datavorous/entrance-exam-dataset huggingface dataset:
If the metal bob of a simple pendulum is replaced by a wooden bob, then its time period will

Options:
A Increase
B Decrease
C Remain the same
D First increase then decrease

The correct answer is:
Remain the same

Solution:
Remains the same because time period of simple pendulum $T$ is independent of mass of the bob.

Topic: Physics, Oscillations, JEE Main
If the middle point of a chord of the circle $x^2+y^2+x-y-1=0$ be $(1,1)$, then the length of the chord is

Options:
A $4$
B $2$
C $5$
D None of these

The correct answer is:
None of these

Solution:
Here radius of circle is
$\begin{aligned}& A C=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2+1}=\sqrt{\frac{3}{2}} \\& \text { and } C B=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{\frac{10}{4}}=\sqrt{\frac{5}{2}} \\& \text { Now } A B^2=A C^2-B C^2=\frac{3}{2}-\frac{5}{2}=-1\end{aligned}$
$\Rightarrow$ There is no possible value for $A B$.
Aliter : Since the point $(1,1)$ lies outside the circle, therefore no such chord exist

Topic: Mathematics, Circle, JEE Main
If the middle points of sides \(\mathrm{BC}, \mathrm{CA} ~\&~ \mathrm{AB}\) of triangle \(A B C\) are respectively \(D, E, F\) then position vector of centre of triangle \(\mathrm{DEF}\), when position vector of \(A, B, C\) are respectively \(\hat{i}+\hat{j}\), \(\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{k}}+\hat{\mathrm{i}}\) is -

Options:
A \((1 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
B \((\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
C \(2(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
D \((2 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)

The correct answer is:
\((2 / 3)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)

Solution:
The position vector of points \(D, E, F\) are respectively
\(\frac{\mathrm{i}+\mathrm{j}}{2}+\mathrm{k}, \mathrm{i}+\frac{\mathrm{k}+\mathrm{j}}{2} \text { and } \frac{\mathrm{i}+\mathrm{k}}{2}+\mathrm{j}\)
So, position vector of centre of \(\triangle \mathrm{DEF}\)
\(=\frac{1}{3}\left[\frac{\mathrm{i}+\mathrm{j}}{2}+\mathrm{k}+\mathrm{i} \frac{\mathrm{k}+\mathrm{j}}{2}+\frac{\mathrm{i}+\mathrm{k}}{2}+\mathrm{j}\right]=\frac{2}{3}[\mathrm{i}+\mathrm{j}+\mathrm{k}]\)

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the middle term in the binomial expansion of 1x+xsinx10 is 638, then the value of 6sin2x+sinx-2 is

Options:

The correct answer is:
0

Solution:
Here, n=10, which is even.
Middle term =102+1th term =6th term
T 6 =   10 C 5 1 x 5 (xsinx) 5
â638=252sinx5
âsinx5=132
âsinx5=125
âsinx=12
â2sinx-1=0
â6sin2x+sinx-2=2sinx-13sinx+2=0

Topic: Mathematics, Binomial Theorem, JEE Main
If the middle term in the expansion of $(1+x)^{2 n}$ is the greatest term, then $x$ lies in the interval

Options:
A $\left(\frac{n}{n+1}, \frac{n+1}{n}\right)$
B $\left(\frac{n+1}{n}, \frac{n}{n+1}\right)$
C $(n-2, n)$
D $(n-1, n)$

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

Solution:
In the expansion of $(1+x)^{2 n}$, middle term is ${ }^{2 n} \mathrm{C}_n x^n$. Since, middle term is the greatest term.
$\therefore{ }^{2 n} \mathrm{C}_n x^n>{ }^{2 n} \mathrm{C}_{n-1} x^{n-1}$
$\text { and }{ }^{2 n} \mathrm{C}_r x^n>{ }^{2 n} \mathrm{C}_{n+1} x^{n+1}$
$\begin{aligned}
& \Rightarrow \quad x>\frac{{ }^{2 n} \mathrm{C}_{n-1}}{{ }^{2 n} \mathrm{C}_n} \text { and } x < \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}} \\
& \text { Hence, } x \in\left(\frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}, \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}\right)
\end{aligned}$
Hence, $x \in\left(\frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}, \frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n} \mathrm{C}_{n+1}}\right)$
$\begin{aligned}
& x \in\left(\begin{array}{c}
\frac{(2 n) !}{(n-1) !(2 n-n+1)} \times \frac{n !(2 n-n) !}{(2 n) !} \\
\frac{(2 n) !}{n !(2 n-n) !} \times \frac{(n-1) !(2 n-n+1)}{(2 n) !}
\end{array}\right) \\
& x \in\left(\frac{n(n-1) ! n !}{n(n-1) !(n+1) n !}, \frac{(n+1) n !(n-1) !}{n(n-1) ! n !}\right) \\
& \therefore \quad x \in\left(\frac{n}{n+1}, \frac{n+1}{n}\right)
\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
If the middle term in the expansion of 1x+xsinx10 is equal to 778, then the number of values of x in 0,2Ï is equal to

Options:

The correct answer is:
2

Solution:
There are 11 terms in the expansion â6th term is the middle term

T6=C5101x5xsinâ¡x5
âC510sinâ¡x5=638
â10Ã9Ã8Ã7Ã65Ã4Ã3Ã2sinâ¡x5=638
âsinâ¡x5=638Ã5Ã4Ã3Ã210Ã9Ã8Ã7Ã6
âsinâ¡x=12

Hence, there are 2 values of x in 0,2Ï

Topic: Mathematics, Binomial Theorem, JEE Main
If the middle term in the expansion of \( \left(\frac{1}{x}+x \sin x\right)^{10} \) is equal to \( 7 \frac{7}{8} \), then the number of values of \( x \) in \( [0,2 \pi] \) is equal to

Options:

The correct answer is:
2

Solution:
There are 11 terms in the expansion â6th term is the middle term

T6=C5101x5xsinâ¡x5
âC510sinâ¡x5=638
â10Ã9Ã8Ã7Ã65Ã4Ã3Ã2sinâ¡x5=638
âsinâ¡x5=638Ã5Ã4Ã3Ã210Ã9Ã8Ã7Ã6
âsinâ¡x=12

Hence, there are 2 values of x in 0,2Ï

Topic: Mathematics, Binomial Theorem, JEE Main
If the middle term in the expansion of $\left(x^2+\frac{1}{x}\right)^n$ is $924 x^6$, then $n=$

Options:
A $10$
B $12$
C $14$
D None of these

The correct answer is:
$12$

Solution:
Since $n$ is even therefore $\left(\frac{n}{2}+1\right)^{t h}$ term is middle term, hence ${ }^n C_{n / 2}\left(x^2\right)^{n / 2}\left(\frac{1}{x}\right)^{n / 2}=924 x^6$
$\Rightarrow \quad x^{n / 2}=x^6 \Rightarrow n=12$

Topic: Mathematics, Binomial Theorem, JEE Main
If the mid-point of the line segment joining the points $A(3,4)$ and $B(k, 6)$ is $P(x, y)$ and $x+y-10=0$, find the value of $k$.

Options:
A 7
B 13
C 10
D 6

The correct answer is:
7

Solution:
Given that the mid-point of the line segment joining the points $A(3,4)$ and $B(k, 6)$ is $\mathrm{P}(\mathrm{x}, \mathrm{y})$.
Using the mid-point formula,
$\begin{aligned} & x=(3+k) / 2 \\ & y=(4+6) / 2 \\ & y=10 / 2=5\end{aligned}$
From the given,
$x+y-10=0$
Substituting $y=5$ in this equation, we get;
$\begin{aligned} & {[(3+k) / 2]+5-10=0} \\ & (3+k) / 2=5 \\ & 3+k=10 \\ & k=10-3=7\end{aligned}$
Therefore, $\mathrm{k}=7$.

Topic: Mathematics, Straight Lines, JEE Main
If the minimum area of the triangle formed by a tangent to the ellipse x2 b2+y24a2=1 and the co-ordinate axis is kab, then k is equal to ___________.

Options:

The correct answer is:
2

Solution:
Equation of tangent at parametric point

b cosθ,2a sin θ to the ellipse is xbcos θ+y2asin θ=1

Co-ordinates of Pbcos θ,0 and Q0,2asin θ respectively.

Therefore, Area of triangle formed by tangent and co-ordinate axes=12×bcos θ×2asin θ=absin θ.cos θ=2absin 2θ

as sin θ≤1 therefore we can say Minimum area of triangle =2ab

⇒k=2

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If the minimum distance of the point (secâ¡Î±,cosecâ¡Î±) from the circle x2+y2=3 can be expressed as a-b, where a,bâN, then the value of ba is

Options:

The correct answer is:
1.5

Solution:
Since, sec2⁡α+cosec2⁡α-3=1+tan2⁡α+1+cot2⁡α-3

=-1+tan⁡α-cot⁡α2+2=1+tan⁡α-cot⁡α2>0

Hence, the point sec⁡α,cos⁡ecα lies outside the circle x2+y2-3=0
Now, the minimum distance
=sec⁡α-02+cosec⁡α-02-3
=4+tan⁡α-cot⁡α2-3 (Minimum is attained, when tan⁡α=cot⁡α)
=2-3≡a-b (Given)
So, a=2,b=3.

Hence, ba=32=1.5

Topic: Mathematics, Circle, JEE Main
If the mirror image of the point 1,3,5 with respect to the plane 4x-5y+2z=8 is α,β,γ, then 5α+β+γ equals :

Options:
A 43
B 47
C 41
D 39

The correct answer is:
47

Solution:
Point Q is image of point P w.r.to plane, M is mid point of P and Q, lies in plane

M1+α2,3+β2,5+γ2

4x-5y+2z=8

41+α2-53+β2+25+γ2=8 ...1

Also PQ perpendicualr to the plane

⇒PQ→‖n→

α-14=β-3-5=γ-52=k (let)

α=1+4kβ=3-5kγ=5+2k ...2

use 2 in 1

21+4k-56-5k2+10+2k=8

k=25

from 2 α=135,β=1,γ=295

5α+β+γ=13+5+29=47

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (26 Feb Shift 2)
If the mirror image of the point 2,4,7 in the plane 3x-y+4z=2 is a,b,c, the 2a+b+2c is equal to

Options:
A 54
B -6
C 50
D -42

The correct answer is:
-6

Solution:
Given, equation of 3x-y+4z=2 and point P2,4,7

Now using the formula of image

x-x1a=y-y1b=z-z1c=-2×ax1+by1+cz1+da2+b2+c2

We get,

x-23=y-4-1=z-74=-26-4+28-226

⇒ x-23=y-4-1=z-74=-2813

⇒ x=-5813,y=8013,z=-2113

Hence, 2a+b+2c=-11613+8013-4213=-6

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (29 Jun Shift 1)
If the molecular weight of two gases are $M_1$ and $M_1$, then at a temperature the ratio of root mean square velocity $v_1$ and $v_2$ will be

Options:
A $\sqrt{\frac{M_1}{M_2}}$
B $\sqrt{\frac{M_2}{M_1}}$
C $\sqrt{\frac{M_1+M_2}{M_1-M_2}}$
D $\sqrt{\frac{M_1-M_2}{M_1+M_2}}$

The correct answer is:
$\sqrt{\frac{M_2}{M_1}}$

Solution:
$v_{r m s}=\sqrt{3 R T / M} \Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{M_2}{M_1}}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
If the molecular wt. of \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\) and \(\mathrm{I}_2\) are \(\mathrm{M}_1\) and \(\mathrm{M}_2\) respectively, then what will be the equivalent wt. of \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\) and \(\mathrm{I}_2\) in the following reaction?
\(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \longrightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\)

Options:
A \(\mathrm{M}_1, \mathrm{M}_2\)
B \(\mathrm{M}_1, \mathrm{M}_2 / 2\)
C \(2 \mathrm{M}_1, \mathrm{M}_2\)
D \(\mathrm{M}_1, 2 \mathrm{M}_2\)

The correct answer is:
\(\mathrm{M}_1, \mathrm{M}_2 / 2\)

Solution:
Hints:
\(\begin{aligned}
& \text { n.f. }\left(\mathrm{S}_2 \mathrm{O}_3^{2-}\right)=1, \text { Equivalent mass }=\frac{\mathrm{M}_1}{1}=\mathrm{M}_1 \\
& \mathrm{n} \mathrm{f} .\left(\mathrm{I}_2\right)=2, \quad \text { Equivalent mass }=\frac{\mathrm{M}_2}{2}
\end{aligned}\)

Topic: Chemistry, Redox Reactions, JEE Main
If the momentum of a body is increased $n$ times, its kinetic energy increases

Options:
A $n$ times
B $2 n$ times
C $\sqrt{n}$ times
D $n^2$ times

The correct answer is:
$n^2$ times

Solution:
$E=\frac{P^2}{2 m} \therefore E \propto P^2$
i.e. if $P$ is increased $n$ times then $E$ will increases $n^2$ times.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
If the momentum of an electron is changed by $p$, then the de-Broglie wavelength associated with it changes by $0.5 \%$. The initial momentum of electron will be

Options:
A $200 p$
B $400 p$
C $\frac{p}{200}$
D $100 p$

The correct answer is:
$200 p$

Solution:
de-Broglie wavelength
Here
$$
\begin{aligned}
\lambda & =\frac{h}{p} \\
\left|\frac{\Delta \lambda}{\lambda}\right| & =\left|\frac{\Delta p}{p}\right| \\
\frac{0.5}{100} & =\frac{p}{p_i} \\
p_i & =\frac{100}{0.5} p \\
p_i & =200 p
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2012 (Mains)
If the $n^{\text {th }}$ term of a series be $3+n(n-1)$, then the sum of $n$ terms of the series is

Options:
A $\frac{n^2+n}{3}$
B $\frac{n^3+8 n}{3}$
C $\frac{n^2+8 n}{5}$
D $\frac{n^2-8 n}{3}$

The correct answer is:
$\frac{n^3+8 n}{3}$

Solution:
Here, $T_n=3+n(n-1)=3+n^2-n$
Now sum $S=\Sigma T_n=\Sigma\left(3+n^2-n\right)$
$=3 n+\frac{1}{6} n(n+1)(2 n+1)-\frac{n(n+1)}{2}$
$=\frac{1}{6} n(n+1)[2 n+1-3]+3 n=\frac{n^3+8 n}{3}$

Topic: Mathematics, Sequences and Series, JEE Main
If the net electric field at point $\mathrm{P}$ along $\mathrm{Y}$ axis is zero, then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$, where $x=$ _____.

Options:

The correct answer is:
5

Solution:
$\begin{aligned} & \frac{\mathrm{Kq}_2}{20} \cos \beta=\frac{\mathrm{Kq}_3}{25} \cos \theta \\ & \frac{\mathrm{Kq}_2}{20} \frac{4}{\sqrt{20}}=\frac{\mathrm{Kq}_3}{25} \frac{4}{\sqrt{25}} \\ & \frac{\mathrm{q}_2}{\mathrm{q}_3}=\frac{20}{25} \sqrt{\frac{20}{25}}=\frac{8}{5 \sqrt{\mathrm{x}}} \\ & \Rightarrow \sqrt{\mathrm{x}}=\frac{8 \times 25 \sqrt{25}}{5 \times 20 \sqrt{20}} \\ & \mathrm{x}=5\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If the normal at P18,12 to the parabola y2=8x cuts it again at Q, then the equation of the normal at point Q on the parabola y2=8x is

Options:
A 27y=99x-3058
B 27y=99x+3058
C 27y=-99x-3058
D None of these

The correct answer is:
27y=99x-3058

Solution:
Let, 2t2,4t=18,12 & Q=2t12,4t1
then t=3 & t1=-t-2t
⇒t1=-3-23=-113
Equation of the normal at Q is
y=-t1x+4t1+2t13
⇒y=113x-443-266227
⇒27y=99x-3058

Topic: Mathematics, Parabola, JEE Main
If the normal drawn at a point $P$ on the curve $3 y=6 x-5 x^3$ passes through $(0,0)$, then the positive integral value of the abscissa of the point $P$ is

Options:
A $1$
B $\frac{2}{3}$
C $\frac{1}{3}$
D $-\frac{2}{3}$

The correct answer is:
$1$

Solution:
Given curve $3 y=6 x-5 x^3$
$\Rightarrow \quad \frac{d y}{d x}=2-5 x^2$
Slope of normal $=\frac{-1}{2-5 x^2}$
Equation of normal with slope $\frac{-1}{2-5 x^2}$
passing through origin $(0,0)$.
$\frac{\left(y-y_1\right)}{x-x_1}=m$
$\Rightarrow \frac{y-0}{x-0}=\frac{-1}{2-5 x^2} \Rightarrow \frac{y}{x}=\frac{-1}{2-5 x^2}$
Since, $(h, k)$ lies on normal
$\frac{k}{h}=\frac{-1}{2-5 h^2}$
$\Rightarrow \frac{k}{h}=\frac{1}{5 h^2-2}$ ...(i)
Since, $(h, k)$ lies on curve
$3 k=6 h-5 h^3$
$\Rightarrow \quad 3 k=h\left(6-5 h^2\right)$
$\Rightarrow \quad \frac{k}{h}=\frac{6-5 h^2}{3}$ ...(ii)
On solving Eqs. (i) and (ii),
$\frac{1}{5 h^2-2}=\frac{6-5 h^2}{3}$
$\begin{aligned} & \Rightarrow \quad 3=\left(6-5 h^2\right)\left(5 h^2-2\right) \\ & \Rightarrow \quad 3=30 h^2-12-25 h^4+10 h^2 \\ & \Rightarrow \quad 15=-25 h^4+40 h^2 \\ & \Rightarrow \quad 5 h^4-8 h^2+3=0\end{aligned}$
For simplicity, let $h^2=z$
The equation becomes $5 z^2-8 z+3=0$
$(5 z-3)(z-1)=0$
Now, $z=\frac{3}{5}, 1$
Putting $h^2=z$
$\Rightarrow \quad h^2=\frac{3}{5} \Rightarrow h= \pm \sqrt{\frac{3}{5}}$
It is not in option.
or $(z-1)=0$
$z=1$
Putting $h^2=z$
$\Rightarrow \quad h^2=1$
$h= \pm 1$
But $h=-1$ is not in option.
$\therefore h=1$ is correct

Topic: Mathematics, Application of Derivatives, JEE Main
If the normal drawn from the origin to the straight line $2 x+7 y+6=0$ makes an angle 0 with the positive $X$-axis, then $0=$

Options:
A $\tan ^{-1} \frac{7}{2}$
B $\pi-\tan ^{-1} \frac{7}{2}$
C $\pi+\tan ^{-1} \frac{7}{2}$
D $\tan ^{-1} \frac{7}{2}$

The correct answer is:
$\pi+\tan ^{-1} \frac{7}{2}$

Solution:
Given equation of straight line,
$2 x+7 y+6=0 \Rightarrow \frac{x}{-3}+\frac{y}{-6 / 7}=1$
So, angle $\theta=\pi+\angle A O M$
$\begin{aligned} & =\pi+\alpha[\text { Let } \angle A O M=\alpha], \\ & \because(\tan \alpha) \times \text { Slope of line }=-1 \\ & \Rightarrow(\tan \alpha) \times\left(-\frac{2}{7}\right)=-1 \\ & \Rightarrow \quad \tan \alpha=7 / 2\end{aligned}$
So, required angle $\theta=\pi+\tan ^{-1} \frac{7}{2}$ Hence, option (c) is correct.

Topic: Mathematics, Straight Lines, JEE Main
If the normal to the curve yx=∫0x2t2-15t+10dt at a point a,b is parallel to the line x+3y=-5,a>1, then the value of a+6b is equal to ________.

Options:

The correct answer is:
406

Solution:
yx=∫0x2t2-15t+10dt

y'xx=a=2x2-15x+10a=2a2-15a+10

Slope of normal =-13

⇒2a2-15a+10=3⇒a=7

and a=12 (rejected)

b=y7=∫072t2-15t+10dt

=2t33-15t22+10t07

⇒6b=4×73-45×49+60×7

a+6b=406

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (16 Mar Shift 1)
If the normal to the ellipse 3x2+4y2=12 at a point P on it is parallel to the line, 2x+y=4 and the tangent to the ellipse at P passes through Q(4,4) then PQ is equal to:

Options:
A 61 2
B 55 2
C 157 2
D 221 2

The correct answer is:
55 2

Solution:
Given the equation of an ellipse is x24+y23=1,where a=2, b=3.
As we know equation of normal to the ellipse x2a2+y2b2=1 can be taken as
axsec⁡θ-bycosec⁡θ=a2-b2
⇒2xsec⁡θ-3ycosec⁡θ=1 for the given ellipse.
Now, this normal is parallel to 2x+y=5.
⇒-2sec⁡θ3cosec⁡θ=-2⇒tan⁡θ=-3⇒θ=2π3,-π3
Hence point Pacos⁡θ, bsin⁡θ=-1,32 or 1, -32
Tangent at -1,32 is x.-14+y.323=1
⇒-x+2y=4 which also passes through Q4, 4
⇒PQ=4±12+4∓322=552.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If the normal to the ellipse x225+y21=1 is at a distance p from the origin, then the maximum value of p is

Options:

The correct answer is:
4

Solution:
Equation of the normal is
5secâ¡Î¸x-cosecÎ¸y=24
So, p=2425sec2â¡Î¸+cosec2Î¸=2436+5tanâ¡Î¸-cotâ¡Î¸2
pmax=246=4

Topic: Mathematics, Ellipse, JEE Main
If the normal to the rectangular hyperbola x2-y2=1 at the point Pπ4 meets the curve again at Qθ, then sec2θ+tanθ=

Options:
A 43
B 57
C 3
D 1

The correct answer is:
57

Solution:
Given,

Equation of hyperbola x2-y2=1

Now parametric point is given by asecθ,btanθ≡2,1 as given θ=45° & a=b=1

Now finding slope of tangent at the given point by differentiating the curve we get,

2x-2ydydx=0

⇒dydx2,1=21

So, slope of normal will be -2,

Now equation of normal will be y=-x2+2

Since normal is intersecting the curve again so,

x2--x2+22=1

⇒x2+42x-10=0

⇒x=2 or -52

So, y=7 for x=-52

Now we know that x=secθ=-52

And y=tanθ=7,

So, the value of sec2θ+tanθ=50+7=57

Topic: Mathematics, Hyperbola, JEE Main
If the normals at two points $P$ and $Q$ of a parabola $y^2=4 a x$ intersect at a third point $R$ on the curve, then the product of ordinates of $P$ and $Q$ is

Options:
A $4 a^2$
B $2 a^2$
C $-4 a^2$
D $8 a^2$

The correct answer is:
$8 a^2$

Solution:
As we know, $t_1 \times t_2=2 \Rightarrow 2 a t_1 \times 2 a t_2=8 a^2$

Topic: Mathematics, Parabola, JEE Main
If the normals at two points P and Q of a parabola y2=4x intersect at a third point R on the parabola y2=4x, then the product of the ordinates of P and Q is equal to

Options:

The correct answer is:
8

Solution:
Let, P=t12,2t1, Q=t22,2t2 and R=t2,2t

then t1,t2 are roots of the equation t=-x-2x
⇒t1,t2 are roots of the equation x2+tx+2=0
⇒t1t2=2
⇒2t12t2=8
⇒ product of the ordinates of P and Q is equal to 8

Topic: Mathematics, Parabola, JEE Main
If the normals at two points x1,y1 and x2,y2 of the parabola y2=4x meets again on the parabola, where x1+x2=8, then y1-y2 is equal to

Options:
A 2
B 3
C 4
D 2

The correct answer is:
4

Solution:
Let, x1,y1≡t12,2t1 & x2,y2≡t22,2t2

Now, x1+x2=8 and both the normals again meets on the parabola,

∴ t12+t22=8 & t1t2=2

On solving, we get,
t1-t22=4⇒t1-t2=2⇒y1-y2=4

Topic: Mathematics, Parabola, JEE Main
If the normals drawn to the hyperbola \(x y=4\) at \(\left(\alpha_i, \beta_i\right)(i=1,2,3,4)\) are concurrent at the point \((a, b)\), then
\(\frac{\left(\alpha_1+\alpha_2+\alpha_3+\alpha_4\right)}{\left(\beta_1+\beta_2+\beta_3+\beta_4\right)}\left(\alpha_1 \alpha_2 \alpha_3 \alpha_4\right)=\)

Options:
A \(\frac{-16 b}{a}\)
B \(\frac{-16 a}{b}\)
C \(\frac{4 b}{a}\)
D \(\frac{4 a}{b}\)

The correct answer is:
\(\frac{-16 a}{b}\)

Solution:
The equation of normal to the given hyperbola \(x y=4\) at point \(\left(2 t, \frac{2}{t}\right)\) is
\(2 t^4-x t^3+y t-2=0\) ...(i)
\(\because\) Normal (i) passes through point \((a, b)\), so \(2 t^4-a t^3+b t-2=0\), the equation having roots are \(\frac{\alpha_1}{2}, \frac{\alpha_2}{2}, \frac{\alpha_3}{2}\) and \(\frac{\alpha_4}{2}\), so
\(\begin{gathered}
\frac{1}{2}\left(\alpha_1+\alpha_2+\alpha_3+\alpha_4\right)=\frac{a}{2} \Rightarrow \alpha_1+\alpha_2+\alpha_3+\alpha_4=a \\
\sum\left(\frac{\alpha_1 \alpha_2}{4}\right)=0 \\
\sum \frac{\alpha_1 \alpha_2 \alpha_3}{8}=\frac{-b}{2} \text { and } \frac{\alpha_1 \alpha_2 \alpha_3 \alpha_4}{16}=-1 \\
\because \beta_1+\beta_2+\beta_3+\beta_4=\frac{4}{\alpha_1}+\frac{4}{\alpha_2}+\frac{4}{\alpha_3}+\frac{4}{\alpha_4}
\end{gathered}\)
\(=4 \frac{\Sigma \alpha_1 \alpha_2 \alpha_3}{\alpha_1 \alpha_2 \alpha_3 \alpha_4}\)
\(=4 \frac{-b / 2 \times 8}{-16}=b\)
\(\begin{aligned}
\frac{\left.\therefore \alpha_1+\alpha_2+\alpha_3+\alpha_4\right)}{\beta_1+\beta_2+\beta_3+\beta_4} & \left(\alpha_1 \alpha_2 \alpha_3 \alpha_4\right) \\
= & \frac{a(-16)}{b}=-16 \frac{a}{b}
\end{aligned}\)
Hence, option (b) is correct.

Topic: Mathematics, Hyperbola, JEE Main
If the normals of the parabola y2=4x drawn at the end points of its latus rectum are tangents to the circle x-32+y+22=r2, then the value of r4 is equal to

Options:

The correct answer is:
4

Solution:
End points of the latus rectum of the parabola y2=4x and 1,±2.
Equation of the normals at points 1,±2 are
y=-x+3 and y=x-3
rr x+y-3=0 and x-y-3=0
These lines are tangent to the circle x-32+y+22=r2
∴3±2-31+1=r=2

Hence, r4=4

Topic: Mathematics, Parabola, JEE Main
If the nucleus 1 3 2 7 Al has a nuclear radius 3.6 fm, then what would be the approximate value of nuclear radius (in fm) of 3 2 12 5 Te?

Options:

The correct answer is:
6

Solution:
Nuclear radius, R=( R 0 ) A 1/3
where A is the mass number.
â´ Râ¡ Teâ¡ Râ¡ Alâ¡ = Aâ¡ Teâ¡ Aâ¡ Alâ¡ 1 / 3 = 1 2 5 2 7 1 / 3 = 5 3
or Râ¡ Te = 5 3 Ã Râ¡ Alâ¡ = 5 3 Ã 3 Â· 6 = 6 fm Given Râ¡ Alâ¡ = 3 Â· 6 fm

Topic: Physics, Nuclear Physics, JEE Main
If the number of 5-element subsets of the set $A=\left\{a_1, a_2, \ldots, a_{20}\right\}$ of 20 distinct elements is $k$ times the number of 5-element subsets containing $a_4$, then $k$ is

Options:
A
5

B
$\frac{20}{7}$

C
4

D
$\frac{10}{3}$

The correct answer is:
4

Solution:
Set $A=\left\{a_1, a_2, \ldots . ., a_{20}\right\}$ has 20 distinct elements.
We have to select 5-element subset.
$\therefore$ Number of 5-element subsets $={ }^{20} \mathrm{C}_5$
According to question
$$
\begin{aligned}
& { }^{20} \mathrm{C}_5=\left({ }^{19} \mathrm{C}_4 \cdot k\right. \\
& \Rightarrow \frac{20 !}{5 ! 15 !}=k \cdot\left(\frac{19 !}{4 ! 15 !}\right) \\
& \Rightarrow \quad \frac{20}{5}=k \Rightarrow k=4
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2012 (07 May Online)
If the number of 7 digit numbers whose sum of the digits is equal to 10 and which is formed by using the digits 1,2 and 3 only is K, then the value of K+46100 is

Options:

The correct answer is:
1.23

Solution:
In a 7 digits number formed by using 1,2 and 3 suppose that 1 appears x times, 2 appears y times and 3 appears z times. Then x+2y+3z=10 and x+y+z=7
⇒y+2z=3
⇒y=1 ,  z=1 or y=3 ,z=0
⇒ x=5,  y=1,  z=1 or x=4 ,  y=3,  z=0
Therefore, the total number is 7!5!+7!4!  3!=42+35=77
⇒k=77 ⇒k+46100=77+46100
=1.23

Topic: Mathematics, Permutation Combination, JEE Main
If the number of σ and π bonds in vinyl acetate are x and y respectively. What is the sum of x + y?

Options:

The correct answer is:
13

Solution:
Vinyl acetate has following structure
Cσ|σ|HH=πσC|σH-σO-σCσ||πO-C|σσ|HH-σH
Total σ bond = x = 11
Total π bond = y = 2
Total sum x + y = 11 + 2 = 13

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
If the number of five digit numbers with distinct digits and 2 at the 10th place is 336k , then k is equal to:

Options:
A 4
B 6
C 7
D 8

The correct answer is:
8

Solution:
2

Total Number of numbers =8×8×7×6=2688=336k⇒k=8

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If the number of five-digit numbers which can be read in the same way from the left and from the right is k, then k100 is equal to

Options:

The correct answer is:
9

Solution:
The number will be of the form XYZYX first place can’t be zero, so it can be any 1 to 9⇒9 ways

Y and Z each can be any of 0 and 9 so, they can be filled in 10 ways each.
The fourth place will be the same as second place and the last place will be the same as the first place.
So, the required number of ways =9×10×10×1×1=900
⇒k=900⇒k100=9.

Topic: Mathematics, Permutation Combination, JEE Main
If the number of integral solutions of x+y+z+w<25 are Cλ23, such that x>-2, y>1, z≥0, w>3, then the value of λ is

Options:
A 3
B 5
C 17
D 19

The correct answer is:
19

Solution:
x+y+z+w<25
Let, x+y+z+w+a=25, such that a>0
⇒x+1+y-2+z+w-4+a-1
=t1+t2+t3+t4+t5=25+1-2-4-1=19
x>-2⇒x≥-1⇒x+1≥0⇒t1≥0,y>1⇒y≥2⇒y-2≥0⇒t2≥0
z≥0⇒t3≥0
w>3⇒w≥4⇒w-4≥0⇒t4≥0
a>0⇒a≥1⇒a-1≥0⇒t5≥0
t1+t2+t3+t4+t5=19 where ti≥0
Number of integral solutions =C5-119+5-1=C4=C192323

Topic: Mathematics, Permutation Combination, JEE Main
If the number of integral solutions x,y,z of the equation xyz=18 is t, then the value of t8 is

Options:

The correct answer is:
9

Solution:
18=1.1.18=1.2.9=1.3.6=2.3.3
Each of 1,2,9,1,3,6 can be permuted in 3! ways
Each of 1,1,18,2,3,3 can be permuted in 3 ways
So, there are 2Ã3!+2Ã3=18 triplets x,y,z
Also, the signs are
+,â,â,â,+,â,â,â,+,+,+,+,
So, the desired number is 18Ã4=72
Hence,
728=9

Topic: Mathematics, Permutation Combination, JEE Main