datavorous/entrance-exam-dataset huggingface dataset:
Suppose $\theta$ and $\phi(\neq 0)$ are such that $\sec (\theta+\phi)$, sec $\theta$ and $\sec (\theta-\phi)$ are in A.P. If $\cos \theta=k \cos \left(\frac{\phi}{2}\right)$ for some $k$, then $k$ is equal to

Options:
A
$\pm \sqrt{2}$

B
$\pm 1$

C
$\pm \frac{1}{\sqrt{2}}$

D
$\pm 2$

The correct answer is:
$\pm \sqrt{2}$

Solution:
Since, $\sec (\theta-\phi), \sec \theta$ and $\sec (\theta+\phi)$ are in A.P.,
$$
\begin{aligned}
& \therefore 2 \sec \theta=\sec (\theta-\phi)+\sec (\theta+\phi) \\
& \Rightarrow \frac{2}{\cos \theta}=\frac{\cos (\theta+\phi)+\cos (\theta-\phi)}{\cos (\theta-\phi) \cos (\theta+\phi)}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 2\left(\cos ^2 \theta-\sin ^2 \phi\right)=\cos \theta[2 \cos \theta \cos \phi \\
& \Rightarrow \cos ^2 \theta(1-\cos \phi)=\sin ^2 \phi=1-\cos ^2 \phi \\
& \Rightarrow \quad \cos ^2 \theta=1+\cos \phi=2 \cos ^2 \frac{\phi}{2} \\
& \therefore \quad \cos \theta=\pm \sqrt{2} \cos \frac{\phi}{2} \\
& \text { But given } \cos \theta=\mathrm{k} \cos \frac{\phi}{2} \\
& \therefore \quad \mathrm{k}=\pm \sqrt{2}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2012 (19 May Online)
Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 orange, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is _____ .

Options:

The correct answer is:
6860

Solution:
We have,

7 Red appleRA

5 white appleWA

8 oranges O

Now, for offering 5 fruits to be selected

Possible selections :- (2O, 1RA, 2WA) or 2O,2RA,1WA, or 3O,1RA,1WA

Required number of ways

=C28C17C25+C28C27C15+C38C17C15

=1960+2940+1960

=6860

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Suppose $\mathrm{BC}$ is a given line segment in the plane and $\mathrm{T}$ is a scalene triangle. The number of points A in the plane such that the triangle with vertices $A, B, C$ (in some order) is similar to triangle $T$ is

Options:
A 4
B 6
C 12
D 24

The correct answer is:
12

Solution:
Let triangle $T$ is DEF possibilities
$\therefore \quad$ A can take two positions if $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}$ We can arrange order of A, B, C in $3 !=6$ ways Total positions which A can take $=6 \times 2=12$

Topic: Mathematics, Permutation Combination, JEE Main
Suppose cos2⁡y⋅dydx=sin⁡x+y+sin⁡x-y, x≤π2 and y≤π2. If yπ3=-π2, then yπ2 is

Options:
A 0
B π4
C -π4
D π2

The correct answer is:
0

Solution:
cos2⁡ydydx=2sin⁡xcos⁡y
⇒cos⁡ydy=2sin⁡xdx
⇒sin⁡y+2cos⁡x=c
Now, yπ3=-π2⇒c=0
So sin⁡y=-2cos⁡x
At x=π2, sin⁡y=0⇒yπ2=0

Topic: Mathematics, Differential Equations, JEE Main
Suppose $\quad f:[-2,2] \rightarrow R \quad$ is defined by
$f(x)=\left\{\begin{array}{ccc}-1, & \text { for } & -2 \leq x \leq 0 \\x-1, & \text { for } & 0 \leq x \leq 2 \\\{x \in[-2,2]: x \leq 0 & \text { and } & f(|x|)=x\end{array}\right.$
is equal tos

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

The correct answer is:
$\left\{-\frac{1}{2}\right\}$

Solution:
We have,
$f(x)=\left\{\begin{array}{c}-1, \text { for }-2 \leq x \leq 0 \\x-1, \text { for } 0 \leq x \leq 2\end{array}\right.$
Now, take $x=-\frac{1}{2}$
$\therefore \quad f\left(\left|-\frac{1}{2}\right|\right)=f\left(\frac{1}{2}\right)=\frac{1}{2}-1=-\frac{1}{2}$
Hence, $\quad f(|x|)=x$
$\therefore \quad$ Domain of $f(x)=\left\{-\frac{1}{2}\right\}$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2003
Suppose $\quad f:[-2,2] \rightarrow R \quad$ is defined by
$f(x)=\left\{\begin{array}{ccc}-1, & \text { for } & -2 \leq x \leq 0 \\x-1, & \text { for } & 0 \leq x \leq 2 \\\{x \in[-2,2]: x \leq 0 & \text { and } & f(|x|)=x\end{array}\right.$
is equal tos

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

The correct answer is:
$\left\{-\frac{1}{2}\right\}$

Solution:
We have,
$f(x)=\left\{\begin{array}{c}-1, \text { for }-2 \leq x \leq 0 \\x-1, \text { for } 0 \leq x \leq 2\end{array}\right.$
Now, take $x=-\frac{1}{2}$
$\therefore \quad f\left(\left|-\frac{1}{2}\right|\right)=f\left(\frac{1}{2}\right)=\frac{1}{2}-1=-\frac{1}{2}$
Hence, $\quad f(|x|)=x$
$\therefore \quad$ Domain of $f(x)=\left\{-\frac{1}{2}\right\}$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2003
Suppose \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=\left\{\begin{array}{cc}1, \quad \text { if } x=1 \\ e^{\left(x^{10}-1\right)}+(x-1)^2 \sin \frac{1}{x-1}\end{array}\right., \text { if } x \neq 1 \text { then }\)

Options:
A \(f^{\prime}(1)\) does not exist
B \(f^{\prime}(1)\) exists and is zero
C \(f^{\prime}(1)\) exist and is 9
D \(f^{\prime}(1)\) exists and is 10

The correct answer is:
\(f^{\prime}(1)\) does not exist

Solution:
Hint: \(f: R \rightarrow R f(x)=\left\{\begin{array}{cc} 1 & \text { if } x=1 & \\ e^{\left(x^{10}-1\right)} &+(x-1)^2 &\cdot \sin \left(\frac{1}{x-1}\right) \text { if } x \neq 1\end{array}\right.\)
\(\begin{aligned}
& f\left(1^{-}\right)=1+0=1 \\
& f\left(1^{+}\right)=1+0=1 \\
& f(1)=1 \\
& \therefore \quad f(x) \text { is continuous at } x=1 \\
& f^{\prime}(x)=\left\{e^{\left(x^{10}-1\right)} \cdot\left(10 x^5\right)+2(x-1) \cdot \sin \left(\frac{1}{x-1}\right)-(x-1)^2 \cdot \cos \left(\frac{1}{x-1}\right) \cdot\left(\frac{1}{x-1}\right)^2\right\}
\end{aligned}\)
which does not exist.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Suppose f:R→0,∞ be a differentiable function such that 5fx+y=fx·fy, ∀ x, y∈R, If f3=320, then ∑n=05fn is equal to:

Options:
A 6875
B 6575
C 6825
D 6528

The correct answer is:
6825

Solution:
Given, equation is:

5fx+y=fx·fy ...1

At x=0, y=0, we get

5f0=f02⇒f0=5

Put y=1 in 1, then we get

5fx+1=fx·f1

⇒fx+1fx=f15

⇒f1f0·f2f1·f3f2=f153

⇒3205=f1353⇒f1=20

∴ 5fx+1=20·fx

⇒fx+1=4fx

So,

f1=4f0=4·5f2=4f1=42·5f3=4f2=43·5f4=4f3=44·5f5=4f4=45·5

∑n=05fn=5+5·4+5·42+5·43+5·44+5·45

⇒∑n=05fn=546-13=6825

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Suppose fx is a polynomial of degree four having critical points at -1, 0, 1. If T=x∈R |fx=f0, then the sum of squares of all the elements of T is :

Options:
A 4
B 6
C 2
D 8

The correct answer is:
4

Solution:
f'x=k.xx+1x-1

=kx3-x

Integrating both sides with respect to x, we get

fx=k x44-x22+c

⇒f0=c

∵ fx=f0

⇒kx4-2x24+c=c

⇒x2x2-2=0

⇒x=0, 2, -2

⇒T=0, 2, -2

Thus, sum of squares of all the elements of T is 02+22+-22=4.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Suppose $f(x)$ is differentiable at $x=1$ and $\lim _{h \rightarrow 0} \frac{1}{h} f(1+h)=5$, then $f^{\prime}(1)$ equals

Options:
A $5$
B $6$
C $3$
D $4$

The correct answer is:
$5$

Solution:
$f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \text {; }$
As function is differentiable so it is continous as it is given that $\lim _{h \rightarrow 0} \frac{f(1+h)}{h}=5$ and hence $f^\prime(1)=0$. Hence $f(1)=\lim _{h \rightarrow 0} \frac{f(1+h)}{h}=5$

Topic: Mathematics, Continuity and Differentiability, JEE Main
Suppose $f(x)$ is differentiable $x=1$ and $\lim _{h \rightarrow 0} \frac{1}{h} f(1+h)=5$, then $f^{\prime}(1)$ equals

Options:
A
3

B
4

C
5

D
6

The correct answer is:
5

Solution:
$f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$; As function is differentiable so it is continuous as it is given that $\lim _{h \rightarrow 0} \frac{f(1+h)}{h}=5$ and hence $f(1)=0$
Hence $f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)}{h}=5$
Hence (C) is the correct answer.

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2005
Suppose $\mathrm{m}, \mathrm{n}$ are positive integers such that $6^{\mathrm{m}}+2^{\mathrm{m}+\mathrm{n}} 3^{\mathrm{w}}+2^{\mathrm{n}}=332 .$ The value of the expression $\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}$ is-

Options:
A 7
B 13
C 19
D 21

The correct answer is:
19

Solution:
$\quad 6^{\mathrm{m}}+2^{\mathrm{m}+\mathrm{n}} \quad 3^{\mathrm{w}}+2^{\mathrm{n}}=332$
maximum possible value of $m$ is 3
checking for $\mathrm{m}=3,2$ and 1
we get $\mathrm{m}=2, \mathrm{n}=3, \mathrm{w}=2$
$$
\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}=4+6+9=19
$$

Topic: Mathematics, Binomial Theorem, JEE Main
Suppose $\mathrm{OABC}$ is a rectangle in the $\mathrm{xy}$-plane where o is the origin and $\mathrm{A}, \mathrm{B}$ lie on the parabola $\mathrm{y}=\mathrm{x}^{2}$. Then $\mathrm{C}$ must lie on the curve.

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

The correct answer is:
$y=x^{2}+2$

Solution:
$m_{O A} \times m_{O B}=-1$
$t_{1} t_{2}=-1$
$t_{1}+t_{2}=h$
$K=\left(t_{1}+t_{2}\right)^{2}-2 t_{1} t_{2}$
$K=h^{2}+2 \Rightarrow y=x^{2}+2$

Topic: Mathematics, Parabola, JEE Main
Suppose $\mathrm{p}, \mathrm{q}, \mathrm{r} \neq 0$ and system of equation $(\mathrm{p}+\mathrm{a}) \mathrm{x}+\mathrm{by}+\mathrm{cz}=0$

$a x+(q+b) y+c z=0$

$a x+b y+(r+c) z=0$ has a non-trivial solution,

then value of $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}$ is

Options:
A -1
B 0
C 1
D 2

The correct answer is:
-1

Solution:
Since the system is homogeneous and has a non-trivial solution

$0=\left|\begin{array}{ccc}p+a & b & c \\ a & q+b & c \\ a & b & r+c\end{array}\right|=\left|\begin{array}{ccc}p & -q & 0 \\ 0 & q & -r \\ a & b & r+c\end{array}\right|$

using $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$ and $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$

Expanding along $\mathrm{C}$, we get $\Rightarrow \mathrm{p}[\mathrm{q}(\mathrm{r}+\mathrm{c})+\mathrm{br}]+\mathrm{aqr}=0$

$\Rightarrow \mathrm{pqr}+\mathrm{pqc}+\mathrm{prb}+\mathrm{qra}=0$

$\Rightarrow \frac{a}{p}+\frac{b}{q}+\frac{c}{r}=-1$

Topic: Mathematics, Determinants, JEE Main
Suppose $S_{1}$ and $S_{2}$ are two unequal circles; $A B$ and $C D$ are the direct common tangents to these circles. A transverse common tangent $\mathrm{PQ}$ cuts $\mathrm{AB}$ in $\mathrm{R}$ and $\mathrm{CD}$ in $\mathrm{S}$. If $\mathrm{AB}=10$, then RS is.

Options:
A 8
B 9
C 10
D 11

The correct answer is:
10

Solution:
$\begin{array}{ll}
\therefore \mathrm{RP}=\mathrm{RA}=10-\alpha & \Rightarrow R S=10-\alpha+\beta \\
\text { Also } \mathrm{SQ}=\mathrm{SD}=10-\beta & \Rightarrow R S=10-\beta+\alpha
\end{array}$
(1) and (2) $\Rightarrow \alpha=\beta$, Hence $\mathrm{RS}=10$

Topic: Mathematics, Circle, JEE Main
Suppose S and Sâ² are foci of the ellipse x225+y216=1. If P is a variable point on the ellipse and if â is the area (in sq. units) of the triangle PSSâ², then the maximum value of â is double of

Options:
A Minimum value of 2a8+2b4a4b2 ∀a, b∈R
B Minimum value of 3a8+3b4a4b2 âa, bâR
C 4a8+4b4a4b2 ∀a, b∈R
D 6a8+6b4a4b2 ∀a, b∈R

The correct answer is:
Minimum value of 3a8+3b4a4b2 âa, bâR

Solution:
Given , a2=25 and b2=16
∴ e=1-b2a2=1-1625= 35
So, the coordinates of foci S and S′ are (3, 0) and (-3, 0) respectively.

Let, P5cos⁡θ,4sin⁡θ be a variable point on the ellipse.
Then, ∆=area of ∆PSS′=12301-3015cos⁡θ4sin⁡θ1=12sin⁡θ
Since the value of sin⁡θ lies between -1 and 1
So, the maximum value of area of ∆PSS′ is 12 sq. units
Also, a8+b4≥2a4b2 (since, A.M≥G.M)

⇒3a8+b4a4b2≥6

Topic: Mathematics, Ellipse, JEE Main
Suppose that 20 pillars of the same height have been erected along the boundary of circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is:

Options:
A 170
B 180
C 210
D 190

The correct answer is:
170

Solution:
Each of the beams connects the top of two non-adjacent pillars, hence we have to find in how many ways we can select the required two pillars which are non-adjacent out of the given 20 pillars.

As, C2 20 is the number of beams connecting any two pillars whereas 20 is the number of adjacent beams.

Thus, the total number of beams =C2 20-20

Using Crn=n!r!n-r!,

=20!2!×18!-20

=20×19×18!2!×18!-20

=190-20

=170.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Suppose that a function f:R→R satisfies fx+y=fx fy for all x, y ε R and f1=3. If ∑i=1nfi=363, then n is equal to ..... .

Options:

The correct answer is:
5

Solution:
fx=ax

⇒f1=a=3

so fx=3x

∑i=1nfi=363

⇒ 3+32+.....3n=363

33n-12=363

3n=243 ⇒ n=5

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Suppose that $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2 \sqrt{x}+x$, then $f(x)$ is

Options:
A $1+2 x^2$
B $2+x^2$
C $1+x$
D $2+x$

The correct answer is:
$2+x^2$

Solution:
$\begin{aligned}
g(x) & =1+\sqrt{x} \text { and } f(g(x))=3+2 \sqrt{x}+x ....(i) \\
& \Rightarrow f(1+\sqrt{x})=3+2 \sqrt{x}+x
\end{aligned}$
Put $1+\sqrt{x}=y \Rightarrow x=(y-1)^2$
then, $f(y)=3+2(y-1)+(y-1)^2=2+y^2$
therefore, $f(x)=2+x^2$.

Topic: Mathematics, Functions, JEE Main
Suppose that gold is being plated on to another metal in an electrolytic cell. The half-cell reaction producing the Au(s) is AuCl4- +3e-âAus+4Cl- . If a 0.30 A current runs for 15.00 minute, what mass of Au(s) will be plated, assume all the electrons are used in the reduction of AuCl4- ? The Faraday constant is 96485 C/mol and molar mass of Au is 197.

Options:
A 0.184 g
B 0.551 g
C 1.84 g
D 0.613 g

The correct answer is:
0.184 g

Solution:
Mass of Au deposited = Number of Faraday passed Ã Eq. mass
=0.30Ã15Ã6096485Ã1973=0.184 g

Topic: Chemistry, Electrochemistry, JEE Main
Suppose that intensity of a laser is 315π W m-2. The rms electric field, in units of V m-1 associated with this source is close to the nearest integer is -ε0=8.86×10-12 C2 N m-2; c=3×108 m s-1

Options:

The correct answer is:
275

Solution:
12C∈0Ems2=315π

Erms=315×2π×3×108×8.86×10-12

=63083.4612×10-4

=7.5484×104

=2.75×102=2.75×102=275

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Suppose that \( \mathrm{f} \) is differentiable for all \( \mathrm{x} \) and that \( \mathrm{f}^{\prime}(\mathrm{x}) \leq 2 \) for all \( \mathrm{x} . \) If \( \mathrm{f}(1)=2 \) and \( \mathrm{f}(4)=8 \), then \( \mathrm{f}(2) \) has the value equal to

Options:
A \( 3 \)
B \( 4 \)
C \( 6 \)
D \( 8 \)

The correct answer is:
\( 4 \)

Solution:
Using Lagrange's mean value theorem for f in [1, 2]
for c ∈ 1 , 2 , f 2 - f 1 2 - 1 = f ′ c ≤ 2
⇒ f 2 - f 1 ≤ 2
⇒ f 2 ≤ 4 (1)
again using Lagrange's mean value theorem in [2, 4]
for d ∈ 1 , 2 , f 4 - f 2 4 - 2 = f ′ d ≤ 2
⇒ f 4 - f 2 ≤ 4
⇒ 8 - f 2 ≤ 4
⇒ f 2 ≥ 4 (2)
from (1) and (2), f(2) = 4.

Topic: Mathematics, Application of Derivatives, JEE Main
Suppose that the angular velocity of rotation of the Earth is increased. Then, as a consequence,

Options:
A weight of the object, everywhere on the earth, will decrease.
B weight of the object, everywhere on the earth, will increase.
C except at poles, weight of the object on the earth, will decrease.
D there will be no change in weight anywhere on the earth.

The correct answer is:
except at poles, weight of the object on the earth, will decrease.

Solution:
Effective acceleration due to gravity, g′=g-ω2Rcos2⁡ϕ, where ϕ is latitude. There will be no change in gravity at poles as ϕ=90o. At other points as ω increases, g' will decrease.

Topic: Physics, Gravitation, JEE Main, JEE Main 2018 (16 Apr Online)
Suppose that the electric field amplitude of electromagnetic wave is $E_{0}=120 \mathrm{NC}^{-1}$ and its frequency $f=50 \mathrm{MHz}$. Then, which of the following value is incorrectly computed?

Options:
A Magnetic field amplitude is $400 \mathrm{nT}$.
B Angular frequency of EM wave is $\pi \times 10^{8} \mathrm{rad} / \mathrm{s}$.
C Propagation constant (angular wave number) is $2.1 \mathrm{rad} / \mathrm{m}$.
D Wavelength of EM wave is $6 \mathrm{~m}$.

The correct answer is:
Propagation constant (angular wave number) is $2.1 \mathrm{rad} / \mathrm{m}$.

Solution:
Given, $E_{0}=120 \mathrm{NC}^{-1}$
$f=50 \mathrm{MHz}=50 \times 10^{6} \mathrm{~Hz}$
(a) magnetic field amplitude,
$\begin{aligned}
B_{0} &=\frac{E_{0}}{c}=\frac{120}{3 \times 10^{8}}=40 \times 10^{-8} \mathrm{~T} \\
&=400 \mathrm{n} \mathrm{T}
\end{aligned}$
(b) Angular frequency, $\omega=2 \pi f=2 \pi \times 50 \times 10^{6}$
$=\pi \times 10^{8} \mathrm{rad} / \mathrm{s}$
(c) Propagation constant,
$k=\frac{\omega}{c}=\frac{\pi \times 10^{8}}{3 \times 10^{8}}=1.047 \mathrm{rad} / \mathrm{m}$
(d) Wavelength, $\lambda=\frac{c}{f}=\frac{3 \times 10^{8}}{50 \times 10^{6}}=\frac{30}{5}=6 \mathrm{~m}$

Topic: Physics, Electromagnetic Waves, JEE Main
Suppose that the electric field part of an electromagnetic wave in vacuum is
\( E=\left\{\left(3.1 \mathrm{~N} \mathrm{C}^{-1}\right) \cos \left[\left(1.8 \mathrm{rad} \mathrm{m}^{-1}\right) y+\left(5.47 \times 10^{6} \operatorname{rad} \mathrm{s}^{-1}\right) t\right]\right\} \hat{\mathrm{i}} . \)
What is the wavelength \( \lambda \) ?

Options:
A \( 3.5 \mathrm{~m} \)
B \( 4.5 \mathrm{~m} \)
C \( 2.5 \mathrm{~m} \)
D \( 3.0 \mathrm{~m} \)

The correct answer is:
\( 3.5 \mathrm{~m} \)

Solution:
The eletric part of electromagnetic wave in vacuum
E = 3.1 cos (1.8 y + 5.4 × 106t) i ^
Comparing with standard equation E = E0 cos (Ky + ωt), we get
Angular frequency ω = 5.4 × 106 rad/s
Wave number K = 1.8 rad/m
The amplitude of the electric field part of the wave
E0 = 3.1 N/C
λ = 2 π Kâ¡ = 2 π 1 · 8 = 3 · 4 9 2 m
λ = 3.5 m

Topic: Physics, Electromagnetic Waves, JEE Main
Suppose that the points h, k, 1, 2 and -3, 4 lie on the line L1. If a line L2 passing through the points h, k and 4, 3 is perpendicular to L1, then kh equals:

Options:
A -17
B 3
C 0
D 13

The correct answer is:
13

Solution:
Slope of line L1 is 4-2-3-1=2-4=-12
Hence slope of L2 is 2.
From line L1, k-2h-1=-12
⇒h+2k=5 …..(1)
From line L2, k-3h-4=2
⇒k=2h-5 ..…(2)
From (1) and (2) h=3 and k=1
⇒kh=13

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Suppose that the side lengths of a triangle are three consecutive integers and one of the angles is twice another. The number of such triangles is/are

Options:
A 1
B 0
C 4
D 2

The correct answer is:
1

Solution:
Let B=2A and BD be the bisector of angle B, then
CD = ab a + c & AD = bc a + c
Now, Î ABC and Î BDC are similar, so
BC AC = CD BC â a 2 = ab a + c b â b 2 = a a + c ......(i)
Since, b>aâââ Either b=a+1 or b=a+2, if b=a+1, then [From Eq.(i)]
a + 1 2 = a + c a â c = 2 + 1 a

c is integer â a=1,b=2,c=3 but then, no triangle will form.
If b=a+2, then obviously c=a+1, and then, [from Eq. (i)]
a+2 2 =a 2a+1
â a 2 â3aâ4=0ââorâa=4
â´ââa=4,âb=6,c=5 is the only possible solution.

Topic: Mathematics, Properties of Triangles, JEE Main
Suppose the cube $x^3-p x+q$ has three distinct real roots where $p>0$ and $q>0$. Then which one of the following holds?

Options:
A
The cubic has minima at $\sqrt{\frac{\mathrm{p}}{3}}$ and maxima at $-\sqrt{\frac{\mathrm{p}}{3}}$

B
The cubic has minima at $-\sqrt{\frac{p}{3}}$ and maxima at $\sqrt{\frac{p}{3}}$

C
The cubic has minima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$

D
The cubic has maxima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$

The correct answer is:
The cubic has minima at $\sqrt{\frac{\mathrm{p}}{3}}$ and maxima at $-\sqrt{\frac{\mathrm{p}}{3}}$

Solution:
Let $f(x)=x^3-p x+q$
Now for maxima/minima
$$
\begin{aligned}
& f^{\prime}(x)=0 \\
& \Rightarrow 3 x^2-p=0 \\
& \Rightarrow x^2=\frac{p}{3} \\
& \therefore x=\pm \sqrt{\frac{p}{3}}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
Suppose the cube $x^3-p x+q$ has three distinct real roots where $p>0$ and $q>0$. Then which one of the following holds?

Options:
A
The cubic has minima at $\sqrt{\frac{\mathrm{p}}{3}}$ and maxima at $-\sqrt{\frac{\mathrm{p}}{3}}$

B
The cubic has minima at $-\sqrt{\frac{p}{3}}$ and maxima at $\sqrt{\frac{p}{3}}$

C
The cubic has minima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$

D
The cubic has maxima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$

The correct answer is:
The cubic has minima at $\sqrt{\frac{\mathrm{p}}{3}}$ and maxima at $-\sqrt{\frac{\mathrm{p}}{3}}$

Solution:
Let $f(x)=x^3-p x+q$
Now for maxima/minima
$$
\begin{aligned}
& f^{\prime}(x)=0 \\
& \Rightarrow 3 x^2-p=0 \\
& \Rightarrow x^2=\frac{p}{3} \\
& \therefore x=\pm \sqrt{\frac{p}{3}}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2008
Suppose the drift velocity vd in a material varied with the applied electric field E as vd ∝E. Then V-I graph for a wire made of such a material is best given by:

Options:
A
B
C
D

The correct answer is:

Solution:
We know that vd=kE and I=n e A vd,
⇒I=n eA kE∵E=Vd⇒I=neAk Vd
⇒I∝V⇒V ∝I2. So,

Topic: Physics, Current Electricity, JEE Main, JEE Main 2015 (10 Apr Online)
Suppose the earth was covered by an ocean of uniform depth h. hâªR. let Ï be density of ocean and Ï be mean density of earth. Let âg be the approximate difference of value of net acceleration due to gravity between the bottom of the ocean and top Îg= g top - g bottom choose the correct option.

Options:
A âg=43ÏGh2Ï-3Ï
B âg=43GÏh 3Ï-2Ï
C âg=43GÏhÂ [Â 2Ï-3Ï]Â
D âg=43GÏhÂ 3Ï-3Ï

The correct answer is:
âg=43GÏh 3Ï-2Ï

Solution:
gA=43GπRρ
gB= G43πρR3+43πR+h3-43πR3σR+h2
≈43GπρR1-2hR+43GπR+h-1-2hRRσ
∆g=gB-gA=43πGh3σ-2ρ

Topic: Physics, Gravitation, JEE Main
Suppose the elements P and Q combine to form two compounds PQ2 and P3Q2 . When 0.1 mole of PQ2 weight 10 g and 0.05 mole of P3Q2 weight 9 g, the atomic weights of P and Q are

Options:
A 40, 30
B 60, 40
C 20, 30
D 30, 20

The correct answer is:
40, 30

Solution:
Let atomic weight of element P is p and that of element Q is q.
For PQ2,n=wMol.wt.
0.1=10p+2q
p+2q=100.1=100 ...(i)
For P3Q2,n=wMol.wt.
0.05=93p+2q
3p+2q=90.05=180 ...(ii)
On solving equation (i) and (ii) we get q = 30
p+2(30)=100
p=100-60=40

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Suppose the family of lines ax+by+c=0 (where a,b,c are in arithmetic progression) be normal to a family of circles. The radius of the circle of the family which intersects the circle x2+y2-4x-4y-1=0 orthogonally is

Options:
A 22 units
B 2 units
C 32 units
D 4 units

The correct answer is:
22 units

Solution:
Since, a,b,c are in A.P

ax+by+c=0 represents a family of lines passing through the point 1,-2.

So, the family of circles (concentric) will be given by x2+y2-2x+4y+c=0 and

2-1×-2+2×-2=-1+c⇒c=-3

⇒radius=1+4+3=22 units

Topic: Mathematics, Circle, JEE Main
Suppose the gravitational force varies inversely as the nth power of the distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to

Options:
A R-n
B Rn-12
C Rn+12
D Rn

The correct answer is:
Rn+12

Solution:
Gravitational force =GMmd2 (Newton's law)
But in this case, F=GMmRn and it provides the necessary centripetal force mv2R
â´ mv2R=GMmRn
â´ v=GMRn-1 and the period T of a planet, T=2ÏRv
â´ T=2ÏRRn-1GM=2ÏRn+1GM=2ÏGMâRn+12
â´ TâRn+12

Topic: Physics, Gravitation, JEE Main
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period planet in circular orbit of radius $\mathrm{R}$ around the sun will be proportional to

Options:
A
$R^{\left(\frac{n+1}{2}\right)}$

B
$R^{\left(\frac{n-1}{2}\right)}$

C
$\mathrm{R}^{\mathrm{n}}$

D
$\mathrm{R}^{\left(\frac{\mathrm{n}-2}{2}\right)}$

The correct answer is:
$R^{\left(\frac{n+1}{2}\right)}$

Solution:
$\mathrm{T} \propto \mathrm{R}^{(\mathrm{n}+1) / 2}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2004
Suppose the height of a pyramid with a square base is decreased by $p \%$ and the lengths of the sides of its square base are increased by $p \%$ (where $p>0)$. If the volume remains the same, then

Options:
A $50 < p < 55$
B $55 < \mathrm{p} < 60$
C $60 < p < 65$
D $65 < p < 70$

The correct answer is:
$60 < p < 65$

Solution:
$\frac{1}{3}\left(x^{2}\right) y=\frac{1}{3} x^{2}\left(\frac{100+p}{100}\right)^{2}\left(\frac{100-p}{100}\right)$
$\Rightarrow p^{2}+100 p-100^{2}=0$
$p=\sqrt{12500}-50$
$60 < p < 65$

Topic: Mathematics, Basic of Mathematics, KVPY, KVPY 2013 (SA)