datavorous/entrance-exam-dataset huggingface dataset:
Let $A$ and $B$ be two non-singular skew symmetric matrices such that $A B=B A$, then $A^{2} B^{2}\left(A^{\top} B\right)^{-1}\left(A B^{-1}\right)^{\top}$ is equal to

Options:
A $\mathrm{A}^{2}$
B $-\mathrm{B}^{2}$
C $-\mathrm{A}^{2}$
D $\mathrm{AB}$

The correct answer is:
$-\mathrm{A}^{2}$

Solution:
$\mathrm{AB}=\mathrm{BA}, \mathrm{A}^{\top}=-\mathrm{A}, \mathrm{B}^{\top}=-\mathrm{B}$
$\because$ matrices are non singular
$\therefore$ order is even
$\mathrm{A}^{2} \mathrm{~B}^{2}\left(\mathrm{~A}^{\top} \mathrm{B}\right)^{-1}\left(\mathrm{AB}^{-1}\right)^{\top}$
$\therefore$ order is even
$A^{2} B^{2}\left(A^{\top} B\right)^{-1}\left(A B^{-1}\right)^{\top}$
$=A^{2} B^{2}(-A B)^{-1}\left(A B^{-1}\right)^{\top}$
$=-A^{2} B^{2} B^{-1} A^{-1}\left(B^{\top}\right)^{-1} A^{\top}$
$=-A^{2} B A^{-1} B^{-1} A$
$=-A^{2} B(B A)^{-1} A$
$=-A^{2} B(A B)^{-1} A=-A^{2} B B^{-1} A^{-1} A=-A^{2}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2021
Let $A$ and $B$ be two square matrices of order 3 and $A B=O_{3}$, where $O_{3}$ denotes the null matrix of order 3. Then,

Options:
A must be $A=O_{3}, B=O_{3}$
B if $A \neq O_{3}$, must be $B \neq O$,
C if $A=O_{3}$. must be $B \neq O_{3}$
D may be $A \neq O_{3} . B \neq O_{3}$

The correct answer is:
may be $A \neq O_{3} . B \neq O_{3}$

Solution:
Since, product of two non-null matrix can be a null matrix.
Therefore, may be
$$
A \neq O_{3}, B \neq O_{3}
$$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2019
Let $a>b>0$ and $f(n)=a^{1 / n}-b^{1 / n}$,$J(n)=(a-b)^{1 / n}$ for all $n \geq 2,$ Then

Options:
A $f(n) < J(n)$
B $f(n)>J(n)$
C $f(n)=J(n)$
D $f(n)+J(n)=0$

The correct answer is:
$f(n) < J(n)$

Solution:
Let $a=4, b=1$ and $n=2$
$\begin{array}{l}
\therefore \quad I(n)=a^{1 / n}-b^{1 / n} \\
=4^{1 / 2}-1^{1 / 2}=2-1=1 \\
\text { and } J(n)=(a-b)^{1 / n} \\
=(4-1)^{1 / 2}=\sqrt{3} \\
\therefore \quad J(n)>I(n)
\end{array}$
Similarly, we can prove for other values of $a, b, n$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2019
Let $A, B$ be two distinct points on the parabola $y^{2}=4 x$. If the axis of the parabola touches a circle of radius $r$ having $A B$ as diameter, the slope of the line $A B$ is

Options:
A $-\frac{1}{r}$
B $\frac{1}{r}$
C $\frac{2}{r}$
D None of the above

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

Solution:
Centre of circle $=\left(\frac{t_{1}^{2}+t_{2}^{2}}{2},\left(t_{1}+t_{2}\right)\right)$
since, circle touch the $x$ -axis, so equation of tangent is $y=0$
$\because$ Radius = Perpendicular distance from centre to the tangent
$\Rightarrow$ Radius $=\left|t_{1}+t_{2}\right|=r$
Slope of $A B=\frac{2}{t_{1}+t_{2}}=\frac{2}{\pm r}$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2018
Let $a, b, c$ and $d$ be any four real numbers. Then. $a^{n}+b^{n}=c^{n}+d^{n}$ holds for any natural number $n$. if

Options:
A $a+b=c+d$
B $a-b=c-d$
C $a+b=c+d \cdot d^{2}+b^{2}=c^{2}+d^{2}$
D $a-b=c-d \cdot d^{2}-b^{2}=c^{2}-d^{2}$

The correct answer is:
$a-b=c-d \cdot d^{2}-b^{2}=c^{2}-d^{2}$

Solution:
From option (d), we have
$\begin{aligned}
a-b &=c-d \\
a^{2}-b^{2} &=c^{2}-d^{2}
\end{aligned}$
Consider, $a^{2}-b^{2}=c^{2}-d^{2}$
$\Rightarrow(a+b)(a-b)=(c-d)(c+d)$
$\Rightarrow \quad a+b=c+d$
On adding Eqs we get
$2 a=2 c \Rightarrow a=c$
$\Rightarrow \quad b=d$
Thus, $a^{n}+b^{n}=c^{n}+d^{n}$ for all $n \in N$

Topic: Mathematics, Mathematical Induction, WBJEE, WBJEE 2015
Let \(A, B, C\) are subsets of set \(X\). Then consider the validity of the following set theoretic statement :

Options:
A \(A \cup(B \backslash C)=(A \cup B) \backslash(A \cup C)\)
B \((A \backslash B) \backslash C=A \backslash(B \cup C)\)
C \((A \cup B) \backslash A=A \backslash B\)
D \(A \backslash C=B \backslash C\)

The correct answer is:
\((A \backslash B) \backslash C=A \backslash(B \cup C)\)

Solution:
Hint:
\(\begin{aligned}
:(A \backslash B) \backslash C & =\left(A \cap B^{\prime}\right) \cap C^{\prime} \\
& =A \cap\left(B^{\prime} \cap C^{\prime}\right) \\
& =A \cap(B \cup C)^{\prime} \\
& =A \backslash(B \cup C)
\end{aligned}\)

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2023
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be real numbers, each greater than 1 , such that $\frac{2}{3} \log _{\mathrm{b}} \mathrm{a}+\frac{3}{5} \log _{\mathrm{c}} \mathrm{b}+\frac{5}{2} \log _{\mathrm{a}} \mathrm{c}=3$. If the value of $b$ is 9 , then the value of ' $a$ ' must be

Options:
A $\sqrt[3]{81}$
B $\frac{27}{2}$
C 18
D 27

The correct answer is:
27

Solution:
$\frac{2 \ln a}{3 \ln b}+\frac{3 \ln b}{5 \ln c}+\frac{5 \ln c}{2 \ln a}=3$
By A.M $=$ G.M
$\frac{2 \ln a}{3 \ln b}=1$
$\Rightarrow a^{2}=b^{3} \Rightarrow a=\left(3^{6}\right)^{1 / 2}=3^{3} \Rightarrow a=27$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2021
Let $a, b, c$ be real numbers such that $a+b+c < 0$ and the quadratic equation $a x^{2}+b x+c=0$ has imaginary roots. Then.

Options:
A a $>0, c>0$
B $a>0 . c < 0$
C $a < 0, c>0$
D $a < 0, c < 0$

The correct answer is:
$a < 0, c < 0$

Solution:
Let $f(x)=a x^{2}+b x+c$
$\Rightarrow \quad f(1)=a+b+c < 0$
Again, $f(x)$ has imaginary zeros. So, $a < 0$. Also, $f(0)=c .$ since $f(x)$ is downward parabola. So, $c < 0$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2019
Let $a, b, c$ be such that $b(a+c) \times 0$
If $\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$
$+\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0$
then the value of $n$ is

Options:
A any integer
B zero
C any even integer
D any odd integer

The correct answer is:
any odd integer

Solution:
We have,
$\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$ + $\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$ $+\left|\begin{array}{ccc}a+1 & a-1 & (-1)^{n+2} a \\ b+1 & b-1 & (-1)^{n+1} b \\ c-1 & c+1 & (-1)^{n} c\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$$+\left|\begin{array}{ccc}(-1)^{n+2} a & a+1 & a-1 \\ (-1)^{n+1} b & b+1 & b-1 \\ (-1)^{n} c & c-1 & c+1\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}a\left(1+(-1)^{n+2}\right) & a+1 & a-1 \\ b\left(-1+(-1)^{n+1}\right) & b+1 & b-1 \\ c\left(1+(-1)^{n}\right) & c-1 & c+1\end{array}\right|=0$
$\therefore n$ is any odd integer.

Topic: Mathematics, Determinants, WBJEE, WBJEE 2017
Let $A, B, C$ be three non-void subsets of set $S$. Let $(A \cap C) \cup\left(B \cap C^{\prime}\right)=\phi$ where $C^{\prime}$ denote the complement of set $C$ in S. Then

Options:
A $A \cap B=\phi$
B $A \cap B \neq \phi$
C $\mathrm{A} \cap \mathrm{C}=\mathrm{A}$
D $\mathrm{A} \cup \mathrm{C}=\mathrm{A}$

The correct answer is:
$A \cap B=\phi$

Solution:
$(A \cap C) \cup\left(B \cap C^{\prime}\right)=\phi$
$\Rightarrow A \cap C=\phi$ and $B \cap C^{\prime}=\phi$ ...(i)
$\Rightarrow \mathrm{B} \subseteq \mathrm{C}$ ...(ii)
from (i) and (ii)
$A \cap B=\phi$

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2021
Let \(a, b, c, p, q\) and \(r\) be positive real numbers such that \(a, b\) and \(c\) are in GP and \(a^{p}=b^{q}=c^{r}\). Then,

Options:
A \(p, q, r\) are in G.P.
B \(p, q, r\) are in A.P.
C \(p, q, r\) are in H.P.
D \(p^{2}, q^{2}, r^{2}\) are in A.P.

The correct answer is:
\(p, q, r\) are in H.P.

Solution:
Let $a^{p}=b^{4}=c^{r}=k$ $\therefore \quad a=k^{1 / p}, b=k^{1 / q}, c=k^{1 / r}$
since, $a, b, c$ are in $\mathrm{GP} .$
$\therefore$
$\frac{b}{a}=\frac{c}{b}$
$\frac{k^{1 / q}}{k^{1 / P}}=\frac{k^{1 / r}}{k^{1 / q}}$
$k^{1 / q-1 / p}=k^{1 / r-1 / 4}$
$\frac{1}{q}-\frac{1}{p}=\frac{1}{r}-\frac{1}{q}$
$\therefore \quad \frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in GP.
$\Rightarrow \quad p, q, r$ in $\mathrm{HP}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2012
Let $A$ be a $3 \times 3$ matrix and $B$ be its adjoint matrix. If $|B|=64,$ then $|A|$ is equal to

Options:
A ±2
B $\pm 4$
C $\pm 8$
D ±12

The correct answer is:
$\pm 8$

Solution:
We have,
$|B|=64$
$\Rightarrow |\operatorname{adj} A \mid=64$
$\Rightarrow|A|^{2}=64$
$|A|=\pm 8$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2016
Let \(A\) be a set containing \(n\) elements. \(A\) subset \(P\) of \(A\) is chosen, and the set \(A\) is reconstructed by replacing the elements of \(P\). A subset \(Q\) of \(A\) is chosen again. The number of ways of choosing \(P\) and \(Q\) such the \(Q\) contains just one element more than \(P\) is

Options:
A \({ }^{2 n} C_{n-1}\)
B \({ }^{2 n} C_n\)
C \({ }^{2 n} C_{n+2}\)
D \(2^{2 n+1}\)

The correct answer is:
\({ }^{2 n} C_{n-1}\)

Solution:
Hint : Required number of ways
\(\begin{aligned}
& ={ }^{\mathrm{n}} \mathrm{C}_0 \cdot{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_1 \cdot{ }^{\mathrm{n}} \mathrm{C}_2+\ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-1} \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \\
& ={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}-1}
\end{aligned}\)

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2023
Let $A$ be a square matrix of order 3 whose all entries are 1 and let $I_{3}$ be the identity matrix of order $3 .$ Then, the matrix $A-3 I_{3}$ is

Options:
A invertible
B orthogonal
C non-invertible
D real Skew Symmetric matrix

The correct answer is:
non-invertible

Solution:
Given, that $A-3 J_{3}$
$$
=\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]-3\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
$=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]-\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$
$=\left[\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right]$
$=[-2(4-1)-1(-2-1)+1(1+2)]$
$=[-2(3)-1(-3)+1(3)]$
$=[-6+3+3]=0$
$\mathrm{Now}, \mathrm{det}=0$
So, matrix $A-3 I_{3}$ is non-invertible matrix.

Topic: Mathematics, Matrices, WBJEE, WBJEE 2019
Let $A$ be the centre of the circle $x^{2}+y^{2}-2 x-4 y-20=0 .$ Let $B(1,7)$ and
$D(4,-2)$ be two points on the circle such that tangents at $B$ and $D$ meet at $C$. The area of the quadrilateral $A B C D$ is

Options:
A $150 \mathrm{sq}$ units
B 50 sq units.
C $75 \mathrm{sq}$ units
D $70 \mathrm{sq}$ units

The correct answer is:
$75 \mathrm{sq}$ units

Solution:
Given, equation of circle is
$$
x^{2}+y^{2}-2 x-4 y-20=0
$$
Center $\left(1,2\right).$ and radius $=\sqrt{(1)^{2}+\left(2)^{2}+20\right.}=5$
Coordinate of intersecting point of tangents at $B$ and $D$ is $C(06,7)$
Area of quadrilateral $A B C D$
$$
=2 \times \operatorname{ar}(\Delta A B C)
$$
$=2 \times \frac{1}{2} \times 15 \times 5=75 \mathrm{sq}$ units

Topic: Mathematics, Circle, WBJEE, WBJEE 2018
Let $A$ be the point $(0,4)$ and $B$ be a moving point on $x$-axis. Let $M$ be the midpoint of $A B$ and let the perpendicular bisector of $A B$ meets the $y$-axis at $R$. The locus of the midpoint $P$ of $M R$ is

Options:
A $y+x^{2}=2$
B $x^{2}+(y-2)^{2}=\frac{1}{4}$
C $(y-2)^{2}-x^{2}=\frac{1}{4}$
D $x^{2}+y^{2}=16$

The correct answer is:
$y+x^{2}=2$

Solution:
Let B $=(2 \alpha, 0)$
$\therefore \mathrm{M}=(\alpha, 2)$
$\mathrm{m}_{\mathrm{AB}}=\frac{4}{-2 \alpha}=-\frac{2}{\alpha}$
$\therefore \mathrm{m}_{\mathrm{MR}}=\frac{\alpha}{2}, \therefore \mathrm{MR} \quad \mathrm{y}-2=\frac{\alpha}{2}(\mathrm{x}-\alpha)$
Put, $\mathrm{x}=0 \Rightarrow \mathrm{y}=2-\frac{\alpha^{2}}{2}$
$\therefore \mathrm{R} \equiv\left(0,2-\frac{\alpha^{2}}{2}\right)$
$\therefore \mathrm{P} \equiv\left(\frac{\alpha}{2}, \frac{2-\alpha^{2} / 2+2}{2}\right) \equiv \left(\frac{\alpha}{2}, 2-\frac{\alpha^{2}}{4}\right)$
$\therefore \mathrm{y}=2-\mathrm{x}^{2} \Rightarrow \mathrm{y}+\mathrm{x}^{2}=2$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2021
Let \(A\) be the point \((0,4)\) in the xy-plane and let \(B\) be the point \((2 t, 0)\). Let \(L\) be the midpoint of \(A B\) and let the perpendicular bisector of \(A B\) meet the \(y\)-axis \(M\). Let \(N\) be the midpoint of LM. Then locus of \(N\) is

Options:
A a circle
B a parabola
C a straight line
D a hyperbola

The correct answer is:
a parabola

Solution:
Hint :
Equation of LM
\(\begin{aligned}
& y-2=\frac{t}{2}(x-t) \\
& \Rightarrow M\left(0, \frac{4-t^2}{2}\right)
\end{aligned}\)
Midpoint of LM : N(h, k)
\(\begin{aligned}
& \Rightarrow \quad 2 h=t, 2 k=4-\frac{t^2}{2} \\
& \Rightarrow \quad x^2=2-y
\end{aligned}\)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2023
Let $a=\min \left\{x^{2}+2 x+3: x \in R\right\}$ and $b=\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^{2}} \cdot$ Then $\sum_{r=0}^{n} a^{r} b^{n-r}$ is

Options:
A $\frac{2^{n+1}-1}{3 \cdot 2^{n}}$
B $\frac{2^{n+1}+1}{3 \cdot 2^{n}}$
C $\frac{4^{n+1}-1}{3 \cdot 2^{n}}$
D $\frac{1}{2}\left(2^{n}-1\right)$

The correct answer is:
$\frac{4^{n+1}-1}{3 \cdot 2^{n}}$

Solution:
$$
\begin{aligned}
&\text { (c) Let } f(x)=x^{2}+2 x+3\\
&\begin{aligned}
a &=f(x)_{\min }=\frac{-D}{4 a}=\frac{-(4-12}{4}=\frac{8}{4}=2 \\
\text { and } b &=\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^{2}}=\lim _{\theta \rightarrow 0} \frac{1-1+2 \sin ^{2} \theta / 2}{\theta^{2}} \\
&=\lim _{\theta \rightarrow 0} \frac{2 \sin ^{2} \theta / 2}{(\theta / 2)^{2} \cdot 4}=\lim _{\theta \rightarrow 0} \frac{1}{2} \cdot\left[\frac{\sin ^{2} \theta / 2}{(\theta / 2)^{2}}\right] \\
&=\frac{1}{2} \cdot \lim _{\theta \rightarrow 0} \frac{\sin ^{2} \theta / 2}{(\theta / 2)^{2}} \\
&=\frac{1}{2} \cdot 1=\frac{1}{2} \\
b &=\frac{1}{2} \\
\text { Now, } & \sum_{r=0}^{n} a^{r} \cdot b^{n-r}
\end{aligned}
\end{aligned}
$$
$=\sum_{r=0}^{n}\left(2^{r}\left(\frac{1}{2}\right)^{n-r}\right)$
$=\sum_{r=0}^{n} 2^{\prime} \cdot 2^{(r-n)}$
$=\sum_{r=0}^{n} 2^{2 r-n}$
$=2^{-n} \sum_{r=0}^{n} 2^{2 r}$
$=2^{-n}\left[1+2^{2}+2^{4}+2^{6}+\ldots+2^{2 n}\right]$
$=2^{-n}\left[\frac{1 \cdot\left(2^{2}\right)^{n+1}-1}{2^{2}-1}\right]\left\{\because S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}\right\}$
$=2^{-n}\left[\frac{4^{n+1}-1}{3}\right]$
$=\frac{4^{n+1}-1}{3 \cdot 2^{n}}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2019
Let $a_n=\left(1^2+2^2+\ldots n^2\right)^n$ and $b_n=n^n(n !)$. Then

Options:
A $a_n < b_n \forall n$
B $a_n>b_n \forall n$
C $a_n=b_n$ for infinitely many $n$
D $a_n < b_n$ if $n$ be even and $a_n>b_n$ if $n$ be odd

The correct answer is:
$a_n>b_n \forall n$

Solution:
Using PMI $\quad a_n>b_n \quad\left[{ }^* a_n>b_n \forall n \geq 2\right.$ as $\left.a_1=b_1\right]$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2022
Let $A=\left(\begin{array}{cc}x+2 & 3 x \\ 3 & x+2\end{array}\right), B=\left(\begin{array}{cc}x & 0 \\ 5 & x+2\end{array}\right)$. Then
all solutions of the equation det $(A B)=0$ is

Options:
A 1,-1,0,2
B 1,4,0,-2
C 1,-1,4,3
D -1,4,0,3

The correct answer is:
1,4,0,-2

Solution:
We know that, $\operatorname{det}(A B)=$ det $(A)$ det $(B)$
$\therefore$ $\operatorname{det}(A B)=0$
$\Rightarrow$ det$(A) \cdot$ det $(B)=0$ $\Rightarrow\left|\begin{array}{cc}x+2 & 3 x \\ 3 & x+2\end{array} \| \begin{array}{cc}x & 0 \\ 5 & x+2\end{array}\right|=0$
$\Rightarrow\left\{(x+2)^{2}-9 x\right\}\{x(x+2)-0\}=0$
$\Rightarrow \quad\left(x^{2}+4 x+4-9 x\right) \times(x+2)=0$
$\Rightarrow \quad x(x+2)\left(x^{2}-5 x+4\right)=0$
$\Rightarrow \quad x(x+2)(x-1)(x-4)=0$
$\Rightarrow \quad x=0,-2,1,4$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2017
Let $\mathrm{A}=\{\mathrm{x} \in \mathbb{R}:-1 \leq \mathrm{x} \leq 1\} \& \mathrm{f}: \mathrm{A} \rightarrow \mathrm{A}$ be a mapping defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}|\mathrm{x}| .$ Then $\mathrm{f}$ is

Options:
A injective but not surjective
B surjective but not injective
C neither injective nor surjective
D bijective

The correct answer is:
bijective

Solution:
Hint:
$f(x)=x|x|=\left\{\begin{array}{c}-x^{2} x \in[-1,0) \\ x^{2} x \in[0,1]\end{array}\right.$

Topic: Mathematics, Functions, WBJEE, WBJEE 2020
Let \(a_1, a_2, a_3, \ldots, a_n\) be positive real numbers. Then the minimum value of \(\frac{a_1}{a_2}+\frac{a_2}{a_3}+\ldots .+\frac{a_n}{a_1}\) is

Options:
A 1
B \(\mathrm{n}\)
C \({ }^n \mathrm{C}_2\)
D 2

The correct answer is:
\(\mathrm{n}\)

Solution:
Hint: \(\frac{\frac{a_1}{a_2}+\frac{a_2}{a_3}+\ldots . .+\frac{a_n}{a_1}}{n} \geq \sqrt[n]{\frac{a_1}{a_2} \times \frac{a_2}{a_3} \times \ldots . \frac{a_n}{a_1}}\)
\(A M \geq G M\) (property)
\(\therefore\) Minimum value \(=\mathrm{n}\)
It is possible when \(a_1=a_2=a_3=\ldots .=a_n\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2023
Let \(A(2 \sec \theta, 3 \tan \theta)\) and \(B(2 \sec \phi, 3 \tan \phi)\) where \(\theta+\phi=\frac{\pi}{2}\) be two points on the hyperbola \(\frac{x^2}{4}-\frac{y^2}{9}=1\). If \((\alpha, \beta)\) is the point of intersection of normals to the hyperbola at \(A\) and \(B\), then \(\beta\) is equal to

Options:
A \(\frac{12}{3}\)
B \(\frac{13}{3}\)
C \(-\frac{12}{3}\)
D \(-\frac{13}{3}\)

The correct answer is:
\(-\frac{13}{3}\)

Solution:
Hint : Equation of Normal at \(A\)
\(\frac{4 x}{2 \sec \theta}+\frac{9 y}{3 \tan \theta}=13\)
at \(B\)
\(\begin{aligned}
& \frac{4 x}{2 \operatorname{cosec} \theta}+\frac{9 y}{3 \cot \theta}=13 \\
& \Rightarrow(2 \cos \theta) x+(3 \cot \theta) y=13 \quad ----(1)\\
& \&~(2 \sin \theta) x+(3 \tan \theta) y=13 \quad ---(2)\\
& \Rightarrow y=-\frac{13}{3}
\end{aligned}\)

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2023
Let $\alpha$ and $\beta$ be iwo distinct roots of $a \cos \theta+b \sin \theta=c,$ where $a, b, c$ are three
real constants and $0 \in|0,2 \pi|$. Then, $\alpha+\beta$ is also a root of the same equation, if

Options:
A $a+b=c$
B $b+c=a$
C $c+a=b$
D $c=a$

The correct answer is:
$c=a$

Solution:
Given equation is $a \cos \theta+b \sin \theta=c$
$\Rightarrow a\left(\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)+\frac{2 b \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=c$
$\left[\because \cos \theta=\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right.$ and $\left.\sin \theta=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right]$
$\Rightarrow a\left(1-\tan ^{2} \frac{\theta}{2}\right)+2 b \tan \frac{\theta}{2}=c\left(1+\tan ^{2} \frac{\theta}{2}\right)$
$\Rightarrow a-\operatorname{atan}^{2} \frac{\theta}{2}+2 b \tan \frac{\theta}{2}-c-c \tan ^{2} \frac{\theta}{2}=0$
$\Rightarrow(c+a) \tan ^{2} \frac{\theta}{2}-2 b \tan \frac{\theta}{2}+(c-a)=0$
Let $\alpha$ and $\beta$ be the roots of the equation $\therefore \alpha+\beta=\frac{2 b}{c+a}$ and $\alpha \beta=\frac{c-a}{c+a}$
Now, $\tan \frac{\alpha+\beta}{2}=\frac{\frac{2 b}{c+a}}{1-\frac{c-a}{c+a}}$
$=\frac{\frac{c+a}{c+a-c+a}}{c+a}=\frac{b}{a}$
since, $\frac{b}{c}$ is a root of the equation. $\therefore \quad(c+a) \frac{b^{2}}{a^{2}}-2 b\left(\frac{b}{a}\right)+c-a=0$
$\Rightarrow b^{2} c+b^{2} a-2 b^{2} a+c a^{2}-a^{3}=0$
$\Rightarrow-b^{2} a+b^{2} c+c a^{2}-a^{3}=0$
$\Rightarrow \quad b^{2} c-b^{2} a+c a^{2}-a^{3}=0$
$\Rightarrow \quad b^{2}(c-a)+a^{2}(c-a)=0$
$\Rightarrow \quad(c-a)\left(b^{2}+a^{2}\right)=0$
$\Rightarrow c-a=0 \quad$ or $\quad b^{2}+a^{2}=0$
$\Rightarrow \quad c=a$ or $b^{2}+a^{2}=0$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2015
Let $\alpha$ and $\beta$ be the roots of $x^{2}+x+1=0 .$ If $n$ be a positive integer, then $\alpha^{n}+\beta^{n}$ is

Options:
A $2 \cos \frac{2 n \pi}{3}$
B $2 \sin \frac{2 n \pi}{3}$
C $2 \cos \frac{n \pi}{3}$
D $2 \sin \frac{n \pi}{3}$

The correct answer is:
$2 \cos \frac{2 n \pi}{3}$

Solution:
We have, $x^{2}+x+1=0$
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{3} i}{2}$
$\Rightarrow \quad \alpha=\frac{-1+\sqrt{3} i}{2}$ and $\beta=\frac{-1-\sqrt{3} i}{2}$
or $\quad \alpha=e^{\frac{i2\pi}{3}}$ and $\beta=e^{\frac{-2\pi i}{3}}$
$\therefore$
$\alpha^{n}+\beta^{n}=e^{\frac{2 n i t}{3}}+e^{\frac{-2\pi i}{3}}$
$=2\left(\frac{e^{\frac{2n\pi i}{3}}+e^{\frac{-2n\pi i}{3}}}{2}\right)$
$=2 \cos \left(\frac{2 n \pi}{3}\right)$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2017
Let $\rho$ be a relation defined on $N$, the set of natural numbers, as $\rho=\{(x, y) \in N \times N: 2 x+y=41\} .$ Then

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

The correct answer is:
$\rho$ is not transitive

Solution:
We have, $\rho=\{(x, y) \in N \times N: 2 x+y=41\}$
For reflexive,
$\begin{array}{l}
x p x \Rightarrow 2 x+x=41 \\
\Rightarrow 3 x=41
\end{array}$
$\Rightarrow x=\frac{41}{3} \notin N$
So, $\rho$ is not reflexive.
For symmetric,
$x \rho y \Rightarrow 2 x+y=41$
and $y \rho x \Rightarrow 2 y+x=41$
$\Rightarrow x p y \neq y \rho x$
So, $\rho$ is not symmetric.
For transitive,
$x p y \Rightarrow 2 x+y=41$
and $y \rho z \Rightarrow 2 y+z=41$
$\Rightarrow x p z$
$\Rightarrow \rho$ is not transitive.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2018
Let $\alpha, \beta$ be the roots of the equation $x^{2}-6 x-2=0$ with $\alpha>\beta$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_{8}}{2 a_{9}}$ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
$x^{2}-6 x-2=0$
$x^{n}-6 x^{n-1}-2 x^{n-2}=0$
$\Rightarrow x^{n}-2 x^{n-2}=6 x^{n-1}$
for $n=10$
$x^{10}-2 x^{8}=6 x^{9}$
$\alpha^{10}-2 \alpha^{8}=6 \alpha^{9}$
$\beta^{10}-2 \beta^{8}=6 \beta^{9}$
$\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^{8}-\beta^{8}\right)=6\left(\alpha^{9}-\beta^{9}\right)$
$\Rightarrow a_{10}-2 a_{8}=6 a_{9}$
$\Rightarrow \frac{a_{10}-2 a_{8}}{2 a_{9}}=3$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2021
Let $\alpha, \beta$ be the roots of $x^{2}-x-1=0$ and $S_{n}=\alpha^{n}+\beta^{n},$ for all integers $n \geq 1 .$ Then, for every integer $n \geq 2$

Options:
A $S_{n}+S_{n-1}=S_{n+1}$
B $S_{n}-S_{n-1}=S_{n+1}$
C $S_{n-1}=S_{n+1}$
D $S_{n}+S_{n-1}=2 S_{n+1}$

The correct answer is:
$S_{n}+S_{n-1}=S_{n+1}$

Solution:
Given, $\alpha$ and $\beta$ are the roots of $x^{2}-x-1=0$
$\therefore \quad \alpha+\beta=1$
and $\alpha \beta=-1$
Now, consider option (a). $S_{n}+S_{n-1}=S_{n+1}$
Put $n=2$, we get, $\quad S_{2}+S_{1}=S_{3}$ $\therefore \quad\left(\alpha^{2}+\beta^{2}\right)+(\alpha+\beta)=\alpha^{3}+\beta^{3}$
$\left[\because \cdot S_{n}=\alpha^{n}+\beta^{n},\right.$ given]
$\Rightarrow(a+\beta)^{2}-2 a \beta+(\alpha+\beta)$
$=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$
$\Rightarrow 1^{2}-2(-1)+1=1^{3}-3(-1)(1)$
$\Rightarrow 2+1=1+3$
$\Rightarrow \quad 1+2+1=1+3$
$\Rightarrow$
$4=4,$ which is true.
Hence, option (a) is correct.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2014
Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be the three unit vectors such that $\vec{\alpha} \cdot \vec{\beta}=\vec{\alpha} \cdot \vec{\gamma}=0$ and the angle between $\vec{\beta}$ and $\vec{\gamma}$
is $30^{\circ} .$ Then $\vec{\alpha}$ is

Options:
A $2(\vec{\beta} \times \vec{\gamma})$
B $-2(\vec{\beta} \times \vec{y})$
C $\pm 2(\vec{\beta} \times \vec{\gamma})$
D $(\vec{\beta} \times \vec{y})$

The correct answer is:
$\pm 2(\vec{\beta} \times \vec{\gamma})$

Solution:
$\vec{\alpha}=\lambda(\vec{\beta} \times \vec{\gamma})=\lambda\left(|\vec{\beta}||\vec{\gamma}|\left|\sin 30^{\circ}\right|\right.$
$\Rightarrow \quad|\vec{\alpha}|=|\lambda|\left(|\beta||\vec{\gamma}| \cdot \frac{1}{2}\right)$
$\begin{array}{lc}\Rightarrow & 1=|\lambda| \cdot 1 \cdot 1 \cdot \frac{1}{2} \\ \Rightarrow & |\lambda|=2 \\ \rightarrow & \lambda-\pm 2\end{array}$
$\therefore \quad \vec{\alpha}=\pm 2(\vec{\beta} \times \vec{\gamma})$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2018