{8}$
$\therefore$ Eccentricity of ellipse, $e=\frac{c}{a}$ $e=\frac{\sqrt{8}}{3}$
$\therefore$ Eccentricity of hyperbola $=\frac{3}{\sqrt{8}}$
$\Rightarrow$
$1+\frac{b^{2}}{a^{2}}=\frac{9}{8}$
$\Rightarrow$
$\frac{b^{2}}{a^{2}}=\frac{1}{8}$
$a^{2}: b^{2}=8: 1$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2018