datavorous/entrance-exam-dataset huggingface dataset:
If $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are the roots of the equation $x^{2}+p x+q=0,$ where $\alpha, \beta, p$ and $q$ are real, then the roots of the equation $\left(p^{2}-4 q\right)\left(p^{2} x^{2}+4 p x\right)-16 q=0$ are

Options:
A $\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)$
B $\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\beta}\right)$ and $\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\beta}\right)$
C $\left(\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)$
D $(\sqrt{\alpha}+\sqrt{\beta})$ and $(\sqrt{\alpha}-\sqrt{\beta})$

The correct answer is:
$\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)$

Solution:
Since, $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are the roots of the equation $x^{2}+p x+q=0$
$\therefore \quad$ Sum of roots $=-p$
$\Rightarrow \quad(\alpha+\sqrt{\beta})+(\alpha-\sqrt{\beta})=-P$
$\Rightarrow$
$\mathbf{2}\alpha=-p \Rightarrow \alpha=-\frac{\underline{p}}{2}$
and produet of roots $=\mathrm{q}$
$(\alpha+\sqrt{\beta})(\alpha-\sqrt{\beta})=q$
$\Rightarrow \alpha^{2}-\beta=q$
$\Rightarrow \quad \beta=\alpha^{2}-q=\left(-\frac{p}{2}\right)^{2}-q=\frac{p^{2}}{4}-q$
$\Rightarrow \quad p^{2}-4 q=4 \beta$
$\therefore$ The given equation $\left(p^{2}-4 q\right)\left(p^{2} x^{2}+4 p x\right)-16 q=0$
$4 \beta\left(p^{2} x^{2}+4 p x\right)-16\left(\alpha^{2}-\beta\right)=0$
$\left[\alpha^{2}-\beta=q\right]$
$\beta\left(4 a^{2} x^{2}-8 x x\right)-4\left(a^{2}-\beta\right)=0$
$\alpha^{2} \beta x^{2}-2 x f\left(x+\beta=\alpha^{2}\right.$
$\Rightarrow \quad(\alpha x \sqrt{\beta}-\sqrt{\beta})^{2}=\alpha^{2}$
$\Rightarrow \quad \operatorname{cox} \sqrt{\beta}-\sqrt{\beta}=\pm \alpha$
$\therefore$
$x=\frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}$
$\Rightarrow x=\left(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}}\right)$ and $\left(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}}\right)$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2012
If $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^{2}+b x+c=0$ and $3 b^{2}=16 a c$ then

Options:
A $\alpha=4 \beta$ or $\beta=4 \alpha$
B $\alpha=-4 \beta$ or $\beta=-4 \alpha$
C $\alpha=3 \beta$ or $\beta=3 \alpha$
D $\alpha=-3 \beta$ or $\beta=-3 \alpha$

The correct answer is:
$\alpha=3 \beta$ or $\beta=3 \alpha$

Solution:
Given that, $(\alpha, \beta)$ are roots of the quadratic
equation
and $\begin{aligned} a x^{2}+b x+c &=0 \\ 3 b^{2} &=16 a c \end{aligned}$
$\therefore$
$\alpha+\beta=\frac{-b}{a}$ and $\alpha \beta=\frac{c}{a}$
From Eq. (i), we get $\quad 3 b \cdot b=16 \cdot a \cdot c$
$\Rightarrow \quad 3\left(\frac{b}{a}\right)^{2}=16\left(\frac{c}{a}\right)$ (divide by $\left.a^{2}\right)$
$\Rightarrow \quad 3(\alpha+\beta)^{2}=16 \alpha \beta[$ from Eq. $(i i)]$
$\Rightarrow 3 \alpha^{2}+3 \beta^{2}+6 \alpha \beta=16 \alpha \beta$
$\Rightarrow \quad 3 \alpha^{2}+3 \beta^{2}=10 \alpha \beta$
$\Rightarrow \quad 3\left(\frac{\alpha}{\beta}\right)+3\left(\frac{\beta}{\alpha}\right)=10$
Let $\quad \frac{\alpha}{\beta}=x$
$\begin{array}{lr}\text { Then, } & 3 x+\frac{3}{x}=10 \\ \Rightarrow & 3 x^{2}-10 x+3=0 \\ \Rightarrow & 3 x^{2}-9 x-x+3=0 \\ \Rightarrow & 3 x(x-3)-1(x-3)=0 \\ \Rightarrow & (x-3)(3 x-1)=0 \\ \Rightarrow & x=\frac{1}{3}, 3 \Rightarrow \frac{\alpha}{\beta}=\frac{1}{3}, 3\end{array}$
$\therefore \quad \beta=3 \alpha$ or $\alpha=3 \beta$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2013
If $\alpha$ and $\beta$ are the roots of $x^{2}-p x+1=0$ and $\gamma$ is a root of $x^{2}+p x+1=0,$ then $(\alpha+\gamma)(\beta+\gamma)$ is

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

The correct answer is:
0

Solution:
Since, $\alpha$ and $\beta$ are the roots of $x^{2}-p x+1=0$.
$\therefore$
$\alpha+\beta=p$ and $\alpha \beta=1$
Also, $\gamma$ is the root of $x^{2}+p x+1=0$ $\therefore \quad \gamma^{2}+p{\gamma}+1=0 \Rightarrow \gamma^{2}=-p \gamma-1$
Now, $(\alpha+\gamma)(\beta+\gamma)=\alpha \beta+\alpha \gamma+\beta \gamma+\gamma^{2}$
$=1+\gamma(\alpha+\beta)-p \gamma-1=\gamma(\alpha+\beta-p)$
$=\gamma \times 0=0$
$$
[\because \alpha+\beta=p]
$$
Alternate Method
Since, $\gamma=-\alpha$ or $-\beta$
$\therefore \quad(\alpha+\gamma)(\beta+\gamma)=0$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2015
If angles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in A.P., then $\frac{\mathrm{a}+\mathrm{c}}{\mathrm{b}}$ is equal to

Options:
A $2 \sin \frac{A-C}{2}$
B $2 \cos \frac{A-C}{2}$
C $\cos \frac{\mathrm{A}-\mathrm{C}}{2}$
D $\sin \frac{\mathrm{A}-\mathrm{C}}{2}$

The correct answer is:
$2 \cos \frac{A-C}{2}$

Solution:
Hints: $2 \mathrm{~B}=\mathrm{A}+\mathrm{C}$
$$
=\frac{\sin A+\sin C}{\sin B}=\frac{2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)}{\sin B}=\frac{2 \sin B}{\sin B} \cos \left(\frac{A-C}{2}\right)=2 \cos \left(\frac{A-C}{2}\right)
$$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2010
If aniline is treated with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ and heated at $200^{\circ} \mathrm{C}$, the product is

Options:
A anilinium sulphate
B benzenesulphonic acid
C maminoberzenesuiphonic acid
D sulphanilic acid

The correct answer is:
sulphanilic acid

Solution:
When aniline is treated with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ followed by heating at $200^{\circ} \mathrm{C}$, the product obtained is sulphanilic acid i.e. $p$ -aminobenzene sulphonic acid

Topic: Chemistry, Amines, WBJEE, WBJEE 2018
If $-\pi < \arg (z) < -\frac{\pi}{2}$ then $\arg \bar{z}-\arg (-\bar{z})$ is

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

The correct answer is:
$\pi$

Solution:
Hints:
$$
\begin{aligned}
& \text { if } \arg (z)=-\pi+\theta \\
& \Rightarrow \arg (\bar{z})=\pi-\theta \\
& \arg (-\bar{z})=-\theta \\
& \arg (\bar{z})-\arg (-\bar{z})=\pi-\theta-(-\theta)=\pi-\theta+\theta=\pi
\end{aligned}
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2010
If $\alpha, \beta$ are the roots of $a x^{2}+b x+c=0(a \neq 0)$ and $\alpha+h, \beta+h$ are the roots of $p x^{2}+q x+r=0$ $(p \neq 0),$ then the ratio of the squares of their discriminants is

Options:
A $a^{2}: p^{2}$
B $a: p^{2}$
C $a^{2}: p$
D $a: 2 p$

The correct answer is:
$a^{2}: p^{2}$

Solution:
Given, $\alpha, \beta$ are the roots of $a x^{2}+b x+c=0$ and
$a+h, \beta+h$ are the roots of $p x^{2}+q x+r=0$
$\therefore \quad \alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$
and $\alpha+h+\beta+h=-\frac{q}{p},(\alpha+h)(\beta+h)=\frac{r}{p}$
Now, $\quad(\alpha+h)-(\beta+h)=\alpha-\beta$
$\Rightarrow \quad[(\alpha+h)-(\beta+h)]^{2}=(\alpha-\beta)^{2}$
$\Rightarrow[(\alpha+h)+(\beta+h)]^{2}-4(\alpha+h)(\beta+h)$
$\Rightarrow \quad \frac{q^{2}-4 p r}{p^{2}}=\frac{b^{2}-4 a c}{a^{2}}$
$\therefore$
$\frac{b^{2}-4 a c}{q^{2}-4 p r}=\frac{a^{2}}{p^{2}}$
Hence, the ratio of the square of their discriminants is $a^{2}: p^{2}$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2014
If $\alpha, \beta$ are the roots of the quadratic equation $x^{2}+a x+b=0,(b \neq 0),$ then the quadratic equation whose roots are $\alpha-\frac{1}{\beta}, \beta-\frac{1}{\alpha},$ is

Options:
A $a x^{2}+a(b-1) x+(a-1)^{2}=0$
B $b x^{2}+a(b-1) x+(b-1)^{2}=0$
C $x^{2}+a x+b v=0$
D $a b x^{2}+b x+a=0$

The correct answer is:
$b x^{2}+a(b-1) x+(b-1)^{2}=0$

Solution:
Given equation is, $x^{2}+a x+b=0,(b \neq 0)$
its roots are $\alpha$ and $\beta$. Then, sum of roots $=\alpha+\beta=-a$
Product of roots $=\alpha \cdot \beta=b$
Now,
$$
\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \beta}\right)
$$
$=-a-\frac{(-a)}{b}[$ from Eqs. (0) and $(i i)]$
$=-a+\frac{a}{b}=\frac{a}{b}(1-b)$
and $\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}$
$=b+\frac{1}{b}-2[$ from $E q \cdot(i i)]$
$=\frac{1}{b}\left(b^{2}-2 b+1\right)=\frac{1}{b}(b-1)^{2}$
Required of quadratic equation whose roots are $\left(\alpha-\frac{1}{\beta}\right)$ and $\left(\beta-\frac{1}{\alpha}\right)$ is
$$
\begin{array}{l}
x^{2}-\left\{\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right\} x \\
+\left\{\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)\right\}=0
\end{array}
$$
On putting the values from Eqs. (i) and (ii). we get $x^{2}-\frac{a}{b}(1-b) x+\frac{1}{b}(b-1)^{2}=0$
$\Rightarrow \quad b x^{2}+a(b-1) x+(b-1)^{2}=0, b \neq 0$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2013
If $\alpha, \beta$ are the roots of the quadratic equation $x^{2}+p x+q=0,$ then the values of $\alpha^{3}+\beta^{3}$ and $\alpha^{4}+\alpha^{3} \beta^{2}+\beta^{4}$ are respectively

Options:
A $3 p q-p^{3}$ and $p^{4}-3 p^{2} q+3 q^{2}$
B $-p\left(3 q-p^{2}\right)$ and $\left(p^{2}-q\right)\left(p^{2}+3 q\right)$
C $p q-4$ and $p^{4}-q^{4}$
D $3 p q-p^{3}$ and $\left(p^{2}-q\right)\left(p^{2}-3 q\right)$

The correct answer is:
$3 p q-p^{3}$ and $\left(p^{2}-q\right)\left(p^{2}-3 q\right)$

Solution:
$\because$ Sum of roots, $\alpha+\beta=-p$ and $\alpha \beta=q$ $\therefore \quad\left(a^{3}+\beta^{3}\right)=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$
$$
\begin{aligned}
=(-p)^{3}-3 q(-p) \\
=&-p^{3}+3 p q \\
&=\left(\alpha^{4}+\beta^{4}\right)+(\alpha \beta)^{2} \\
&=\left(\alpha^{2}+\beta^{2}\right)^{2}-(\alpha \beta)^{2} \\
&=\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]^{2}-(\alpha \beta)^{2} \\
&=\left[(-p)^{2}-2 q\right]^{2}-3(q)^{2} \\
&=\left[p^{2}-2 q\right]^{2}-3 q^{2} \\
&=p^{4}-4 p^{2} q+4 q^{2}-q^{2} \\
&=p^{4}-4 p^{2} q+3 q^{2} \\
&=p^{4}-3 p^{2} q-p^{2} q+3 q^{2} \\
&=p^{2}\left(p^{2}-3 q\right)-q\left(p^{2}-3 q\right) \\
&=\left(p^{2}-q\right)\left(p^{2}-3 q\right)
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2014
If $\mathrm{b}=\int_{0}^{1} \frac{\mathrm{e}^{t}}{\mathrm{t}+1} d t$, then $\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{-1}}{\mathrm{t}-\mathrm{a}-1}$ is

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

The correct answer is:
$-\mathrm{be}^{-\mathrm{a}}$

Solution:
$\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} d t$
$=\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{-(2 \mathrm{a}-1-\mathrm{t})}}{2 \mathrm{a}-1-\mathrm{t}-\mathrm{a}-1} \mathrm{dt}$
$=\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{\mathrm{t}-2 \mathrm{a}+1}}{\mathrm{a}-2-\mathrm{t}} \mathrm{dt}$
$t-(a-1)=x \quad \Rightarrow d t=d x$
$=\int_{0}^{1} \frac{e^{x} \cdot e^{-a}}{-x-1} d x$
$=-e^{-a} \int_{0}^{1} \frac{e^{x}}{x+1} d x=-b e^{-a}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2021
If $b_{1} b_{2}=2\left(c_{1}+c_{2}\right)$ and $b_{1}, b_{2}, c_{1}, c_{2}$ are all real numbers, then at least one of the equations $\bar{x}^{2}+h_{1} x+c_{1}=0$ and $x^{2}+b_{2} x+c_{2}=0$ has

Options:
A real roots
B purely imagnary roots
C roots of the form $a+i b(a, b \in R, a b \neq 0)$
D rational roots

The correct answer is:
real roots

Solution:
We have equations
$$
\begin{aligned}
x^{2}+b_{1} x+c_{1} &=0 \\
D_{1} &=b_{1}^{2}-4 c_{1} \\
x^{2}+b_{2} x+c_{2} &=0
\end{aligned}
$$
$$
\begin{array}{c}
D_{2}=b_{2}^{2}-4 c_{2} \\
\text { Now, } D_{1}+D_{2}=b_{1}^{2}+b_{2}^{2}-4\left(c_{1}+c_{2}\right) \\
=b_{1}^{2}+b_{2}^{2}-2 b_{1} b_{2} \quad\left[\because b_{1} b_{2}=2\left(c_{1}+c_{2}\right)\right] \\
=\left(b_{1}-b_{2}\right)^{2} \geq 0
\end{array}
$$
$\Rightarrow$ At least one of $D_{1}$ and $D_{2}$ are non-negative
real roots.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2018
If $\mathrm{B}$ and $\mathrm{B}^{\prime}$ are the ends of minor axis and $\mathrm{S}$ and $\mathrm{S}^{\prime}$ are the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, then the area of the rhombus
$\mathrm{SBS}^{\prime} \mathrm{B}^{\prime}$ will be

Options:
A 12 sq. unit
B 48 sq. unit
C 24 sq. unit
D 36 sq. unit

The correct answer is:
24 sq. unit

Solution:
Hint:
$\mathrm{B}=(0,3), \mathrm{B}^{\prime}=(0,-3)$
$\mathrm{S} \equiv(\mathrm{ae}, 0) \equiv\left(5 \times \sqrt{1-\frac{9}{25}}, 0\right) \equiv(4,0) \quad \therefore$ area $\mathrm{SBS}^{\prime} \mathrm{B}^{\prime}=2 \times$ area $\triangle \mathrm{BSB}^{\prime}=2 \times \frac{1}{2} \times 4 \times 6=24$ sq. unit

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2020
If $\alpha, \beta$ be the roots of $x^2-a(x-1)+b=0$, then the value of $\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}$

Options:
A $\frac{4}{a+b}$
B $\frac{1}{a+b}$
C 0
D -1

The correct answer is:
0

Solution:
$$
\begin{aligned}
& \text { Hints }: x^2-a x=a+3 \quad \alpha \beta=a+b \\
& \alpha+\beta=a \\
& \alpha^2-a \alpha=-(a+b) \\
& \beta^2-a \alpha=-(a+b) \\
&-\frac{1}{a+b}-\frac{1}{a+b}+\frac{2}{a+b}=0
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2009
If $\mathrm{c}_{0}, \mathrm{c}_{1}, \mathrm{c}_{2}, \ldots \ldots \mathrm{c}{15}$ are the Binomial co-efficients in the expansion of $(1+\mathrm{x})^{15}$, then the value of $\frac{\mathrm{C}{1}}{\mathrm{c}{0}}+2 \frac{\mathrm{C}{2}}{\mathrm{c}{1}}+3 \frac{\mathrm{c}{3}}{\mathrm{c}{2}}+\ldots \ldots+15 \frac{\mathrm{c}{15}}{\mathrm{c}{14}}$ is

Options:
A 1240
B 120
C 124
D 140

The correct answer is:
120

Solution:
Hint:
$S_{n}=\sum_{r=1}^{15} r \frac{{ }^{15} C_{r}}{{ }^{15} c_{r-1}}=\sum_{r=1}^{15}(15-r+1)=16 \times 15-\frac{15 \times 16}{2}=120$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2020
If $\mathrm{c}_0, \mathrm{c}_1, \mathrm{c}_2, \ldots \ldots \ldots \ldots \ldots . \mathrm{c}_{\mathrm{n}}$ denote the co-efficients in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$ then the value of $\mathrm{c}_1+2 \mathrm{c}_2+3 \mathrm{c}_3+\ldots . .+\mathrm{nc}_{\mathrm{n}}$ is

Options:
A $\mathrm{n} \cdot 2^{\mathrm{n}-1}$
B $(\mathrm{n}+1) 2^{\mathrm{n}-1}$
C $(n+1) 2^n$
D $(n+2) 2^{n-1}$

The correct answer is:
$\mathrm{n} \cdot 2^{\mathrm{n}-1}$

Solution:
Hints : $(1+\mathrm{x})^{\mathrm{n}}=\dot{c}_0+\mathrm{xc}_1+\mathrm{x}^2 \mathrm{c}_2+\ldots \ldots . \mathrm{x}^{\mathrm{n}} \mathrm{c}_{\mathrm{n}}$
$n(1+x)^{n-1}=c_1+2 x c_2+\ldots \ldots . . n x^{n-1} c_n$
Put $x=1$
$$
\mathrm{n}(2)^{\mathrm{n}-1}=\mathrm{c}_1+2 \mathrm{c}_2+3 \mathrm{c}_2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\mathrm{nc}_{\mathrm{n}}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2009
If $C$ is a point on the line segment joining $A(-3,4)$ and $B(2,1)$ such that $A C=2 B C$, then the coordinate of $C$ is

Options:
A $\left(\frac{1}{3}, 2\right)$
B $\left(2, \frac{1}{3}\right)$
C $(2,7)$
D $(7,2)$

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

Solution:
Hints :
$$
\begin{aligned}
& C\left(\frac{4-3}{3}, \frac{2+4}{3}\right) \\
& C\left(\frac{1}{3}, 2\right)
\end{aligned}
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2009
If $\mathrm{C}$ is the reflecton of $\mathrm{A}(2,4)$ in $\mathrm{x}$-axis and $\mathrm{B}$ is the reflection of $\mathrm{C}$ in $\mathrm{y}$-axis, then $|\mathrm{AB}|$ is

Options:
A 20
B $2 \sqrt{5}$
C $4 \sqrt{5}$
D 4

The correct answer is:
$4 \sqrt{5}$

Solution:
$$
\begin{aligned}
& \text { Hints: } \mathrm{A} \equiv(2,4) ; \mathrm{C} \equiv(2,-4) ; \mathrm{B} \equiv(-2,-4) \\
& |A B|=\sqrt{(2-(-2))^2+(4-(-4))^2}=\sqrt{4^2+8^2} \\
& =\sqrt{16+64}=\sqrt{80}=\sqrt{16 \times 5}=4 \sqrt{5}
\end{aligned}
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2009
If $\mathrm{Cl}_{2}$ is passed through hot aqueous $\mathrm{NaOH}$, the products formed have Cl in different oxidation states. These are indicated as

Options:
A -1 and +1
B -1 and +5
C 1 and 5
D -1 and +3

The correct answer is:
-1 and +5

Solution:
If $\mathrm{Cl}_{2}$ is passed through hot aqueous $\mathrm{NaOH}$, it results in the formation of sodium chloride and sodium chlorate as
$$
6 \mathrm{NaOH}+3 \mathrm{Cl}_{2} \longrightarrow 5 \mathrm{NaCl}+\mathrm{NaClO}_{3}+3 \mathrm{H}_{2} \mathrm{O}
$$
In $\mathrm{NaCl}$, let oxidation state of $\mathrm{Cl}$ be $\mathrm{x}$
$$
+1+x=0 \text { or } \quad x=-1
$$
In $\operatorname{NaClO}_{3}$, let oxidation state of $\mathrm{Cl}$ be $y$
$+1+y+(-2) 3=0$
$y-5=0$
$\therefore y=+5$
Thus, the obtained compounds have Cl in -1 and +5 oxidation states.

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2014
If $\cos x$ and $\sin x$ are solutions of the differential equation $a_{0} \frac{d^{2} y}{d x^{2}}+a_{1} \frac{d y}{d x}+a_{2} y=0$ where $a_{0}, a_{1}$ and $a_{2}$ are real constants, then which of the following is/are always true?

Options:
A $A \cos x+B \sin x$ is a solution, where $A$ and $B$ are feal constants
B $A \cos \left(x+\frac{\pi}{4}\right)$ is a solution, where $A$ is a real constant
C $A \cos x \sin x$ is a solution, where $A$ is a real constant
D $A \cos \left(x+\frac{\pi}{4}\right)+B \sin \left(x -\frac{\pi}{4}\right)$ is a solution, where A and $B$ are real constants

The correct answers are:
$A \cos x+B \sin x$ is a solution, where $A$ and $B$ are feal constants, $A \cos \left(x+\frac{\pi}{4}\right)$ is a solution, where $A$ is a real constant, $A \cos \left(x+\frac{\pi}{4}\right)+B \sin \left(x -\frac{\pi}{4}\right)$ is a solution, where A and $B$ are real constants

Solution:
Let $f(x)=\cos x$ and $g(x)=\sin x$ Consider the Wronskian of $f(x)$ and $g(x)$.
$$
\begin{aligned}
W &=\left|\begin{array}{ll}
f(x) & g(x) \\
f^{\prime}(x) & g^{\prime}(x)
\end{array}\right| \\
&=\left|\begin{array}{ll}
\cos x & \sin x \\
-\sin x & \cos x
\end{array}\right| \\
&=\cos ^{2} x+\sin ^{2} x=1 \neq 0
\end{aligned}
$$
Thus, the functions are linearly independent. So, the general solution of given differential equation is given by $y=A \cos x+B \sin x,$ where $A$ and $B$ are real constants.
[$\because$ if $y_{1}$ and $y_{2}$ are linearly independent solutions of the differential equation $a y^{\prime \prime}+b y^{\prime}+c=0,$ then the general solution is $y=c_{1} y_{1}+c_{2} y_{2},$ where $c_{1}$ and $c_{2}$ are constants] Hence, option (a) is true.
(b) Let $y=A \cos \left(x+\frac{\pi}{4}\right)$
$$
\begin{array}{l}
=A\left(\cos x \cdot \cos \frac{\pi}{4}-\sin x \cdot \sin \frac{\pi}{4}\right) \\
=\frac{A}{\sqrt{2}}(\cos x-\sin x) \\
=\frac{A}{\sqrt{2}} \cos x+\left(-\frac{A}{\sqrt{2}}\right) \sin x
\end{array}
$$
which is in the form of general solution.
Hence, option (b) is true.
(c) Let $y=A \cos x \sin x$
which cannot be
expressed in the form of general solution.
(d) Let $y=A \cos \left(x+\frac{\pi}{4}\right)+B \sin \left(x-\frac{\pi}{4}\right)$
$$
\begin{array}{l}
=A\left(\cos x \cdot \frac{1}{\sqrt{2}}-\sin x \cdot \frac{1}{\sqrt{2}}\right) \\
\quad+B\left(\sin x \cdot \frac{1}{\sqrt{2}}-\cos x \cdot \frac{1}{\sqrt{2}}\right) \\
=\cos x\left(\frac{A}{\sqrt{2}}-\frac{B}{\sqrt{2}}\right)+\sin x\left(\frac{B}{\sqrt{2}}-\frac{A}{\sqrt{2}}\right) \text { which is }
\end{array}
$$
in the form of general solution. Hence, option (d) is true.

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2015
If $\int \cos x \log \left(\tan \frac{x}{2}\right) d x$$=\sin x \log \left(\tan \frac{x}{2}\right)+f(x),$ then $f(x)$ is equal to (assuming $c$ is a arbitrary real constant)

Options:
A $c$
B $c-x$
C $c+x$
D $2 x+c$

The correct answer is:
$c-x$

Solution:
Let $I=\int \cos x \log \left(\tan \frac{x}{2}\right)$
$=\log \left(\tan \frac{x}{2}\right) \cdot \sin x-\int \sin x \cdot \frac{1}{\tan \frac{x}{2}} \cdot \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} d x$
$=\sin x \log \left(\tan \frac{x}{2}\right)-\int \sin x \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} d x$
$=\sin x \cdot \log \left(\tan \frac{x}{2}\right)-\int \frac{\sin x}{\sin x} d x$
$=\sin x \cdot \log \left(\tan \frac{x}{2}\right)-\int 1 d x$
$=\sin x \log \left(\tan \frac{x}{2}\right)-x+c$
$\therefore f(x)=c-x$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2019
If $\left(\cot \alpha_1\right)\left(\cot \alpha_2\right) \ldots \ldots .\left(\cot \alpha_n\right)=1,0 < \alpha_1, \alpha_2, \ldots . . \alpha_n < \pi / 2$, then the maximum value of $\left(\cos a_1\right)\left(\cos \alpha_2\right) \ldots \ldots$ $\left(\cos \alpha_n\right)$ is given by

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

The correct answer is:
$\frac{1}{2^{n / 2}}$

Solution:
$\frac{1}{\left(\cos \alpha_1 \cos \alpha_2 \ldots . \cos \alpha_n\right)^2}$
$\begin{aligned}
&=\left(1+\tan ^2 \alpha_1\right)\left(1+\tan ^2 \alpha_2\right) \ldots \ldots\left(1+\tan ^2 \alpha_n\right) \\
&\geq\left(2 \tan \alpha_1\right)\left(2 \tan \alpha_2\right) \ldots \ldots .\left(2 \tan \alpha_n\right)[\text { AM }-\text { GM inequality } \\
&=2^n \times \frac{1}{\cot \alpha_1 \cot \alpha_2 \ldots . . \cot \alpha_n}=2^n \\
&\therefore \cos \alpha_1 \cos \alpha_2 \ldots . . \cos \alpha_n \leq \frac{1}{2^{n / 2}}
\end{aligned}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2022
If $\cot \frac{2 x}{3}+\tan \frac{x}{3}=\operatorname{cosec} \frac{k x}{3},$ then the value of $k$ is

Options:
A 1
B 2
C 3
D -1

The correct answer is:
2

Solution:
Given, $\cot \frac{2 x}{3}+\tan \frac{x}{3}=\operatorname{cosec} \frac{k x}{3}$
Let $\theta=\frac{x}{3},$ then
$\therefore \quad \cot 2 \theta+\tan \theta=\operatorname{cosec} k \theta$
$\Rightarrow \quad \frac{\cos 2 \theta}{\sin 2 \theta}+\frac{\sin \theta}{\cos \theta}=\operatorname{cosec} k \theta$
$\Rightarrow \frac{2 \cos ^{2} \theta-1}{2 \sin \theta \cos \theta}+\frac{\sin \theta}{\cos \theta}=\operatorname{cosec} k \theta$
$\Rightarrow \frac{2 \cos ^{2} \theta-1+2 \sin ^{2} \theta}{2 \sin \theta \cos \theta}=\operatorname{cosec} k \theta$
$\Rightarrow \quad \frac{1}{2 \sin \theta \cos \theta}=\operatorname{cosec} k \theta$
$\Rightarrow \quad \operatorname{cosec} 2 \theta=\operatorname{cosec} k \theta$
$\Rightarrow \quad k=2$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2015
If $\frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{f}(\mathrm{x})\}=\mathrm{g}(\mathrm{x})$, then $\int_a^b \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx}$ is equal to

Options:
A $\frac{1}{2}\left[\mathrm{f}^2(\mathrm{~b})-\mathrm{f}^2(\mathrm{a})\right]$
B $\frac{1}{2}\left[\mathrm{~g}^2(\mathrm{~b})-\mathrm{g}^2(\mathrm{a})\right]$
C $f(b)-f(a)$
D $\frac{1}{2}\left[\mathrm{f}\left(\mathrm{b}^2\right)-\mathrm{f}\left(\mathrm{a}^2\right)\right]$

The correct answer is:
$\frac{1}{2}\left[\mathrm{f}^2(\mathrm{~b})-\mathrm{f}^2(\mathrm{a})\right]$

Solution:
Hints $: f(x)=\int g(x) d x$
$$
\begin{aligned}
& \int_a^b f(x) \cdot g(x) \cdot d x=(f(x) f(x))_a^b-\int_a^b g(x) f(x) d x \\
& I=f^2(b)-f^n(a)^{-1} \\
& I=\frac{1}{2}\left(f^2(b)-f^2(a)\right)
\end{aligned}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2010
If $E$ and $B$ are the magnitudes of electric and magnetic fields respectively in some region of space, then the possibilities for which a charged particle may move in that space with a uniform velocity of magnitude $v$ are

Options:
A $E=v B$
B $E \neq 0, B=0$
C $E=0, B \neq 0$
D $E \neq 0, B \neq 0$

The correct answers are:
$E=v B$, $E=0, B \neq 0$, $E \neq 0, B \neq 0$

Solution:
The charged particle will move with uniform velocity in that space if $E=0$ and $B \neq 0,$ since a moving charge in a magnetic field experience a force. Also if $E = 0, B \neq 0,$ the charged particle will move with uniform velocity in that space.
If $E=v B$, then particle again moves with uniform velocity.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2013
If $\int e^{\sin x} \cdot\left[\frac{x \cos ^{3} x-\sin x}{\cos ^{2} x}\right] d x=e^{\sin x} f(x)+c$
where c is constant of integration, then $f(x)$ is
equal to

Options:
A $\sec x-x$
B $x-\sec x$
C $\tan x-x$
D $x-\tan x$

The correct answer is:
$x-\sec x$

Solution:
We have. $\int e^{\operatorname{cin} x}\left(\frac{x \cos ^{3} x-\sin x}{\cos ^{2} x}\right) d x=e^{\sin x} f(x)+c$
$\int e^{\operatorname{sin} x}(x \cos x-\sec x \tan x) d x=e^{\sin x} f(x)+c$
$\int e^{\operatorname{sin} x}(x \cos x-1+1-\sec x \tan x) d x$
$=e^{\sin x} f(x)+c$
$\left.\int[ e^{\sin x} \cos x(x-\sec x)+e^{\sin x}(1-\sec x \tan x)\right] d x$
$=e^{\sin x} f(x)+c$
$\int \frac{d}{d x}\left\{e^{\sin x}(x-\sec x)\right\} d x=e^{\sin x} f(x)+c$
$e^{\operatorname{cin} x}(x-\sec x)=e^{\sin x} f(x)+c$
$f(x)=x-\sec x$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2018