datavorous/entrance-exam-dataset huggingface dataset:
If ∑k=110kk4+k2+1=mn, where m and n are co-prime, then m+n is equal to

Options:

The correct answer is:
166

Solution:
Given,

∑k=110kk4+k2+1=mn

⇒12∑k=110k2+k+1-k2-k+1k2+k+1k2-k+1=mn

by using the formula x4+x2+1=x2+x+1x2-x+1

⇒12∑k=1101k2-k+1-1k2+k+1=mn

⇒121-13+13-17+17-.........-1111=mn

⇒121-1111=mn

⇒55111=mn

So, m+n=166

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2022 (26 Jul Shift 2)
If K1 and K2 are maximum kinetic energies of photoelectrons emitted when lights of wavelength Î»1 and Î»2 respectively incident on a metallic surface. If Î»1=3Î»2, then

Options:
A K1>K23
B K1<K23
C K1=3K2
D K2=3K1

The correct answer is:
K1<K23

Solution:
Eienstein's photoelectric equation is given as,
K1=hc1λ1-1λ0 ...(i)
and K2=hc1λ2-1λ0 ...(ii)
Dividing Equation (i) by Equation (ii), we get
K1K2=1λ1-1λ01λ2-1λ0
Since, λ1=3λ2
K1K2=13λ2-1λ01λ2-1λ0
=13λ2 1-3λ0/λ21λ21-1λ0
K1K2=13 1-3λ0/λ21-1λ0/λ2
3K1K2=1-3/λ0/λ21-1/λ0λ2<1
∴ 3K1K2<1⇒K1<K23

Topic: Physics, Dual Nature of Matter, JEE Main
If K=11C2+210C2+9C2+8C2+...+2C2, then the value of K100 is equal to

Options:

The correct answer is:
3.85

Solution:
We know that nC2=nn-12
⇒2nC2=nn-1=n2-n
⇒2∑n=210nC2=∑n=210n2-n
=22+32+..+92+102-2+3+..+9+10
=12+22+..+92+102-1+2+3+..+10
=10×11×216-10×112
=385-11C2
⇒11C2+210C2+9C2+..+2C2=385
⇒K100=3.85

Topic: Mathematics, Binomial Theorem, JEE Main
If $k$ is a scalar and $/$ is a unit matrix of order 3 , then $\operatorname{adj}(k I)=$

Options:
A $k^3 I$
B $k^2 I$
C $-k^3 I$
D $-k^2 I$

The correct answer is:
$k^2 I$

Solution:
Let $I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$, then $k I=\left[\begin{array}{lll}k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k\end{array}\right]$
$\Rightarrow \operatorname{adj}(k I)=\left[\begin{array}{ccc}k^2 & 0 & 0 \\ 0 & k^2 & 0 \\ 0 & 0 & k^2\end{array}\right]=k^2 I$.

Topic: Mathematics, Matrices, JEE Main
If $K_{s p}$ of $\mathrm{CaF}_2$ at $25^{\circ} \mathrm{C}$ is $1.7 \times 10^{-10}$, the combination amongst the following which gives a precipitate of $\mathrm{CaF}_2$ is

Options:
A
$1 \times 10^{-2} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-3} \mathrm{M} \mathrm{F}^{-}$

B
$1 \times 10^{-4} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-4} \mathrm{M} \mathrm{F}^{-}$

C
$1 \times 10^{-2} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-5} \mathrm{M} \mathrm{F}^{-}$

D
$1 \times 10^{-3} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-5} \mathrm{M} \mathrm{F}^{-}$

The correct answer is:
$1 \times 10^{-2} \mathrm{M} \mathrm{Ca}^{2+}$ and $1 \times 10^{-3} \mathrm{M} \mathrm{F}^{-}$

Solution:
When ionic product i.e. the product of the concentration of ions in the solution exceeds the value of solubility product, formation of precitpiate occurs.
$\begin{aligned} & \mathrm{CaF}_2 \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \\ & \text {Ionic product }=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\ & \text { when, }\left[\mathrm{Ca}^{2+}\right]=1 \times 10^{-2} \mathrm{M} \\ & {\left[\mathrm{F}^{-}\right]^2=\left(1 \times 10^{-3}\right)^2 \mathrm{M} } \\ & =1 \times 10^{-6} \mathrm{M} \\ \therefore & {\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^2=\left(1 \times 10^{-2}\right)\left(1 \times 10^{-6}\right)=1 \times 10^{-8} } \\ & \text { In this case, } \\ & \text { Ionic product }\left(1 \times 10^{-8}\right)> \\ & \text { solubility product }\left(1.7 \times 10^{-10}\right)\end{aligned}$
$\therefore$ Hence (a) is correct option.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2012 (19 May Online)
If $k \leq \sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x \leq K$, then

Options:
A $k=0, K=\pi$
B $k=0, K=\frac{\pi}{2}$
C $k=\frac{\pi}{2}, K=\pi$
D None of these

The correct answer is:
$k=0, K=\pi$

Solution:
We have
$\sin ^{-1} x+\cos ^{-1} x+\tan ^{-1} x=\frac{\pi}{2}+\tan ^{-1} x$
$\begin{aligned}
& \text {Since } \frac{-\pi}{2} \leq \tan ^{-1} x \leq \frac{\pi}{2} \Rightarrow 0 \leq \frac{\pi}{2}+\tan ^{-1} x \leq \pi \\
& \therefore K=\pi, k=0 .
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If Ksp of Ag2CO3 is 8×10-12 , the molar solubility of Ag2CO3 in 0.1 M AgNO3 is:

Options:
A 8x10-13M
B 8x10-11M
C 8x10-12M
D 8x10-10M

The correct answer is:
8x10-10M

Solution:
(i)
(ii)
K sp =8× 10 −12 = ( 2 S ′ +0.1 ) 2 S ′
or S ′ =8× 10 −10 M

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2019 (12 Jan Shift 2)
If KE of the particle of mass $m$ performing UCM in a circle of radius $r$ is $E$. Find the acceleration of the particle

Options:
A $\frac{2 E}{m r}$
B $\left(\frac{2 E}{m r}\right)^{2}$
C $2 \mathrm{Emr}$
D $\frac{4 E}{m r}$

The correct answer is:
$\frac{2 E}{m r}$

Solution:
Kinetic energy
$$
\begin{aligned}
\frac{1}{2} m v^{2}=E \\
\frac{1}{2} m r \frac{v^{2}}{r} &=E \\
\frac{1}{2} m r a &=E \\
a &=\frac{2 E}{m r}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
If L1 is the line of intersection of the planes 2x-2y+3z-2=0, x-y+z+1=0 and L2 is the line of intersection of the planes x+2y-z-3=0, 3x-y+2z-1=0, then the distance of the origin from the plane, containing the lines L1 and L2 is

Options:
A 12
B 142
C 132
D 122

The correct answer is:
132

Solution:
Direction Ratio's of the line L1
ijk2-231-11
=i1-j -1+k(0)
= 1, 1, 0
At y=0
2x+3z=2x+z=-1→ (-5,0,4) is point
Equation of L1
x+51=y1=z-40=α
Similarly for L2
x-753=y-5=z+85-7=β
Point of intersection of L1 and L2
(-1,4,4)
Direction Ratio's of normal to plane passing through L1 and L2 =ijk1103-5-7=i-7-j-7+k-5-3=-7i+7j-8k=-7, 7, -8
Equation of plane
-7x+7y-8z=c .
c=3 (Since plane passes through(-1,4,4) )
Equation of plane
7x-7y+8 z+3=0
Distance from origin
=372+-72+82=3162=392
=132

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (08 Apr)
If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines, will be

Options:
A $\left(m_1 n_2-m_2 n_1\right),\left(n_1 l_2-n_2 l_1\right),\left(l_1 m_2-l_2 m_1\right)$
B $\left(l_1 l_2-m_1 m_2\right),\left(m_1 m_2-n_1 n_2\right),\left(n_1 n_2-l_1 l_2\right)$
C $\frac{1}{\sqrt{l_1^2+m_1^2+n_1^2}}, \frac{1}{\sqrt{l_2^2+m_2^2+n_2^2}}, \frac{1}{\sqrt{3}}$
D $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$

The correct answer is:
$\left(m_1 n_2-m_2 n_1\right),\left(n_1 l_2-n_2 l_1\right),\left(l_1 m_2-l_2 m_1\right)$

Solution:
Let lines are $l_1 x+m_1 y+n_1 z+d=0$
and $l_2 x+m_2 y+n_2 z+d=0$
If $l x+m y+n z+d=0$ is perpendicular to (i) and (ii), then, $\left\|_1+m m_1+n n_1=0,\right\|_2+m m_2+n n_2=0$
$\Rightarrow \frac{l}{m_1 n_2-m_2 n_1}=\frac{m}{n_1 l_2-l_1 n_2}=\frac{n}{l_1 m_2-l_2 m_1}=d$
Therefore, direction cosines are
$\left(m_1 n_2-m_2 n_1\right),\left(n_1 l_2-l_1 n_2\right),\left(l_1 m_2-l_2 m_1\right) .$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If $L$ and $R$ are respectively the inductance and resistance, then the dimensions of $\frac{L}{R}$ will be

Options:
A $M^0 L^0 T^{-1}$
B $M^0 L T^0$
C $M^0 L^0 T$
D Cannot be represented in terms of $M, L$ and T

The correct answer is:
$M^0 L^0 T$

Solution:
We know that;
Dimensional formula of \(\mathrm{L}=\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\)
Dimensional formula for \(\mathrm{R}=\mathrm{R}=\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\)
\(\frac{\mathrm{L}}{\mathrm{R}}=\frac{\mathrm{ML}^2}{\mathrm{~T}^2 \mathrm{~A}^2} \times \frac{\mathrm{T}^2 \mathrm{~A}^2}{\mathrm{ML}^2}=\mathrm{M}^{\mathrm{O}} \mathrm{L}^{\mathrm{O}} \mathrm{T}\)
Hence,
option (C) is correct answer.

Topic: Physics, Units and Dimensions, JEE Main
If L,C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?

Options:
A LC
B LR
C CR
D LC

The correct answer is:
LC

Solution:
Dimensions of L is ML2T-2A-2

Dimensions of C is M-1L-2T4A2

And dimensions of R is ML2T-3A-2

Clearly, dimensions of LC=T,

Dimensions of LR=T,

Dimensions of CR=T and

Dimensions of LC=ML2T-2A-2M-1L-2T4A2=M2L4T-6A-4=R2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2022 (27 Jun Shift 2)
If L, C and R denote the inductance, capacitance and resistance respectively, the dimensional formula for C 2 LR is

Options:
A [ML−2T−1I0]
B M0L0T3
C [M−1L−2T6I2]
D [M0L0T2I0]

The correct answer is:
M0L0T3

Solution:
C2LR=C2L2×RL

As LR and LC both have the dimensions of time, we get

C2LR=M0L0T4×M0L0T-1=M0L0T3

Topic: Physics, Units and Dimensions, JEE Main
If $L$ denotes the inductance of an inductor through which a current $i$ is flowing, the dimensions of $L i^2$ are

Options:
A $M {L}^2 T^{-2}$
B Not expressible in $MLT$
C $M {L} T^{-2}$
D $M^2 L^2 T^{-2}$

The correct answer is:
$M {L}^2 T^{-2}$

Solution:
$\frac{1}{2} L i^2=$ Stored energy in an inductor $=$ $\left[M {L}^2 T^{-2}\right]$

Topic: Physics, Units and Dimensions, JEE Main
If limx→0(1+px+qx2)cosecx=e5, then

Options:
A p=5,q∈R
B p=5,q>0
C q=5,p∈R
D q=5,p=0

The correct answer is:
p=5,q∈R

Solution:
∵ given limit is of the form 1∞
∴ elimx→0cosecx1+px+qx2-1=e5
limx→0px+qx2sin⁡x=5
limx→0px+qx2x⋅xsin⁡x=5
limx→0p+qx1=5 ⇒p=5

Topic: Mathematics, Limits, JEE Main
If limxâ0sinâ¡2x-asinâ¡xx3 exists finitely, then the value of a is

Options:
A 0
B 2
C 1
D 4

The correct answer is:
2

Solution:
limx→0sin⁡x2cos⁡x-ax⋅x2=limx→0sin⁡xx2cos⁡x-ax2
For this limit to exists finitely,
limx→02cos⁡x-ax2=finite
∴ It must be 00 form
∴ 2cos⁡0-a=0⇒a=2

Topic: Mathematics, Limits, JEE Main
If limx→0x1+acos⁡x-bsin⁡xx3=1, then the value of ab is equal to

Options:

The correct answer is:
3.75

Solution:
Using the expansion, we get,
limx→0x+ax1-x22!+x44!-......-bx-x33!+......x3=1
limx→01+a-bx+-a2+b6x3+.....x3=1
For limit to exist numerator & denominator must be of the same degree
∴1+a-b=0 … 1
Also, -a2+b6=1⇒3a-b+6=0 … 2
By equations 1 & 2, we get,
a=-52 and b=-32

⇒ab=154=3.75

Topic: Mathematics, Limits, JEE Main
If limx→1x4-1x-1=limx→kx3-k3x2-k2 , then k is

Options:
A 32
B 43
C 38
D 83

The correct answer is:
83

Solution:
Given limits can be written as,

limx→1x2+1x+1x-1x-1=limx→kx-kx2+k2+xkx-kx+k
⇒4=3k22k
⇒k=83.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If limxââaex+bcosx+c+dxxsin2x=3, then the value of 272abdc3 is equal to

Options:

The correct answer is:
34

Solution:
Using expansions, we get,
limx→0a1+x+x22!+x33!+..+b1-x22!+...+c+dxxx-x33!+.....2=3
limx→0a+b+c+a+dx+a-b2x2+a6x3+…x31-x23!+.....2=3
∵ in the denominator lowest power of x is 3
For the limit to be finite, the numerator should also have the least power of x as 3
∴a+b+c=0  ...1
a+d=0...2
a-b2=0...3
Now, a61=3⇒a=18
From 1,2,3, we get,

a=18, b=18, c=–36, d=–18
abdc3=-183-8183=18

Topic: Mathematics, Limits, JEE Main
If L=limx→0e-x22-cos⁡xx2tan2x, then the value of 3L is equal to ______. If your answer is K, mark 4K as answer after rounding off to nearest integer.

Options:

The correct answer is:
1

Solution:
Using expansion series, we get,

L=limx→01-x22+x442!...-1-x22!+x44!-...x2x+x33+...2
L=limx→01-1+-x22+x22+x48-x424+....x41+x23+....2
⇒L=18-124=112

Hence, 3L=0.25

Topic: Mathematics, Limits, JEE Main
If L=limx→0e-x22-cos⁡xx2tan2x, then the value of 3L is equal to

Options:

The correct answer is:
0.25

Solution:
Using expansion series, we get,

L=limx→01-x22+x442!...-1-x22!+x44!-...x2x+x33+...2
L=limx→01-1+-x22+x22+x48-x424+....x41+x23+....2
⇒L=18-124=112

Hence, 3L=0.25

Topic: Mathematics, Limits, JEE Main
If $l=\lim _{x \rightarrow 0} \frac{x}{|x|+x^2}$, then the value of $l$ is

Options:
A $1$
B $-1$
C $2$
D non-existant

The correct answer is:
non-existant

Solution:
Let $\mathrm{f}(x)=\frac{x}{|x|+x^2}$
$\begin{aligned}
& \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0} \frac{x}{-x+x^2}=\lim _{x \rightarrow 0} \frac{1}{-1+x}=-1 \\
& \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)=\lim _{x \rightarrow 0} \frac{x}{x+x^2}=\lim _{x \rightarrow 0} \frac{1}{1+x}=1
\end{aligned}$
Here, $\lim _{x \rightarrow 0^{-}} \mathrm{f}(x) \neq \lim _{x \rightarrow 0^{+}} \mathrm{f}(x)$
$\therefore \quad$ Value of $l$ is non-existant

Topic: Mathematics, Limits, JEE Main
If L=limxâÏ41âtanx1âsin2x1+tanxÏâ4x3, then the value of 40L is equal to ______. If your answer is K, mark 4K as answer after rounding off to nearest integer.

Options:

The correct answer is:
4

Solution:
L=limx→π4tanπ4−x1−sin2x4π4−xπ−4x2
=limx→π4141-sin⁡2xπ-4x2
=limx→π414-cos⁡2x⋅22π-4x-4 (applying L-Hospital rule)
=limx→π4116cos2xπ-4x
=limx→π4116-sin⁡2x×2-4 (again applying L-Hospital rule)
=116-1-4×2⇒L=132

Topic: Mathematics, Limits, JEE Main
If L=limxâÏ41âtanx1âsin2x1+tanxÏâ4x3, then the value of 40L is equal to

Options:

The correct answer is:
1.25

Solution:
L=limx→π4tanπ4−x1−sin2x4π4−xπ−4x2
=limx→π4141-sin⁡2xπ-4x2
=limx→π414-cos⁡2x⋅22π-4x-4 (applying L-Hospital rule)
=limx→π4116cos2xπ-4x
=limx→π4116-sin⁡2x×2-4 (again applying L-Hospital rule)
=116-1-4×2⇒L=132

Topic: Mathematics, Limits, JEE Main
If l, m, n are the direction cosines of a line which makes angles α, β and γ with the coordinate axes X, Y, Z, respectively, then lm+mn+nl takes the maximum value when

Options:
A α,β,γ are in arithmetic progression
B α=β=γ
C any two of α, β, γ are the same
D one of α, β, γ is zero and the remaining two are non-zero and unequal.

The correct answer is:
α=β=γ

Solution:
As we know,

l=cosα

m=cosβ

n=cosγ

Also, cos2α+cos2β+cos2γ=1

Now

lm+mn+nl=cosαcosβ+cosβcosγ+cosγcosα

lm+mn+nl will take the maximum value only when α=β=γ, that will be lm+mn+nl=3cos2α=1

Topic: Mathematics, Three Dimensional Geometry, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
If \( l, \mathrm{~m}, \mathrm{n} \) are in arithmetic progression, then the straight line \( l \mathrm{x}+\mathrm{my}+\mathrm{n}=0 \) will pass through the point

Options:
A \( (-1,2) \)
B \( (1,-2) \)
C \( (1,2) \)
D \( (2,1) \)

The correct answer is:
\( (1,-2) \)

Solution:
Since, l , m, n are in AP
∴ 2 m = l + n
Given equation of line is
l x + my + n = 0
âµ l - 2 m + n = 0 , x = 1 & y = - 2 satisfy l n + my + n = 0

Topic: Mathematics, Sequences and Series, JEE Main
If l n x + y = 2 x y , then y ′ 0 is equal to

Options:
A 1
B −1
C 2
D 0

The correct answer is:
1

Solution:
Given equation is ln( x+y )=2xy ...(1)
For x=0
ln( 0+y )=2.0.y=0
⇒lny=0
⇒y=1
Now, differentiating (1), we get,
1x+y1+dydx=2y+2xdydx
At point (0,1), we get,
10+11+dydx=21+2.0.dydx
⇒1+dydx=2
⇒dydx=1=y'(0)

Topic: Mathematics, Differentiation, JEE Main
If L=sin2π16-sin2π8 and M=cos2π16-sin2π8

Options:
A L=-122+12cosπ8
B L=142-14cosπ8
C M=142+14cosπ8
D M=122+12cosπ8

The correct answer is:
M=122+12cosπ8

Solution:
L=sinπ16+π8sinπ16-π8
=sin3π16·sin-π16

=12cos3π16+π16-cos3π16-π16=1212-cosπ8
M=cosπ16+π8cosπ16-π8

=cos3π16·cos-π16
=12cos3π16+π16+cos3π16-π16=1212+cosπ8

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If $l x+m y=1$ is a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $a^2 m^2-b^2 l^2=$

Options:
A $\frac{m^2}{l^2}\left(a^2+b^2\right)^2$
B $\left(l^2+m^2\right)\left(a^2+b^2\right)^2$
C $\frac{l^2}{m^2}\left(a^2+b^2\right)^2$
D $l^2 m^2\left(a^2+b^2\right)^2$

The correct answer is:
$l^2 m^2\left(a^2+b^2\right)^2$

Solution:
Given equation of normal, $l x+m y=1$
$$
y=\frac{-l}{m}+\frac{1}{m}
$$
Which is of the form $y=m x+c$
Since, the lines $y=m x+c$ will be normal to the hyperbola
$$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { if } c^2=\frac{m^2\left(a^2+b^2\right)^2}{a^2-b^2 m^2} \\
& \therefore \quad\left(\frac{1}{m}\right)^2=\frac{\left(-\frac{l}{m}\right)^2\left(a^2+b^2\right)^2}{a^2-b^2\left(-\frac{l}{m}\right)^2}
\end{aligned}
$$
$\begin{aligned} & \frac{1}{m^2}=\frac{\frac{l^2}{m^2}\left(a^2+b^2\right)^2}{\frac{m^2 a^2-l^2 b^2}{m^2}} \\ & \frac{1}{m^2}=\frac{l^2\left(a^2+b^2\right)^2}{m^2 a^2-l^2 b^2} \\ & a^2 m^2-b^2 l^2=l^2 m^2\left(a^2+b^2\right)^2\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
If $l x+m y+n=0$ is tangent to the parabola $x^2=y$, then condition of tangency is

Options:
A $I^2=2 m n$
B $I=4 m^2 n^2$
C $m^2=4 / n$
D $I^2=4 m n$

The correct answer is:
$I^2=4 m n$

Solution:
Given that $l x+m y+n+0 \ldots(i)$
$x^2=y \ldots(ii)$
The point of intersection of the line and parabola are obtained by solving (i) and (ii) simultaneously
substituting the values of $x$ from (i) in (ii), we get $\left(\frac{m y+n}{l}\right)^2=y \quad \Rightarrow m^2 y^2+n^2+2 m n y=y l^2$
$\Rightarrow m^2 y^2+\left(2 m n-l^2\right) y+n^2=0 \ldots(iii)$
If lines (iii) touches the parabola (ii), then discriminant
$=0 \Rightarrow\left(2 m n-l^2\right)^2=4 m^2 n^2$
$\Rightarrow 4 m^2 n^2+l^4-4 m n l^2=4 m^2 n^2 \Rightarrow l^2=4 m n$.

Topic: Mathematics, Parabola, JEE Main
If $l x+m y+n=0$ represents a chord of the ellipse $b^2 x^2+a^2 y^2=a^2 b^2$ whose eccentric angles differ by $90^{\circ}$, then

Options:
A $a^2 l^2+b^2 m^2=n^2$
B $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{\left(a^2-b^2\right)^2}{n^2}$
C $a^2 l^2+b^2 m^2=2 n^2$
D None of these

The correct answer is:
$a^2 l^2+b^2 m^2=2 n^2$

Solution:
Equation of chord joining points $P(a \cos \alpha, b \sin \alpha)$ and $Q(a \cos \beta, b \sin \beta)$ is $\frac{x}{a} \cos \left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b} \sin \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{\alpha-\beta}{2}\right)$
Now, $\beta=\alpha+90^{\circ}$
$\frac{x}{a} \cos \left(\frac{2 \alpha+90^{\circ}}{2}\right)+\frac{y}{b} \sin \left(\frac{2 \alpha+90^{\circ}}{2}\right)=\frac{1}{\sqrt{2}}$
now, compare it with, $l x+m y=-n$,
we get
$\begin{aligned}
& \frac{\cos \left(\frac{2 a n a 0^{\circ}}{2}\right)}{a l}=\frac{\sin \left(\frac{2 \alpha a 0^{\circ}}{20^{\circ}}\right)}{b m}=-\frac{1}{\sqrt{2} n} \\
& \because \cos ^2 \theta+\sin ^2 \theta=1 \\
& \Rightarrow a^2 l^2+b^2 m^2=2 n^2
\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
If \( \lambda_{1} \) and \( \lambda_{2} \) be wavelengths of \( \gamma \)-rays and \( X \)-rays respectively then-

Options:
A \( \lambda_{1}>\lambda_{2} \)
B \( \lambda_{1} < \lambda_{2} \)
C \( \lambda_{1}=\lambda_{2} \)
D None of the above

The correct answer is:
\( \lambda_{1} < \lambda_{2} \)

Solution:
Gamma rays have the highest frequency in the electromagnetic spectrum. X-rays, on the other hand, fall just below the gamma rays in the spectrum.
In terms of wavelength, the X-rays have a high wavelength compared to gamma rays.

Topic: Physics, Electromagnetic Waves, JEE Main
If last three digits of the expression \( 17^{256} \) are \( A_{1}, A_{2} \& A_{3} \) respectively, then the value of expression \( 2 A_{2}+A_{3}-4 A_{1} \) equals

Options:
A \( 0 \)
B \( -4 \)
C \( 5 \)
D \( -7 \)

The correct answer is:
\( -7 \)

Solution:
We can have 17256=(290−1)128

=1000I+128C2(290)2−128C1(290)+1, where I is an integer

=1000I+128(290)(18415-1)+1

=1000m+681

∴A1=6,A2=8,A3=1

Topic: Mathematics, Binomial Theorem, JEE Main
If latent heat of fusion of ice is 80 cals per g at 0oC , calculate molal depression constant for water.

Options:
A 18.63
B 186.3
C 1.863
D 0.1863

The correct answer is:
1.863

Solution:
Kf=RTf21000ÃlV Here R=2 cals ,
Tf=0+273=273, lf=80 cals
Kf=2Ã273Ã2731000Ã80=1.863

Topic: Chemistry, Solutions, JEE Main
If \( \left|z^{2}-1\right|=|z|^{2}+1 \), then \( z \) lies on

Options:
A the real axis
B the imaginary axis
C a circle
D an ellipse

The correct answer is:
the imaginary axis

Solution:
Using the relation, if

z1+z2=z1+z2

Then, argz1=argz2

since, z2+-1=z2+-1

Then, argz2=arg-1

⇒2argz=π ∵arg-1=π

⇒argz=π2

⇒z lies on Y-axis (imaginary axis).

Topic: Mathematics, Complex Number, JEE Main
If length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ whose middle point is $\left(\frac{1}{2}, \frac{2}{5}\right)$, is $7 \sqrt{\frac{k}{25}}$ unit, then $\mathrm{k}=$

Options:
A 41
B 42
C 43
D 44

The correct answer is:
41

Solution:
Equation of the ellipse is :
$\frac{x^2}{25}+\frac{y^2}{16}=1$
Equation of the chord having $\left(\frac{1}{2} \frac{3}{5}\right)$ as midpoint is:
$\begin{aligned}
& T=S_1 \\
& \Rightarrow \frac{\left(\frac{1}{4}\right)}{25}+\frac{\left(\frac{4}{25}\right)}{16}-1=\frac{\left(\frac{1}{2}\right) x}{25}+\frac{\left(\frac{2}{5}\right) y}{16}-1 \\
& \Rightarrow 4 x+5 y=4 \\
& \Rightarrow 5 y=4(1-x) \ldots(2)
\end{aligned}$
Solving with equation (1), we get
$\begin{aligned}
& 16 x^2+16(1-x)^2=400 \\
& \Rightarrow x^2-x-12=0 \\
& \Rightarrow x=4,-3
\end{aligned}$
From equation ( 2 ),
For $x=4, y=-\frac{12}{5}$
For $x=-3, y=\frac{16}{5}$
Therefore, length of the chord
$=\sqrt{7^2+\left(\frac{28}{5}\right)^2}=7 \sqrt{\frac{41}{25}} \text { unit }$
Hence, $\mathbf{k}=41$

Topic: Mathematics, Ellipse, JEE Main
If letters of the word REGULATIONS are randomly arranged, then the probability that there are exactly four letters between R and S is

Options:
A 655
B 355
C 511
D 611

The correct answer is:
655

Solution:
Total cases =11!
Number of favourable cases =2!×9C4×4!×6!
Hence, required probability =2!×9C4×4!×6!11!
=2!×9!×4!×6!11!×5!×4!
=2×611×10=655

Topic: Mathematics, Probability, JEE Main
If light travels a distance $x$ in $t_1 \sec$ in air and $10 x$ distance in $t_2 \sec$ in a medium, the critical angle of the medium will be

Options:
A $\tan ^{-1}\left(\frac{t_1}{t_2}\right)$
B $\sin ^{-1}\left(\frac{t_1}{t_2}\right)$
C $\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)$
D $\tan ^{-1}\left(\frac{10 t_1}{t_2}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)$

Solution:
From the formula $\sin C=\frac{1}{{ }_1 \mu_2} \Rightarrow \sin C={ }_2 \mu_1$
$=\frac{u_1}{u_2}=\frac{v_2}{v_1} \Rightarrow \sin C=\frac{10 x / t_2}{x / t_1}$
$\Rightarrow \sin C=\frac{10 t_1}{t_2} \Rightarrow C=\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)$

Topic: Physics, Ray Optics, JEE Main
If light travels a distance x in time t1 in air and 10x distance in time t2 in a certain medium, then find the critical angle for total internal reflection, of the medium.

Options:
A sin-120t1t2
B sin-110t1t2
C sin-1t1t2
D sin-1t210t1

The correct answer is:
sin-110t1t2

Solution:
μ=xt110xt2=t210t1
sin⁡θc=1t2/10t1=10t1t2⇒θc=sin-110t1t2

Topic: Physics, Ray Optics, JEE Main
If limn→∞⁡1a+2a+…+nan+1a-1na+1+na+2+…+na+n=160 for some positive real number a, then a is equal to

Options:
A 172
B 152
C 7
D 8

The correct answer is:
7

Solution:
limn→∞na(∑r=1n(rn)a)(n+1)a−1.n2(a+1+1n2)=160

⇒limn→∞na∑r=1rnana+11+1na-1a+1+1n2=160

⇒limn→∞1n∑r=1n(rn)a(1+1n)a-1a+1+1n2=160

⇒∫01xadx(a+12)=160
⇒ 1a+1a+12=160
⇒ a+1 2a+1=120
⇒2a2+3a-119=0
⇒2a2+17a-14a-119=0
⇒ a-7 2a+17=0
a=7, -172.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2017 (09 Apr Online)
If limnâââr=12n3r2n3er3n3=ea-eb, then a+b is equal to

Options:

The correct answer is:
8

Solution:
The given limit is

limn→∞∑r=12n3r2n3er3n3=∫023x2ex3dx
=ex302
=e8-1=e8-e0

Hence, a+b=8

Topic: Mathematics, Definite Integration, JEE Main
If $\lim _{n \rightarrow \infty} \sum_{r-1}^n \frac{4 r^3}{r^4+n^4}=p$, then $e^p=$

Options:
A 4
B 3
C 2
D 1

The correct answer is:
2

Solution:
$\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{r^4+n^4}=p$
$$
e^p=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 r^3}{n^4 \cdot\left[1+\frac{r^4}{n^4}\right]} \Rightarrow \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{4 \frac{r^3}{n^3}}{\left[n\left[1+\frac{r^4}{n^4}\right]\right.}
$$
Let $x=\frac{r}{n}$ and $d x=\frac{1}{n}$, so limit value will be from ' 0 ' to 1 .
So,
$$
\begin{aligned}
4 \int_0^1 \frac{x^3}{1+x^4} d x & \Rightarrow 1+x^4=t \\
4 x^3 d x=d t & \Rightarrow \int_1^2 \frac{d t}{t}
\end{aligned}
$$
Now, $x=0$, then $t=1 x=1$, then $t=2$
$$
\begin{array}{ll}
& {[\ln t]_1^2} \\
& \\
\text { So, } & \ln 2-\ln (1) \Rightarrow \ln 2=P \\
e^p=e^{\ln 2} \Rightarrow 2
\end{array}
$$

Topic: Mathematics, Definite Integration, JEE Main
If limx→0⁡1+px+qx2cosec x=2048, then the value of p11 is equal to (take ln⁡2=0.69)

Options:

The correct answer is:
0.69

Solution:
The given limit is of the form 1∞
∴elimx→0⁡cosec x1+px+qx2-1=2048
limx→0⁡px+qx2sin⁡x=ln⁡2048
limx→0⁡px+qx2x⋅xsin⁡x=11ln⁡2
limx→0⁡p+qx1=11ln⁡2⇒p=11ln⁡2

Hence, p11=ln2=0.69

Topic: Mathematics, Limits, JEE Main
If limx→01x81−cosx22−cosx24+cosx22cosx24=2−k then the value of k is

Options:

The correct answer is:
8

Solution:
limx→01x81−cosx22−cosx24+cosx22cosx24

=limx→01−cosx221−cosx24x8

=limx→01−cosx22x4×1−cosx24x4

=limx→01−cosx224×x44×1−cosx2416×x416

Using limx→01-cosxx2=12

=12×14×12×116=1256=2-8

⇒2-8=2-k ⇒k=8

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (03 Sep Shift 1)