datavorous/entrance-exam-dataset huggingface dataset:
If $A=\left\{5^{n}-4 n-1: n \in N\right\}$ and
$B=\{16(n-1): n \in N],$ then

Options:
A $A=B$
B $A \cap B=\phi$
C $A \subseteq B$
D $B \subseteq A$

The correct answer is:
$A \subseteq B$

Solution:
We have,
$\begin{aligned} A &=5^{n}-4 n-1=(1+4)^{n}-4 n-1 \\ &=\left({ }^{n} C_{0}+{ }^{n} C_{1} \times 4+{ }^{n} C_{2} \times 4^{2}+\ldots+{ }^{n} C_{n} \times 4^{n}\right) \\ &=4^{2}\left({ }^{n} C_{2}+{ }^{n} C_{3} \times 4+\ldots+{ }^{n} C_{n} \times 4^{n-2}\right) \end{aligned}$
$\therefore$ A contains some multiples of 16
Clearly, $B$ contains all multiples of 16 including $0 .$
$A \subseteq B$

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2016
If $a$ and $b$ are arbitrary positive real numbers, then the least possible value of $\frac{6 a}{5 b}+\frac{10 b}{3 a}$ is

Options:
A 4
B $\frac{6}{5}$
C $\frac{10}{3}$
D $\frac{68}{15}$

The correct answer is:
4

Solution:
Hint:
$\frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \sqrt{\frac{6 a}{5 b} \times \frac{10 b}{3 a}}, \quad \frac{6 a}{5 b}+\frac{10 b}{3 a} \geq 2 \times 2 \geq 4$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2020
If \(A\) and \(B\) are coefficients of \(x^{\mathrm{n}}\) in the expansions of \((1+x)^{2 \mathrm{n}}\) and \((1+x)^{2 \mathrm{n}-1}\) respectively, then \(\mathrm{A} / \mathrm{B}\) is equal to

Options:
A 4
B 2
C 9
D 6

The correct answer is:
2

Solution:
\(\begin{array}{r}\text {Hints: }
A={ }^{2 n} C_n \\
B={ }^{2 n-1} C_n
\end{array}\)
\(\frac{\mathrm{A}}{\mathrm{B}}=\frac{{ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}}}{{ }^{2 \mathrm{n}-1} \mathrm{C}_{\mathrm{n}}}=\frac{2 \mathrm{n}}{\mathrm{n}}=2\)

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2011
If $A$ and $B$ are two matrices such that $A B=B$ and $B A=A,$ then $A^{2}+B^{2}$ equals

Options:
A 248
B $2 \mathrm{BA}$
C $A+B$
D $A B$

The correct answer is:
$A+B$

Solution:
Given, $A B=B$ and $B A=A$ $\begin{aligned} \text { Now, } A^{2}+B^{2} &=A \cdot A+B \cdot B \\ &=A(B A)+B(A B)=(A B) A+(\ln ) \\ &=B A+A B=A+B \end{aligned}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2015
If $A$ and $B$ square matrices of the same order and $A B=3 I$, then $A^{-1}$ is equal to

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

The correct answer is:
$\frac{1}{3} \mathrm{~B}$

Solution:
Hints: $\mathrm{AB}=3 \mathrm{I}, \mathrm{A}^{-1} \cdot \mathrm{AB}=3 \cdot \mathrm{A}^{-1} \mathrm{I} \Rightarrow \mathrm{B}=3 \mathrm{~A}^{-1} \Rightarrow A^{-1}=\frac{1}{3} \mathrm{~B}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2009
If $a, b \in\{1,2,3\}$ and the equation $a x^{2}+b x+1=0$ has real roots, then

Options:
A $a>b$
B $a \leq b$
C number of possible ordered pairs $(a, b)$ is 3
D $a < b$

The correct answers are:
number of possible ordered pairs $(a, b)$ is 3, $a < b$

Solution:
We have,
$$
a x^{2}+b x+1=0
$$
For real roots, $D \geq 0$
$$
\therefore \quad b^{2}-4 a \geq 0
$$
$\Rightarrow \quad b^{2} \geq 4 a$
$\therefore(a, b)=(1,2),(1,3),(2,3)$
$\therefore$ Number of ordered pairs $(a, b)=3$ and $a$ is always less than $b$.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2017
If $a, b$ and $c$ are in $\mathrm{AP}$, then the straight line $a x+2 b y+c=0$ will always pass through a fixed point whose coordinates are

Options:
A (1,-1)
B (-1,1)
C (1,-2)
D (-2,1)

The correct answer is:
(1,-1)

Solution:
Given that, $a$, $b$ and $c$ are in $\mathrm{AP}$
$\therefore \quad 2 b=a+c$
$\Rightarrow \quad c=2 b-a$
and equation of straight line is
$a x+2 b y+c=0$
$\Rightarrow \quad a x+2 b y+(2 b-a)=0$
$\Rightarrow \quad a(x-1)+b(2 y+2)=a \cdot 0+b \cdot 0$
On comparing, we get
$x-1=0 \Rightarrow x=1$
and $\quad 2 y+2=0$
$\Rightarrow \quad y=-1$
Hence, the required fixed point is (1,-1)
$$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2013
If $a, b$ and $c$ are in arithmetic progression, then the roots of the equation $a x^{2}-2 h x+c=0$ are

Options:
A 1 and $\frac{c}{a}$
B $-\frac{1}{a}$ and $-c$
C -1 and $-\frac{c}{a}$
D -2 and $-\frac{c}{2 a}$

The correct answer is:
1 and $\frac{c}{a}$

Solution:
Since, $a, b$ and $c$ are in $\mathrm{AP}$.
$\therefore$
$2 b=a+c$
Given, quadratic equation, $a x^{2}-2 b x+c=0$
$\Rightarrow \quad a x^{2}-(a+c) x+c=0$
$2 b=a+c)$
$\Rightarrow \quad a x^{2}-a x-c x+c=0$
$\begin{array}{lr}\Rightarrow & a x(x-1)-c(x-1)=0 \\ \Rightarrow & (x-1)(a x-c)=0 \\ \Rightarrow & x=1, \frac{c}{a}\end{array}$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2012
If $a, b$ and $c$ are positive numbers in a GP, then the roots of the quadratic equation $\left(\log _{e} a\right) x^{2}-\left(2 \log _{e} b\right) x+\left(\log _{e} c\right)=0$ are

Options:
A -1 and $\frac{\log _{e} c}{\log _{e} a}$
B 1 and $-\frac{\log _{e} c}{\log _{e} a}$
C 1 and $\log _{a} c$
D -1 and $\log _{c} a$

The correct answer is:
1 and $\log _{a} c$

Solution:
Since, $a, b$ and $c$ are in $\mathrm{GP}$.
$\therefore b^{2}=a c$
Given equation is
$(\log _{e} \text { a) } x^{2}-\left(2 \log _{e} \text { b) } x+\log _{e} c\right)=0$
Put $x=1,$ we get
$\begin{aligned}
& \log _{e} a-2 \log _{e} b+\log _{e} c=0 \\
\Rightarrow & 2 \log _{e} b=\log _{e} a+\log _{0} c \\
\Rightarrow \quad & \log _{e} b^{2}=\log _{e} a c
\end{aligned}$
$\Rightarrow \quad b^{2}=a c,$ which is true.
Hence, one of the root of given equation is $1 .$
Let another root be $\alpha$.
$\therefore$ Sum of roots, $1+\alpha=\frac{2 \log _{e} b}{\log _{e} a}=\frac{\log _{e} b^{2}}{\log _{e} a}$
$\begin{aligned}
\Rightarrow \quad a &=\frac{\log _{e} a c}{\log _{e} a}-1 \\
&=\frac{\left(\log _{e} a+\log _{e} c\right)}{\log _{e} a}-1 \\
&=\frac{\log _{e} c}{\log _{c} a}=\log _{a} c
\end{aligned}$
Hence, roots are 1 and $\log _{a} c .$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2014
If $a, b$ are odd integers, then the roots of the equation $2 a x^2+(2 a+b) x+b=0, a \neq 0$ are

Options:
A rational
B irrational
C non-real
D equal

The correct answer is:
rational

Solution:
Discriminant is a perfect square $\Rightarrow$ Rational roots

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2022
If $A, B$ are two events such that $P(A \cup B) \geq \frac{3}{4}$ and $\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8},$ then

Options:
A $P(A)+P(B) \leq \frac{11}{8}$
B $P(A) \cdot P(B) \leq \frac{3}{8}$
C $P(A)+P(B) \geq \frac{7}{8}$
D None of the above

The correct answers are:
$P(A)+P(B) \leq \frac{11}{8}$, $P(A)+P(B) \geq \frac{7}{8}$

Solution:
We know,
$$
P(A \cup B)=P(A)+P(B)-P(A \cap B)
$$
$\Rightarrow P(A)+P(B)=P(A \cup B)+P(A \cap B)$
Given, $\quad \frac{3}{4} \leq P(A \cup B) \leq 1$
and $\quad \frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}$
So. $\quad \frac{7}{8} \leq P(A \cup B)+P(A \cap B) \leq \frac{11}{8}$
$\therefore P(A)+P(B) \geq \frac{7}{8}$ and $P(A)+P(B) \leq \frac{11}{8}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2016
If $\mathrm{a}(\vec{\alpha} \times \vec{\beta})+\mathrm{b}(\vec{\beta} \times \vec{\gamma})+\mathrm{c}(\vec{\gamma} \times \vec{\alpha})=\overrightarrow{0}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are non-zero scalars, then the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are

Options:
A parallel
B non-coplanar
C coplanar
D mutually perpendicular

The correct answer is:
coplanar

Solution:
$\therefore \vec{\alpha} \times \vec{\beta}, \vec{\beta} \times \vec{\lambda}, \vec{\gamma} \times \vec{\alpha}$ are coplanar $\Rightarrow \vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are coplanar

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2021
If $\vec{a}+\vec{b}=\vec{c}$ and $a+b=c$, then the angle included between $\vec{a}$ and $\vec{b}$ is

Options:
A $90^{\circ}$
B $180^{\circ}$
C $120^{\circ}$
D Zero

The correct answer is:
Zero

Solution:
Hints: Here $\vec{a}+\vec{b}=\vec{c} \& c=a+b$
Now, $c=\sqrt{a^2+b^2+2 a b \cos \theta}$
$\begin{aligned}
& (a+b)=\sqrt{a^2+b^2+2 a b \cos \theta} \\
& a^2+b^2+2 a b=a^2+b^2+2 a b \cos \theta \\
& \cos \theta=1, \theta=0^{\circ}
\end{aligned}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2010
If $a, b, c$ are in G. P. and $\log a .-\log 2 b, \log 2 b-\log 3 c, \log 3 c-\log a$ are in A. P., then $a, b, c$ are the lengths of the sides of a triangle which is

Options:
A acute angled
B obtuse angled
C right angled
D equilateral

The correct answer is:
obtuse angled

Solution:
$$
\because 2 \log \frac{2 b}{3 c}=\log \frac{a}{2 b}+\log \frac{3 c}{a} \Rightarrow \frac{4 b^2}{9 c^2}=\frac{3 c}{2 b} \Rightarrow 2 b=3 c \Rightarrow \frac{c}{b}=\frac{2}{3} \text { (common ratio) }
$$
$\therefore$ Sides are a, $\frac{2 \mathrm{a}}{3}, \frac{4 \mathrm{a}}{9}$
Using Cosine rule
$\Rightarrow \cos \mathrm{A} < 0 \Rightarrow$ obteuse angled tiangle

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2022
If $a, b, c$ are in G.P. $(a>1, b>1, c>1)$, then for any real number $x($ with $x>0, x \neq 1), \log _{\mathrm{a}} x, \log _{\mathrm{b}} x, \log _{\mathrm{c}} x$ are in

Options:
A G.P.
B A.P.
C H.P.
D G.P. but not in H.P.

The correct answer is:
H.P.

Solution:
Hints : $a, b, c$ are in G.P.
$\Rightarrow \log _x a, \log _x b, \log _x c$ are in A.P.
$\Rightarrow \frac{1}{\log _x a}, \frac{1}{\log _x b}, \frac{1}{\log _x c}$ arein H.P.
$\Rightarrow \log _a X, \log _b X, \log _c X$ are in H.P.

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2009
If $a, b, c$ are real, then both the roots of the equation $(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$ are always

Options:
A positive
B negative
C real
D imaginary

The correct answer is:
real

Solution:
Hints : $3 x^2-2 x(a+b+c)+a b+b c+c a=0$
$$
\begin{aligned}
& \mathrm{D}=4(a+b+c)^2-4.3(a b+b c+c a) \\
& =4\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\
& =\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\
& \geq 0
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2009
If $\mathbf{A}=\mathbf{B}+\mathbf{C}$ have scalar magnitudes of 5,4,3 units respectively, then the angle between $A$ and $\mathbf{C}$ is

Options:
A $\cos ^{-1}(3 / 5)$
B $\cos ^{-1}(4 / 5)$
C $\pi / 2$
D $\sin ^{-1}(3 / 4)$

The correct answer is:
$\cos ^{-1}(3 / 5)$

Solution:
Here, triangle is

$\theta=\cos ^{-1}\left(\frac{|\mathbf{C}|}{|\mathbf{A}|}\right)$
$\theta=\cos ^{-1}\left(\frac{3}{5}\right)$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2012
If $\int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x=\alpha\left[\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}\right]+c$, then $\alpha=$

Options:
A $\frac{2}{\mathrm{~b}^{2}}$
B $\frac{2}{a^{2}}$
C $-\frac{2}{b^{2}}$
D $-\frac{2}{\mathrm{a}^{2}}$

The correct answer is:
$-\frac{2}{b^{2}}$

Solution:
$\int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x$
Put $a+b \cos x=t$
$-b \sin x d x=d t$
$=\int \frac{2 \sin x \cos x}{t^{2}} \cdot \frac{d t}{-b \sin x}=\int \frac{-2}{b} \cdot\left(\frac{t-a}{b}\right) \cdot \frac{1}{t^{2}} d t=\frac{-2}{b^{2}} \int\left(\frac{1}{t}-\frac{a}{t^{2}}\right) d t \quad \therefore \alpha=\frac{-2}{b^{2}}$
$=-\frac{2}{b^{2}}\left[\ln |t|+\frac{a}{t}\right]+c$
$=\frac{-2}{b^{2}}\left[\ln |a+b \cos x|+\frac{a}{a+b \cos x}\right]+c$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2021
If [a] denote the greatest integer which is less than or equal to a. Then, the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}[\sin x \cos x] d x$ is

Options:
A $\frac{\pi}{2}$
B $\pi$
C $-\pi$
D $-\frac{\pi}{2}$

The correct answer is:
$-\frac{\pi}{2}$

Solution:
Let $/=\int_{-\pi / 2}^{\pi / 2}[\sin x \cdot \cos x] d x=\int_{-\pi / 2}^{\pi / 2}\left[\frac{1}{2} \sin 2 x\right] d x$

Put $\theta=2 x \Rightarrow d \theta=2 d x$
Also, when $\quad x=-\pi / 2,$ then $\theta=-\pi$ when $\quad x=\frac{\pi}{2},$ then $\theta=\pi$
$\begin{aligned} \text { Then, } & I=\frac{1}{2} \int_{\pi}^{\pi}\left[\frac{1}{2} \sin \theta\right] d \theta \\ &=\frac{1}{2}\left[\int_{\pi}^{0}(-1) d x+\int_{0}^{\pi}(0) d x\right] \\ &=\frac{1}{2}[-x]_{\pi}^{2}+0=-\frac{\pi}{2} \end{aligned}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2013
If a hyperbola passes through the point \(\mathrm{P}(\sqrt{2}, \sqrt{3})\) and foci at \(( \pm 2,0)\) then the tangent to this hyperbola at \(\mathrm{P}\) is

Options:
A \(\mathrm{y}=\mathrm{x} \sqrt{6}-\sqrt{3}\)
B \(\mathrm{y}=\mathrm{x} \sqrt{3}-\sqrt{6}\)
C \(\mathrm{y}=\mathrm{x} \sqrt{6}+\sqrt{3}\)
D \(y=x \sqrt{3}+\sqrt{6}\)

The correct answer is:
\(\mathrm{y}=\mathrm{x} \sqrt{6}-\sqrt{3}\)

Solution:
Hint: \(a e=2, a^2+b^2=4\)
\(\Rightarrow \frac{2}{a^2}-\frac{3}{b^2}=1 \Rightarrow b^2=3 ; a^2=1\)
\(\frac{x^2}{1}-\frac{y^2}{3}=1\). Tangent at \(P(\sqrt{2}, \sqrt{3})\) is
\(\sqrt{6 x}-y=\sqrt{3}\)

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2023
If $\vec{a}=\hat{i}+\hat{j}-\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ is unit vector perpendicular to $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$, then unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is

Options:
A $\pm \frac{1}{\sqrt{6}}(2 \hat{i}-\hat{j}+\hat{k})$
B $\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$
C $\pm \frac{1}{\sqrt{6}}(\hat{i}-2 \hat{j}+\hat{k})$
D $\pm \frac{1}{\sqrt{2}}(\hat{j}-\hat{k})$

The correct answer is:
$\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$

Solution:
$\because \overrightarrow{\mathrm{d}}$ is normal to the plane of $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$
$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\pm 2(\hat{\mathrm{j}}+\hat{\mathrm{k}}) \therefore$ unit vector $\overrightarrow{\mathrm{d}}=\pm \frac{1}{\sqrt{2}}(\hat{\mathrm{j}}+\hat{\mathrm{k}})$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2022
If $A$ is a square matrix then,

Options:
A $\mathrm{A}+\mathrm{A}^{\mathrm{T}}$ is symmetric
B $\mathrm{AA}^{\mathrm{T}}$ is skew - symmetric
C $\mathrm{A}^{\mathrm{T}}+\mathrm{A}$ is skew-symmetric
D $\quad \mathrm{A}^{\mathrm{T}} \mathrm{A}$ is skew symmetric

The correct answer is:
$\mathrm{A}+\mathrm{A}^{\mathrm{T}}$ is symmetric

Solution:
Hints: $\left(\mathrm{A}+\mathrm{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}+\left(\mathrm{A}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}+\mathrm{A}=\mathrm{A}+\mathrm{A}^{\mathrm{T}}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2009
If $a_{n}(>0)$ be the $n^{\text {th }}$ term of a G.P. then $\left|\begin{array}{lll}\log a_{n} & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8}\end{array}\right|$ is equal to

Options:
A 1
B 2
C $-2$
D 0

The correct answer is:
0

Solution:
$\log a_{n}=\log \left(a r^{n-1}\right)=\log a+(n-1) \log r$

$=\left|\begin{array}{ccc}\log \mathrm{a}+(\mathrm{n}-1) \log \mathrm{r} & \log \mathrm{a}+\mathrm{nlogr} & \log \mathrm{a}+(\mathrm{n}+1) \log \mathrm{r} \\ 3 \log \mathrm{r} & 3 \log \mathrm{r} & 3 \log \mathrm{r} \\ 6 \log r & 6 \log r & 6 \log r\end{array}\right|$
= 0

Topic: Mathematics, Determinants, WBJEE, WBJEE 2021
If a person can throw a stone to maximum height of \(\mathrm{h}\) metre vertically, then the maximum distance through which it can be thrown horizontally by the same person is

Options:
A \(\frac{\mathrm{h}}{2}\)
B \(\mathrm{h}\)
C \(2 \mathrm{~h}\)
D \(3 \mathrm{~h}\)

The correct answer is:
\(2 \mathrm{~h}\)

Solution:
Hints:
\(\mathrm{h}=\frac{\mathrm{u}^2}{2 \mathrm{~g}} \Rightarrow \mathrm{u}^2=2 \mathrm{gh}\)
Maximum horizontal distance
\(\begin{aligned}
& \mathrm{R}_{\max }=\frac{\mathrm{u}^2}{\mathrm{~g}}\left(\text { when } \theta=45^{\circ}\right) \\
& \mathrm{R}_{\max }=2 \mathrm{~h}
\end{aligned}\)

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2011
If $a_{r}=(\cos 2 r \pi+i \sin 2 r \pi)^{1 / 9}$, then the value of
$\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is equal to

Options:
A 1
B -1
C 0
D 2

The correct answer is:
0

Solution:
We have,
$a_{r}=(\cos 2 r \pi+i \sin 2 r \pi)^{1 / 9}=e^{\frac{2 r \pi i}{9}}$
Now, $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|=\left|\begin{array}{ccc}\ e^{\frac{2 \pi i}{9}} & e^{\frac{4 \pi i}{9}} & e^{\frac{6 \pi i}{9}} \\ e^{\frac{8 \pi i}{9}} & e^{\frac{10 \pi i}{9}} & e^{\frac{12 \pi i}{9}} \\ e^{\frac{14 \pi i}{9}} & e^{\frac{16 \pi i}{9}} & e^{\frac{18 \pi i}{9}}\end{array}\right|$
$=e^{\frac{2 \pi i}{9}} \times e^{\frac{8 \pi i}{9}}\left|\begin{array}{ccc}1 & e^{\frac{2 \pi i}{9}} & e^{\frac{4 \pi i}{9}} \\ 1 & e^{\frac{2 \pi i}{9}} & e^{\frac{4 \pi i}{9}} \\ e^{\frac{14 \pi i}{9}} & e^{\frac{16 \pi i}{9}} & e^{\frac{18 \pi i}{9}}\end{array}\right|$
$=e^{\frac{2 \pi i}{9}} \times e^{\frac{8 \pi}{9}} \times 0 \quad\left(\because R_{1}\right.$ and $R_{2}$ are identical $)$
$=0$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2018
If a species has 16 protons, 18 electrons and 16 neutrons, find the species and its charge

Options:
A $\mathrm{S}^{1-}$
B $\mathrm{Si}^{2-}$
C $\mathrm{P}^{3-}$
D $\mathrm{S}^{2-}$

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

Solution:
Hints : $16 \mathrm{p}$ means $\mathrm{z}=16$
$18 \mathrm{e}^{-}$means, 2 unit negative charge is present.
Hence species is $\mathrm{S}^{-2}$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2010
If a straight line passes through the point $(\alpha, \beta)$ and the portion of the line intercepted between the axes is divided equally at that point, then $\frac{x}{a}+\frac{y}{\beta}$ is

Options:
A 0
B 1
C 2
D 4

The correct answer is:
2

Solution:
Let the line be $A B=\frac{x}{a}+\frac{y}{b}=1$
$\ldots(0$
$\therefore$ The coordinates of $A$ and $B$ are respectively
$(a, 0)$ and $(0, b)$

Since, $(\alpha, \beta)$ is the mid-point of $A B$
$\begin{array}{ll}\therefore \quad & \alpha=\frac{a}{2}, \beta=\frac{b}{2} \\ \therefore & a=2 x \text { and } b=2 \beta\end{array}$
Crom Eq. (i), $\frac{x}{2 x}+\frac{y}{2 \beta}=1 \Rightarrow \frac{x}{\alpha}+\frac{y}{\beta}=2$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2012
If a string, suspended from the ceilling is given a downward force $F_1$, its length becomes $L_1$, Its length is $L_2$, if the downward force is $\mathrm{F}_2$. What is its actual length?

Options:
A $\frac{L_1+L_2}{2}$
B $\sqrt{L_1 L_2}$
C $\frac{\mathrm{F}_2 \mathrm{~L}_1+\mathrm{F}_1 \mathrm{~L}_2}{\mathrm{~F}_2+\mathrm{F}_1}$
D $\frac{\mathrm{F}_2 \mathrm{~L}_1-\mathrm{F}_1 \mathrm{~L}_2}{\mathrm{~F}_2-\mathrm{F}_1}$

The correct answer is:
$\frac{\mathrm{F}_2 \mathrm{~L}_1-\mathrm{F}_1 \mathrm{~L}_2}{\mathrm{~F}_2-\mathrm{F}_1}$

Solution:
$\begin{aligned} &\left(L_1-L\right)=\frac{F_1 L}{A y} ; L_2-L=\frac{F_2 L}{A y} \\ \Rightarrow & \frac{L_1-L}{L_2-L}=\frac{F_1}{F_2} \Rightarrow F_2 L_1-F_2 L=F_1 L_2-F_1 L \Rightarrow L=\frac{F_2 L_1-F_1 L_2}{F_2-F_1} \end{aligned}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2022
If $a, x$ are real numbers and $|a| < 1,|x| < 1$ then $1 (1+a) x+\left(1+a+a^{2}\right) x^{2}+\ldots \infty \quad$ is equal to

Options:
A $\frac{1}{(1-a)(1-a x)}$
B $\frac{1}{(1-a)(1-x)}$
C $\frac{1}{(1-x)(1-a x)}$
D $\frac{1}{(1+a x)(1-a)}$

The correct answer is:
$\frac{1}{(1-x)(1-a x)}$

Solution:
We have. $1+(1+a) x+\left(1+a+a^{2}\right) x^{2}+\ldots \infty$
$=\sum_{n=1}^{\infty}\left(1+a+a^{2}+\ldots+a^{n-1}\right) x^{n-1}$
$=\sum_{n=1}^{\infty}\left(\frac{1-a^{n}}{1-a}\right) x^{n-1}$
$=\sum_{n=1}^{\infty} \frac{x^{n-1}}{1-a}-\sum_{n=1}^{\infty} \frac{a^{n} x^{n-1}}{1-a}$
$=\frac{1}{1-a} \sum_{n=1}^{\infty} x^{n-1}-\frac{a}{1-a} \sum_{n=1}^{\infty}(a x)^{n-1}$
$=\frac{1}{1-a}\left[\left(1+x+x^{2}+\ldots \infty\right)\right]$ $-\frac{a}{1-a}\left[\left\{1+a x+(a x)^{2}+\ldots \infty\right\}\right]$
$=\frac{1}{(1-a)} \cdot \frac{1}{(1-x)}-\frac{a}{(1-a)(1-a x)}$
$=\frac{1}{(1-a x)(1-x)}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2016
If all the nucleophilic substitution reactions at saturated carbon atoms in the above sequence of reactions follow \(\mathrm{S}_{\mathrm{N}} 2\) mechanism, then \(\underline{E}\) and \(\underline{E}\) will be respectively,

Options:
A

B

C

D

The correct answer is:

Solution:
Hint :

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2023
If \(\alpha\) and \(\beta\) are the roots of the equation \(x^2+x+1=0\), then the equation whose roots are \(\alpha^{19}\) and \(\beta^7\) is

Options:
A \(x^2-x-1=0\)
B \(x^2-x+1=0\)
C \(x^2+x-1=0\)
D \(x^2+x+1=0\)

The correct answer is:
\(x^2+x+1=0\)

Solution:
Hints: \(\alpha\) and \(\beta\) are the roots of \(x^2+x+1=0\)
\(\begin{array}{ll}
\alpha=\omega & \beta=\omega^2 \\
\alpha^{19}=\omega & \beta^7=\omega^2 \\
x^2-\left(\alpha^{19}+\beta^7\right) x+\alpha^{19} \beta^7=0
\end{array}\)
Thou,
\(\begin{aligned}
& x^2-\left(\omega+\omega^2\right) x+\omega \cdot \omega^2=0 \\
& x^2+x+1=0
\end{aligned}\)

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2011
If $\alpha$ and $\beta$ are roots of $a x^{2}+b x+c=0,$ then
the equation whose roots are $\alpha^{2}$ and $\beta^{2},$ is

Options:
A $a^{2} x^{2}-\left(b^{2}-2 a c\right) x+c^{2}=0$
B $a^{2} x^{2}+\left(b^{2}-a c\right) x+c^{2}=0$
C $a^{2} x^{2}+\left(b^{2}+a c\right) x+c^{2}=0$
D $a^{2} x^{2}+\left(b^{2}+2 a c\right) x+c^{2}=0$

The correct answer is:
$a^{2} x^{2}-\left(b^{2}-2 a c\right) x+c^{2}=0$

Solution:
Given. $\alpha$ and $\beta$ are roots of $a x^{2}+b x+c=0$
$\therefore \quad \alpha+\beta=\frac{-b}{a}$
and $\quad a \beta=\frac{c}{a}$
Now, if the roots are $\alpha^{2}$ and $\beta^{2}$, then
$$
\begin{aligned}
a^{2}+\beta^{2} &=(\alpha+\beta)^{2}-2 \alpha \beta \\
&=\left(\frac{-b}{a}\right)^{2}-\frac{2c}{a}=\frac{b^{2}}{a^{2}}-\frac{2 c}{a} \\
\text { And } \quad \alpha^{2} \beta^{2} &=(\alpha \beta)^{2}=\left(\frac{c}{a}\right)^{2}=\frac{c^{2}}{a^{2}}
\end{aligned}
$$
The equation whose roots are $\alpha^{2}$ and $\beta^{2}$. is
$$
x^{2}-\left(\frac{b^{2}}{a^{2}}-\frac{2 c}{a}\right) x+\frac{c^{2}}{a^{2}}=0
$$
$\therefore \quad a^{2} x^{2}-\left(b^{2}-2 a c\right) x+c^{2}=0$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2016
If $\alpha$ and $\beta$ are roots of $x^{2}-x+1=0,$ then the value of $\alpha^{2013}+\beta^{2013}$ is

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

The correct answer is:
-2

Solution:
Given equation is $x^{2}-x+1=0$
$$
\begin{aligned}
x &=\frac{1 \pm \sqrt{1-4}}{2} \quad\left(\because x=\frac{b \pm \sqrt{b^{2}-4 a c}}{2 a}\right) \\
&=\frac{1 \pm i \sqrt{3}}{2}=\frac{1+i \sqrt{3}}{2}, \frac{1-i \sqrt{3}}{2} \\
\Rightarrow-x &=\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2} \\
\Rightarrow+x &=\omega,-\omega^{2}
\end{aligned}
$$
Since, $(\alpha, \beta)$ are the roots of given equation. Then, $\alpha=-\omega$ and $\beta=-\omega^{2}$
$$
\begin{aligned}
\therefore \alpha^{2013}+\beta^{2013} &=-(\omega)^{2013}+\left(-\omega^{2}\right)^{2013} \\
&=-\omega^{2013}-\omega^{4026} \\
&=-\left(\omega^{3}\right)^{671}-\left(\omega^{3}\right)^{1342} \\
&=-(1)^{671}-(1)^{1342} \\
&=-1-1=-2
\end{aligned}
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2013