datavorous/entrance-exam-dataset huggingface dataset:
The equation of the tangent at the point $(a \sec \theta, b \tan \theta)$ of the conic $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, is

Options:
A $x \sec ^2 \theta-y \tan ^2 \theta=1$
B $\frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1$
C $\frac{x+a \sec \theta}{a^2}-\frac{y+b \tan \theta}{b^2}=1$
D None of these

The correct answer is:
$\frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1$

Solution:
$\frac{x(a \sec \theta)}{a^2}-\frac{y(b \tan \theta)}{b^2}=1 \Rightarrow \frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1$ as tangent equation is, \(\frac{x \cdot x_1}{a^2}-\frac{y \cdot y_1}{b^2}\)

Topic: Mathematics, Hyperbola, JEE Main
The equation of the tangent line at the point \( (4,2) \) to the curve with parametric equation given by \( x=t^{2} \) and \( y=t^{3}-3 t \)
where \( t \) is parameter is

Options:
A \( y=15 x-58 \)
B \( y=2 \)
C \( y=x-2 \)
D \( y=\frac{9 x}{4}-7 \)

The correct answer is:
\( y=\frac{9 x}{4}-7 \)

Solution:
Slope M=dydx=dy/dtdx/dt

∴ M=3t2-32t at t=2⇒M=94

Equation of tangent at 4,2 is M=94

∴ y-2=94x-4⇒y=94x-7.

Topic: Mathematics, Application of Derivatives, JEE Main
The equation of the tangent parallel to $y-x+5=0$ drawn to $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$ is

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

The correct answer is:
$x-y-1=0$

Solution:
Given hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$ ...(i)
Equation of tangent parallel to $y-x+5=0$ is
$$
\begin{aligned}
& & y-x+\lambda &=0 & \\
\Rightarrow & & y &=x-\lambda ...(ii)
\end{aligned}
$$
If line (ii) is a tangent to hyperbola (i), then
$$
\begin{array}{rc}
-\lambda=\pm \sqrt{3 x-2} \\
\left(\text { from } c=\pm \sqrt{a^{2} m^{2}-b^{2}}\right) \\
\Rightarrow & -\lambda=\pm 1 \\
\Rightarrow & \lambda=-1,+1
\end{array}
$$
Put the values of $\lambda$ in Eq. (ii), we get $x-y-1=0 \quad$ and $\quad x-y+1=0$ are the required tangents.

Topic: Mathematics, Application of Derivatives, JEE Main
The equation of the tangent to the curve $y=4 x e^{x}$ at $\left(-1, \frac{-4}{e}\right)$ is

Options:
A $y=-1$
B $y=-\frac{4}{e}$
C $x=-1$
D $x=\frac{-4}{e}$

The correct answer is:
$y=-\frac{4}{e}$

Solution:
iven curve is $y=4 x e^{x}$ $\frac{d y}{d x}=4 e^{x}+4 x e^{x}$
At $\left(-1,-\frac{4}{e}\right)\left(\frac{d y}{d x}\right)_{(-1,-4 / e)}=4 e^{-1}+4(-1) e^{-1}=0$
$\therefore$ Equation of tangent is $\begin{aligned} &\left(y+\frac{4}{e}\right) &=0(x+1) \\ \Rightarrow & y &=-\frac{4}{e} \end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
The equation of the tangent to the curve $y=x+\frac{4}{x^2}$, that is parallel to the $x$-axis, is

Options:
A
$y=1$

B
$y=2$

C
$y=3$

D
$y=0$

The correct answer is:
$y=3$

Solution:
Parallel to $x$-axis $\quad \Rightarrow \frac{d y}{d x}=0 \quad \Rightarrow 1-\frac{8}{x^3}=0$ $\Rightarrow x=2 \quad \Rightarrow y=3$
Equation of tangent is $y-3=0(x-2) \quad \Rightarrow y-3=0$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2010
The equation of the tangent to the ellipse $x^2+16 y^2-16$ making an angle of $60^{\circ}$ with $x$-axis is

Options:
A $\sqrt{3} x-y+7=0$
B $\sqrt{3} x-y-7=0$
C $\sqrt{3} x-y \pm 7=0$
D None of these

The correct answer is:
$\sqrt{3} x-y \pm 7=0$

Solution:
We have, $x^2+16 y^2=16$
$\Rightarrow \frac{x^2}{4^2}+\frac{y^2}{1^2}=1$
This is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a^2=16$ and $b^2=1$
So, the equation of the tangents with slope $m$ are
$y=m \times \pm \sqrt{a^2 m^2+b^2}$
Given, $m=\tan 60^{\circ}$
$\Rightarrow m=\sqrt{3}$
From equation (1), we get
$\begin{aligned} & y=\sqrt{3} x \pm \sqrt{16 \times 3+1} \\ & \Rightarrow y=\sqrt{3} x \pm 7\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
The equation of the tangent to the hyperbola $2 x^2-3 y^2=6$ which is parallel to the line $y=3 x+4$, is

Options:
A $y=3 x+5$
B $y=3 x-5$
C $y=3 x+5$ and $y=3 x-5$
D None of these

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

Solution:
Let tangent be $y=3 x+c$
$c= \pm \sqrt{a^2 m^2-b^2}= \pm \sqrt{3.9-2}= \pm 5 \Rightarrow y=3 x \pm 5 .$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the tangent to the hyperbola $4 y^2=x^2-1$ at the point $(1,0)$ is

Options:
A $x=1$
B $y=1$
C $y=4$
D $x=4$

The correct answer is:
$x=1$

Solution:
The equation of the tangent to $4 y^2=x^2-1$ at $(1,0)$ is $4(y \times 0)=x \times 1-1$ or $x-1=0$ or $x=1$.

Topic: Mathematics, Hyperbola, JEE Main
The equation of the tangent to the parabola $y^2=4 a x$ at point $\left(a / t^2, 2 a / t\right)$ is

Options:
A $t y=x t^2+a$
B $t y=x+a t^2$
C $y=t x+a t^2$
D $y=t x+\left(a / t^2\right)$

The correct answer is:
$t y=x t^2+a$

Solution:
Equation of the tangent to the parabola,
$y^2=4 a x$ is $y y_1=2 a\left(x+x_1\right)$
$\Rightarrow \quad y \cdot \frac{2 a}{t}=2 a\left(x+\frac{a}{t^2}\right)$
$\Rightarrow \frac{\underline{y}}{t}=\left(x+\frac{a}{t^2}\right) \Rightarrow \frac{y}{t}=\frac{t^2 x+a}{t^2}$ $\Rightarrow t y=t^2 x+a$

Topic: Mathematics, Parabola, JEE Main
The equation of the tangent to the parabola $y^2=4 x+5$ parallel to the line $y=2 x+7$ is

Options:
A $2 x-y-3=0$
B $2 x-y+3=0$
C $2 x+y+3=0$
D None of these

The correct answer is:
$2 x-y+3=0$

Solution:
Equation of parabola is $Y^2=4 X$, where $X=x+\frac{5}{4}$
Tangent parallel to $Y=2 X+7$ is $Y=2 X+\frac{a}{m}$ where $a=1, m=2$
$\Rightarrow y=2\left(x+\frac{5}{4}\right)+\frac{1}{2} \Rightarrow y=2 x+3$ i.e., $2 x-y+3=0$

Topic: Mathematics, Parabola, JEE Main
The equation of the tangent to the parabola y2=4x+5 which is parallel to the line y=2x+7 is

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

The correct answer is:
y=2x+3

Solution:
y2=4x+5 is same as y2=4x+54
The equation of the tangent to y2=4x+54 which is parallel to y=2x+7 is

y=2x+54+12
ây=2x+3

Topic: Mathematics, Parabola, JEE Main
The equation of the tangent to the parabola y2=4x whose slope is positive and which also touches x2+y2=12 is

Options:
A y=x+1
B y=2x+1
C x+y=2
D y=4x+12

The correct answer is:
y=x+1

Solution:
Tangent to parabola is y=mx+1m
Tangent to circle is y=mxÂ±121+m2
For common tangent
1m2=121+m2⇒m4+m2-2=0
⇒m2=1⇒m=1

Topic: Mathematics, Parabola, JEE Main
The equation of the tangent to the parabola $y=x^2-x$ at the point where $x=1$, is

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

The correct answer is:
$y=x-1$

Solution:
Point $(1,0) \Rightarrow\left(\frac{d y}{d x}\right)_{(1,0)}=2-1=1$
Hence the tangent is $y-0=x-1$.

Topic: Mathematics, Parabola, JEE Main
The equation of the tangents drawn at the ends of the major axis of the ellipse $9 x^2+5 y^2-30 y=0$, are

Options:
A $y= \pm 3$
B $x= \pm \sqrt{5}$
C $y=0, y=6$
D None of these

The correct answer is:
$y=0, y=6$

Solution:
Given equation of ellipse is
$9 x^2+5 y^2-30 y=0$
The above equation can be rewritten as
$\begin{aligned} & 9 x^2+5\left(y^2-6 y\right)=0 \\ & \Rightarrow 9 x^2+5(y-6 y+9)=45 \\ & \Rightarrow 9 x^2+5(y-3)^2=45 \\ & \Rightarrow \frac{x^2}{5}+\frac{(y-3)^2}{9}=1\end{aligned}$
Since, $\mathrm{a}^2 \lt \mathrm{b}^2$, so axis of ellipse is $\mathrm{Y}$-axis.
To find vertex at $\mathrm{Y}$-axis, put $\mathrm{x}=0$.
$\begin{aligned} & \therefore 5(y-3)^2=45 \\ & \Rightarrow(y-3)^2=9 \\ & \Rightarrow y-3= \pm 3 \\ & \Rightarrow y=0 \text { or } y=6\end{aligned}$
Therefore, tangents at vertex are $\mathrm{y}=0, \mathrm{y}=6$.

Topic: Mathematics, Ellipse, JEE Main
The equation of the tangents to the conic $3 x^2-y^2=3$ perpendicular to the line $x+3 y=2$ is

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

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

Solution:
Tangent to $\frac{x^2}{1}-\frac{y^2}{3}=1$ and perpendicular to $x+3 y-2=0$ is given by $y=3 x \pm \sqrt{9-3}=3 x \pm \sqrt{6}$
As, \(m=3\) (for perpendicular) So, tangent en is,
\(\begin{aligned}
& y=m x \pm \sqrt{a^2 m^2-b^2} \\
& y=3 x \pm \sqrt{1 \cdot 9-(\sqrt{3})^2} \Rightarrow y = 3 \times \pm \sqrt{6}
\end{aligned}\)

Topic: Mathematics, Hyperbola, JEE Main
The equation of the tangents to the hyperbola $3 x^2-4 y^2=12$ which cuts equal intercepts from the axes, are

Options:
A $y+x= \pm 1$
B $y-x= \pm 1$
C $3 x+4 y= \pm 1$
D $3 x-4 y= \pm 1$

The correct answer is:
$y-x= \pm 1$

Solution:
The tangent at $(h, k)$ is $\frac{x}{4 / h}-\frac{y}{3 / k}=1$
$\therefore \frac{4}{h}=\frac{3}{k} \Rightarrow \frac{h}{k}=\frac{4}{3}$ ....(i)
and $3 h^2-4 k^2=12$ ....(ii)
As point ${ }^{(h, k)}$ lies on it, using (i) and (ii), we get the tangent as $y-x= \pm 1$.

Topic: Mathematics, Hyperbola, JEE Main
The equation of the transverse and conjugate axis of the hyperbola $16 x^2-y^2+64 x+4 y+44=0$ are

Options:
A $x=2, y+2=0$
B $x=2, y=2$
C $y=2, x+2=0$
D None of these

The correct answer is:
$y=2, x+2=0$

Solution:
$\begin{aligned}
& (4 x+8)^2-(y-2)^2=-44+64-4 \\
& \Rightarrow \quad \frac{16(x+2)^2}{16}-\frac{(y-2)^2}{16}=1
\end{aligned}$
Transverse and conjugate axes are $y=2, x=-2$.

Topic: Mathematics, Hyperbola, JEE Main
The equation of wave is given by Y=10-2sin2π160t-0.5x+π4, where x and Y are in m and t in s. The speed of the wave is _______ km h-1.

Options:

The correct answer is:
1152

Solution:
The equation of the wave is

Y=10-2sin2π(160t-0.5x+π4)

The velocity of a wave is given by v=ωk ...(i).

A wave equation is given by

Y=Asinωt-kx+ϕ

Using equation (i),

v=2π×1602π×0.5=320 m s-1

⇒v=320×185

⇒v=1152 km h-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The equation $\left(P+\frac{a}{V^2}\right)(V-b)$ constant. The units of a are

Options:
A Dyne $\times \mathrm{cm}^5$
B Dyne $\times \mathrm{cm}^4$
C Dyne $/ \mathrm{cm}^3$
D Dyne $/ \mathrm{cm}^2$

The correct answer is:
Dyne $\times \mathrm{cm}^4$

Solution:
Units of $a$ and $P V^2$ are same and equal to $dyne \times \mathrm{cm}^4$.

Topic: Physics, Units and Dimensions, JEE Main
The equation, \( \pi^{x}=-2 x^{2}+6 x-9 \) has

Options:
A no solution
B one solution
C two solutions
D infinite solutions

The correct answer is:
no solution

Solution:
πx=-2x2+6x-9

For -2x2+6x-9, leading coefficient is negative implies that graph will be concave down.

Maximum value of -2x2+6x-9 occurs at x=-b2a=-6-2=3.

Maximum values is -232+6×3-9=-18+18-9=-9.

This implies range of -2x2+6x-9 is (-∞,-9].

πx is increasing exponential function which is always positive.

πx∈0,∞

Left-hand side is always positive and right-hand side is always negative implies both cannot be equal.

⇒No solution.

Topic: Mathematics, Quadratic Equation, JEE Main
The equation $\frac{1}{r}=\frac{1}{8}+\frac{3}{8} \cos \theta$ represents

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

The correct answer is:
a hyperbola

Solution:
We have, $\frac{1}{r}=\frac{1}{8}+\frac{3}{8} \cos \theta$
$$
\begin{array}{ccrl}
\Rightarrow & & 8=r+3 r \cos \theta \\
\Rightarrow & r & =8-3 x \\
\Rightarrow & r^2 & =(8-3 x)^2 \\
\Rightarrow & & x^2+y^2=64+9 x^2-48 x \\
\Rightarrow & & 8 x^2-y^2-48 x+64=0
\end{array}
$$
Which represents a hyperbola.

Topic: Mathematics, Ellipse, JEE Main
The equation tan4x-2sec2x+a2=0 will have at least one solution, if

Options:
A a≤4
B a≤2
C a≤3
D a>2

The correct answer is:
a≤3

Solution:
tan4x-2sec2x+a2=0
tan4x-21+tan2x+a2=0
tan4x-2tan2x+1=3-a2
tan2x-12≥0⇒3-a2≥0
⇒a2-3≤0⇒a≤3

Topic: Mathematics, Trigonometric Equations, JEE Main
The equation that is incorrect is:

Options:
A Λm0KBr-Λm0NaCl=Λm0KBr-Λm0KCl
B Λm0KG-Λm0KQ=Λm0NaG-Λm0NaQ
C Îm0H2O=Îm0HCl+Îm0NaOH-Îm0NaCl
D Λm0NaBr-Λm0Nal=Λm0KBr-Λm0KI

The correct answer is:
Λm0KBr-Λm0NaCl=Λm0KBr-Λm0KCl

Solution:
Option (1) is incorrect.
According to Kohlrausch's law correct expression is
Λ m 0 NaBr − Λ m 0 Nal = Λ m 0 KBr − Λ m 0 KI
The other statements are correct.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The equation that represents magnetic field of a plane electromagnetic wave which is propagating along $x$-direction with wavelength $10 \mathrm{~mm}$ and maximum electric field $60 \mathrm{Vm}^{-1}$ in $y$-direction is (where, $c=$ speed of light)

Options:
A $\left(6 \times 10^{-7}\right) \sin [0.2 \pi(c t-x)] \hat{k} T$
B $\left(2 \times 10^{-7}\right) \sin [200 \pi(c t-x)] \hat{k} T$
C $\left(2 \times 10^{-7}\right) \sin [200 \pi(c t-x)] \hat{i} \mathrm{~T}$
D $\left(6 \times 10^{-7}\right) \sin [02 \pi(c t-x)] \hat{i} T$

The correct answer is:
$\left(2 \times 10^{-7}\right) \sin [200 \pi(c t-x)] \hat{k} T$

Solution:
Equation of magnetic field of an electromagnetic wave is given by
$B=B_0 \sin (\omega t-k x) \hat{k}$
$=B_0 \sin \left(2 \pi f t-\frac{2 \pi}{\lambda} \cdot x\right) \hat{k} \quad\left[\because \omega=2 \pi f\right.$ and $\left.k=\frac{2 \pi}{\lambda}\right]$
$=B_0 \sin \frac{2 \pi}{\lambda}[f \lambda t-x] \hat{k}$
$B=B_0 \sin \frac{2 \pi}{\lambda}[c t-x] \hat{k}$ ...(i)
Also direction of $\mathbf{B}$ vector is perpendicular to $\mathbf{E}$ and also to direction of propagation
And $B_0=\frac{E_0}{c}$
Here, $E_0=60 \mathrm{Vm}^{-1}$
$\therefore \quad B_0=\frac{60^{}}{3 \times 10^8}=2 \times 10^{-7}$
Also $k=\frac{2 \pi}{\lambda}$
Here $k=200 \pi=\frac{2 \pi}{\lambda}$
$\Rightarrow \quad \lambda=\frac{1}{100} \mathrm{~m}=\frac{100}{100} \mathrm{~cm}=10 \mathrm{~mm}=10^{-2} \mathrm{~m}$
Putting the values is eq. (i), we get
$B=2 \times 10^{-7} \sin \left[\frac{2 \pi}{10^{-2}}(c t-x)\right] \hat{k}$
$=2 \times 10^{-7} \sin 200 \pi(t-x) \hat{k}$

Topic: Physics, Electromagnetic Waves, JEE Main
The equation that represents the water-gas shift reaction is :

Options:
A CH4g+H2Og1270 KNiCOg+3H2g
B 2Cs+O2g+4N2g⟶1273K2COg+4N2g
C Cs+H2Og⟶1270 KCOg+H2g
D COg+H2Og →catalyst673K CO2g + H2g

The correct answer is:
COg+H2Og →catalyst673K CO2g + H2g

Solution:
COg+H2Og →catalyst673K CO2g + H2g

Reaction is called water gas shift reaction

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The equation to the normal to the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ at $(-4,0)$ is

Options:
A $2 x-3 y=1$
B $x=0$
C $x=1$
D $y=0$

The correct answer is:
$y=0$

Solution:
We have,
$$
\frac{x^{2}}{16}-\frac{y^{2}}{9}=1
$$
Since, $(-4,0)$ is vertex of the hyperbola, so equation of normal at $(-4,0)$ is $y=0$.

Topic: Mathematics, Hyperbola, JEE Main
The equation which is balanced and represents the correct product(s) is

Options:
A Li 2 O + 2 KCl → 2 LiCl + K 2 O
B CoClNH35++5H+→Co2++5NH4++Cl-
C MgH2O62++EDTA4-→excess NaOHMgEDTA2++6H2O
D CuSO4+4KCN→K2CuCN4+K2SO4

The correct answer is:
CoClNH35++5H+→Co2++5NH4++Cl-

Solution:
Li 2 O + 2 KCl → 2 LiCl + K 2 O is incorrect since, KCl is less reactive than LiCl and will not form LiCl.
MgH2O62++EDTA4-→excess NaOHMgEDTA2++6H2O is incorrect since, the charge on the complex will be -2, not +2.
CuSO4+4KCN→K2CuCN4+K2SO4 is incorrect since, the final complex formed is K3CuCN4, not K2CuCN4.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2014 (06 Apr)
The equation $x^2-2 x y+y^2+3 x+2=0$ represents

Options:
A A parabola
B An ellipse
C A hyperbola
D A circle

The correct answer is:
A parabola

Solution:
$\Delta=(1)(1)(2)+2\left(\frac{3}{2}\right)(0)(-1)-(1)(0)^2-$ (1) $\left(\frac{3}{2}\right)^2-2(-1)^2$
$=2-\frac{9}{4}-2 \lt 0$ and $h^2-a b=1-1=0$
i.e., $h^2=a b \Rightarrow$ a parabola.

Topic: Mathematics, Parabola, JEE Main
The equation $x^2-2 x y+y^2+3 x+2=0$ represents

Options:
A A parabola
B An ellipse
C A hyperbola
D A circle

The correct answer is:
A parabola

Solution:
$\begin{aligned} & \left. \Delta=(1)(1)(2)+2\left(\frac{3}{2}\right)(0)(-1)-(1)(0)^2-(1)\left(\frac{3}{2}\right)^2-2(-1)^2\right. \\ & =2-\frac{9}{4}-2 \lt 0 \text { and } h^2-a b=1-1=0 . \\ & \text { i.e., } h^2=a b \Rightarrow \text { a parabola. }\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
The equation $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$, where $-2 < x < 2$, represents a

Options:
A Circle
B Pair of lines
C Parabola
D Line segment

The correct answer is:
Line segment

Solution:
Given,
$$
\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4
$$
According above equation
Let
$$
\begin{aligned}
P & =(x, y) \\
A & =(2,0) \\
B & =(-2,0)
\end{aligned}
$$
From Eq. (i)
$$
P A+P B=A B
$$
$\Rightarrow$ Eq. $A, P, B$ are collinear
$\Rightarrow$ Eq. (i) represents a line segment
Hence, option (4) is correct.

Topic: Mathematics, Straight Lines, JEE Main
The equation $x^2+4 x y+y^2+2 x+4 y+2=0$ represents

Options:
A An ellipse
B A pair of straight lines
C A hyperbola
D None of these

The correct answer is:
A hyperbola

Solution:
Obviously $h^2 \gt a b$ and $\Delta=(1)(1)(2)+2(2)(1)(2)-(1)(2)^2-(1)(1)^2-2(2)^2 \lt 0$ Hence it is a hyperbola.

Topic: Mathematics, Hyperbola, JEE Main
The equation $x^2-16 x y-11 y^2-12 x+6 y+21=0$ represents

Options:
A Parabola
B Ellipse
C Hyperbola
D Two straight lines

The correct answer is:
Hyperbola

Solution:
$\Delta \neq 0, h^2 \gt a b$, condition for hyperbola, So equation is hyperbola.

Topic: Mathematics, Hyperbola, JEE Main
The equation x3+3x2+6x+3-2cosx=0 has n solution(s) in 0,1, then the value of n+2 is equal to

Options:

The correct answer is:
2

Solution:
Let fx=x3+3x2+6x+3-2cosx
f'x=3x2+6x+6+2sinx
f ′ x =3 x 2 +2x+2 +2sinx
f ′ x is always positive as the minimum value of 3x2+2x+2 is 3 and that of 2sin⁡x is -2, so fx is increasing in 0,1
f0=1, f1=13-2cos⁡x>0
fx=0 has no solution in 0,1
n=0⇒n+2=2

Topic: Mathematics, Application of Derivatives, JEE Main
The equation $x^3+x-1=0$ has

Options:
A no real root.
B exactly two real roots.
C exactly one real root.
D more than two real roots.

The correct answer is:
exactly one real root.

Solution:
Let $\mathrm{f}(x)=x^3+x-1$
A root of $\mathrm{f}(x)$ exists, if $\mathrm{f}(x)=0$ for at least one value of $x$.
$$
\begin{aligned}
& \mathrm{f}(0)=-1 < 0 \\
& \mathrm{f}(1)=1>0
\end{aligned}
$$
$\therefore \quad$ By intermediate value theorem, there has to be a point ' $c$ ' between 0 and 1 such that $\mathrm{f}(x)=0$.
$\therefore \quad$ The given equation has exactly one real root.
Alternate Method:
Let $\mathrm{f}(x)=x^3+x-1$
$$
\begin{array}{ll}
\therefore \quad \mathrm{f}^{\prime}(x)=3 x^2+1 \\
\quad \Rightarrow \mathrm{f}^{\prime}(x)>0 \quad \forall x \in \mathrm{R}
\end{array}
$$
$\Rightarrow \mathrm{f}(x)$ is an increasing function.
$\Rightarrow \mathrm{f}(x)$ intersects $\mathrm{X}$-axis at only one point.
$\therefore \quad$ The given equation has exactly one real root.

Topic: Mathematics, Quadratic Equation, JEE Main
The equation \( x^{2}-2 x y+y^{2}+3 x+2 \) represents

Options:
A A parabola
B An ellipse
C A hyperbola
D A circle

The correct answer is:
A parabola

Solution:
Given equation is x2-2xy+y2+3x+2=0

Comparing the above equation with ax2+2hxy+by2+2gx+2fy+c=0
Here, a=1, b=1, h=-1, g=32, f=0, c=2
Now, Δ=abc+2fgh-af2-bg2-ch2
⇒Δ=112+2032-1--102-1322-2-12⇒Δ=-94
i.e., Δ≠0 and h2-ab=1-1=0
i.e., h2=ab

given equation represents a parabola

Topic: Mathematics, Parabola, JEE Main
The equation y=sinxsin⁡x+2-sin2⁡(x+1) represents a straight line lying in:

Options:
A first, third and fourth quadrants
B second and third quadrants only
C first, second and fourth quadrants
D third and fourth quadrants only

The correct answer is:
third and fourth quadrants only

Solution:
y=sin⁡(x+2)sin⁡x-sin2⁡x+1
=122sin⁡x+2sin⁡x-2sin2⁡x+1

∵ 2sin⁡Asin⁡B =cos⁡A-B-cos⁡A+B & cos⁡2A=1-2sin2⁡A
=12cos⁡x+2-x-cos⁡2x+2-1-cos⁡2x+1
=12cos⁡2-cos⁡2x+2-1+cos⁡2x+2
=121-cos⁡2=-122sin2⁡1
=-sin2⁡1 ( constant and negative )
Hence, the straight line lies in the third and fourth quadrants only.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The equation |z-i|=|z-1|,i=-1, represents:

Options:
A a circle of radius 12
B a circle of radius 1
C the line through the origin with slope 1
D the line through the origin with slope -1

The correct answer is:
the line through the origin with slope 1

Solution:
Let z=x+iy

z-i=z-1⇒x2+(y-1)2=(x-1)2+y2

⇒y=x
Hence, the equation represents a line through the origin with slope 1 .

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (12 Apr Shift 1)
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, JEE Main
The equations $2 x+3 y+4=0 ; 3 x+4 y+6=0$ and $4 x+5 y+8=0$ are

Options:
A consistent with unique solution
B inconsistent
C consistent with infinitely many solutions
D None of the above

The correct answer is:
consistent with unique solution

Solution:
Consider first two equations :
$2 x+3 y=-4$ and $3 x+4 y=-6$
We have $\Delta=\left|\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right|=-1 \neq 0$
$$
\Delta_{\mathrm{x}}=\left|\begin{array}{ll}
-4 & 3 \\
-6 & 4
\end{array}\right|=2 \quad \text { and } \quad \Delta_{\mathrm{y}}=\left|\begin{array}{ll}
2 & -4 \\
3 & -6
\end{array}\right|=0
$$
$$
\therefore \mathrm{x}=-2 \text { and } \mathrm{y}=0
$$
Now this solution satisfies all the equations, so the equations are consistent with unique solution.

Topic: Mathematics, Determinants, JEE Main
The equations kx2+x+k=0 and kx2+kx+1=0 have exactly one root in common for

Options:
A k=-12,1
B k=1
C k=-12
D k=12

The correct answer is:
k=-12

Solution:
Let α be the common root
⇒ kα2+α+k=0
kα2+kα+1=0
On solving, we get,

α21-k2=αk2-k=1k2-k
α2α=1-k2k2-k and α1=k2-kk2-k
⇒ α=1-k2k2-k=1⇒k2-k=1-k2
⇒ 2k2-k-1=0⇒k=-12,1
For k=1, equations are identical, thus not possible
Hence, k=-12

Topic: Mathematics, Quadratic Equation, JEE Main
The equations of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the $x$ and $y$-axes respectively are

Options:
A $x^{2}+y^{2} \pm 4 x \pm 8 y=0$
B $x^{2}+y^{2} \pm 2 x \pm 4 y=0$
C $x^{2}+y^{2} \pm 8 x \pm 16 y=0$
D $x^{2}+y^{2} \pm x \pm y=0$

The correct answer is:
$x^{2}+y^{2} \pm 4 x \pm 8 y=0$

Solution:
$\ln \Delta O A C, O C^{2}=2^{2}+4^{2}=20$
$\therefore$ Required equation of circle is
$$
\begin{array}{l}
(x \pm 2)^{2}+(y \pm 4)^{2}=20 \\
\Rightarrow x^{2}+y^{2} \pm 4 x \pm 8 y=0
\end{array}
$$

Topic: Mathematics, Circle, JEE Main
The equations of the lines L1 and L2 are y=mx and y=nx, respectively. Suppose L1 makes twice as large an angle with the horizontal (measured counterclockwise from the positive x-axis) as does L2 and m=4n, then the value of m2+4n2m2-6n2 is equal to (where, nâ 0)

Options:
A 3
B -3
C 2
D -2

The correct answer is:
2

Solution:
Let, m=tan⁡2θ and n=tan⁡θ
And m=4n and m≠0,

∴tan⁡2θ≠0; θ≠0
tan⁡2θ=4tan⁡θ
2tan⁡θ1-tan2⁡θ=4tan⁡θ
∴1=2-2tan2⁡θ⇒2tan2⁡θ=1

⇒tan2⁡θ=12=n2
⇒m2=16n2=8⇒m2+4n2m2-6n2=105=2

Topic: Mathematics, Straight Lines, JEE Main
The equations of the lines passing through the point $(1,0)$ and at a distance $\frac{\sqrt{3}}{2}$ from the origin, are

Options:
A $\sqrt{3} x+y-\sqrt{3}=0, \sqrt{3} x-y-\sqrt{3}=0$
B $\sqrt{3} x+y+\sqrt{3}=0, \sqrt{3} x-y+\sqrt{3}=0$
C $x+\sqrt{3} y-\sqrt{3}=0, x-\sqrt{3} y-\sqrt{3}=0$
D None of these

The correct answer is:
$\sqrt{3} x+y-\sqrt{3}=0, \sqrt{3} x-y-\sqrt{3}=0$

Solution:
The equation of lines passing through $(1,0)$ are given by $y=m(x-1)$. Its distance from origin is $\frac{\sqrt{3}}{2}$
$\Rightarrow\left|\frac{-m}{\sqrt{1+m^2}}\right|=\frac{\sqrt{3}}{2} \Rightarrow m= \pm \sqrt{3}$. Hence the lines are $\sqrt{3} x+y-\sqrt{3}=0$ and
$\sqrt{3} x-y-\sqrt{3}=0$

Topic: Mathematics, Straight Lines, JEE Main
The equations of the lines which cuts off an intercept -1 from $y$-axis are equally inclined to the axes are

Options:
A $x-y+1=0, x+y+1=0$
B $x-y-1=0, x+y-1=0$
C $x-y-1=0, x+y+1=0$
D None of these

The correct answer is:
$x-y-1=0, x+y+1=0$

Solution:
Here $c=-1$ and $m=\tan \theta=\tan 45^{\circ}=1$
(Since the line is equally inclined to the axes, so $\theta=45^{\circ}$ )
Hence equation of straight line is $y= \pm(1 . x)-1$
$\Rightarrow x-y-1=0$ and $x+y+1=0$.

Topic: Mathematics, Straight Lines, JEE Main
The equations of the sides AB, BC & CA of a triangle ABC are 2x+y=0, x+py=21a a≠0 and x-y=3 respectively. Let P2,a be the centroid of the triangle ABC, then BC2 is equal to

Options:

The correct answer is:
122

Solution:
We have,

Since, G2,a, so

1+α+β+33=2 and -2-2α+β3=a

⇒α+β=2 and -2α+β=3a+2

So, α=-a, β=2+a

So, B≡-a,2a, C≡5+a,2+a

Now, B and C lies on the line x+py=21a, so

-a+2pa=21a

⇒p=11

Also,

5+a+112+a=21a

⇒27=9a⇒a=3

So,

B≡-3,6, C≡8,5

So,

BC2=121+1=122

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (24 Jan Shift 2)