datavorous/entrance-exam-dataset huggingface dataset:
If the tangent at \( P \) and \( Q \) on the parabola meet in \( T \), then \( S P, S T \) and \( S Q \) are in

Options:
A AP
B GP
C HP
D None of these

The correct answer is:
GP

Solution:
Since, tangent at P and Q on the parabola meet in T

Taking parametric co-ordinates of the point of tangency, let the coordinates of P and Q are at12, 2at1 and at22, 2at2 respectively, then coordinates of T are at1t2, at1+t2

Now S being the focus of the standard parabola y2=4ax i.e. co-ordinates of S being a,0.

By distance formula,

∴SP=a-at122+0-2at122=a2+a2t14-2a2t12+4a2t12=a2+a2t14+2a2t12=a1+t12

Similarly, SQ=a-at222+0-2at222=a2+a2t24-2a2t22+4a2t22=a2+a2t24+2a2t22=a1+t22

and ST=a21-t1t22+a2t1+t22
=a21+t12+t22+t12t22
=a1+t12a1+t22=SP×SQ
Thus, SP, ST, SQ are in GP

Topic: Mathematics, Parabola, JEE Main
If the tangent at the point x1,y1 on the curve y=x3+3x2+5 passes through the origin, then x1,y1 does NOT lie on the curve

Options:
A x2+y281=2
B y29-x2=8
C y=4x2+5
D x3-y2=2

The correct answer is:
x3-y2=2

Solution:
Given y=x3+3x2+5

Now dydx=3x2+6x

So at point x1,y1

dydxat x1,y1=3x12+6x1 ... i

Also slope of line in two point form will be y1-0x1-0 ... ii

Also x1,y1 lie on curve

We get y1=x13+3x12+5 ... iii

From equation i and ii we get 3x13+6x12=y1 (as slope is equal)

Now y1=3x13+6x12=x13+3x12+5

⇒2y1=3x12+152

i.e. 2y=3x2+152 does not intersect the curve x3-y2=2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (24 Jun Shift 1)
If the tangent drawn at $A(2,1)$ to the curve $\mathrm{x}=1+\frac{1}{\mathrm{y}^2}$ meets the curve again at $\mathrm{B}$, then

Options:
A the tangent drawn at $\mathrm{B}$ coincides with the tangent drawn at $\mathrm{A}$
B the angle between the tangents drawn at A and B is neither 0 nor $\frac{\pi}{2}$
C the tangent drawn at $\mathrm{A}$ and the tangent drawn at $\mathrm{B}$ are perpendicular to each other
D the tangent drawn at $\mathrm{A}$ is parallel to the tangent drawn at B

The correct answer is:
the angle between the tangents drawn at A and B is neither 0 nor $\frac{\pi}{2}$

Solution:
The given curve is : $x=1+\frac{1}{y^2}$ ...(i)
$\Rightarrow 1=-2 \cdot \frac{1}{y^3} \cdot y^{\prime} \Rightarrow y^{\prime}=\frac{-y^2}{2}$
$\Rightarrow \quad\left(y^{\prime}\right)_{(2,1)}=-\frac{1}{2}$.
The equation of tangent at point $(2,1)$
$(y-1)=-\frac{1}{2}(x-2) \Rightarrow y-2=-\frac{1}{2} x$
$\Rightarrow \quad x=4-2 y$ ...(ii)
We can get the point $B$ by solving (i) and (ii)
$4-2 y=1+\frac{1}{y^2} \Rightarrow 2 y^3-3 y^2+1=0$
$\begin{aligned} & \Rightarrow \quad(y-1)^2(2 y+1)=0 \\ & \Rightarrow y=-\frac{1}{2}, 1,1\end{aligned}$
For point $B, y=-\frac{1}{2}$
At $y=-\frac{1}{2}, x=1+4=5$
$\therefore$ Co-ordinate of $B\left(5,-\frac{1}{2}\right)$
Let us find the equation of tangent at $B\left(5,-\frac{1}{2}\right)$
$\Rightarrow\left(y+\frac{1}{2}\right)=-\frac{1}{8}(x-5)$
$\Rightarrow y=-\frac{1}{8} x+\frac{1}{8}$ ...(iii)
$\because \quad$ Eqns. (ii) and (iii) are not same. So, option (1) is incorrect.
Now, let slope of tangent line at point $A$ and $B$ are $m_1$ and $m_2$ respectively. Then from equations (ii) and (iii), we get :
$m_1=-\frac{1}{2}, m_2=-\frac{1}{8}$
Angle between the tangents drawn at $A$ and $B$ is
$\theta=\tan ^{-1}\left(\frac{m_1+m_2}{1-m_1 m_2}\right)$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{-\frac{1}{2}-\frac{1}{8}}{1-\frac{1}{16}}\right)$
$\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{2}{3}\right) \neq 0$ or $\frac{\pi}{2}$

Topic: Mathematics, Application of Derivatives, JEE Main
If the tangent drawn to the hyperbola 4y2=x2+1 intersect the co-ordinates axes at the distinct points A and B, then the locus of the midpoint of AB is :

Options:
A x2-4y2+16x2y2=0
B 4x2-y2+16x2y2=0
C x2-4y2-16x2y2=0
D 4x2-y2-16x2y2=0

The correct answer is:
x2-4y2-16x2y2=0

Solution:
Let tangent drawn at point x1, y1 to the hyperbola

4y2=x2+1 is 4yy1=xx1+1.

This tangent intersect coordinate axes at A and B respectively, then A-1x1, 0& B0, 14y1 .

Let mid point is Mh, k.

2h=-1x1⇒x1=-12h ...1

2k=14y1⇒y1=18k ...2

Since, point x1, y1, lies on the hyperbola.

So, 4y12=x12+1

From equations 1 and 2, we get

418k2=-12h2+1⇒116k2=14h2+1

⇒4h2=16k21+4h2

⇒x2=4y2+16x2y2

⇒x2-4y2-16x2y2=0⇒ Locus of M.

Topic: Mathematics, Hyperbola, JEE Main
If the tangent drawn to the hyperbola 4y2=x2+1 intersect the co-ordinates axes at the distinct points A and B, then the locus of the midpoint of AB is :

Options:
A x2-4y2+16x2y2=0
B 4x2-y2+16x2y2=0
C x2-4y2-16x2y2=0
D 4x2-y2-16x2y2=0

The correct answer is:
x2-4y2-16x2y2=0

Solution:
Let tangent drawn at point x1, y1 to the hyperbola

4y2=x2+1 is 4yy1=xx1+1.

This tangent intersect coordinate axes at A and B respectively, then A-1x1, 0& B0, 14y1 .

Let mid point is Mh, k.

2h=-1x1⇒x1=-12h ...1

2k=14y1⇒y1=18k ...2

Since, point x1, y1, lies on the hyperbola.

So, 4y12=x12+1

From equations 1 and 2, we get

418k2=-12h2+1⇒116k2=14h2+1

⇒4h2=16k21+4h2

⇒x2=4y2+16x2y2

⇒x2-4y2-16x2y2=0⇒ Locus of M.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2018 (15 Apr)
If the tangent to the conic, y-6=x2 at (2,10) touches the circle, x2+y2+8x-2y=k (for some fixed k) at a point α, β; then (α, β) is

Options:
A -717,Â 617
B -817, 217
C -617,Â 1017
D -417,117

The correct answer is:
-817, 217

Solution:
The equation of tangent (T=0) would be 12 y+10- 6=2x
4x-y+2=0
The center of the circle is (-4,1) and the point of contact would be the foot of perpendicular from (-4,1) on 4x-y+2=0
α+44=β-1-1= --16-1+242+12
α+44=1517 and β-1-1=1517
α= -817, β= -1517+1=217
Hence option -817, 217 is correct.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2015 (10 Apr Online)
If the tangent to the curve y=ex at a point c, ec and the normal to the parabola y2=4x at the point 1, 2 intersect at the same point on the x-axis, then the value of c is .....

Options:

The correct answer is:
4

Solution:
For the parabola y2=4x, a=1

Let at2, 2at=1, 2

⇒t=1

So, equation of normal is tx+y=2at+at3.

⇒x+y=3 intersects x-axis at 3, 0

Given curve y=ex⇒dydx=ex

So, equation of tangent to the curve at c, ec is y-ec=ecx-c.

At 3, 0⇒0-ec=ec3-c⇒c=4.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If the tangent to the curve,y=fx=x logex,x>0 at a point c, fc is parallel to the line-segment joining the points 1, 0 and e, e,then c is equal to :

Options:
A e-1e
B e1e-1
C e11-e
D 1e-1

The correct answer is:
e1e-1

Solution:
f'c=1+lnc=ee-1

lnc=1e-1

c=e1e-1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (06 Sep Shift 2)
If the tangent to the curve, y=x3+ax–b at the point (1, –5) is perpendicular to the line, –x+y+4=0, then which one of the following points lies on the curve?

Options:
A 2, –2
B 2, –1
C –2, 1
D –2, 2

The correct answer is:
2, –2

Solution:
Given y=x3+ax-b

∵ it passes through 1,-5

∴ 1+a-b=-5

⇒a-b=-6 ...1

Slope of the line ax+by+c=0 is -ab, hence, slope of the line -x+y+4=0 is 1

We know that if two lines are perpendicular, then the product of their slopes is -1.

∴ Slope of tangent of y, which is perpendicular to -x+y+4=0 is -1

Also, we know that the slop of tangent to a curve y=fx at a point x1, y1 is given by f'x1.

∴ f'1=-1

And, f'x=3x2+a

⇒3+a=-1

⇒a=-4 ...2

From 1 and 2, we get a=-4, b=2

∴ fx=y=x3-4x-2

At, x=2, y=23-4×2-2=-2 and at x=-2, y=-23-4×2-2=-18.

Thus, out of the given options only 2,-2 satisfy the curve, hence this point will lie on the curve.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If the tangent to the curve y=x3 at the point Pt,t3 meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1: 2 is:

Options:
A 0
B -2t3
C -t3
D 2t3

The correct answer is:
-2t3

Solution:
Slope of tangent at Pt,t3=dydxt,t3

=3x2x=t=3t2

So equation tangent at Pt,t3:

y-t3=3t2x-t
for point of intersection with y=x3

x3-t3=3t2x-3t3

⇒ x-tx2+xt+t2=3t2x-t

for x≠t

x2+xt+t2=3t2

⇒ x2+xt-2t2=0⇒x-tx+2t=0

So for Q:x=-2t,Q-2t,-8t3
ordinate of required point :2t3-8t32+1=-2t3

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (24 Feb Shift 1)
If the tangent to the curve y=x3-x2+x at the point a,b is also tangent to the curve y=5x2+ 2x-25 at the point 2,-1, then 2a+9b is equal to ______.

Options:

The correct answer is:
195

Solution:
Given,

Curve y=5x2+2x-25

Now differentiating to find slope at point P2,-1 we get,

y'=10x+2 ⇒yP'=22

Now tangent to curve at P will be,

y+1=22x-2

⇒y=22x-45

Now the tangent is also tangent to the curve y=x3-x2+x and passing through a,b so plottnig the diagram we get,

Now differentiating the curve y=x3-x2+x we get,

dydx=3x2-2x+1

dydxQ=a,b=3a2-2a+1

So, equating with tangent slope we get,

3a2-2a+1=22

⇒3a2-2a-21=0

⇒3a2-9a+7a-21=0

⇒3a+7a-3=0

Now from curve b=a3-a2+a

So, at a=3 ⇒b=21

Also a=3 & b=21 are satisfying the line y=22x-45

So, 2a+9b=2×3+9×21=195

Note at a=-7/3 tangent will be parallel so it is rejected

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (29 Jul Shift 2)
If the tangent to the curve y=x+siny at a point a,b is parallel to the line joining 0, 32 and 12, 2, then

Options:
A b=a
B b-a=1
C a+b=1
D b=π2+a

The correct answer is:
b-a=1

Solution:
y=x+siny

dydx=1+cosydydx

dydx=11-cosy

Now for a, b, dydx=11-cosb

As the tangent is parallel to the line joining 0, 32 and 12, 2

⇒11-cosb=2-3212-0

⇒ cosb=0 ⇒ b=2n+1π2, n∈I

Point lie on curve b=a+sinb

b-a=sinb

|b-a|=1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (02 Sep Shift 1)
If the tangent to the curve y=xx2-3, x∈R, x≠±3, at a point α, β≠0, 0 on it is parallel to the line 2x+6y-11=0, then:

Options:
A 2α+6β=19
B 2α+6β=11
C 6α+2β=19
D 6α+2β=9

The correct answer is:
6α+2β=19

Solution:
The slope of a line ax+by+c=0 is -ab, hence the slope of the given line 2x+6y-11=0 is -13.

Also, we know that the slope of tangent to a curve y=fx at a point x1, y1 is dydxx1, y1

We have y=xx2-3

Applying quotient rule for differentiation, i.e. ddxuv=v·dudx-u·dvdxu2, we get

dydx=1·x2-3-x·2xx2-32

⇒dydx=x2-3-2x2x2-32

⇒dydx=-x2+3x2-32

Also, we have ⇒dydxα, β=-13

⇒-α2+3α2-32=-13

⇒3α2+9=α2-32

⇒3α2+9=α4-6α2+9

⇒α4-9α2=0

⇒α2α2-9=0

⇒α=0, 3,-3

But α≠0, ⇒α=3, -3

For α=3, β=332-3=12

⇒6α+2β=19 and 2α+6β=9

For α=-3, β=-36=-12

⇒6α+2β=19 and 2α+6β=9.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (10 Apr Shift 2)
If the tangent to the ellipse x2+4y2=16 at P4cos⁡θ,2sin⁡θ is also a normal to the circle x2+y2=8x+4y, then the sum of all possible values of θ is equal to ∀θ∈0,2π

Options:
A π2
B π4
C 5π2
D 7π4

The correct answer is:
π2

Solution:
Tangent to the ellipse is xcosâ¡Î¸+2ysinâ¡Î¸=4 which is a normal to the given circle
â it satisfies 4,2
âsinâ¡Î¸+cosâ¡Î¸=1
â´Î¸=0,Ï2

Topic: Mathematics, Ellipse, JEE Main
If the tangent to the parabola y2=x at a point α,β,β>0 is also a tangent to the ellipse, x2+2y2=1 then α is equal to:

Options:
A 2-1
B 22+1
C 2+1
D 22-1

The correct answer is:
2+1

Solution:
Since α, β lies on the parabola ⇒α=β2

⇒α, β=β2, β

Tangent to a parabola y2=4ax at a point x1, y1 is yy1=2ax+x1

Therefore, tangent to the parabola y2=x at β2, β isy⋅β=12x+β2

⇒y=x2β+β2 ...i

Tangent to parabola is also a tangent to ellipse

x2+2y2=1 ⇒x2+y212=1

And, a line y=mx+c is a tangent to the ellipse x2a2+y2b2=1 if c2=a2m2+b2

Using the condition c2=a2m2+b2, we get

β24=1212β2+12

⇒β24=14β2+12

⇒β4-2β2-1=0

Using Sridharacharya's Formula the roots of a quadratic equation ax2+bx+c=0, a≠0 are x=-b±b2-4ac2a

⇒β2=--2±-22-4×1×-12×1

⇒β2=2±82

⇒β2=1+2 or β2=1-2 (Neglected as β2 is positive)

⇒β2=2+1

⇒α=2+1.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If the tangents are drawn from any point on the line \( x+y=3 \) to the circle \( x^{2}+y^{2}=9 \), then the chord of contact always passes through a fixed point. Find that point

Options:
A \( (3,5) \)
B \( (3,3) \)
C \( (5,3) \)
D none of these

The correct answer is:
\( (3,3) \)

Solution:
Let α,3-α be any point on x+y=3

∴ Equation of chord of contact is αx+3-αy=9

i.e. αx-y+3y-9=0

The chord passes through the point 3,3 for all values of α.

Topic: Mathematics, Circle, JEE Main
If the tangents at the points P and Q on the circle x2+y2-2x+y=5 meet at the point R94, 2, then the area of the triangle PQR is

Options:
A 54
B 138
C 58
D 134

The correct answer is:
58

Solution:
Given,

The tangents at the points P and Q on the circle x2+y2-2x+y=5 meet at the point R94, 2,

Now plotting the diagram of the above data we have,

Now we know that,

Length of tangent is given by, PR=QR=S1

So, L=S1=942+22-2×94+2-5

⇒L=54

Now distance between CR will be, CR=94-12+2+122=2516+254=12516=554

Now finding PQ by equating area, we get

12×52×L=12×PQ2×CR

⇒12×52×54=12×PQ2×554

⇒PQ=5

Now finding area of triangle PQR by heron's formula,

s=25+54

So, area will be ∆=ss-as-bs-c

⇒∆=25+5425+54-5425+54-5425+54-5

⇒∆=25+542542545-254

⇒∆=5425-2016=58

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (06 Apr Shift 2)
If the tangents drawn at the point O0,0 and P1+5,2 on the circle x2+y2-2x-4y=0 intersect at the point Q, then the area of the triangle OPQ is equal to

Options:
A 3+52
B 4+252
C 5+352
D 7+352

The correct answer is:
5+352

Solution:
Tangent at Rx1,y1 when point R is on the circle x2+y2+2gx+2fy+c=0 can be written as

xx1+yy1+gx+x1+fy+y1+c=0.

Tangent at O will be

-x+0-2y+0=0 ⇒x+2y=0

Tangent at P will be

x1+5+2y-x+1+5-2y+2=0

⇒5x=5+5⇒x=5+1 and y=-5+12

So Q5+1,-5+12

Length of tangent L=OQ=6+25+3+52=52+522=5+52

We know area of triangle OPQ =RL3R2+L2

Here, radius of circle R=5

So area=5×5+5235+5+522

=52×125+75+755+5520+25+105+5=52200+80550+105

=5220+855+55-55-5=52100-205+405-4020=53+52=5+352

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (28 Jun Shift 1)
If the tangents drawn at the points P and Q on the parabola y2=2x-3 intersect at the point R0,1, then the orthocentre of the triangle PQR is

Options:
A 0,1
B 2,-1
C 6,3
D 2,1

The correct answer is:
2,-1

Solution:
On plotting the diagram of given value we get,

Now given parabola is y2=2x-3⇒y2=2x-32

Let a point on parabola be P=32+t22,t

Equation of tangent at P will be yt=x+32+t22-3

∵ R0,1 lies on this tangent

Then t=t22-32

i.e. t=3 or -1.

So P≡6,3 & Q≡2,-1

Now triangle PQR is right angled at Q, so 2,-1 is the orthocentre of the triangle.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (28 Jul Shift 1)
If the tangents drawn to the hyperbola $4 y^2=x^2+$ 1 intersect the co-ordinate axes at the distinct points $A$ and $B$, then the locus of the mid point of $A B$ is

Options:
A
$x^2-4 y^2+16 x^2 y^2=0$

B
$4 x^2-y^2+16 x^2 y^2=0$

C
$4 x^2-y^2-16 x^2 y^2=0$

D
$x^2-4 y^2-16 x^2 y^2=0$

The correct answer is:
$x^2-4 y^2-16 x^2 y^2=0$

Solution:
Equation of hyperbola is :
$$
\begin{aligned}
& 4 y^2=x^2+1 \Rightarrow-x^2+4 y^2=1 \\
\Rightarrow &-\frac{x^2}{1^2}+\frac{y^2}{\left(\frac{1}{2}\right)^2}=1 \\
\therefore & a=1, b=\frac{1}{2}
\end{aligned}
$$
Now, tangent to the curve at point $\left(x_1, y_1\right)$ is given by.
$$
\begin{aligned}
&4 \times 2 y_1 \frac{d y}{d x}=2 x_1 \\
&\Rightarrow \frac{d y}{d x}=\frac{2 x_1}{8 y_1}=\frac{x_1}{4 y_1}
\end{aligned}
$$
Equation of tangent at $\left(x_1, y_1\right)$ is
$$
\begin{aligned}
&y=m x+c \\
&\Rightarrow y=\frac{x_1}{4 y_1} \cdot x+c
\end{aligned}
$$
As tangent passes through $x_1, y_1$
$$
\begin{aligned}
&\therefore y_1=\frac{x_1 x_1}{4 y_1}+c \\
&\Rightarrow C=\frac{4 y_1^2-x_1^2}{4 y_1}=\frac{1}{4 y_1}
\end{aligned}
$$
Therefore, $y=\frac{x_1}{4 y_1} x+\frac{1}{4 y_1}$
$$
\Rightarrow 4 y_1 y=x_1 x+1
$$
which intersects $x$ axis at $A\left(\frac{-1}{x_1}, 0\right)$ and $y$ axis at $B\left(0, \frac{1}{4 y_1}\right)$
Let midpoint of $A B$ is $(h, k)$
$$
\begin{aligned}
&\therefore h=\frac{-1}{2 x_1} \\
&\Rightarrow x_1=\frac{-1}{2 h} \& y_1=\frac{1}{8 k}
\end{aligned}
$$
Thus, $4\left(\frac{1}{8 k}\right)^2=\left(\frac{-1}{2 h}\right)^2+1$
$$
\begin{aligned}
&\Rightarrow \frac{1}{16 k^2}=\frac{1}{4 h^2}+1 \\
&\Rightarrow 1=\frac{16 k^2}{4 h^2}+16 k^2 \quad \Rightarrow h^2=4 k^2+16 h^2 k .
\end{aligned}
$$
So, required equation is
$$
x^2-4 y^2-16 x^2 y^2=0
$$

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If the tangents on the ellipse 4x2+y2=8 at the points 1, 2 and (a, b) are perpendicular to each other, then a2 is equal to

Options:
A 217
B 417
C 6417
D 12817

The correct answer is:
217

Solution:
4x2+y2=8
⇒8x+2ydydx=0
dydx=-8x2y=-4xy
∴ Slope at 1, 2, m1=-2
Slope at a,b, m2=-4ab
∴ -2×-4ab=-1
b=-8a
Again,
4a2+b2=8
⇒4a2+64a2=8
⇒68a2=8
⇒a2=868

⇒a2=217

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (08 Apr Shift 1)
If the tangents PQ and PR are drawn from a variable point P to the ellipse x216+y29=1, such that the fourth point S of the parallelogram PQSR lies on the circumcircle of the ÎPQR, then the area (in sq. units) of the locus of P is

Options:
A 7π
B 16π
C 25π
D 9π

The correct answer is:
25π

Solution:
Parallelogram PQSR is a cyclic quadrilateral
⇒ Parallelogram is a rectangle
⇒∠QPR=90o
Let, the equation of the tangent from Ph,k to the ellipse is y=mx±16m2+9⇒k-mh2=16m2+9
∵m1m2=-1⇒h2-16=-k2+9
⇒ locus of P is x2+y2=25

Hence, the area of locus of P is 25π sq. units

Topic: Mathematics, Ellipse, JEE Main
If the temperature increases, then what happens to the frequency of the sound produced by the organ pipe

Options:
A Increases
B Decreases
C Unchanged
D Not definite

The correct answer is:
Increases

Solution:
When the temperature increases, then the frequency of the sound produced by the organ pipe will: Increases.

Topic: Physics, Waves and Sound, JEE Main
If the temperature increases, then what happens to the frequency of the sound produced by the organ pipe?

Options:
A increases
B decreases
C unchanged
D not definite

The correct answer is:
increases

Solution:
Due to rise in temperature, the speed of sound increases. Since frequency $n=\frac{v}{\lambda}$ and $\lambda$ remains unchanged. Hence frequency $n$ increases.

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2011
If the temperature is raised by 1 K from 300 K, the percentage change in the speed of sound in the gaseous mixture is [R=8.314 J mol-1 K-1]

Options:
A 0.167%
B 0.334%
C 1%
D 2%

The correct answer is:
0.167%

Solution:
From v=Î³RTM
âvv=12âTT
âvvÃ100=12âTTÃ100
=12Ã1300Ã100=0.167 %

Topic: Physics, Waves and Sound, JEE Main
If the temperature of an ideal gas in a sealed, rigid container is increased to 1.5 times the initial value (in K) , the density of gas

Options:
A becomes 1.5 times the initial value
B becomes 1/1.5 times the initial value
C becomes 2.25 times the initial value
D remains same

The correct answer is:
remains same

Solution:
Density=massvolume
For a sealed rigid container mass as well as volume remains a constant hence the density remains same even if the temperature changes.

Topic: Chemistry, States of Matter, NEET
If the temperature of cold junction decreases, then neutral temperature

Options:
A increases
B decreases
C remains same
D may increases or may decreases

The correct answer is:
remains same

Solution:
If the temperature of cold junction decreases then neutral temperature remains same.

Topic: Physics, Thermal Properties of Matter, MHT CET, MHT CET 2011
If the temperature of cold junction of thermocouple is lowered, then the neutral temperature

Options:
A increases
B approaches inversion temperature
C decreases
D remains the same

The correct answer is:
remains the same

Solution:
Neutral temperature is independent of temperature of cold junction.

Topic: Physics, Thermal Properties of Matter, MHT CET, MHT CET 2008
If the temperature of the sun were to increase from $T$ to $2 T$ and its radius from $R$ to $2 R$, then the ratio of the radiant energy received on earth to what it was previously will be

Options:
A
4

B
16

C
32

D
64

The correct answer is:
64

Solution:
According to stefan's law,
$P \propto A T^4$ and $A \propto r^2$
$P \propto r^2 T^4$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2004
If the tenth term of the sequence S=1+5+13+29+.……. is k, then k500 is equal to

Options:

The correct answer is:
4.09

Solution:
S=1+5+13+29+.â¦â¦+t9+t10
S=.â¦â¦1+5+13+.â¦â¦.+t8+t9+t10
Subtracting we get 0=1+4+8+16+.â¦â¦9 terms-t10
ât10=1+4+8+16+.â¦..9 terms=1+41-291-2
=1+429-1=1+4512-1
=2045
âk=2045âk500=4.09

Topic: Mathematics, Sequences and Series, JEE Main
If the term independent of x in the expansion of 32x2-13x9 is k, then 18k is equal to:

Options:
A 11
B 5
C 9
D 7

The correct answer is:
7

Solution:
Tr+1=9Cr3x229-r-13xr

=9Cr329-r-13rx18-3r

For the term independent of x, put r=6

⇒T7=9C6323-136=9C3163=9×8×73×2×1163=718

⇒k=T7⇒k=718⇒18k=7

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If the term without x in the expansion of x23+αx322 is 7315 , then α is equal to _____ .

Options:

The correct answer is:
1

Solution:
General term in the binomial expansion of x23+αx322 is

Tr+1=Cr22x2322-rαx3r

⇒Tr+1=Cr22x44-11r3αr

For term independent of x,

44-11r3=0

⇒r=4

So,

T5=C422α4=7315

⇒22!4!×18!α4=7315

⇒7315α4=7315

⇒α4=1

⇒α=1

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (01 Feb Shift 2)
If the terminal speed of a sphere of gold (density $=19.5 \mathrm{~kg} / \mathrm{m}^3$ ) is $0.2 \mathrm{~m} / \mathrm{s}$ in a viscous liquid (density $=1.5 \mathrm{~kg} / \mathrm{m}^3$ ) of the same size in the same liquid.

Options:
A
$0.2 \mathrm{~m} / \mathrm{s}$

B
$0.4 \mathrm{~m} / \mathrm{s}$

C
$0.133 \mathrm{~m} / \mathrm{s}$

D
$0.1 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$0.1 \mathrm{~m} / \mathrm{s}$

Solution:
$\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}_{\mathrm{g}}}=\frac{\left(\rho_{\mathrm{s}}-\rho_{\ell}\right)}{\left(\rho_{\mathrm{g}}-\rho_{\ell}\right)}$
$\mathrm{v}_{\mathrm{s}}=0.1 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2006
If the third term in the binomial expansion of 1+xlog2x5 equals 2560, then a possible value of x is

Options:
A 42
B 18
C 2 2
D 14

The correct answer is:
14

Solution:
The general term in the expansion of a+bn is Tr+1=Crnan-rbr.

Given, in the expansion of 1+xlog2x5, third term is 2560

⇒T3= 5C2xlog2x2=2560

Using, Crn=n!r!·n-r!, we get

⇒5!2!·3!xlog2x2=2560

⇒5·4·3!2×1·3!x2log2x=2560

⇒x2log2x=256

Taking logarithm to the base 2 on both sides

⇒log2x2log2x=log2256

Now, using logamn=nlogam & logaa=1,

⇒2log2xlog2x=log228

⇒ 2log2x2=8

⇒log2x=±2

Using, logax=b, ⇒x=ab

x=22, 2-2

⇒x=4, 14

Here, only x=14 is given in the options.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If the third term in the binomial expansion of $(1+x)^m$ is $-\frac{1}{8} x^2$ then the rational value of $m$ is

Options:
A $2$
B $1 / 2$
C $3$
D $4$

The correct answer is:
$1 / 2$

Solution:
We have $(1+x)^m=1+m x+\frac{m(m-1)}{2!} x^2+\ldots$
By hypothesis, $\frac{m(m-1)}{2} x^2=-\frac{1}{8} x^2$
$\Rightarrow 4 m^2-4 m=-1 \Rightarrow(2 m-1)^2=0 \Rightarrow m=\frac{1}{2}$

Topic: Mathematics, Binomial Theorem, JEE Main
If the third term in the binomial expansion of $(1+x)^m$ is $-\frac{1}{8} x^2$, then the rational value of $m$ is

Options:
A $2$
B $1/2$
C $3$
D $4$

The correct answer is:
$1/2$

Solution:
We have $(1+x)^m=1+m x+\frac{m(m-1)}{2!} x^2+\ldots$
By hypothesis, $\frac{m(m-1)}{2} x^2=-\frac{1}{8} x^2$
$\Rightarrow 4 m^2-4 m=-1 \Rightarrow(2 m-1)^2=0 \Rightarrow m=\frac{1}{2}$

Topic: Mathematics, Binomial Theorem, JEE Main
If the third term of a G.P. is 4 then the product of its first 5 terms is

Options:
A $4^3$
B $4^4$
C $4^5$
D None of these

The correct answer is:
$4^5$

Solution:
Given that $a r^2=4$
Then product of first 5 terms
$=a(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)=a^5 r^{10}=\left[a r^2\right]^5=4^5$

Topic: Mathematics, Sequences and Series, JEE Main