datavorous/entrance-exam-dataset huggingface dataset:
The number of solution(s) of the equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4 x-1}$ is/are

Options:
A 2
B 0
C 3
D 1

The correct answer is:
0

Solution:
Given equation is
$$
\sqrt{x+1}-\sqrt{x-1}=\sqrt{4 x-1}
$$
On squaring both sides, we get
$$
(x+1)+(x-1)-2 \sqrt{x^{2}-1}=4 x-1
$$
$\Rightarrow \quad 2 x-2 \sqrt{x^{2}-1}=4 x-1$
$\Rightarrow \quad-2 \sqrt{x^{2}-1}=2 x-1$
Again, squaring both sides, we get
$$
4\left(x^{2}-1\right)=4 x^{2}+1-4 x
$$
$\Rightarrow \quad-4=+1-4 x$
$\Rightarrow \quad 4 x=5$
$\Rightarrow$
$$
x=\frac{5}{4}
$$
But, when we put $x=\frac{5}{4}$ in Eq. (i), we get
$$
\sqrt{\frac{5}{4}+1}-\sqrt{\frac{5}{4}-1}=\sqrt{4 \times \frac{5}{4}-1}
$$
$\Rightarrow$
$\sqrt{\frac{9}{4}}-\sqrt{\frac{1}{4}}=\sqrt{5-1} \Rightarrow \frac{3}{2}-\frac{1}{2}=2$
$\Rightarrow 1=2,$ which is not true.
Hence, no value of $x$ satisfy the given equation.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2014
The number of solutions of the equation $x+y+z=10$ where $x, y$ and $z$ are positive integers

Options:
A 36
B 55
C 72
D 45

The correct answer is:
36

Solution:
Given equation, is $x+y+z=10$
where, $x, y$ and $z$ are positive integers.
$\therefore$ Required number of solutions $={ }^{(10-1)} \mathrm{C}_{(3-1)}$
$$
={ }^{9} C_{2}=\frac{9 \times 8}{2}=36
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2013
The number of unpaired electron in $\mathrm{Mn}^{2+}$ ion is

Options:
A 2
B 3
C 5
D 6

The correct answer is:
5

Solution:

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2022
The number of unpaired electrons in Ni (atomic number $=28$ ) are

Options:
A 0
B 2
C 4
D 8

The correct answer is:
2

Solution:
The number of unpaired electrons in $\mathrm{Ni}$ (atomic number $=28$ ) are 2 . It can be easily concluded from the electronic configuration of $\mathrm{kL}$. Electronic configuration of $$
\mathrm{Ni}=\mathrm{l} s^{2} 2 \mathrm{s}^{2} 2 p^{6} 3 \mathrm{s}^{2} 3 p^{6} 4 \mathrm{s}^{2} \mathrm{W}^{8}
$$
Orbitals of $\mathrm{Ni}$ in $3 d=$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2018
The number of unpaired electrons in
$\quad\left[\mathrm{NiCl}_{4}\right]^{2-}, \mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2}$
respectively are

Options:
A 2,2,1
B 2,0,1
C 0,2,1
D 2,2,0

The correct answer is:
2,0,1

Solution:
Configuration for metal (M) in various complex.

$\mathrm{Ni}^{2 *}$ has 2 unpaired electrons

$\mathrm{Ni}^{0}$ has zero unpaired electrons with
$s p^{3}$ -hybridisation.
$\mathrm{Cu}^{2}$ "has one unpaired electron with $d s p^{2}$ -hybridisation.

$\therefore$ (b) is the correct answer, i.e.
Number of unpaired electron (2,0,1)

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2017
The number of unpaired electrons in the uranium $\left({ }_{92} \mathrm{U}\right)$ atom is:

Options:
A 4
B 6
C 3
D 1

The correct answer is:
4

Solution:
Hint:
$_{92} \mathrm{U}:\left[\mathrm{R}_{\mathrm{n}}\right]^{\mathrm{g} 8} 5 \mathrm{f}^{3} 6 \mathrm{~d}^{1} 7 \mathrm{~s}^{2}$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2020
The number of values of $k,$ for which the equation $x^{2}-3 x+k=0$ has two distinct roots lying in the interval $(0,1),$ are

Options:
A three
B two
C infinitely many
D no value of $k$ satisfies the requirement

The correct answer is:
no value of $k$ satisfies the requirement

Solution:
Given equation is
$$
x^{2}-3 x+k=0
$$
As.
$$
-\frac{b}{2 a}=\frac{3}{2} \notin (0,1)
$$
Since, both roots cannot lie between 0 and
So, no value of $k$ is possible.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2016
The number of ways in which the letters of the word ARRANGE can be permuted such that the R's occur together, is

Options:
A $\frac{7 !}{2!2 !}$
B $\frac{7!}{2!}$
C $\frac{6 !}{2!}$
D $5 ! \times 2!$

The correct answer is:
$\frac{6 !}{2!}$

Solution:
Given word is 'ARRANGE'.
If R's occur together, number of letters in word = 6
To arrange R's together number of ways $=\frac{2 !}{2 !}=1$
$\therefore$ Number of ways to permuted 'ARRANGE' = $\frac{6!}{2 !}$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2016
The number of words that can be written using all the letters of the word "IRRATIONAL' is

Options:
A $\frac{10 !}{(2 !)^{3}}$
B $\frac{10 !}{(21)^{2}}$
C $\frac{10 !}{2 !}$
D 10 !

The correct answer is:
$\frac{10 !}{(2 !)^{3}}$

Solution:
These are 10 letters in the word IRRATIONAL in which these are $2^{4} I, 2 \cdot R^{\prime}$ and $2 \cdot A^{\prime}$
$\therefore$ The total number of word $s=\frac{10 !}{2 ! 2 ! 2 !}=\frac{10 !}{(21)^{3}}$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2012
The number of zeros at the end of 100! is

Options:
A 21
B 22
C 23
D 24

The correct answer is:
24

Solution:
$\mathrm{E}_5(100 !)=\left[\frac{100}{5}\right]+\left[\frac{100}{25}\right]+\left[\frac{100}{125}\right]=20+4+0=24$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2022
The optically active molecule is

Options:
A
B
C
D

The correct answer is:

Solution:
Only molecule (c) is optically active. Other moelcules are meso, due to presence of plane of symmetry.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2013
The orange solid on heating gives a colourless gas and a green solid which can be reduced to metal by aluminium powder. The orange and the green solids are, respectively

Options:
A $\mathrm{NH}_{4} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$
B $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$
C $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{CrO}_{3}$
D $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CrO}_{4}$ and $\mathrm{CrO}_{3}$

The correct answer is:
$\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$

Solution:
Orange solid is ammonium dichromate.
$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2} \uparrow+\mathrm{Cr}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}$
$$
\mathrm{Cr}_{2} \mathrm{O}_{3}+2 \mathrm{Al} \longrightarrow 2 \mathrm{Cr}+\mathrm{Al}_{2} \mathrm{O}_{3}
$$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2013
The order of decreasing ease of abstraction of hydrogen atoms in the following molecule

Options:
A $\mathrm{H}_{a}>\mathrm{H}_{b}>\mathrm{H}_{c}$
B $\mathrm{H}_{a}>\mathrm{H}_{c}>\mathrm{H}_{\mathrm{b}}$
C $\mathrm{H}_{b}>\mathrm{H}_{a}>\mathrm{H}_{c}$
D $\mathrm{H}_{\mathrm{c}}>\mathrm{H}_{\mathrm{b}}>\mathrm{H}_{a}$

The correct answer is:
$\mathrm{H}_{a}>\mathrm{H}_{c}>\mathrm{H}_{\mathrm{b}}$

Solution:
Abstraction of $\left(\mathrm{H}_{a}\right)$ hydrogen atom results in the formation of $3^{\circ}$ allylic carbocation wheres abstraction of $\mathrm{H}_{c}$ gives $2^{\circ}$ allylic carbocation (which is less stable as compared to former one). However, abstraction of $\mathrm{H}_{b}$ gives only a $2^{\circ}$ carbocation which is less stable as compared to former two. More the stability of generated carbocation, higher is the tendency of abstraction of hydrogen. Thus, the order of decreasing ease of abstraction of hydrogen atoms from the given molecule is
$$
\mathrm{H}_{a}>\mathrm{H}_{c}>\mathrm{H}_{b}
$$

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
The order of equivalent conductances at infinite dilution for LiCl, $\mathrm{NaCl}$ and $\mathrm{KCl}$ is

Options:
A $\mathrm{LCl}>\mathrm{NaCl}>\mathrm{KCl}$
B $\mathrm{KCl}>\mathrm{NaCl}>\mathrm{LiCl}$
C $\mathrm{NaCl}>\mathrm{KCl}>\mathrm{LiCl}$
D $\mathrm{LCl}>\mathrm{KCl}>\mathrm{NaCl}$

The correct answer is:
$\mathrm{KCl}>\mathrm{NaCl}>\mathrm{LiCl}$

Solution:
KCl is more ionic than NaCl while NaCl is more ionic than LiCl
Also, more the ionic nature, more is the conductivity thus, order of equivalent conductance at infinite dilution is $\mathrm{KCl}>\mathrm{NaCl}>\mathrm{LiCl}$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2016
The order of the differential equation $\frac{d^2 y}{d x^2}=\sqrt{1-\left(\frac{d y}{d x}\right)^2}$ is

Options:
A 3
B 2
C 1
D 4

The correct answer is:
2

Solution:
No Solution Available

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2009
The order of the differential equation of all parabolas whose axis of symmetry along $X$ -axis is

Options:
A 2
B 3
C 1
D None of these

The correct answer is:
2

Solution:
The general equation of the parabola is $x=a y^{2}+b$.
where a and $b$ are arbitrary constants.
So, the differential equation is of order 2.

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2016
The ore chromite is

Options:
A $\mathrm{FeCr}_{2} \mathrm{O}_{4}$
B $\mathrm{CoCr}_{2} \mathrm{O}_{3}$
C $\mathrm{CrFe}_{2} \mathrm{O}_{4}$
D $\mathrm{FeCr}_{2} \mathrm{O}_{3}$

The correct answer is:
$\mathrm{FeCr}_{2} \mathrm{O}_{4}$

Solution:
Ore chomite is $\mathrm{FeCr}_{2} \mathrm{O}_{4}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2013
The output Y of the logic circuit given below is

Options:
A $\bar{A}+B$
B $\bar{A}$
C $\overline{(\overline A+B)} \cdot \bar{A}$
D $\overline{(\overline A+B)} \cdot A$

The correct answer is:
$\bar{A}$

Solution:
By absorption laws
$$
Y=\bar{A}+\bar{A} \cdot B
$$
or $\quad Y=(\bar{A} \cdot B)+\bar{A}=\bar{A} \cdot(B+1)=\bar{A} \cdot 1=\bar{A}$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2014
The oxidant which is used as an antiseptic is :

Options:
A $\mathrm{KBrO}_3$
B $\mathrm{KMnO}_4$
C $\mathrm{CrO}_3$
D Merrifield resin

The correct answer is:
$\mathrm{KMnO}_4$

Solution:
No Solution Available

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2009
The ozone layer forms naturally by

Options:
A the interaction of CFC with oxygen
B the interaction of UV radiation with oxygen
C the interaction of IR radiation with oxygen
D the interaction of oxygen and water vapour.

The correct answer is:
the interaction of UV radiation with oxygen

Solution:
No Solution Available

Topic: Chemistry, Environmental Chemistry, WBJEE, WBJEE 2011
The paramagnetic behaviour of $\mathrm{B}_{2}$ is due to the presence of

Options:
A two unpaired electrons in $\pi^{*}$ MO
B two unpaired electrons in $\pi$ MO
C two unpaired electrons in $\dot{\sigma}$ MO
D two unpaired electrons in $\sigma^{*}$ MO

The correct answer is:
two unpaired electrons in $\pi^{*}$ MO

Solution:
$B_{2}=5+5=10 e^{-}$
$$
=\sigma 1 s^{2} \dot{\sigma} 1 s^{2}, \sigma 2 s^{2} \dot{\sigma} 2 s^{2}, \pi 2 p_{x}^{1}, \pi 2 p_{y}^{\prime}
$$
Due to the presence of 2 unpaired electrons, in $\pi$ bonding orbitals, $B_{2}$ shows paramagnetic behaviour.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2012
The perfect gas equation for $4 \mathrm{g}$ of hydrogen gas is

Options:
A $p V=R T$
B $p V=2 R T$
C $p V=\frac{1}{2} R T$
D $p V=4 R T$

The correct answer is:
$p V=2 R T$

Solution:
Mass of gas $=4 \mathrm{g} \quad$ (given)
From pertect gas equation
$$
p V=n R T
$$
where $n=$ number of moles here, $\left(n=\frac{4}{2}=2\right)$
$\therefore \quad p V=2 R T$
It will be the perfect gas equation for $4 \mathrm{g}$ of hydrogen.

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2016
The period of oscillation of a simple pendulum of length \(l\) suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination \(\alpha\), is given by

Options:
A \(2 \pi \sqrt{\frac{1}{g \cos \alpha}}\)
B \(2 \pi \sqrt{\frac{1}{g \sin \alpha}}\)
C \(2 \pi \sqrt{\frac{1}{g}}\)
D \(2 \pi \sqrt{\frac{1}{\mathrm{~g} \tan \alpha}}\)

The correct answer is:
\(2 \pi \sqrt{\frac{1}{g \cos \alpha}}\)

Solution:
Hints:
\(\begin{aligned}
& \mathrm{g}_{\text {eff }}=\mathrm{g} \cos \alpha \\
& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}_{\text {eff }}}}
\end{aligned}\)

Topic: Physics, Oscillations, WBJEE, WBJEE 2011
The period of the function \(f(x)=\cos 4 x+\tan 3 x\) is

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

The correct answer is:
\(\pi\)

Solution:
Hints: \(\operatorname{LCM}\left(\frac{2 \pi}{4}, \pi / 3\right)=\pi\)

Topic: Mathematics, Functions, WBJEE, WBJEE 2011
The pH of $10^{-4} \mathrm{M}$ KOH solution will be

Options:
A 4
B 11
C 10.5
D 10

The correct answer is:
10

Solution:
$10^{-4} \mathrm{M} \mathrm{KOH}=10^{-4} \mathrm{M}\left[\mathrm{OH}^{-}\right]$
We know that
$$
\left.\left[\mathrm{H}^{+}\right] \mathrm{IOH}^{-1}\right]=1 \times 10^{-14}
$$
$$
\begin{array}{c}
{\left[\mathrm{H}^{+}\right] \times 10^{-4}=1 \times 10^{-14}} \\
{\left[\mathrm{H}^{+}\right]=\frac{1 \times 10^{-14}}{1 \times 10^{-4}}=1 \times 10^{-10} \mathrm{M}} \\
\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-10}\right) \\
=10
\end{array}
$$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2014
The plane $\ell \mathrm{x}+\mathrm{my}=0$ is rotated about its line of intersection with the plane $\mathrm{z}=0$ through an angle $\alpha$. The equation changes to

Options:
A $\ell \mathrm{x}+\mathrm{my} \pm \tan \alpha \sqrt{\ell^{2}+\mathrm{m}^{2}}=0$
B $\ell \mathrm{x}+\mathrm{my} \pm \mathrm{z} \tan \alpha \sqrt{\ell^{2}+\mathrm{m}^{2}+1}=0$
C $\ell \mathrm{x}+\mathrm{my} \pm \mathrm{z} \tan \alpha \sqrt{\ell^{2}+1}=0$
D $\ell \mathrm{x}+\mathrm{my} \pm \mathrm{z} \tan \alpha \sqrt{\ell^{2}+\mathrm{m}^{2}}=0$

The correct answer is:
$\ell \mathrm{x}+\mathrm{my} \pm \mathrm{z} \tan \alpha \sqrt{\ell^{2}+\mathrm{m}^{2}}=0$

Solution:
$P_{1}: \ell x+m y=0, \quad P_{2}=z=0$
Plane through common line of $P_{1}$ and $P_{2}$
$\mathrm{P}_{3}: \ell \mathrm{x}+\mathrm{my}+\mathrm{nz}=0$
angle between $P_{1}$ and $P_{3}=\alpha$
$\therefore \cos \alpha=\hat{n}_{1}, \hat{n}_{2}$
$=\frac{\ell^{2}+\mathrm{m}^{2}}{\sqrt{\ell^{2}+\mathrm{m}^{2}} \sqrt{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}}}=\sqrt{\frac{\ell^{2}+\mathrm{m}^{2}}{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}}}$
$\Rightarrow \cos ^{2} \alpha=\frac{\ell^{2}+\mathrm{m}^{2}}{\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}}$
$\Rightarrow \mathrm{n}^{2}=\left(\ell^{2}+\mathrm{m}^{2}\right) \tan ^{2} \alpha$
$\Rightarrow \mathrm{n}=\pm \sqrt{\ell^{2}+\mathrm{m}^{2}} \tan \alpha$
$\mathrm{P}_{3}: \ell \mathrm{x}+\mathrm{my} \pm \sqrt{\ell^{2}+\mathrm{m}^{2}}$ tan $\alpha=0$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2021
The point $(-4,5)$ is the vertex of a square and one of its diagonals is $7 x-y+8=0$. The equation of the other diagonal is

Options:
A $7 x-y+23=0$
B $7 y+x=30$
C $7 y+x=31$
D $\quad x-7 y=30$

The correct answer is:
$7 y+x=31$

Solution:
$$
\begin{aligned}
\text { Hints } & : x+7 y=k \\
& -4+35=k \\
& 31=k \\
& x+7 y-31=0
\end{aligned}
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2009
The point in the interval $[0,2 \pi]$, where $f(x)=e^x \sin x$ has maximum slope, is

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

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

Solution:
Hints: $f^{\prime}(x)=e^x(\sin x+\cos x)$
$$
\begin{aligned}
& f^{\prime \prime}(x)=e^x(\sin x+\cos x+\cos x-\sin x) \Rightarrow f^{\prime \prime}(x)=e^x \cos x=0 \\
& \Rightarrow x=\frac{\pi}{2}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
The point of contact of the tangent to the parabola $y^2=9 x$ which passes through the point $(4,10)$ and makes an angle $\theta$ with the positive side of the axis of the parabola where $\tan \theta>2$, is

Options:
A $\left(\frac{4}{9}, 2\right)$
B $(4,6)$
C $(4,5)$
D $\left(\frac{1}{4}, \frac{1}{6}\right)$

The correct answer is:
$\left(\frac{4}{9}, 2\right)$

Solution:
$$
\tan \theta=\frac{1}{t}>2
$$
Equation of tangent at $\mathrm{t}$
$$
\begin{aligned}
&\left.\mathrm{yt}=\mathrm{x}+\frac{9}{4} \mathrm{t}^2 \text { (passes through }(4,10)\right) \Rightarrow 10 \mathrm{t}=4+\frac{9}{4} \mathrm{t}^2 \Rightarrow 9 \mathrm{t}^2-40 \mathrm{t}+16=0 \Rightarrow \mathrm{t}=4, \frac{4}{9}\left(\frac{1}{\mathrm{t}}>2\right) \Rightarrow \mathrm{t}=\frac{4}{9} \\
&\therefore \text { point }\left(\mathrm{at}^2, 2 \mathrm{at}\right) \Rightarrow\left(\frac{4}{9}, 2\right)
\end{aligned}
$$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2022
The point on the parabola $y^{2}=64 x$ which is nearest to the line $4 x+3 y+35=0$ has coordinates

Options:
A (9,-24)
B (1,81)
C (4,-16)
D (-9,-24)

The correct answer is:
(9,-24)

Solution:
Given equation of parabola is
$$
y^{2}=64 x
$$
The point at which the tangent to the curve is parallel to the line is the nearest point on the curve.
On differentiating both sides of Eq. (i), we get $2 y \frac{d y}{d x}=64$
$\Rightarrow \quad \frac{d y}{d x}=\frac{32}{y}$
Also, slope of the given line is $-\frac{4}{3}$.
$\therefore$
$$
-\frac{4}{3}=\frac{32}{y} \Rightarrow y=-24
$$
From Eq. (i), $(-24)^{2}=64 x \Rightarrow x=9$
$\therefore$ Hence, the required point is (9,-24)

Topic: Mathematics, Parabola, WBJEE, WBJEE 2014
The point $P(3,6)$ is first reflected on the line $y=x$ and then the image point $Q$ is again reflected on the line $y=-x$ to get the image point $Q^{\prime} .$ Then, the circumcentre of the $\Delta P Q Q^{\prime}$ is

Options:
A (6,3)
B (6,-3)
C (3,-6)
D (0,0)

The correct answer is:
(0,0)

Solution:
The coordinates of $Q$ and $Q^{\prime}$ will be (6,3)
and $(-3,-6),$ respectively. Now, slope of $P Q=\frac{6-3}{3-6}=-1$ and Slope of $Q Q^{\prime}=\frac{-6-3}{-3-6}=1$
$\therefore$ Slope of $P Q \times$ Slope of $Q Q^{\prime}=-1 \times 1=-1$
$\therefore \Delta P Q Q^{\prime}$ is right angled triangle at $Q .$
$\therefore$ Circumcentre will be the mid-point of hypotenuse $P Q^{\prime}=\left(\frac{-3+3}{2}, \frac{-6+6}{2}\right)$
$$
=(0,0)
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2017
The point $Q$ is the image of the point $P(1,5)$ about the line $y=x$ and $R$ is the image of the point $Q$ about the line $y=-x .$ The circumcentre of the $\Delta P Q R$ is

Options:
A (5,1)
B (-5,1)
C $(1,-5)$
D $(0,0)$

The correct answer is:
$(0,0)$

Solution:
Given, point P(1,5) image of the point $P(1,5)$ about the line $y=x$ is $Q(5,1)$ and image of the point $Q$ on line $y=-x$ is $R(-1,-5)$
$\therefore$ Required circumcentre $=$ Mid-point of $P$ and $R$ $=\left(\frac{1-1}{2}, \frac{5-5}{2}\right)=(0,0$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2018
The points $(-a,-b),(a, b),(0,0)$ and $\left(a^{2}, a b\right), a \neq 0, b \neq 0$ are always

Options:
A collinear
B vertices of a parallelogram
C vertices of a rectangle
D lie on a circle

The correct answer is:
collinear

Solution:
Let the four points be $A(-a,-b), B(a, b), C(0,0)$ and $D\left(a^{2}, a b\right)$
If $A,B$ and $C$ are collinear.
Then, $\left|\begin{array}{ccc}-a & -b & 1 \\ a & b & 1 \\ 0 & 0 & 1\end{array}\right|=0$
$\Rightarrow \quad-a(b-0)+b(a-0)+1(0)=0$
$-a b+a b=0$
Hence, $A, B$ and $C$ ae collinear.
Again, if $B, C$ and $D$ are collinear.
Then, $\left|\begin{array}{ccc}a & b & 1 \\ 0 & 0 & 1 \\ a^{2} & a b & 1\end{array}\right|=0$
$\Rightarrow \quad a(0-a b)-b\left(0-a^{2}\right)+1(0) ;=0$
$-a^{2} b+a^{2} b=0$
$\Rightarrow$
Hence, $B, C$ and $D$ are collinear. So, the points $A, B, C$ and $D$ are akways collinear.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2016
The points of intersection of two ellipses $x^{2}+2 y^{2}-6 x-12 y+20=0$ and $2 x^{2}+y^{2}-10 x-6 y+15=0$ lie on a circle. The centre of the circle is

Options:
A $(8,3)$
B $(8,1)$
C $\left(\frac{8}{3}, 3\right)$
D $(3,8)$

The correct answer is:
$\left(\frac{8}{3}, 3\right)$

Solution:
$x^{2}+2 y^{2}-6 x-12 y+20+\lambda\left(2 x^{2}+y^{2}-10 x-6 y+15\right)=0$
$\begin{array}{l}
\Rightarrow(1+2 \lambda) \mathrm{x}^{2}+(2+\lambda) \mathrm{y}^{2}-(10 \lambda+6) \mathrm{x}-(6 \lambda+12) \mathrm{y}+15 \lambda+20=0 \\
2+\lambda=1+2 \lambda \\
\Rightarrow \lambda=1
\end{array}$
$\therefore$ Centre $:\left(\frac{8}{3}, 3\right)$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2021
The points of the ellipse $16 x^{2}+9 y^{2}=400$ at which the ordinate decreases at the same rate at which the abscissa increases is/are given by

Options:
A $\left(3, \frac{16}{3}\right)$ and $\left(-3,-\frac{16}{3}\right)$
B $\left(3,-\frac{16}{3}\right)$ and $\left(-3, \frac{16}{3}\right)$
C $\left(\frac{1}{16}, \frac{1}{9}\right)$ and $\left(-\frac{1}{16},-\frac{1}{9}\right)$
D $\left(\frac{1}{16},-\frac{1}{9}\right)$ and $\left(-\frac{1}{16}, \frac{1}{9}\right)$

The correct answer is:
$\left(3, \frac{16}{3}\right)$ and $\left(-3,-\frac{16}{3}\right)$

Solution:
Given that, the ordinate decreases at the same rate at which the abscissa increases, therefore
$$
\frac{d y}{d t}=-\frac{d x}{d t}
$$
Also, equation of ellipse
$$
16 x^{2}+9 y^{2}=400
$$
On differentiating wrt. $t$, we get
$16 \times 2 x \frac{d x}{d t}+9 \times 2 y \frac{d y}{d t}=0$
$\Rightarrow \quad 16 x \frac{d x}{d t}+9 y \frac{d y}{d t}=0$
$\Rightarrow \quad 9 y \frac{d y}{d t}=-16 x \frac{d x}{d t}$
$\Rightarrow \quad 9 y \frac{d y}{d t}=-16 x\left(-\frac{d y}{d t}\right) \quad$ [using $\left.\mathrm{Eq.}(i)\right]$
$\Rightarrow \quad 9 y=16 x \Rightarrow y=\frac{16}{9} x$
$16 x^{2}+9\left(\frac{16}{9} x\right)^{2}=400$
$\Rightarrow \quad 16 x^{2}+\frac{(16)^{2}}{9} x^{2}=400$
$\Rightarrow \quad 16 x\left[x+\frac{16}{9} x\right]=400$
$\Rightarrow \quad 16 x\left[\frac{25}{9} x\right]=400 \Rightarrow x^{2}=\frac{400 \times 9}{25 \times 16}$
$\Rightarrow \quad x^{2}=9 \Rightarrow x=\pm 3$
$\therefore$ Required points are $\left(3, \frac{16}{3}\right)$ and $\left(-3-\frac{16}{3}\right)$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2016
The points representing the complex number z for which arg $\left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$ lie on

Options:
A a circle
B a straight line
C an ellipse
D a parabola

The correct answer is:
a circle

Solution:
Let $z=x+i y$ $\therefore \quad \frac{z-2}{z+2}=\frac{x+iy-2}{x+i y+2}$
$=\frac{(x-2)+i y}{(x+2)+6 y} \times \frac{(x+2)-i y}{(x+2)-i y}$
$=\frac{(x-2)(x+2)+i y(x+2)-i y(x-2)-i^{2} y^{2}}{(x+2)^{2}-(i y)^{2}}$
$=\frac{x^{2}-4+i x y+2 i y-i x y+2 i y+y^{2}}{(x+2)^{2}+y^{2}}$
$x^{2}+y^{2}-4-\frac{4}{\sqrt{3}} y=0,$ which represents a
circle.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2012
The poisson's ratio of a material is 0.5 . If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by $4 \%$. The percentage increase in the length is :

Options:
A $1 \%$
B $2 \%$
C $2.5 \%$
D $4 \%$

The correct answer is:
$4 \%$

Solution:
Poisson ratio $=0.5$
Therefore density is constant hence change in volume is zero we have $\mathrm{V}=\mathrm{A} \times \ell=$ constant
$\log \mathrm{V}=\log \mathrm{A}+\log \ell$ or $\frac{d A}{A}+\frac{d \ell}{\ell}=0 \Rightarrow \frac{d \ell}{\ell}=-\frac{d A}{A}$
That is $4 \%$

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2009
The polar coordinate of a point $P$ is $\left(2,-\frac{\pi} {4}\right)$ The polar coordinate of the point $Q$ which is such that line joining $P Q$ is bisected perpendicularly by the initial line, is

Options:
A $\left(2, \frac{\pi}{4}\right)$
B $\left(2, \frac{\pi}{6}\right)$
C $\left(-2, \frac{\pi}{4}\right)$
D $\left(-2, \frac{\pi}{6}\right)$

The correct answer is:
$\left(2, \frac{\pi}{4}\right)$

Solution:
Since, $Q$ is the point for which $P Q$ is bisected perpendicularly by the initial line $(X$ -axis). Therefore, $Q$ will be the image of $P$ in $X$ -axis
$\therefore$ Coordinates of $Q$ is $(2, \pi / 4)$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2019
The portion of the tangent to the curve \(x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}, a > 0\) at any point of it, intercepted between the axes

Options:
A varies as abscissa
B varies as ordinate
C is constant
D varies as the product of abscissa and ordinate

The correct answer is:
is constant

Solution:
Hint : for the given curve \(x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}, a > 0\)
Parametric Coordinates are \(x=a \cos ^3 \theta\)
\(y=a \sin ^3 \theta\)
\(\begin{gathered}
\frac{d y}{d x}=\text { slope of tangent at any }=\frac{d y / d \theta}{d x / d \theta} \\
\text { point }(x, y)
\end{gathered}\)
\(=\frac{3 a \sin ^2 \theta \cdot \cos \theta}{3 a \cos ^2 \theta(-\sin \theta)}=-\tan \theta\)
Equation of tangent at \(\left(a \cos ^3 \theta, a \sin ^3 \theta\right)\)
\(y-a \sin ^3 \theta=-\tan \theta\left(x-a \cos^3 \theta\right)\)
its \(x\)-intercept \(=a \cos \theta\)
its \(y\)-intercept \(=a \sin \theta\)
So, the tangent cuts the axes at \(A(a \cos \theta, 0)\) and \(B(0, a \sin \theta)\) respectively.
\(A B=\sqrt{(a \cos \theta)^2+(a \sin \theta)^2}=a\), which is constant

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2023