datavorous/entrance-exam-dataset huggingface dataset:
If $P, Q$ and $R$ are angles of $\Delta P Q R$, then the value of $\left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$ is equal to

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

The correct answer is:
0

Solution:
$$
\begin{aligned}
&\text { Multiplying } C_{1} \text { by } p \text { and then doing }\\
&\begin{aligned}
C_{1} & \rightarrow C_{1}+q C_{2}+r C_{3} \text { , we get } \\
\Delta &=\frac{1}{p}\left[\begin{array}{ccc}
-p & \cos R & \cos Q \\
p \cos R & -1 & \cos P \\
p \cos Q & \cos P & -1
\end{array} \mid\right.\\
&=\frac{1}{p}\left|\begin{array}{ccc}
-p+q \cos R+r \cos Q & \cos R & \cos Q \\
p \cos R-q+r \cos P & -1 & \cos P \\
p \cos Q+q \cos P-r & \cos P & -1
\end{array}\right| \\
=\frac{1}{p}\left|\begin{array}{ccc}
0 & \cos R & \cos Q \\
0 & -1 & \cos P \\
0 & \cos P & -1
\end{array}\right|=0
\end{aligned}
\end{aligned}
$$
Using the prejection formula $p=q \cos R+r \cos Q$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2012
If $p, q$ are odd integers, then the roots of the equation $2 p x^{2}+(2 p+q) x+q=0$ are

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

The correct answer is:
rational

Solution:
Given equation.
$$
\begin{array}{l}
2 p x^{2}+(2 p+q) x+q=0 \\
\therefore D=(2 p+q)^{2}-4(2 p)(q) \\
\quad=4 p^{2}+q^{2}+4 p q-8 p q \\
\quad=4 p^{2}+q^{2}-4 p q \\
\quad=(2 p-q)^{2}
\end{array}
$$
$=$ a perfect square
Given equation has rational roots

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2017
If $p, q$ are the roots of the equation $x^{2}+p x+q=0,$ then

Options:
A $p=1, q=-2$
B $p=0, q=1$
C $p=-2, q=0$
D $p=-2, q=1$

The correct answer is:
$p=1, q=-2$

Solution:
Given, $\rho, q$ are roots of equation $x^{2}+p x+q=0$
$\therefore \quad p+q=-p \Rightarrow 2 p+q=0$
$\begin{array}{lr}\text { and } & p q=q \\ \Rightarrow & \rho=1\end{array}$
Put $p=1$ in Eq, we get $2 \times 1+q=0$
$\begin{array}{lr}\Rightarrow & q+2=0 \\ \Rightarrow & q=-2 \\ \text { Hence, } p=1, q=-2\end{array}$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2016
If $\mathrm{P}(\mathrm{x})=a \mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$ and $\mathrm{Q}(\mathrm{x})=-\mathrm{ax}^{2}+\mathrm{dx}+\mathrm{c}$, where $\mathrm{ac} \neq 0 \quad[\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$ are all real], then $\mathrm{P}(\mathrm{x}) \cdot \mathrm{Q}(\mathrm{x})=0$ has

Options:
A at least two real roots
B two real roots
C four real roots
D no real root

The correct answer is:
at least two real roots

Solution:
Hint:
If $P(x)=a x^{2}+b x+c, Q(x)=-a x^{2}+d x+c$
$\mathrm{D}_{1}=\mathrm{b}^{2}-4 \mathrm{ac}$
$\mathrm{D}_{2}=\mathrm{d}^{2}+4 \mathrm{ac}$
$\Rightarrow \mathrm{D}_{1}+\mathrm{D}_{2}>0$
Atleast two real roots.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2020
If $P Q$ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that $\Delta O P Q$ is equilateral. $O$ being the centre. Then, the eccentricity $e$ satisfies

Options:
A $1 < e < \frac{2}{\sqrt{3}}$
B $e=\frac{2}{\sqrt{2}}$
C $e=\frac{\sqrt{3}}{2}$
D $e>\frac{2}{\sqrt{3}}$

The correct answer is:
$e>\frac{2}{\sqrt{3}}$

Solution:
Given equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $\Delta O P Q$ is an equlateral triangle. $P Q$ is double ordinate of the hyperbola.

Let the coordinates of $P$ and $Q$ be $(a \sec \theta, b$ tan $\theta)$ and $(a \sec \theta,-b$ tane $),$ respectively.
In$\Delta O P Q$,
$O P=P Q$
$\Rightarrow \quad a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta=(2 b \tan \theta)^{2}$
$\Rightarrow \quad a^{2} \sec ^{2} \theta=3 b^{2} \tan ^{2} \theta$
$\Rightarrow \quad \sin ^{2} \theta=\frac{a^{2}}{3 b^{2}}$
Now, $\sin ^{2} \theta < 1$
$\Rightarrow \quad \frac{a^{2}}{3 b^{2}} < 1$
$\frac{b^{2}}{a^{2}}>\frac{1}{3}$
$\Rightarrow \quad 1+\frac{b^{2}}{a^{2}}>\frac{4}{3}$
$\therefore$
$e^{2}>\frac{4}{3} \Rightarrow e>\frac{2}{\sqrt{3}}$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2016
If pressure of real gas $\mathrm{O}_{2}$ in a container is given by $\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$, then the mass of the gas in the container is

Options:
A $32 \mathrm{gm}$
B $16 \mathrm{gm}$
C $4 \mathrm{gm}$
D $64 \mathrm{gm}$

The correct answer is:
$16 \mathrm{gm}$

Solution:
$\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$
$\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}$
$2\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]\left[\mathrm{V}-\frac{\mathrm{b}}{2}\right]=\mathrm{RT}$
$\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]\left[\mathrm{V}-\frac{\mathrm{b}}{2}\right]=\frac{\mathrm{RT}}{2}$
$\therefore \mathrm{n}=1 / 2 \mathrm{~mole}$
$\therefore$ Mass of $\mathrm{O}_{2}=\frac{1}{2} \times 32=16 \mathrm{gm}$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2021
If $\theta \in \mathbb{R}$ and $\frac{1-i \cos \theta}{1+2 i \cos \theta}$ is real number, then $\theta$ will be (when $I:$ Set of integers)

Options:
A $(2 n+1) \frac{\pi}{2}, n \in l$
B $\frac{3 n \pi}{2}, n \in I$
C $n \pi, n \in I$
D $2 n \pi, n \in l$

The correct answers are:
$(2 n+1) \frac{\pi}{2}, n \in l$

Solution:
Let $Z=\frac{1-i \cos \theta}{1+2 i \cos \theta}$
$\therefore \quad \bar{z}=\frac{1+i \cos \theta}{1-2 i \cos \theta}$
Since, $z$ is a real number, then $z-\bar{z}=0$ $\Rightarrow \frac{1-i \cos \theta}{1+2 i \cos \theta}=\frac{1+i \cos \theta}{1-2 i \cos \theta}$
$\Rightarrow(1-i \cos \theta)(1-2 i \cos \theta)=(1+i \cos \theta)(1+2 i \cos \theta)$
$\Rightarrow 1-2 i \cos \theta-i \cos \theta-2 \cos ^{2} \theta$ $=1+2 i \cos \theta+i \cos \theta-2 \cos ^{2} \theta$
$\Rightarrow \quad 6 i \cos \theta=0$
$\Rightarrow \quad \cos \theta=0$
$\Rightarrow \quad \theta=(2 n+1) \frac{\pi}{2}, n \in I$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2019
If \(R\) and \(\mathrm{R}^1\) are equivalence relations on a set \(A\), then so are the relation

Options:
A \(\mathrm{R}^{-1}\)
B \(\mathrm{R} \cup \mathrm{R}^1\)
C \(\mathrm{R} \cap \mathrm{R}^1\)
D All of these

The correct answers are:
\(\mathrm{R}^{-1}\), \(\mathrm{R} \cap \mathrm{R}^1\)

Solution:
Hint: If \(R ~\&~ R^1\) are equivalence relation
\(\mathrm{R}^{-1} \Rightarrow\) Equivalence
\(\mathrm{R} \cap \mathrm{R}^1 \Rightarrow\) Equivalence
\(R \cup R^1\) may or may not equivalence

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2023
If $R$ be the set of all real numbers and $f: R \rightarrow R$ is given by $f(x)=3 x^{2}+1$. Then, the set $f^{-1}([1,6])$ is

Options:
A $\left\{-\sqrt{\frac{5}{3}}, 0, \sqrt{\frac{5}{3}}\right\}$
B $\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]$
C $\left[-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right]$
D $\left(-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right)$

The correct answer is:
$\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]$

Solution:
Given, $f(x)=3 x^{2}+1$
Let $\quad y=3 x^{2}+1$
$\Rightarrow \quad 3 x^{2}=y-1 \Rightarrow x^{2}=\frac{y-1}{3}$
$\Rightarrow \quad x=\pm \sqrt{\frac{y-1}{3}}$
$\therefore \quad f^{-1}(x)=\pm \sqrt{\frac{x-1}{3}}$
When $x \in[1,6]$
Then, $\quad f^{-1}(x) \in\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]$

Topic: Mathematics, Functions, WBJEE, WBJEE 2014
If $\mathrm{R}$ is the Rydberg Constant in $\mathrm{cm}^{-1}$, then hydrogen atom does not emit any radiation of wave-length in the range of

Options:
A $\frac{1}{\mathrm{R}}$ to $\frac{4}{3 \mathrm{R}} \mathrm{cm}$
B $\frac{7}{5 \mathrm{R}}$ to $\frac{19}{5 \mathrm{R}} \mathrm{cm}$
C $\frac{4}{\mathrm{R}}$ to $\frac{36}{5 \mathrm{R}} \mathrm{cm}$
D $\frac{9}{\mathrm{R}}$ to $\frac{144}{7 \mathrm{R}} \mathrm{cm}$

The correct answer is:
$\frac{7}{5 \mathrm{R}}$ to $\frac{19}{5 \mathrm{R}} \mathrm{cm}$

Solution:
Hint:
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}^{2}}}-\frac{1}{\mathrm{n}_{\mathrm{i}^{2}}}\right]$
For range of wavelengths:$\mathrm{n}_{\mathrm{i}}=1,2,3, \ldots . .$ for Lyman, Balmer, Paschen, $\ldots \ldots$
$\mathrm{n}_{\mathrm{f}}=\mathrm{n}_{\mathrm{i}}+1$ and $\mathrm{n}_{\mathrm{f}}=\infty$ for upper and lower range
Thus, Lyman : $\left[\frac{1}{\mathrm{R}}\right.$ to $\left.\frac{4}{3 \mathrm{R}}\right]$, Balmer : $\left[\frac{4}{\mathrm{R}}\right.$ to $\left.\frac{36}{5 \mathrm{R}}\right]$, Paschen : $\left[\frac{9}{\mathrm{R}} \mathrm{to} \frac{144}{36 \mathrm{R}}\right]$, Bracket : $\left[\frac{16}{\mathrm{R}}\right.$ to $\left.\frac{400}{9 \mathrm{R}}\right]$, Pfund : $\left[\frac{25}{\mathrm{R}}\right.$ to $\left.\frac{900}{11 \mathrm{R}}\right]$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2020
If radium and chlorine combine to form radium chloride, the compound would be

Options:
A half as radioactive as radium
B twice as radioactive
C as radioactive as radium
D not radioactive

The correct answer is:
as radioactive as radium

Solution:
Radium in isolated form and in its chloride (i.e. $Ra^{+}$ ) only differ in the number of orbital electron. But, radioactivity is independent of chemical environment of an ion or atom, i.e. radiouctivity is the phenomenon which does not depend on the orbital electrons but depends only on the composition of nucleus

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2016
If $S_{r}=\left|\begin{array}{ccc}2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1)\end{array}\right|$, then the
value of $\sum_{r=1}^{n} S_{r}$ is independent of

Options:
A only $x$
B only $y$
C only $n$
D $x y, z$ and $n$

The correct answer is:
$x y, z$ and $n$

Solution:
We have, $S_{r}=\left|\begin{array}{ccc}2 r & x & n(n+1) \\ 6 r^{2}-1 & y & n^{2}(2 n+3) \\ 4 r^{3}-2 n r & z & n^{3}(n+1)\end{array}\right|$
$\Rightarrow \quad \sum_{r=1}^{n} S_{r}=\left|\begin{array}{ccc}2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n}\left(6 r^{2}-1\right) & y & n^{2}(2 n+3) \\ \sum_{r=1}^{n}\left(4 r^{3}-2 n r\right) & z & n^{3}(n+1)\end{array}\right|$
$=\left|\begin{array}{ccc}n(n+1) & x & n(n+1) \\ n^{2}(2 n+3) & y & n^{2}(2 n+3) \\ n^{3}(n+1) & z & n^{3}(n+1)\end{array}\right|$
$=0 \quad\left(\because C_{1}\right.$ and $C_{3}$ are identical $)$
Hence, $\sum_{r=1}^{n} S_{r}$ is independent of $x, y, z$ and $n$.

Topic: Mathematics, Determinants, WBJEE, WBJEE 2018
If $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2},$ then the value of $x$ is

Options:
A 5
B 4
C 12
D 11

The correct answer is:
5

Solution:
Given,
$\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2} ..(i)$
Let $\operatorname{cosec}^{-1} \frac{13}{12}=y$
Then. $\operatorname{cosec} y=\frac{13}{12} \Rightarrow \sin y=\frac{12}{13}$
$\therefore \cos y=\sqrt{1-\sin ^{2} y}$
$\begin{array}{l}
=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{1-\frac{144}{169}} \\
=\sqrt{\frac{25}{169}}=\frac{5}{13} \Rightarrow y=\cos ^{-1} \frac{5}{13}
\end{array}$
Eq. (i) becomes, $\sin ^{-1}\left(\frac{x}{13}\right)+\cos ^{-1}\left(\frac{5}{13}\right)=\frac{\pi}{2}$
We know that
$\sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2}$
$\therefore$ Both angles of Eq. (i) should be same.
$\therefore x=5$

Topic: Mathematics, Inverse Trigonometric Functions, WBJEE, WBJEE 2014
If $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\frac{x^{4}}{8}+\ldots\right)=\frac{\pi}{6},$ where $|x| < 2$ then the value of $x$ is

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

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

Solution:
We have, $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\frac{x^{4}}{8}+\ldots\right)=\frac{\pi}{6}$
$\Rightarrow \sin ^{-1}\left(\frac{x}{1-\left(\frac{-x}{2}\right)}\right)=\frac{\pi}{6} \quad\left[\because S_{\infty}=\frac{a}{1-r}\right]$
$\Rightarrow \quad \sin ^{-1}\left(\frac{2 x}{2+x}\right)=\frac{\pi}{6}$
$\Rightarrow \frac{2 x}{2+x}=\sin \frac{\pi}{6} \Rightarrow \frac{2 x}{2+x}=\frac{1}{2}$
$\Rightarrow \quad 4 x=2+x \quad \Rightarrow \quad 3 x=2 \quad \Rightarrow \quad x=\frac{2}{3}$

Topic: Mathematics, Inverse Trigonometric Functions, WBJEE, WBJEE 2015
If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2},$ then the value of $x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}}$ is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
We know that $\left|\sin ^{-1} x\right| \leq \frac{\pi}{2}$
Hence, from the given relation we observe that each of $\sin ^{-1} x, \sin ^{-1} y$ and $\sin ^{-1} x$ will be $\frac{\pi}{2}$
so that $x=y=x=\sin \frac{\pi}{2}=1$
$\begin{aligned} \therefore \quad & x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}} \\ &=(1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}} \\ &=1+1+1-\frac{1}{1 \times 1 \times 1} \\ &=3-1=2 \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, WBJEE, WBJEE 2012
If $\sin ^{2} \theta+3 \cos \theta=2,$ then $\cos ^{3} \theta+\sec ^{3} \theta$ is equal out to

Options:
A 1
B 4
C 9
D 18

The correct answer is:
18

Solution:
Given, $\sin ^{2} \theta+3 \cos \theta=2$
$\begin{array}{lr}\Rightarrow & 1-\cos ^{2} \theta+3 \cos \theta=2 \\ \Rightarrow & \cos ^{2} \theta-3 \cos \theta+1=0\end{array}$
$\Rightarrow \quad \cos \theta=\frac{3 \pm \sqrt{9-4}}{2}$ (by quadratic formula)
$\Rightarrow \quad \cos \theta=\frac{3 \pm \sqrt{5}}{2}$
Here, $\cos \theta=\frac{3-\sqrt{5}}{2}$
$\left(\because \frac{3+\sqrt{5}}{2} > 1 \text{ and } -1 \leq \cos \theta \leq 1\right)$
and $\sec \theta=\frac{2}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3-\sqrt{5}}$
$=\frac{2(3+\sqrt{5})}{4}=\frac{3+\sqrt{5}}{2}$
Now, $\cos ^{3} \theta+\sec ^{3} \theta=(\cos \theta+\sec \theta)$
$\quad\left(\cos ^{2} \theta+\sec ^{2} \theta-\cos \theta \cdot \sec \theta\right)$
$\begin{array}{l}
=\left(\frac{3-\sqrt{5}+3+\sqrt{5}}{2}\right) \\
\quad \quad\left\{(\cos \theta+\sec \theta)^{2}-3 \cos \theta \cdot \sec \theta\right\} \\
=3 \cdot\left\{(3)^{2}-3\right\}=3(9-3)=3 \times 6=18
\end{array}$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2013
If $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$, then general
value of $\theta$ is

Options:
A $\frac{n \pi}{4}, n \pi \pm \frac{\pi}{3}$
B $\frac{n \pi}{4}, n \pi \pm \frac{\pi}{6}$
C $\frac{n \pi}{4}, 2 n \pi \pm \frac{\pi}{3}$
D $\frac{n \pi}{4}, 2 n \pi \pm \frac{\pi}{6}$

The correct answer is:
$\frac{n \pi}{4}, n \pi \pm \frac{\pi}{3}$

Solution:
We have, $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$
$\Rightarrow \quad \sin 6 \theta+\sin 2 \theta+\sin 4 \theta=0$
$\Rightarrow \quad 2 \sin 4 \theta \cdot \cos 2 \theta+\sin 4 \theta=0$
$\Rightarrow \quad \sin 4 \theta(2 \cos 2 \theta+1)=0$
$\Rightarrow \quad \sin 4 \theta=0$ or $2 \cos 2 \theta+1=0$
$\Rightarrow \quad 4 \theta=n \pi$ or $\cos 2 \theta=\frac{-1}{2}=\cos \frac{2 \pi}{3}$
$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $2 \theta=2 n \pi \pm \frac{2 \pi}{3}$
$\Rightarrow \quad \theta=\frac{n \pi}{4}$ or $\theta=n \pi \pm \frac{\pi}{3},$ where $n \in z$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2018
If $\sin 6 \theta+\sin 4 \theta+\sin 2 \theta=0$ then the general value of $\theta$ is

Options:
A $\frac{\mathrm{n} \pi}{4}, \mathrm{n} \pi \pm \frac{\pi}{3}$
B $\frac{\mathrm{n} \pi}{4}, \mathrm{n} \pi \pm \frac{\pi}{6}$
C $\frac{\mathrm{n} \pi}{4}, 2 \mathrm{n} \pi \pm \frac{\pi}{3}$
D $\frac{\mathrm{n} \pi}{4}, 2 \mathrm{n} \pi \pm \frac{\pi}{6}$

The correct answer is:
$\frac{\mathrm{n} \pi}{4}, \mathrm{n} \pi \pm \frac{\pi}{3}$

Solution:
Hints : $2 \sin 4 \theta \cos 2 \theta+\sin 4 \theta=0$
$\sin 4 \theta=0$
$2 \cos 2 \theta=-1$
$4 \theta=n \pi$
$\cos 2 \theta=-\frac{1}{2}=\cos \frac{2 \pi}{3}$
$\theta=\frac{n \pi}{4}$
$2 \theta=2 \mathrm{n} \pi \pm \frac{2 \pi}{3}, \Rightarrow \theta=\mathrm{n} \pi \pm \frac{\pi}{3}$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2010
If $\sin \alpha, \cos \alpha$ be the roots of the equation $x^{2}-b x+c=0 .$ Then, which of the following statements is/are correct?

Options:
A $c \leq \frac{1}{2}$
B $b \leq \sqrt{2}$
C $c>\frac{1}{2}$
D $b>\sqrt{2}$

The correct answers are:
$c \leq \frac{1}{2}$, $b \leq \sqrt{2}$

Solution:
Given equation is $x^{2}-b x+c=0$
and roots are $\sin \alpha$ and $\cos \alpha$
Also $\sin \alpha+\cos \alpha=b$
and $\sin \alpha \cdot \cos \alpha=c$
$$
\begin{array}{r}
\because \sin ^{2} \alpha+\cos ^{2} \alpha=(\sin \alpha+\cos \alpha)^{2} \\
-2 \sin \alpha \cdot \cos \alpha
\end{array}
$$
$\Rightarrow \quad 1=b^{2}-2 c$
[from Eqs. (ii) and (iii)]
$-\sqrt{1+1} \leq(\sin \alpha+\cos \alpha) \leq \sqrt{1+1}$
$\Rightarrow \quad-\sqrt{2} \leq b \leq \sqrt{2}$
and $2 \sin \alpha ; \cos \alpha=x 2 c$
[from Eq. (iii)] $\begin{array}{ll}\Rightarrow & \sin 2 \alpha=2 c(\because-1 \leq \sin 2 \alpha \leq 1) \\ \Rightarrow & 2 c \leq 1 \Rightarrow c \leq \frac{1}{2}\end{array}$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2013
If \(\sin \theta\) and \(\cos \theta\) are the roots of the equation \(a x^2-b x+c=0\), then \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) satisfy the relation

Options:
A \(a^2+b^2+2 a c=0\)
B \(\mathrm{a}^2-\mathrm{b}^2+2 \mathrm{ac}=0\)
C \(\mathrm{a}^2+\mathrm{c}^2+2 \mathrm{ab}=0\)
D \(\mathrm{a}^2-\mathrm{b}^2-2 \mathrm{ac}=0\)

The correct answer is:
\(\mathrm{a}^2-\mathrm{b}^2+2 \mathrm{ac}=0\)

Solution:
Hints: \(\sin \theta+\cos \theta=\frac{b}{a}\)
\(\begin{aligned}
& \sin \theta \cdot \cos \theta=\frac{c}{a} \\
& \left(\frac{b}{a}\right)^2=1+\frac{2 c}{a} \\
& b^2=a^2+2 a c \\
& a^2-b^2+2 a c=0
\end{aligned}\)

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2011
If \(\sin \theta+\cos \theta=0\) and \(0 < \theta < \pi\), then \(\theta\)

Options:
A 0
B \(\frac{\pi}{4}\)
C \(\frac{\pi}{2}\)
D \(\frac{3 \pi}{4}\)

The correct answer is:
\(\frac{3 \pi}{4}\)

Solution:
Hints: \(\sin \theta+\cos \theta=0\)
\(\Rightarrow \tan \theta=-1 \quad \theta=\frac{3 \pi}{4}\)

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2011
If \(\sin \theta=\frac{2 t}{1+t^2}\) and \(\theta\) lies in the second quadrant, then \(\cos \theta\) is equal to

Options:
A \(\frac{1-t^2}{1+t^2}\)
B \(\frac{t^2-1}{1+t^2}\)
C \(\frac{-\left|1-t^2\right|}{1+t^2}\)
D \(\frac{1+t^2}{\left|1-t^2\right|}\)

The correct answer is:
\(\frac{-\left|1-t^2\right|}{1+t^2}\)

Solution:
Hints: \(\theta\) in 2nd quad \(\cos \theta < 0\)
\(\begin{aligned}
& |\cos \theta|=\left|\frac{1-t^2}{1+t^2}\right|=\frac{\left|1-t^2\right|}{1+t^2} \\
& \cos \theta=-\frac{\left|1-t^2\right|}{1+t^2}
\end{aligned}\)

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2011
If $\chi$ stands for the magnetic susceptibility of a substance, $\mu$ for its magnetic permeability and $\mu_{0}$ for the permeability of free space, then

Options:
A for a paramagnetic substance: $\chi < 0$, $\mu>0$
B for a paramagnetic substance: $\chi>0$, $\mu>\mu_{0}$
C for a diamagnetic substance: $\chi>0, \mu < 0$
D for a ferromagnetic substance: $\chi>1$, $\mu>\mu_{0}$

The correct answers are:
for a paramagnetic substance: $\chi>0$, $\mu>\mu_{0}$, for a ferromagnetic substance: $\chi>1$, $\mu>\mu_{0}$

Solution:
We know that magnetic susceptibility
$\chi=\mu_{r}-1$
and \mu_r=\frac{\mu}{\mu_{0}}$
=Relative permeability
For paramagnetic substance,
$\begin{array}{ll}
& \chi>0 \\
\therefore & \mu_{r}>1 \\
\therefore \mu>\mu_{0}
\end{array}$
For diamagnetic substance,
$\chi< 0$
$\therefore \mu_r < 1$
$\therefore \mu < \mu_{0}$
For ferromagnetic substance,
$\begin{array}{l}
\chi>>1 \\
\mu>>M_{0}
\end{array}$
If
$0 < \mu < \mu_{0}$
Then substance will not be paramagnetic. Hence option (a) is incorrect and option (b) and (d) are correct.

Topic: Physics, Magnetic Properties of Matter, WBJEE, WBJEE 2017
If sum of an infinite geometric series is $\frac{4}{5}$ and its 1 st term is $\frac{3}{4}$, then its common ratio is

Options:
A $\frac{7}{16}$
B $\frac{9}{16}$
C $\frac{1}{9}$
D $\frac{7}{9}$

The correct answer is:
$\frac{7}{16}$

Solution:
Hints: $\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{4}{3} \quad$ Then $\frac{\frac{3}{4}}{1-\mathrm{r}}=\frac{4}{3} \Rightarrow \mathrm{r}=1-\frac{9}{16}=\frac{7}{16}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2010
If $t_1$ and $t_2$ be the parameters of the end points of a focal chord for the parabola $y^2=4 a x$, then which one is true?

Options:
A $\mathrm{t}_1 \mathrm{t}_2=1$
B $\frac{t_1}{t_2}=1$
C $\mathrm{t}_1 \mathrm{t}_2=-1$
D $\mathrm{t}_1+\mathrm{t}_2=-1$

The correct answer is:
$\mathrm{t}_1 \mathrm{t}_2=-1$

Solution:
Hints : $\mathrm{t}_1 \mathrm{t}_2=-1$ Fact

Topic: Mathematics, Parabola, WBJEE, WBJEE 2010
If $\phi(t)=\left\{\begin{array}{l}1, \text { for } 0 \leq t < 1 \\ 0, \text { otherwise }\end{array},\right.$ then $\int_{-3000}^{3000}\left(\sum_{r^{\prime}=2014}^{2016} \phi(t-r') \phi(t-2016)\right) d t$ is

Options:
A a real number
B 1
C 0
D does not exist

The correct answers are:
a real number, 1

Solution:
$\int_{-3000}^{3000}\left(\sum_{r'=2014}^{2016} \phi\left(t-r^{\prime}\right) \phi(t-2016)\right) d t$
$=\int_{-3000}^{3000}[\phi(t-2014) \phi(t-2016)+\phi(t-2015) \phi(t-2016) +\phi(t-2016) \phi(t-2016)] \mathrm{dt}$
$=\int_{-3000}^{3000} \phi(t-2016)[\phi(t-2014)+\phi(t-2015)$ $+\phi(t-2016)] d t$
$=\int_{-3000}^{2016} 0 d t+\int_{2016}^{2017} 1 \cdot(0+0+1) dt+\int_{2017}^{3000} 0 d t$
$=0+1+0=1$ which is a real number.

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2016
If the algebraic sum of the distances from the points $(2,0),(0,2)$ and $(1,1)$ to a variable straight line be zero, then the line passes through the fixed point

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

The correct answer is:
$(1,1)$

Solution:
$\left(\frac{2+0+1}{3}, \frac{0+2+1}{3}\right)$ i.e. $(1,1)$.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2022
If the charge on a capacitor is increased by 2 coulomb, the energy stored in it increases by $21 \%$. The original charge on the capacitor is

Options:
A $10 \mathrm{C}$
B $20 \mathrm{C}$
C $30 \mathrm{C}$
D $40 \mathrm{C}$

The correct answer is:
$20 \mathrm{C}$

Solution:
$\frac{\frac{q_f^2}{2 C}-\frac{q_i^2}{2 C}}{\frac{q_i^2}{2 C}} \times 100=21$ and $q_f-q_i=2$
solving we get $q_i=20$ coulomb

Topic: Physics, Capacitance, WBJEE, WBJEE 2009
If the circle $x^{2}+y^{2}+2 g x+2 f y+c=0$ cuts the three circles $x^{2}+y^{2}-5=0$ $x^{2}+y^{2}-8 x-6 y+10=0$ and $x^{2}+y^{2}-4 x+2 y-2=0$ at the extremities of their diameters, then

Options:
A $c=-5$
B $f g=147 / 25$
C $g+2 t=c+2$
D $4 f=3 g$

The correct answers are:
$4 f=3 g$

Solution:
Centres and constant terms in the circles $x^{2}+y^{2}-5=0, x^{2}+y^{2}-8 x-6 y+10=0$ and
$x^{2}+y^{2}-4 x+2 y-2=0$ are $C_{1}^{\prime}(0,0), c_{1}=-5$
$C_{2}^{\prime}(4,3), c_{2}=10$ and $C_{3}^{\prime}(2,-1), c_{3}=-2$
Also, centre and constant of circle $x^{2}+y^{2}+2 g x+2 f y+c=0$
is $\quad C_{4}^{\prime}(-g,-f)$ and $c_{4}=c$
since, the first circle intersect all the three at the
extremities of diameter, therefore they are orthogonal to each other.
$\therefore \quad 2\left(g_{1} g_{2}+f_{1} f_{2}\right)=c_{1}+c$
$\therefore \quad 2[-g(0)+(-f)(0)]=c-5$
$\Rightarrow \quad c=5$
$$
2[-g(4)+(-f)(3)]=c+10
$$
$\Rightarrow \quad-2(4 g+3 f)=5+10$
$\Rightarrow \quad 4 g+3 f=-\frac{15}{2}$
and $2[-g(2)+(-f)(-1)]=c-2$
$\Rightarrow \quad 2[-2 g+f]=5-2$
$\Rightarrow \quad-2 g+f=\frac{3}{2}$
On solving Eqs. (i) and (ii), we get $f=-\frac{9}{10}$ and $g=-\frac{12}{10}$
$\therefore \quad f g=\frac{-9}{10} \times-\frac{12}{10}=\frac{27}{25}$
and $\quad 4 f=4 \times \frac{-9}{10}$
$\Rightarrow \quad 4 f=\frac{-36}{10}$
$\Rightarrow \quad 4 f=3 g$

Topic: Mathematics, Circle, WBJEE, WBJEE 2014
If the circles $x^{2}+y^{2}+2 x+2 k y+6=0$ and $x^{2}+y^{2}+2 k y+k=0$ intersect orthogonally. then $k$ is equal to

Options:
A 2 or $-\frac{3}{2}$
B -2 or $-\frac{3}{2}$
C 2 or $\frac{3}{2}$
D -2 or $\frac{3}{2}$

The correct answer is:
2 or $-\frac{3}{2}$

Solution:
Two circles are orthogonally if and only if $\quad 2\left(g_{1} g_{2}+f_{1} f_{2}\right)=c_{1}+c_{2}$
$\Rightarrow \quad 2[(1 \times 0+(k) k]=6+k$
$\Rightarrow \quad 2 k^{2}=6+k$
$\Rightarrow \quad 2 k^{2}-k-6=0$
$\Rightarrow \quad 2 k^{2}-4 k+3 k-6=0$
$\Rightarrow \quad \quad 2 k(k-2)+3(k-2)=0$
$\Rightarrow \quad(k-2)(2 k+3)=0$
$\Rightarrow \quad k=2-\frac{3}{2}$

Topic: Mathematics, Circle, WBJEE, WBJEE 2012