datavorous/entrance-exam-dataset huggingface dataset:
The Entropy (S) of a black hole can be written as $S=\beta k_B A$, where $k_B$ is the Boltzmann constant and $A$ is the area of the black hole. Then $\beta$ has dimension of

Options:
A $\mathrm{L}^2$
B $\quad \mathrm{ML}^2 \mathrm{~T}^{-1}$
C $\mathrm{L}^{-2}$
D dimensionless

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

Solution:
$\begin{array}{ll} \mathrm{S}= \beta \mathrm{K}_\beta \mathrm{A} \\ \frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\theta]}=[\beta]\left[\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\theta}\right]\left[\mathrm{L}^2\right] \quad[\beta]=\left[\mathrm{L}^{-2}\right] \end{array}$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2022
The equation $2 x^{2}+5 x y-12 y^{2}=0$ represents a

Options:
A circle
B pair of non-perpendicular intersecting straight lines
C pair of perpendicular straight lines
D hyperbola

The correct answer is:
pair of non-perpendicular intersecting straight lines

Solution:
Given equation is
$$
\begin{array}{r}
2 x^{2}+5 x y-12 y^{2}=0 \\
\Rightarrow \quad 2 x^{2}+8 x y-3 x y-12 y^{2}=0 \\
\Rightarrow \quad 2 x(x+4 y)-3 y(x+4 y)=0 \\
\Rightarrow \quad(x+4 y)(2 x-3 y)=0 \\
\therefore \quad x+4 y=0 \text { and } 2 x-3 y=0
\end{array}
$$
which represent a pair of straight lines.
Compair Eq. (i) with $a x^{2}+2 h x y+b y^{2}=0$
we get $a=2$ and $b=-12$
$\because a+b \neq 0$
So, lines are not perpendicular to each other. Hence, it is a pair of non-perpendicular intersecting straight lines.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2013
The equation $\sqrt{3} \sin x+\cos x=4$ has

Options:
A only one solution
B two solutions
C infinitely many solutions
D no solution

The correct answer is:
no solution

Solution:
Hints : $\sqrt{3} \sin x+\cos x=2 \sin \left(x+\frac{\pi}{6}\right) \leq 2$. Therefore
$\sqrt{3} \sin x+\cos x=4 \quad$ cannot have a solution

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2009
The equation $6^{x}+8^{x}=10^{x}$ has

Options:
A no real root
B infinitely many rational roots
C exactly one real root
D two distinct real roots

The correct answer is:
exactly one real root

Solution:
$6^{x}+8^{x}=10^{x}\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1 \Rightarrow\left(\frac{3}{5}\right)^{x}=1-\left(\frac{4}{5}\right)^{x}$
$\therefore \quad 6^{x}+8^{x}=10^{x}$ has exactly one real root

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2021
The equation \(8 x^2+12 y^2-4 x+4 y-1=0\) represents

Options:
A an ellipse
B a hyperbola
C a parabola
D a circle

The correct answer is:
an ellipse

Solution:
Hints: \(a x^2+b y^2+2 h x y+2 g x+2 f y+c=0\)
represents ellipse if \(\mathrm{h}^2-a b < 0\)
\(\begin{aligned}
& 3 x^2+12 y^2-4 x+4 y-1=0 \\
& h=0, a=3, b=12 \\
& h^2-a b < 0
\end{aligned}\)

Topic: Mathematics, Pair of Lines, WBJEE, WBJEE 2011
The equation of a line parallel to the line $3 x+4 y=0$ and touching the circle $x^{2}+y^{2}=9$ in the first quadrant, is

Options:
A $3 x+4 y=15$
B $3 x+4 y=45$
C $3 x+4 y=9$
D $3 x+4 y=27$

The correct answer is:
$3 x+4 y=15$

Solution:
Equation of circle is $x^{2}+y^{2}=9$ ...(i)
And equation of line is $3 x+4 y=0$ ....(ii)
Equation of line parallel to the line (ii) .
$=$
$$
\begin{aligned}
3 x+4 y &=k \\
y &=\frac{-3 x}{4}+\frac{k}{4} ...(iii)
\end{aligned}
$$
$\because$ Circle touches the line (iii), so by the condition of tangency
$$
c^{2}=a^{2}\left(1+m^{2}\right)
$$
$\Rightarrow \quad\left(\frac{k}{4}\right)^{2}=9\left[1+\left(\frac{-3}{4}\right)^{2}\right]$
$\left[\because c=\frac{k}{4}, a=3, m=\frac{-3}{4}\right]$
$\Rightarrow$
$$
k=\pm 15
$$
Hence, $3 x+4 y=15$ touches the circle in the first quadrant.

Topic: Mathematics, Circle, WBJEE, WBJEE 2016
The equation of a progressive wave is $y=4 \sin (4 \pi t-0.04 x+\pi / 3)$ where $x$ is in meter and $t$ is in second. The velocity of the wave is

Options:
A $100 \pi \mathrm{m} / \mathrm{s}$
B $50 \pi \mathrm{m} / \mathrm{s}$
C $25 \pi \mathrm{m} / \mathrm{s}$
D $\pi \mathrm{m} / \mathrm{s}$

The correct answer is:
$100 \pi \mathrm{m} / \mathrm{s}$

Solution:
Hints : Velocity of wave $=\frac{\omega}{\mathrm{K}}=\frac{4 \pi}{0.04}=100 \pi \mathrm{m} / \mathrm{sec}$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2010
The equation of auxiliary circle of the ellipse $16 x^{2}+25 y^{2}+32 x-100 y=284$ is

Options:
A $x^{2}+y^{2}+2 x-4 y-20=0$
B $x^{2}+y^{2}+2 x-4 y=0$
C $(x+1)^{2}+(y-2)^{2}=400$
D $(x+1)^{2}+(y-2)^{2}=225$

The correct answer is:
$x^{2}+y^{2}+2 x-4 y-20=0$

Solution:
Given equation of ellipse is
$$
16 x^{2}+25 y^{2}+32 x-100 y=284
$$
Simplifying the given equation, we have ellipse as
$$
\frac{(x+1)^{2}}{25}+\frac{(y-2)^{2}}{16}=1
$$
So, the auxiliary equation of circle is
$$
\begin{array}{r}
(x+1)^{2}+(y-2)^{2}=25 \\
\Rightarrow \quad x^{2}+2 x+1+y^{2}-4 y+4=25
\end{array}
$$
$\therefore$
$x^{2}+y^{2}+2 x-4 y-20=0$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2016
The equation of circle of radius $\sqrt{17}$ unit, with centre on the positive side of $x$ -axis and through the point $(0,1)$ is

Options:
A $x^{2}+y^{2}-8 x-1=0$
B $x^{2}+y^{2}+8 x-1=0$
C $x^{2}+y^{2}-9 y+1=0$
D $2 x^{2}+2 y^{2}-3 x+2 y=4$

The correct answer is:
$x^{2}+y^{2}-8 x-1=0$

Solution:
Hint:
Let centre be $\equiv(a, 0)(a>0)$
So, circle : $(x-a)^{2}+y^{2}=17 ;$ as it passes through $(0,1)$, so $a^{2}+1=17 \quad \Rightarrow a=4(a \neq-4, a>0)$
$\therefore$ Equation is : $x^{2}+y^{2}-8 x-1=0$

Topic: Mathematics, Circle, WBJEE, WBJEE 2020
The equation of hyperbola whose coordinates of the foci are (±8,0) and the length of latusrectum is 24 units, is

Options:
A $3 x^{2}-y^{2}=48$
B $4 x^{2}-y^{2}=48$
C $x^{2}-3 y^{2}=48$
D $x^{2}-4 y^{2}=48$

The correct answer is:
$3 x^{2}-y^{2}=48$

Solution:
Let equation of hyperbola be
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
$$
Given, foci, $(\pm 8,0)=(\pm a e, 0)$
$\Rightarrow \quad a e=8$
and length of latusrectum $=\frac{2 b^{2}}{a}$
$\therefore$
$24=\frac{2 b^{2}}{a}$
$\Rightarrow \quad b^{2}=12 a$
$\therefore$ From Eq. (ii).
$$
a^{2} e^{2}=64
$$
$\Rightarrow$
$a^{2}\left(\frac{a^{2}+b^{2}}{a^{2}}\right)=64$
$\Rightarrow \quad a^{2}+b^{2}=64$
$\Rightarrow \quad a^{2}+12 a=64$
$\Rightarrow \quad a^{2}+12 a-64=0$
$\Rightarrow \quad a^{2}+16 a-4 a-64=0$
$\Rightarrow \quad a(a+16)-4(a+16)=0$
$\Rightarrow \quad(a+16)(a-4)=0$
$\Rightarrow \quad a=4$
[$\because$ a cannot be negative] On putting $a=4$ in Eq. (iii), we get
$$
b^{2}=12 \times 4 \Rightarrow b^{2}=48
$$
$\therefore$ From Eq. (i). $\frac{x^{2}}{16}-\frac{y^{2}}{48}=1$
$\Rightarrow \quad 3 x^{2}-y^{2}=48$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2014
The equation of normal of $x^2+y^2-2 x+4 y-5=0$ at $(2,1)$ is

Options:
A $y=3 x-5$
B $2 y=3 x-4$
C $y=3 x+4$
D $y=x+1$

The correct answer is:
$y=3 x-5$

Solution:
Hints: $0(1,-2) \mathrm{A}(2,1)$
Slope $A \rightarrow \frac{y-1}{-2-1}=\frac{x-2}{1-2}, \quad \frac{y-1}{-3}=\frac{x-2}{-1}=1, y-1=3(x-2)$ $y=3 x-5$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
The equation of one of the curves whose slope at any point is equal to $\mathrm{y}+2 \mathrm{x}$ is

Options:
A $y=2\left(e^x+x-1\right)$
B $y=2\left(e^x-x-1\right)$
C $y=2\left(e^x-x+1\right)$
D $y=2\left(e^x+x+1\right)$

The correct answer is:
$y=2\left(e^x-x-1\right)$

Solution:
$$
\begin{aligned}
& \text { Hints: } \frac{d y}{d x}=y+2 x \quad \text { Put } y+2 x=z \Rightarrow \frac{d y}{d x}+z=\frac{d z}{d x} \\
& \frac{d z}{d x}-2=z, \quad \frac{d z}{d x}=z+2 \Rightarrow \int \frac{d z}{z+2}=\int d x \\
& \log (z+2)=x+c, \log (y+2 x+2)=x+c \\
& y+2 x+2=x+c, y=2\left(e^x-x-1\right)
\end{aligned}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2010
The equation of state for $\mathrm{n}$ moles of an ideal gas is $\mathrm{PV}=\mathrm{nRT}$, where $\mathrm{R}$ is a constant. The SI unit for $\mathrm{R}$ is

Options:
A $\mathrm{JK}^{-1}$ per molecule
B $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
C $\mathrm{J} \mathrm{Kg}^{-1} \mathrm{~K}^{-1}$
D $\mathrm{JK}^{-1} \mathrm{~g}^{-1}$

The correct answer is:
$\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

Solution:
$\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2009
The equation of state of a gas is given by $\left(p+\frac{a}{v^{3}}\right)\left(V-b^{2}\right)=c T,$ where $p, V, T$ are pressure, volume and temperature respectively and $a, b, c$ are constants. The dimensions of $a$ and $b$ are respectively

Options:
A $\left[\mathrm{ML}^{8} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{3 / 2}\right]$
B $\left[\mathrm{ML}^{5} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{3}\right]$
C $\left[\mathrm{ML}^{5} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{6}\right]$
D $\left[\mathrm{ML}^{6} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{3 / 2}\right]$

The correct answer is:
$\left[\mathrm{ML}^{8} \mathrm{T}^{-2}\right.$ and $\left.\mathrm{L}^{3 / 2}\right]$

Solution:
$[P]=\left[\frac{a}{V^{3}}\right] \Rightarrow\left[M L^{-1} T^{-2}\right]=\frac{a}{\left[L^{3}\right]^{3}}$
$\begin{aligned} \Rightarrow \quad a &=\left[M L^{8} T^{-2}\right] \\ &[V]=\left[b^{2}\right] \Rightarrow\left[L^{3}\right]=b^{2} \Rightarrow b=\left[L^{3 / 2}\right] \end{aligned}$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2013
The equation of the chord of the circle $x^2+y^2-4 x=0$ whose mid point is $(1,0)$ is

Options:
A $y=2$
B $y=1$
C $x=2$
D $x=1$

The correct answer is:
$x=1$

Solution:
Hints:
$$
\text { Equation: } X=1
$$

Topic: Mathematics, Circle, WBJEE, WBJEE 2009
The equation of the circle passing through the point (1,1) and the points of intersection of $x^{2}+y^{2}-6 x-8=0$ and $x^{2}+y^{2}-6=0$ is

Options:
A $x^{2}+y^{2}+3 x-5=0$
B $x^{2}+y^{2}-4 x+2=0$
C $x^{2}+y^{2}+6 x-4=0$
D $x^{2}+y^{2}-4 y-2=0$

The correct answer is:
$x^{2}+y^{2}+3 x-5=0$

Solution:
Let $\quad S_{1}=x^{2}+y^{2}-6 x-8=0$
and $\quad S_{2}=x^{2}+y^{2}-6=0$
Now, the equation of the circle passing through the point (1,1) and the point of intersection of $S_{1}$ and $S_{2}$ is
$$
S_{1}+\lambda S_{2}=0
$$
$\Rightarrow \quad\left(x^{2}+y^{2}-6 x-8\right)+\lambda\left(x^{2}+y^{2}-6\right)=0$
$\Rightarrow \quad(1+\lambda) x^{2}+(1+\lambda) y^{2}-6 x$
$$
+(-8-6 \lambda)=0 \ldots 0
$$
Since, Eq. (i) passes through the point (1,1)
$\because$ Put $x=1, y=1$ in Eq. (ii), we get
$$
\begin{array}{r}
(1+\lambda)+(1+\lambda)-6+(-8-6 \lambda)=0 \\
\Rightarrow \quad-4 \lambda-12=0
\end{array}
$$
$\Rightarrow \quad \lambda=-3$
On putting the value of ' $\lambda$ ' in Eq. ( 0 , we get
$$
\begin{array}{r}
-2 x^{2}-2 y^{2}-6 x+10=0 \\
\Rightarrow \quad x^{2}+y^{2}+3 x-5=0
\end{array}
$$

Topic: Mathematics, Circle, WBJEE, WBJEE 2013
The equation of the common tangent with positive slope to the parabola $y^{2}=8 \sqrt{3} x$ and the hyperbola $4 x^{2}-y^{2}=4$ is

Options:
A $y=\sqrt{6} x+\sqrt{2}$
B $y=\sqrt{6} x-\sqrt{2}$
C $y=\sqrt{3} x+\sqrt{2}$
D $y=\sqrt{3} x-\sqrt{2}$

The correct answer is:
$y=\sqrt{6} x+\sqrt{2}$

Solution:
Equation of tangent in slope form of parabola $y^{2}=8 \sqrt{3} x$ is
$$
y=m x+c
$$
where, $\quad c=\frac{a}{m}$
$\therefore$
$c=\frac{2 \sqrt{3}}{m}$
Also, tangent to the hyperbola
$$
4 x^{2}-y^{2}=4
$$
or $\frac{x^{2}}{1}-\frac{y^{2}}{4}=1$ is
$$
\begin{array}{l}
c^{2}=a^{2} m^{2}-b^{2} \\
c^{2}=1 m^{2}-4
\end{array}
$$
$\Rightarrow \quad\left(\frac{2 \sqrt{3}}{m}\right)^{2}=m^{2}-4$
$\Rightarrow \quad \frac{12}{m^{2}}=m^{2}-4$
$\Rightarrow \quad m^{4}-4 m^{2}-12=0$
$\Rightarrow \quad m^{4}-6 m^{2}+2 m^{2}-12=0$
$\Rightarrow \quad m^{2}\left(m^{2}-6\right)+2\left(m^{2}-6\right)=0$
$\Rightarrow \quad\left(m^{2}+2\right)\left(m^{2}-6\right)=0$
$\Rightarrow \quad m^{2}-6=0$ and $m^{2}+2 \neq 0$
$\Rightarrow \quad m^{2}=6$
$\Rightarrow \quad m=\pm \sqrt{6}$
i.e $m=\sqrt{6}$ as $m$ is positive slope.
$\therefore$
$y=\sqrt{6} x+\frac{2 \sqrt{3}}{\sqrt{6}}$
$\Rightarrow \quad y=\sqrt{6} x+\sqrt{2}$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2014
The equation of the directrices of the hyperbola $3 x^{2}-3 y^{2}-18 x+12 y+2=0$ is

Options:
A $x=3 \pm \sqrt{\frac{13}{6}}$
B $x=3 \pm \sqrt{\frac{6}{13}}$
C $x=6 \pm \sqrt{\frac{13}{3}}$
D $x=6 \pm \sqrt{\frac{3}{13}}$

The correct answer is:
$x=3 \pm \sqrt{\frac{13}{6}}$

Solution:
Given equation is
$$
3 x^{2}-3 y^{2}-18 x+12 y+2=0
$$
It can be written as
$\frac{(x-3)^{2}}{\left(\sqrt{\frac{13}{3}}\right)^{2}}-\frac{(y-2)^{2}}{\left(\sqrt{\frac{13}{3}}\right)^{2}}=1$
$\begin{aligned} \text { here, } a &=b=\sqrt{\frac{13}{3}} \\ \therefore \quad e &=\sqrt{1+\frac{b^{2}}{a^{2}}} \\ &=\sqrt{1+1} \\ &=\sqrt{2} \end{aligned}$
$\therefore$ Equation of the directrix are
$$
x=3 \pm \sqrt{\frac{13}{6}}
$$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2019
The equation of the latus rectum of a parabola is $x+y=8$ and the equation of the tangent at the vertex is $x+y=12$. Then the length of the latus rectum is

Options:
A $4 \sqrt{2}$ unit
B $2 \sqrt{2}$ unit
C 8 unit
D $8 \sqrt{2}$ unit

The correct answer is:
$8 \sqrt{2}$ unit

Solution:
Hint:
The distance between latus rectum and equation of tangent at vertex is 'a'. Here $\mathrm{a}=\frac{4}{\sqrt{1+1}}=2 \sqrt{2}$
So, length of latus rectum $=4 \mathrm{a}=8 \sqrt{2}$ unit

Topic: Mathematics, Parabola, WBJEE, WBJEE 2020
The equation of the locus of the point of intersection of the straight lines \(x \sin \theta+(1-\cos \theta) y=a \sin \theta\) and \(x\) sin \(\theta-(1+\) \(\cos \theta) y+a \sin \theta=0\) is

Options:
A \(y \pm a x\)
B \(x= \pm\) ay
C \(y^2=4 x\)
D \(x^2+y^2=a^2\)

The correct answer is:
\(x^2+y^2=a^2\)

Solution:
Hints: \(y=a \sin \theta\)
\(x=a \cos \theta \text {. }\)
\(x^2+y^2=a^2\)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2011
The equation of the plane through (1,2,-3) and (2,-2,1) and parallel to $X$ -axis is

Options:
A $y-z+1=0$
B $y-z-1=0$
C $y+z-1=0$
D $y+z+1=0$

The correct answer is:
$y+z+1=0$

Solution:
Equation of the plane will be $\left|\begin{array}{ccc}x-1 & y-2 & z+3 \\ 2-1 & -2-2 & 1-(-3) \\ 1 & 0 & 0\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}x-1 & y-2 & z+3 \\ 1 & -4 & 4 \\ 1 & 0 & 0\end{array}\right|=0$
$\Rightarrow \quad 1[4(y-2)+4(z+3)]=0$
[Expanding along $\left.R_{j}\right]$
$\Rightarrow \quad 4 y-8+4 z+12=0$
$\Rightarrow \quad 4 y+4 z+4=0$
$\Rightarrow \quad y+z+1=0$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2017
The equation of the plane through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4$ and parallel to the $x$-axis is

Options:
A $y+3 z+6=0$
B $y+3 z-6=0$
C $y-3 z+6=0$
D $y-3 z-6=0$

The correct answer is:
$y-3 z+6=0$

Solution:
$\quad(2 x+3 y-z+4)+\lambda(x + y+z-1)=0$
$\overrightarrow{\mathrm{n}}=(2+\lambda) \hat{\mathrm{i}}+(3+\lambda) \hat{\mathrm{j}}+(-1+\lambda) \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{n}} \cdot \overrightarrow{\mathrm{i}}=0 \Rightarrow \lambda=-2$
$\Rightarrow \mathrm{y}-3 z+6=0$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2022
The equation of the plane through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

Options:
A $8 x+14 y+13 z+37=0$
B $8 x-14 y-13 z-37=0$
C $8 x-14 y-13 z+37=0$
D None of the above

The correct answer is:
$8 x+14 y+13 z+37=0$

Solution:
The equation of plane passing through \((2,-1,-3)\) is \(a(x-2)+b(y+1)+c(z+3)=0\) Also, as the plane is parallel is the given two lines,
\(\therefore 3 a+2 b-4 c=0\) and \(2 a-3 b+2 c=0\)
\(\Rightarrow \mathrm{a}=-8, \mathrm{~b}=-14, \mathrm{c}=-13\)
\(\therefore\) The equation of the required plane is
\(-8(x-2)-14(y+1)-13(z+3)=0\)
\(\Rightarrow 8 x+14 y+13 z+37=0\)
So the answer is (a).

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2020
The equation of the plane, which bisects the line joining the points (1,2,3) and (3,4,5) at right angles is

Options:
A $x+y+z=0$
B $x+y-z=9$
C $x+y+z=9$
D $x+y-z+9=0$

The correct answer is:
$x+y+z=9$

Solution:
The given points are $A(1,2,3)$ and $B(3,4,5)$ The direction ratios of line segment $A B$ is given by
$$
\begin{array}{l}
\left(x_{2}-x_{1}\right),\left(y_{2}-y_{1}\right),\left(z_{2}-z_{1}\right) \\
(3-1),(4-2),(5-3)=(2,2,2)=(1,1,1)
\end{array}
$$
Since, the plane bisects $A B$ at right angles, $\mathbf{A B}$ is the normal to the plane. Which is n Therefore, $\mathbf{n}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Let $c$ be the mid-point of $A B$
$$
\begin{aligned}
\left(\frac{x_{1}+x_{2}}{2},\right.&\left.\frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right) \\
=& c\left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{3+5}{2}\right) \\
=& c(2,3,4)
\end{aligned}
$$
Let this vector be $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
Hence, the required equation of vector
$$
\{\mathbf{r}-(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \mathbf{k})\}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k})=0
$$
We know that, $\mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
$$
\begin{array}{l}
\Rightarrow\{(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(2\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})\}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \\
\Rightarrow \quad(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
=(2\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
\Rightarrow x+y+z=2+3+4 \\
\Rightarrow x+y+z=9
\end{array}
$$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2019
The equation of the straight line passing through the point $(4,3)$ and making intercepts on the co-ordinate axes whose sum is $-1$ is

Options:
A $\frac{x}{2}-\frac{y}{3}=1$
B $\frac{x}{-2}+\frac{y}{1}=1$
C $-\frac{x}{3}+\frac{y}{2}=1$
D $\frac{x}{1}-\frac{y}{2}=1$

The correct answers are:
$\frac{x}{2}-\frac{y}{3}=1$, $\frac{x}{-2}+\frac{y}{1}=1$

Solution:
Hint:
Let line be $\frac{x}{a}+\frac{y}{b}=1$, then $a+b=-1$
$\frac{4}{a}-\frac{3}{1+a}=1 \Rightarrow a=\pm 2$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2020
The equation of the tangent to the conic $x^2-y^2-8 x+2 y+11=0$ at $(2,1)$ is

Options:
A $x+2=0$
B $2 x+1=0$
C $x+y+1=0$
D $x-2=0$

The correct answer is:
$x-2=0$

Solution:
Hints : Equation of tangent at $\left(x_1, y_1\right)$ is
$$
\begin{aligned}
& x x_1-y y_1-4\left(x+x_1\right)+\left(y+y_1\right)+11=0 \\
& x_1=2 ; y=1
\end{aligned}
$$
$\therefore$ Equation of tangent is
$$
\begin{aligned}
& 2 x-y-4(x+2)+(y+1)+11=0 \\
& \text { or } \quad-2 x-8+12=0
\end{aligned}
$$
$$
\begin{array}{ll}
\text { or } & -2 x+4=0 \\
\text { or } & 2 x=4 \\
\text { or } & x=2 \\
\text { or } & x-2=0
\end{array}
$$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2009
The equation $r \cos \left(\theta-\frac{\pi}{3}\right)=2$ represents

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

The correct answer is:
a straight line

Solution:
Hint:
$r \cos \left(\theta-\frac{\pi}{3}\right)=2 \quad \Rightarrow r \cos \theta \times \frac{1}{2}+r \sin \theta \times \frac{\sqrt{3}}{2}=2 \Rightarrow x+\sqrt{3} y=4(x=r \cos \theta, y=r \sin \theta)$
$\therefore$ a straight line

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2020
The equation \(r^2 \cos ^2\left(\theta-\frac{\pi}{3}\right)=2\) represents

Options:
A a parabola
B a hyperbola
C a circle
D a pair of straight lines

The correct answer is:
a pair of straight lines

Solution:
\(\begin{aligned}
& \text {Hint: } r^2\left(\frac{\cos \theta}{2}+\frac{\sqrt{3}}{2} \sin \theta\right)^2=2 \\
& \Rightarrow \quad(r \cos \theta+\sqrt{3} r \sin \theta)^2=8 \\
& \Rightarrow \quad(x+y \sqrt{3})^2=8 \\
& \Rightarrow \quad(x+y \sqrt{3}-2 \sqrt{2})(x+y \sqrt{3}+2 \sqrt{2})=0 \\
& \Rightarrow \quad \text { a pair of straight lines }
\end{aligned}\)

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2023
The equation $\sin x(\sin x+\cos x)=k$ has real solutions, where k is a real number. Then,

Options:
A $0 \leq k \leq \frac{1+\sqrt{2}}{2}$
B $2-\sqrt{3} \leq k \leq 2+\sqrt{3}$
C $0 \leq k \leq 2-\sqrt{3}$
D $\frac{1-\sqrt{2}}{2} \leq k \leq \frac{1+\sqrt{2}}{2}$

The correct answer is:
$\frac{1-\sqrt{2}}{2} \leq k \leq \frac{1+\sqrt{2}}{2}$

Solution:
Let, $f(x)=\sin x(\sin x+\cos x)$
$=\sin ^{2} x+\sin x \cos x$
$=\frac{1-\cos 2 x}{2}+\frac{2 \sin x \cos x}{2}$
$=\frac{1}{2}-\frac{1}{2} \cos 2 x+\frac{1}{2} \sin 2 x$
$=\frac{1}{2}+\frac{1}{2}(\sin 2 x-\cos 2 x)$
$\begin{aligned} \text { Now, }-\sqrt{(1)^{2}+(1)^{2}} & \leq \sin 2 x-\cos 2 x \\ & \leq \sqrt{(1)^{2}+(1)^{2}} \end{aligned}$
$\left[\because \sqrt{a^{2}+b^{2}} \leq a \sin x+b \cos x \leq \sqrt{a^{2}+b^{2}}\right]$
$\Rightarrow \quad-\sqrt{2} \leq \sin 2 x-\cos 2 x \leq \sqrt{2}$
$\Rightarrow \quad-\frac{\sqrt{2}}{2} \leq \frac{\sin 2 x-\cos 2 x}{2} \leq \frac{\sqrt{2}}{2}$
$\Rightarrow \frac{1}{2}-\frac{\sqrt{2}}{2} \leq \frac{1}{2}+\frac{\sin 2 x-\cos 2 x}{2} \leq \frac{1}{2}+\frac{\sqrt{2}}{2}$
$\Rightarrow \frac{1-\sqrt{2}}{2} \leq f(x) \leq \frac{1+\sqrt{2}}{2}$
$\therefore \quad \frac{1-\sqrt{2}}{2} \leq k \leq \frac{1+\sqrt{2}}{2}$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2017
The equation $x^{3}-y x^{2}+x-y=0$ represents

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

The correct answer is:
a straight line

Solution:
$x^{3}-y x^{2}+x-y=0$
$x^{2}(x-y)+(x-y)=0$
$\left(x^{2}+1\right)(x-y)=0$
$x^{2}+1 \neq 0$
$x-y=0$
$x=y$
So, the equation represents a straight line.

Topic: Mathematics, Pair of Lines, WBJEE, WBJEE 2016
The equation $x^{\left(\log _{3} x\right)^{2}-\frac{9}{2} \log _{3} x+5}=3 \sqrt{3}$ has

Options:
A at least one real root
B exactly one real root
C exactly one irrational root
D complex roots

The correct answers are:
at least one real root, exactly one irrational root

Solution:
Hint $: 2 t^{3}-9 t^{2}+10 t-3=0, t=\log _{3} x$
$\Rightarrow(t-3)(t-1)\left(t-\frac{1}{2}\right)=0$
$\Rightarrow x=3^{3}, 3^{1}, 3^{1 / 2}$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2020
The equation $x \log x=3-x$

Options:
A has no root in (1,3)
B has exactly one root in (1,3)
C $x \log x-(3-x)>0$ in [1,3]
D $x \log x-(3-x) < 0$ in [1,3]

The correct answer is:
has exactly one root in (1,3)

Solution:
Let $\quad f(x)=x \log x+x-3$
$\Rightarrow \quad f^{\prime}(x)=x \cdot \frac{1}{x}+\log x+1$
$\Rightarrow \quad f^{\prime}(x)=\log x+2>0$
$\Rightarrow f(1)=-2$ and $f(3)=3 \log 3, f(1) \cdot f(3 < 0)$
Hence, exactly one root in $x \in(1,3)$ as $f(x)>0$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2018
The equation $y^{2}+4 x+4 y+k=0$ represents a parabola whose latusrectum is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
$y^{2}+4 x+4 y+k=0$ $y^{2}+2 \times 2 y+4-4+4 x+k=0$
$(y+2)^{2}=-4 x-k+4$
$(y+2)^{2}=-4\left(x-\frac{4+k}{4}\right)$
$\therefore$ Latusrectum $=4$ units

Topic: Mathematics, Parabola, WBJEE, WBJEE 2012
The equation $z \bar{z}+(2-3 i) z+(2+3 i) \bar{z}+4=0$ represents a circle of radius

Options:
A 2 unit
B 3 unit
C 4 unit
D 6 unit

The correct answer is:
3 unit

Solution:
Hint:
Centre and radius of $z \bar{z}+\bar{a} z+a \bar{z}+b=0$ are $-a$ and $\sqrt{|a|^{2}-b} \therefore$ radius $=\sqrt{13-4}=3$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2020
The equations of the circles, which touch both the axes and the line $4 x+3 y=12$ and have centres in the first quadrant, are

Options:
A $x^{2}+y^{2}+x-y+1=0$
B $x^{2}+y^{2}-2 x-2 y+1=0$
C $x^{2}+y^{2}-12 x-12 y+36=0$
D $x^{2}+y^{2}-6 x-6 y+36=0$

The correct answers are:
$x^{2}+y^{2}-2 x-2 y+1=0$, $x^{2}+y^{2}-12 x-12 y+36=0$

Solution:
Radius $(r)=$ perpendicular distance on line $4 x+3 y=12$ from centre

$\Rightarrow \quad r=\frac{|4 r+3 r-12|}{\sqrt{16+9}}$
$\Rightarrow \quad|7 r-12|=5 r$
$\Rightarrow \quad 7 r-12=\pm 5 r$
$\therefore \quad 2 r=12 \Rightarrow r=6$
and $12 r=12 \Rightarrow r=1$
(1) When centre is (1,1) and radius is $1,$ then equation of circle is
$$
(x-1)^{2}+(y-1)^{2}=1
$$
$\Rightarrow \quad x^{2}+y^{2}-2 x-2 y+1=0$
(ii) When centre is (6,6) and radius is 2 , then equation of circle is
$(x-6)^{2}+(y-6)^{2}=36$
$\Rightarrow \quad x^{2}+y^{2}-12 x-12 y+36=0$

Topic: Mathematics, Circle, WBJEE, WBJEE 2013