datavorous/entrance-exam-dataset huggingface dataset:
If the apertaure of a telescope is decreased the resolving power will

Options:
A increases
B decreases
C remain same
D zero

The correct answer is:
decreases

Solution:
The resolving power $=\frac{a}{1.22 \lambda}$
where $a=$ aperture of telescope
$\therefore$ Resolving power $\propto$ aperture of telescope If resolving power is decreased, the resolving power will decrease.

Topic: Physics, Ray Optics, MHT CET, MHT CET 2010
If the area bounded by fx=tan3⁡x+tan⁡x from x=0 to x=π4 is k square units, then the maximum value of gx=ksin⁡x is ∀x∈0,π4

Options:
A 2
B 122
C 4
D 8

The correct answer is:
122

Solution:
The required area is
A=â«0Ï4tan3â¡x+tanâ¡xdx
=â«0Ï4tanâ¡xtan2â¡x+1dx
= â« 0 Ï 4 tanxâ sec 2 xdx
Let tanâ¡x=t
âsec2â¡xdx=dt
âI=â«01tdt=t2201
=12
∴g x = 1 2 sinx≤ 1 2⋅ 2

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by \( f(x)=\max (\sin x, \cos x) ; 0 \leq x \leq \frac{\pi}{2}, x=\frac{\pi}{2} \) and the coordinates axes is equal to \( k \), then \( [k+3] \) is equal to (where, [.] denotes the greatest integer function

Options:
A \( 2 \)
B \( 8 \)
C \( 4 \)
D \( 6 \)

The correct answer is:
\( 4 \)

Solution:
f x = { cos x for 0 ≤ x ≤ π / 4 sin x for π / 4 < x ≤ π / 2
∴ Required Area = 2 ∫ 0 π / 4 cos xdx = 2 sin x 0 π / 4
= 2 sq units
⇒ k = 2
⇒ k + 3 = 2 + 3 = 4

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by the curve 2y2=3x, lines x+y=3, y=0 and outside the circle x-32+y2=2 is A, then 4π+ 4A is equal to __________.

Options:

The correct answer is:
42

Solution:
Given:

2y2=3x, x+y=3slope is -1 & x-32+y2=2

Now plotting the diagram of the above function we have,

So, required area is

A=∫0323-y-2y23dy-π22×45°360°

⇒A=3y-y22032-23y33032-π4

⇒A=218-π4

So,

A=14212-π

⇒π+4A=212

⇒4π+4A=42

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (15 Apr Shift 1)
If the area bounded by the curve y+x2=8x and the line y=12 is K sq. units, then the value of 3K10 is

Options:

The correct answer is:
3.2

Solution:
The curves are y=-x2+8x=-xx-8 and y=12
The intersection points are A and B where
-x2+8x=12⇒x2-8x+12=0
⇒x-2x-6=0
⇒x=2,6
i.e. A2,12 and B(6,12)
Thus, the required area is
A=∫26-x2+8x-12dx
=-x3326+4x226-12x26
=-216-83+436-4-126-2
=-2083+128-48=80-2083=323 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by the curves x2+y2â¤4, x+yâ¥2, and yâ¥1 is 2ÏK-K2-12 sq. units, then the value of K is equal to

Options:

The correct answer is:
3

Solution:
The required area is A=∫124-y2dy-Area∆APQ
⇒A=y4-y22+42sin-1⁡y212-12×1×1
⇒A=0+2sin-1⁡1-32-2sin-1⁡12-12
=π-32-π3-12
=2π3-32-12 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by the curves x2+y≤2 and y≥x is k2 sq. units, then 2k is equal to

Options:
A 9
B 27
C 18
D 32

The correct answer is:
18

Solution:
Required area is A=∫-212-x2-xdx
=2x-x33-x22-21
=2x-x33-x22-21
=6-93+32=92 sq. units
∴2k=18

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by the curves x,y|x2-y+1≥0 and x,y|x+y-3≥0 is k square units, then the value of 3k is equal to

Options:
A 27
B 9
C 92
D 272

The correct answer is:
272

Solution:
The required area is k sq. units
⇒k=∫-213-x-x2+1dx
=2x-x22-x33-21
⇒k=92
⇒3k=272

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by the parabola y=2-x2 and the line y=-x is k2 sq. units, then the value of 2k is equal to

Options:
A 9
B 27
C 18
D 32

The correct answer is:
18

Solution:
Solving y=2-x2 & y=-x; we get -x=2-x2
⇒x=2,-1
⇒A-1,1 & B2,-2
Thus, the required area
A=∫-122-x2--xdx
=2x-x33+x22-12
=92
⇒k=9∴2k=18

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by y2=4ax and x2=4ay is 643 square units, then the positive value of a is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Point of intersection of the two given curves are 0,0 and 4a,4a

The area bounded =2â«04ax-x24adx
=2x22-x312a04a
=28a2-16a23=16a23 sq. units
Given, 16a23=643

Hence, the positive value of a=2

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by yâ¤e-x-e and yâ¥0 is A sq. units, then logeâ¡A is equal to

Options:

The correct answer is:
2

Solution:
The required area is A=122ee=e2 sq. units

â´logeâ¡A=logeâ¡e2=2

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by y=x2-4x+3 and the x-axis from x=1 to x=3 is pq (where, p & q are coprime), then the value of p+q is

Options:

The correct answer is:
97

Solution:
Area =∫134x-x2+12dx =943

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by y=x2 and y=21+x2 is K1π-K23 sq. units (where K1,K2∈Z), then the value of K1+K2 is equal to

Options:
A 3
B 1
C –1
D –2

The correct answer is:
3

Solution:
Drawing graphs of y=x2 and y=2x2-1

For points of intersection,

x2=2x2+1⇒x4+x2-2=0
x2=1⇒x=±1
Required area =2∫0121+x2-x2dx
=π-23sq.units
K1=1, K2=2⇒K1+K2=3

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by y+x-Ïâ¤Ï and yâ¥Ï2 is KÏ2 sq. units, then the value of 8K is equal to

Options:

The correct answer is:
2

Solution:
Given curve is y=π-x-π
Solving with y=π2, we get,

A as π2,π2 and B as 3π2,π2
Thus, the required area is =12⋅3π2-π2⋅π2
=12⋅π⋅π2=π24 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
If the area bounded by y=x, y=sinx and x=π2 is π2k-1 sq. units, then the value of k is equal to

Options:
A 2
B 4
C 6
D 8

The correct answer is:
8

Solution:
Required area =arΔAOB-∫0π2sin⁡xdx
=12⋅π2⋅π2⏟12⋅base⋅height-1
=π28-1 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
If the area covered by y=2x and y=22x-1 from x=1 to x=e is ln⁡a sq. units, then 2e-12a2 is equal to

Options:
A e2
B e4
C 1
D e

The correct answer is:
e4

Solution:
As 2x>22x-1 ∀x∈1,e
∴ the required area is A=∫1e2x-22x-1dx
=2ln⁡x1e-12ln⁡2x-11e
=ln⁡e2-ln⁡2e-1
=ln⁡e22e-1 sq. units

⇒ln⁡e22e-1=lna

⇒a2e-1=e2⇒a22e-12=e4

Topic: Mathematics, Area Under Curves, JEE Main
If the area enclosed between the curves y=kx2 and x=ky2, k>0, is 1 sq. unit. Then k is

Options:
A 3
B 13
C 32
D 23

The correct answer is:
13

Solution:
The given curves are, y=kx2, x=ky2

To find the point of intersection of the curves, put y from the first curve into second, to get

x=kk2x4

⇒x=0 or x3=1k3⇒x=1k

Hence, y=0 or y=k1k2=1k

Therefore, point of intersection is 1k, 1k.

Hence, the required area is ∫01ky2-y1dx=1

⇒∫01kxk-kx2dx=1

⇒1kx3/23/2-kx3301k=1

⇒23k2-13k2=1

⇒k2=13

⇒k=±13

But, given k>0
So, k=13.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If the area enclosed by the parabolas P1:2y=5x2 and P2:x2-y+6=0 is equal to the area enclosed by P1 and y=αx,α>0, then α3 is equal to _____ .

Options:

The correct answer is:
600

Solution:
Given:

P1:2y=5x2

P2:x2-y+6=0

On solving both equations, we have

2y=5y-6

⇒3y=30

⇒y=10

⇒x=±2

So, points of intersection are 2,10 & -2,10.

Also, solving P1 and y=αx, we have

x=0, 2α5

According to question 2∫02x2+6-5x22dx=∫02α5αx-5x22dx

⇒2x33+6x-5x3602=αx22-5x3602α5

⇒212+83-203=2α325-8α36×25

⇒2α375=16

⇒ α3=600

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (25 Jan Shift 1)
If the area enclosed by y2=2x and x2+4+4x=4y2 is k square units, then the value of 3k is equal to

Options:

The correct answer is:
8

Solution:
Given equation is x+22=4y2 or x+2=2y and x+2=-2y
Solving x+2=2y with the equation of the parabola, we get,
y2=22y-2
⇒y2-4y+4=0
⇒y=2
Hence, the pair of lines are tangent to the parabola.
So, k=2∫02y22dy-∫022y-2dy
=2y36-y2+2y02
=286=83
∴3k=8

Topic: Mathematics, Area Under Curves, JEE Main
If the area (in sq. units) bounded by the parabola y2=4λx and the line y=λx, λ>0, is 19, then λ is equal to

Options:
A 43
B 26
C 48
D 24

The correct answer is:
24

Solution:
The given equations are y2=4λx and y=λx

Using above equations, we can write

λx2=4λx⇒λ2x2-4λx=0

⇒λxλx-4=0

⇒x=0 & x=4λ

Area =19=∫04λ4λx-λxdx

⇒4λx3232-λx2204λ=19

⇒λ=24

Hence, the required value is 24.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (12 Apr Shift 2)
If the area (in sq. units) of the region x, y: y2≤4x, x+y≤1, x≥0, y≥0 is a2+b, then a-b is equal to

Options:
A 6
B 103
C -23
D 83

The correct answer is:
6

Solution:
The given equations are,

y2=4x & x+y=1

⇒y2=41-y

⇒y2+4y-4=0 ⇒y=22-2

∴ Required Area =∫022-21-y-y24dx=y-y22-y312022-2

=22-2-22-222-22-2312-0

=22-21-2-1-2-123=-103+823

⇒-103+823=a2+b ⇒a=83, b=-103
∴ a-b=83+103=6

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If the area (in sq. units) of the triangle formed by the intersection of a line parallel to the x-axis and passing through the point P(h, k) with the line y=x and x+y=2 is 4h2, then the locus of the point P is

Options:
A 2 x = Â± y - 1
B 3 x = ± y - 1
C 5 x = ± y - 1
D 7 x = ± y - 1

The correct answer is:
2 x = ± y - 1

Solution:
Here, the triangle formed by a line parallel to the x-axis and passing through P (h, k) and the straight line y=x and y=2-x is shown below.

Since, area of Δ ABC = 4 h 2
∴ 1 2 · AB · AC = 4 h 2
Where AB = 2 k - 1 and AC = 2 k - 1
⇒ 1 2 · 2 k - 1 2 = 4 h 2
⇒ 4 h 2 = k - 1 2
⇒ 2 h = ± k - 1
Hence, the locus of the point P is 2 x = ± y - 1

Topic: Mathematics, Straight Lines, JEE Main
If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4 x=0$, the parabola $y^2=x$ and the $X$-axis is $A$, then $6 A-9 \sqrt{3}=$

Options:
A $\pi$
B $2 \pi$
C $3 \pi$
D $4 \pi$

The correct answer is:
$4 \pi$

Solution:
Given curve
$x^2+y^2-4 x=0 \Rightarrow(x-2)^2+y^2=4 \text { and } y^2=x$
intersecting point of circle and parabola is $(0,0)$ and $(3, \sqrt{3})$
Required area
$A=\int_0^3 \sqrt{x} d x+\int_3^4 \sqrt{4-(x-2)^2} d x$
$\begin{aligned} & A=\left[\frac{2}{3}(x)^{3 / 2}\right]_0^3+\left[\frac{x-2}{2} \sqrt{4-(x-2)^2}+2 \sin ^{-1} \frac{x-2}{2}\right]_3^4 \\ & A=2 \sqrt{3}+\left[\left(0+2 \sin ^{-1}(1)\right)-\left(\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{1}{2}\right)\right)\right] \\ & A=2 \sqrt{3}+\pi-\frac{\sqrt{3}}{2}-\frac{\pi}{3}=\frac{3 \sqrt{3}}{2}+\frac{2 \pi}{3} \\ & 6 A-9 \sqrt{3}=4 \pi\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
If the area of an equilateral triangle inscribed in the circle x2+y2+10x+12y+c=0 is 273sq.units, then c is equal to:

Options:
A 25
B 13
C -25
D 20

The correct answer is:
25

Solution:
Given, area of the equilateral triangle inscribed in the circle is 273, hence
34 a2=27 3 ⇒a=63 ...1

Again, rcos30°=a2

⇒r=6

Hence, the radius of a circle, 52+62-c=6
⇒25+36-c=36
⇒c=25

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (10 Jan Shift 2)
If the area of the auxillary circle of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$ is twice the area of the ellipse, then the eccentricity of the ellipse is

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

The correct answer is:
$\frac{\sqrt{3}}{2}$

Solution:
Equation of ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Then, equation of auxillary circle will be
$$
x^{2}+y^{2}=a^{2} \quad(a>b)
$$
Now, it is given that
Area of auxillary circle $=2 \times$ Area of ellipse
$$
\begin{array}{ll}
\Rightarrow & \pi a^{2}=2 \times \pi a b \\
\Rightarrow & a=2 b
\end{array}
$$
Now, cocentricity of the ellipse,
$$
\begin{aligned}
e &=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{b^{2}}{(2 b)^{2}}} \\
&=\sqrt{1-\frac{b^{2}}{4 b^{2}}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main
If the area of the circle \(x^2+y^2=2\) is divided into two parts by the parabola \(y=x^2\), then the area (in sq units) of the larger part is

Options:
A \(\frac{3 \pi}{2}-\frac{1}{3}\)
B \(6 \pi-\frac{4}{3}\)
C \(\frac{4 \pi}{3}-\frac{2}{3}\)
D \(4 \pi-\frac{1}{4}\)

The correct answer is:
\(\frac{3 \pi}{2}-\frac{1}{3}\)

Solution:
Given equation of curves
\(x^2+y^2=2 \text { and } y=x^2\)
For point of intersection, on solving the given curves, we get
\(\begin{aligned}
& y^2+y-2=0 \Rightarrow y^2+2 y-y-2=0 \\
\Rightarrow & y(y+2)-1(y+2) =0 \\
\Rightarrow & y =1 \qquad {[\because y > 0] }
\end{aligned}\)
So, the required area
\(\begin{aligned}
& =\pi+2 \int_0^1\left(\sqrt{\left.2-y^2\right)}-\sqrt{y}\right) d y \\
& =\pi+2\left[\frac{y}{2} \sqrt{2-y^2}+\frac{2}{2} \sin ^{-1} \frac{y}{\sqrt{2}}-\frac{2}{3} y^{3 / 2}\right]_0^1 \\
& =\pi+2\left[\frac{1}{2}+\frac{\pi}{4}-\frac{2}{3}\right]=\frac{3 \pi}{2}+1-\frac{4}{3}
\end{aligned}\)
\(=\frac{3 \pi}{2}-\frac{1}{3} \text { sq. units }\)
Hence, option (1) is correct.

Topic: Mathematics, Area Under Curves, JEE Main
If the area of the region bounded by the curves y2-2y=-x and x+y=0 is A, then 8A=

Options:

The correct answer is:
36

Solution:
Given,

Equations of curve y2-2y=-x and x+y=0

Now solving above equations we get, x=0,-3

Now plotting the area we get,

Now from above diagram required area will be

A=∫032y-y2+ydy

⇒A=∫033y-y2dy

⇒A=3y22-y3303

⇒A=272-273

⇒A=276=92 sq. units

So, A=92⇒8A=9×4=36.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (24 Jan Shift 2)
If the area of the region bounded by the curves, y=x2, y=1x and the lines y=0 and x=tt>1 is 1 sq. unit, then t is equal to

Options:
A e23
B e32
C 32
D 43

The correct answer is:
e23

Solution:
Area =∫01x2dx+∫1t1xdx=1

⇒13+lnt=1⇒t=e23

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2018 (16 Apr Online)
If the area of the region S=(x,y):2y-y2≤x2≤2y,x≥y is equal to n+2n+1-πn-1, then the natural number n is equal to _______

Options:

The correct answer is:
5

Solution:
Given,

S=(x,y):2y-y2≤x2≤2y,x≥y,

Now plotting the diagram, we get

Now from above diagram, given region are,

2y-y2≤x2, x2≤2y and x≥y

Now the intersecting point of x2+y2-2y=0 & x=y will be 1,1 and x2=2y & x=y will be 2,2

So, required area will be,

A=∫011+1-x2-x22+∫12x-x22dx

⇒A=x-x21-x2-12·sin-1x-x3601+x22-x3612

⇒A=76-π4=n+2n+1-πx-1

So, on comparing we get, n=5

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (06 Apr Shift 1)
If the area of the region (x,y):0≤y≤min2x,6x-x2 is A, then 12 A is equal to _______.

Options:

The correct answer is:
304

Solution:
Solving, y=2x and y=6x-x2

⇒2x=6x-x2

⇒4x-x2=0

⇒x4-x=0

⇒x=0,4

⇒y=0,8

⇒x,y≡0,0, 4,8

So, the required area is given by,

A=12×4×8+∫466x-x2dx

⇒A=16+3x2-x3346

⇒A=763

⇒12A=304

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (27 Jan Shift 2)
If the area of the region x,y:x2-2≤y≤x is A, then 6A+162 is equal to ______________

Options:

The correct answer is:
27

Solution:
Given,

x2-2≤y & y≤x

Now on finding the intersection point we get, x,y≡1,1 & 2,2

Now plotting the curve of the above function we get,

Now from above diagram required bounded area will be,

A=∫12x-2-x2dx+∫22x-x2-2dx

⇒A=x22-2x+x3312+x22-x33+2x22

⇒A=1-22+223-12-2+13+2-83+4-1-223+22

⇒A=92-823

⇒6A=27-162

⇒ 6A+162=27

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (10 Apr Shift 2)
If the area of the region $\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0 < \mathrm{a} < 1\right\}$ is $\left(\log _{\mathrm{e}} 2\right)-\frac{1}{7}$ then the value of $7 \mathrm{a}-3$ is equal to:

Options:
A 0
B 2
C -1
D 1

The correct answer is:
-1

Solution:
$\begin{aligned} & \text { area } \int_1^2\left(\frac{1}{x}-\frac{a}{x^2}\right) d x \\ & {\left[\ell n x+\frac{a}{x}\right]_1^2} \\ & \ell \operatorname{nn} 2+\frac{a}{2}-a=\log _e 2-\frac{1}{7} \\ & \frac{-a}{2}=-\frac{1}{7} \\ & a=\frac{2}{7} \\ & 7 a=2 \\ & 7 a-3=-1\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If the area of the rhombus enclosed by the lines x±y±n=0 be 2 square units, then

Options:
A n2=4
B n2=2
C n2=12
D n2=1

The correct answer is:
n2=1

Solution:
By solving the equation of the sides of the rhombus, its vertices are
0,-n1,-n1, 0,0,  n1 and n1,0
∴ area =12×2n1×2n1=2
⇒n2=1

Topic: Mathematics, Straight Lines, JEE Main
If the area of the triangle formed by the x-axis, the normal and the tangent to the circle x-22+y-32=25 at the point 5,7 is A, then 24A is equal to _________.

Options:

The correct answer is:
1225

Solution:
Equation of normal

4x-3y+1=0

and equation of tangents

3x+4y-43=0

Area of triangle=12433+14×7

=12172+312×7

A=122524 Sq. unit

24 A=1225

Note: As positive x-axis was given in the question so question was bonus in JEE Mains.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (24 Feb Shift 2)
If the area of the triangle whose one vertex is at the vertex of the parabola, $y^{2}+4\left(x-a^{2}\right)=0$ and the other two vertices are the points of intersection of the parabola and $y$ -axis, is 250 sq. units, then a value of 'a' is :

Options:
A $5 \sqrt{5}$
B $5\left(2^{1 / 3}\right)$
C $(10)^{33}$
D 5

The correct answer is:
5

Solution:
$y^{t}=-4\left(x-a^{2}\right)$

$\mathrm{Area}=\frac{1}{2}(4 a)\left(a^{2}\right)=2 a^{3}$

Since $2 a^{3}=250 \Rightarrow a=5$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (11 Jan Shift 2)
If the area of triangle formed by the points \( (2 a, b),(a+b, 2 b+a) \) and \( (2 b, 2 a) \) is \( 2 \) sq units, then the area of the triangle whose vertices are \( (a+b, a-b),(3 b-a, b+3 a) \) and \( (3 a-b, 3 b-a) \) will be

Options:

The correct answer is:
8

Solution:
Given Aa+b,a-b, B-a+3b,3a+b & C3a-b,-a+3b

Midpoint of AB=2b,2a=P

Midpoint of BC=a+b,a+2b=Q

Midpoint of CA=2a,b=R

Given the area of triangle PQR=2 sq. units

Since, triangle PQR is formed by joining the mid points of the sides of triangle ABC,

Therefore, area of triangle ABC=4×area of triangle PQR=8 sq. units

Topic: Mathematics, Straight Lines, JEE Main
If the area to be covered for TV telecast is doubled, then the height of transmitting antenna (TV tower) will have to be

Options:
A Doubled
B Halved
C Quadrupled
D 16 times

The correct answer is:
Doubled

Solution:
We know that
d= 2hRe
A=πd2=2πhRe
area proportional to the height of the antenna
⇒A∝h

Topic: Physics, Communication System, JEE Main
If the arguments of 1-i3+i1+3i and Z-2ZÂ¯-1 are equal, then the locus of Z is part of a circle with centre a,b. The value of 1a+b is

Options:

The correct answer is:
0.5

Solution:
Let, Z=x+iy
Now,
Z-2Z¯-1=ZZ¯-2Z¯-Z+2
=x2+y2-3x+2+iy
argument of 1-i3+i1+3i=-π4+π6+π3=π4

Now, since both arguments are equal
⇒ x2+y2-3x+2=y (where, y>0)
⇒ x2+y2-3x-y+2=0
Hence, the centre of the circle is 32,12=a,b
⇒ 1a+b=132+12=12=0.5

Topic: Mathematics, Complex Number, JEE Main
If the arithmetic and geometric means of $a$ and $b$ be $A$ and $G$ respectively, then the value of $A-G$ will be

Options:
A $\frac{a-b}{a}$
B $\frac{a+b}{2}$
C $\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\right]^2$
D $\frac{2 a b}{a+b}$

The correct answer is:
$\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\right]^2$

Solution:
Arithmetic mean of $a$ and $b=A=\frac{a+b}{2}$
and geometric mean $G=\sqrt{a b}$
Then, $A-G=\frac{a+b}{2}-\sqrt{a b}$ $=\frac{a+b-2 \sqrt{a b}}{2}$
$=\frac{(\sqrt{a})^2+(\sqrt{b})^2-2(\sqrt{a})(\sqrt{b})}{2}$ $=\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\right]^2$

Topic: Mathematics, Sequences and Series, JEE Main
If the arithmetic and geometric means of $a$ and $b$ be $A$ and $G$ respectively, then the value of $A-G$ will be

Options:
A $\frac{a-b}{a}$
B $\frac{a+b}{2}$
C $\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\right]^2$
D $\frac{2 a b}{a+b}$

The correct answer is:
$\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\right]^2$

Solution:
Arithmetic mean of $a$ and $b=A=\frac{a+b}{2}$
and geometric mean $G=\sqrt{a b}$
Then, $A-G=\frac{a+b}{2}-\sqrt{a b}=$ $=\frac{a+b-2 \sqrt{a b}}{2}$
$=\frac{(\sqrt{a})^2+(\sqrt{b})^2-2(\sqrt{a})(\sqrt{b})}{2}=\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\right]^2$

Topic: Mathematics, Sequences and Series, JEE Main
If the arithmetic, geometric and harmonic means between two distinct positive real numbers be $A, G$ and H respectively, then the relation between them is

Options:
A $A \gt G \gt H$
B $A \gt G \lt H$
C $H \gt G \gt A$
D $G \gt A \gt H$

The correct answer is:
$A \gt G \gt H$

Solution:
$\mathrm{A}=\frac{(\mathrm{a}+\mathrm{b})}{2}, G=\sqrt{a b}$ and $H=\frac{2 a b}{(a+b)}$.
$A-G=\frac{(a+b)}{2}-\sqrt{a b}$
$=\frac{(a+b-2 \sqrt{a b})}{2}$
$=\frac{(\sqrt{a}-\sqrt{b})^2}{2} \geq 0$
So $A \geq G$
Step 2: Relation between $G ~\&~ H$
$G-H=\sqrt{a b}-\frac{2 a b}{(a+b)}$
$=\frac{\sqrt{a b}[a+b-2 \sqrt{(a b)]}}{(a+b)}$
$=$ $\frac{\sqrt{(a b)}(\sqrt{a}-\sqrt{b})^2}{(a+b)}$ $\geq 0$
$G \geq H$
Thus $\mathrm{A} \geqslant \mathrm{G} \geq \mathrm{H}$.

Topic: Mathematics, Sequences and Series, JEE Main
If the arithmetic, geometric and harmonic means between two positive real numbers be $A, G$ and $H$, then

Options:
A $A^2=G H$
B $H^2=A G$
C $G=A H$
D $G^2=A H$

The correct answer is:
$G^2=A H$

Solution:
$A=\frac{a+b}{2}, G=\sqrt{a b}$ and $H=\frac{2 a b}{a+b}$
Then $G^2=a b\ldots(i)$
and $A H=\left(\frac{a+b}{2}\right) \cdot \frac{2 a b}{a+b}=a b \ldots(ii)$
From (i) and (ii), we have $G^2=A H$

Topic: Mathematics, Sequences and Series, JEE Main
If the arithmetic mean and the geometric mean of the pth and qth terms of the sequence -16,8,-4,2,… satisfy the equation 4x2-9x+5=0, then p+q is equal to _______.

Options:

The correct answer is:
10

Solution:
Let tp & tq is the pth & qth terms of the series -16,8,-4,2,...

Here a=-16, r=-12

Then, tp=-16-12p-1

tq=-16-12q-1

The roots of the given quadratic equation 4x2-9x+5=0 is x=1,54.

AM=54, GM=1 ∵AM≥GM

Now, given 54=tp+tq2,1=tptq

So, 1=256-12p+q-2⇒2p+q-2=-1p+q-228

⇒2p+q-10=-1p+q-2

It is equal when 2p+q-10=1=-1p+q-2

Hence, p+q-10=0⇒p+q=10

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (26 Feb Shift 2)
If the arithmetic mean of the numbers $x_1, x_2, x_3, \ldots . ., x_n$ is $\bar{x}$, then the arithmetic mean of numbers $a x_1+b, a x_2+b, a x_3+b, \ldots \ldots . ., a x_n+b$, where $a, b$ are two constants would be

Options:
A $\bar{x}$
B $n a \bar{x}+n b$
C $a \bar{x}$
D $a \bar{x}+b$

The correct answer is:
$a \bar{x}+b$

Solution:
Required mean $=\frac{\left(a x_1+b\right)+\left(a x_2+b\right)+\ldots . .+\left(a x_n+b\right)}{n}$
$\begin{aligned} &=\frac{a\left(x_1+x_2+\ldots .+x_n\right)+n b}{n}=a \bar{x}+b, \\ & \quad\left(\because \frac{x_1+x_2+\ldots .+x_n}{n}=\bar{x}\right) .\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If the arithmetic mean of two numbers a and b, a>b>0 , is five times their geometric mean, then a+ba-b is equal to:

Options:
A 7312
B 324
C 62
D 5612

The correct answer is:
5612

Solution:
Arithmetic Mean =a+b2
Geometric Mean =ab

⇒a+b2=5ab
⇒a+b=10ab

⇒(a+b)2=100ab

⇒(a-b)2=(a+b)2-4ab=96ab

Now,

⇒a+ba-b=(a+b)2(a-b)2=10096

=5612

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2017 (08 Apr Online)
If the arithmetic mean of two numbers be $A$ and geometric mean be $G$, then the numbers will be

Options:
A $A \pm\left(A^2-G^2\right)$
B $\sqrt{A} \pm \sqrt{A^2-G^2}$
C $A \pm \sqrt{(A+G)(A-G)}$
D $\frac{A \pm \sqrt{(A+G)(A-G)}}{2}$

The correct answer is:
$A \pm \sqrt{(A+G)(A-G)}$

Solution:
A.M. $=\frac{a+b}{2}=A$ and G.M. $=\sqrt{a b}=G$
On solving $a$ and $b$ are given by the values
$A \pm \sqrt{(A+G)(A-G)}$
Trick : Let the numbers be 1,9 . Then $A=5$ and $G=3$. Now put these values in options.
Here (3) $\Rightarrow 5 \pm \sqrt{8 \times 2}$ i.e. 9 and 1 .

Topic: Mathematics, Sequences and Series, JEE Main
If the average kinetic energy of a molecule of hydrogen gas at 300 K is E, then the average kinetic energy of a molecule of nitrogen gas at the same temperature is

Options:
A 7E
B 14E
C E7
D E

The correct answer is:
E

Solution:
We know that,
Kinetic energy ∝T
In the question, temperature is same in both conditions. So, the kinetic energy in the second condition is E .

Topic: Physics, Kinetic Theory of Gases, JEE Main
If the average of the first $n$ numbers in the sequence $148,146,144, \ldots$, is 125 , then $n$ is equal to

Options:
A 18
B 24
C 30
D 36

The correct answer is:
24

Solution:
Given sequence $148,146,144, \ldots$ and average is 125.
$$
\begin{aligned}
\therefore \quad S_n & =\frac{n}{2}[2 \times 148+(n-1)(-2)] \quad[\because d=-2] \\
& =\frac{n}{2}[296-2 n+2] \\
& =n[148-n+1]=n[149-n]
\end{aligned}
$$
Hence, $\quad 125=\frac{n[149-n]}{n}$
$$
\begin{aligned}
& \Rightarrow \quad 125=149-n \\
& \therefore \quad n=149-125=24 \\
&
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Main