datavorous/entrance-exam-dataset huggingface dataset:
The area (in sq.units) of the region {x, y:y2≥2x and x2+y2≤4x, x ≥0, y ≥0} is

Options:
A π-423
B π2-223
C π-43
D π-83

The correct answer is:
π-83

Solution:
y2≥2x (Area outside the parabola)
x2+y2≤4x (Area inside the circle)
Finding point of intersection of the curves, we get,
x2+y2=4x and y2=2x
⇒x2+2x=4x
⇒x2=2x
⇒x=0, x=2

If x=0,then y=0 and if x=2, then y=±2.
Coordinates of A0, 0, B2, 2
As x ≥0 ,y ≥0 only area above x-axis would be considered.

Hence, area
=∫024x-x2 dx- 2 ∫02x dx
=∫024-x-22 dx- 2 ∫02x dx
=x-22 4x-x2+42 .sin-1x-22- 2 x323202
=0-2sin-1-1- 2 .2322=π-83 sq.units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2016 (03 Apr)
The area (in sq. units) of the region {x, y:y2â¥2x and x2+y2â¤4x, x â¥0, y â¥0} is

Options:
A π-423
B π2-223
C π-43
D π-83

The correct answer is:
π-83

Solution:
y2≥2x (Area outside the parabola)
x2+y2≤4x (Area inside the circle)
Finding point of intersection of the curves, we get,
x2+y2=4x and y2=2x
⇒x2+2x=4x
⇒x2=2x
⇒x=0, x=2
If x=0, then y=0 and if x=2, then y= ±2.
Coordinates of A0, 0 and B2, 2
As x ≥0 y ≥0 only area above x-axis would be considered.

Hence, area
=∫024x-x2 dx- 2 ∫02x dx
=∫024-x-22 dx- 2 ∫02x dx
=x-22 4x-x2+42 .sin-1⁡x-2202- 2 x323202
=0-2sin-1⁡-1- 2 .2322] =π-83 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) of the smaller of the two circles that touch the parabola, y2=4x at the point 1, 2 and the x -axis is

Options:
A 8π3-22
B 8π2-2
C 4π3+2
D 4π2-2

The correct answer is:
8π3-22

Solution:
The equation of the tangent to the parabola y2=4ax at x1, y1 is yy1=2ax+x1.

Equation of tangent to the parabola y2=4x at 1, 2 is
y⋅2=2x+1

⇒x-y+1=0 ...i

The equation of the family of circles touching a line l=0 at x1, y1 is x-x12+y-y12+λl=0.

Equation of family of circle touching the parabola at 1, 2 is

x-12+y-22+λx-y+1=0 ...ii

Since circle touches x-axis, hence by putting y=0, we get

x2+λ-2x+λ+5=0

As the circle touches the x-axis, hence this equation will have only one root.

And we know that a quadratic equationax2+bx+x=0, a≠0, has only one root when its discriminant D=b2-4ac is zero.

Now D=0

⇒λ-22-4⋅1⋅λ+5=0

⇒λ2-4λ+4-4λ-20=0

⇒λ2-8λ-16=0

Now, applying the Sridharacharya formula, we get

λ=--8±-82-4×1×-162×1

⇒λ=4±42 for smaller circle λ=4-42.

Hence, the equation of the circle is x-12+y-22+4-42x-y+1=0

⇒x2+y2-2x1-22-2y4-22+9-42=0

Since, the circle touches the x-axis, hence the absolute value of the y-co-ordinate of the circle is the radius of the circle.

Thus, the radius of smaller circle r=4-22 units.

Area of circle πr2=π4-222

=π16-162+8=π24-162

=8π3-22 sq units.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The area (in sq. units) of the smaller portion enclosed between the curves, x2+y2=4 and y2=3x, is:

Options:
A 13+4π3
B 13+2π3
C 123+π3
D 123+2π3

The correct answer is:
13+4π3

Solution:
To find point of intersection, eliminate y from both the equation, we get
x2+3x-4=0
⇒ x+4 x-1=0
x=-4, x=1
Area =∫013·x·dx+∫124-x2·dx×2
=3x323201+x24-x2+2sin-1⁡x212×2
=323+2·π2-32+π3×2
=23-32+2π3×2
=123+2π3×2=13+4π3 sq. units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2017 (08 Apr Online)
The area (in sq. units) of the triangle formed by the latus rectum and the tangents at the end points of the latus rectum of x216-y29=1 is equal to

Options:

The correct answer is:
4.05

Solution:
For the given hyperbola, e=54

From the figure,
PS1=b2a=94
Meeting point of tangents at P & R is Qae,0≡165,0
Hence, the area of the triangle, A= 1 2 ⋅ 9 2 ⋅ 9 5 =4.05 sq. units

Topic: Mathematics, Hyperbola, JEE Main
The area (in sq. units) of the triangle formed by the lines y=2x,y=-2x and the tangent at the point 5,4 on 4x2-y2=4 is equal to

Options:
A 4
B 2
C 1
D 3

The correct answer is:
2

Solution:
Equation of the tangent at point 5,4 on 4x2-y2=4 is 4x5-4y=4⇒5x-y=1.
Now, point of intersection of y=5x-1 & y=2x is 15-2,25-2.
Point of intersection of y=5x-1 and y=-2x is 15+2,-25+2.
Point of intersection of y=2x & y=-2x is 0,0
Now, area of the triangle formed by the points 0,0,15-2,25-2,15+2,-25+2
=1200115-225-2115+2-25+21=12×1×-2-2=2 sq.units

Topic: Mathematics, Hyperbola, JEE Main
The area (in square unit) of the circle which touches the lines $4 x+3 y=15$ and $4 x+3 y=5$ is

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

The correct answer is:
$\pi$

Solution:
Given lines are parallel.
$d=\frac{15-5}{\sqrt{4^{2}+3^{2}}}=\frac{10}{5}$ $\Rightarrow d=2=$ diameter of the circle $\therefore$ Radius of circle $=1$ $\therefore$ Area of circle $=\pi \mathrm{r}^{2}=\pi$ sq unit

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square unit) of the triangle formed by $x+y+1=0$ and the pair of straight lines $x^{2}-3 x y+2 y^{2}=0$ is

Options:
A $\frac{7}{12}$
B $\frac{5}{12}$
C $\frac{1}{12}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{12}$

Solution:
$$
\begin{array}{l}
x^{2}-2 x y-x y+2 y^{2}=0 \\
\Rightarrow(x-2 y)(x-y)=0 ...(i)\\
\Rightarrow x=2 y, x=y...(ii)
\end{array}
$$
Also, $x+y+1=0$ On solving Eqs. (i) and (ii), we get
(ii)
$$
\begin{array}{l}
A\left(-\frac{2}{3},-\frac{1}{3}\right), B\left(-\frac{1}{2},-\frac{1}{2}\right), C(0,0) \\
\therefore \text { Area of } \triangle A B C=\frac{1}{2}\left|\begin{array}{rrr}
-\frac{2}{3} & -\frac{1}{3} & 1 \\
1 & -\frac{1}{2} & 1 \\
0 & 0 & 1
\end{array}\right| \\
=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{6}\right]=\frac{1}{2}\left[\frac{1}{6}\right]=\frac{1}{12}
\end{array}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) bounded by the curves $|x|=2,|y|=2$ and $x y \leq \frac{1}{2}$ is

Options:
A $9+3 \log 2$
B $3+\log 2$
C $2+\log 3$
D $8+2 \log 3$

The correct answer is:
$9+3 \log 2$

Solution:
We have,
$|x|=2,|y|=2, x y \leq \frac{1}{2}$
$\begin{aligned} \text { Required area } & =2[2 \times 2]+2\left[\int_0^{1 / 4} 2 d x+\int_{1 / 4}^2 \frac{1}{2 x} d x\right] \\ & =8+2\left[2[x]_0^{1 / 4}+\frac{1}{2}[\log x]_{1 / 4}^2\right] \\ & =8+4\left[\frac{1}{4}-0\right]+\left[\log 2-\log \frac{1}{4}\right] \\ & =8+1+\log 2-\log \left(2^{-2}\right) \\ & =9+\log 2+2 \log 2=9+3 \log 2\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) bounded by the curves $y^2=4 x$ and $x^2=4 y$ in the plane is

Options:
A $\frac{8}{3}$
B $\frac{16}{3}$
C $\frac{32}{3}$
D $\frac{64}{3}$

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

Solution:
Given curves are
The intersecting points of Eqs. (i) and (ii) are $(0,0)$ and $(4,4)$
$\therefore$ Required area $=$ Area of shaded portion $O A B C$
$\begin{aligned}
& =\int_0^4\left(\sqrt{4 x}-\frac{x^2}{4}\right) d x \\
& =\left[\frac{2 \cdot x^{3 / 2}}{3 / 2}-\frac{x^3}{12}\right]_0^4 \\
& =\left[\frac{4}{3} \cdot 8-\frac{64}{12}\right]=\left[\frac{32}{3}-\frac{16}{3}\right]
\end{aligned}$
$=\frac{16}{3}$ sq. units.

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) bounded by the curves y=x, 2y-x+3=0, X-axis and lying in the first quadrant is

Options:
A 18 sq. units
B 274 sq. units
C 9 sq. units
D 36 sq. units

The correct answer is:
9 sq. units

Solution:
First draw the given curves and represent the required area.
Eliminate x to find point of intersection of y=x and 2y-x+3=0
⇒2y-y2+3=0
⇒y2-2y-3=0
⇒y-3y+1=0
⇒y=3,-1
∴Required area A=∫09xdx-12×6×3
A=x3232 09-9
∴A=-9

Hence, area is 9 sq. units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2013 (07 Apr)
The area in square units bounded by the normal at $(1,2)$ to the parabola $y^{2}=4 x, X$-axis and the curve is given by

Options:
A $\frac{10}{3}$
B $\frac{7}{3}$
C $\frac{4}{3}$
D None of these

The correct answer is:
$\frac{10}{3}$

Solution:
We have, $y^{2}=4 x$
Differentiating w.r.t. $x$, we get
$$
\begin{aligned}
&2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y} \\
&\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2}=1
\end{aligned}
$$
Equation of normal to the curve at $(1,2)$ is
$$
\begin{aligned}
&\begin{aligned}
y-y_{1}=& \frac{1}{\left(-\frac{d y}{d x}\right)}\left(x-x_{1}\right) \Rightarrow(y-2)=-\frac{1}{1}(x-1)
\end{aligned} \\
&\Rightarrow \quad y-2=-x+1 \Rightarrow x+y=3 \\
&\text { The line } x+y=3 \text { meets the X-axis at } x=3 \\
&\therefore \text { Required area }=\int_{0}^{1} \sqrt{4 x} d x+\int_{1}^{3}(3-x) d x \\
&=2\left[\frac{x^{3 / 2}}{3 / 2}\right]_{0}^{1}+\left[3 x-\frac{x^{2}}{2}\right]_{1}^{3} \\
&=\frac{4}{3}(1)+\left[9-\frac{9}{2}-3+\frac{1}{2}\right]=\frac{4}{3}+\left(\frac{9}{2}-\frac{5}{2}\right) \\
&=\frac{4}{3}+\frac{4}{2}=\frac{4}{3}+2=\frac{10}{3} \text { sq units }
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) bounded by y=x2+x+1 and x+y=2 is

Options:
A 22
B 423
C 823
D 23

The correct answer is:
823

Solution:
Area =∫x1x22-x-x2+x+1dx

where x1 and x2 are roots of the equation 2-x=x2+x+1
⇒x2+2x-1=0
So, x1+x2=-2, x1x2=-1, x1-x2=4+4=22
Hence, the required area =∫x1x21-2x-x2dx=x2-x11-x2+x1-x22+x2x1+x123
=221--2-4--13
=223-53=823 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) enclosed by y-x=1 and x2+y2=1 is

Options:
A 2
B zero
C infinite
D None of these

The correct answer is:
zero

Solution:
Graph of x2 + y2=1 is
and graph of y- x=1 is
So area enclosed by two graph is zero.

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) of the quadrilateral formed by joining the focii of the two ellipses x29+y25=1 and x25+y29=1 is

Options:
A 4
B 2
C 6
D 8

The correct answer is:
8

Solution:
Major axis of ellipse x29+y25=1 is x-axis.

Major axis of ellipse x25+y29=1 is y-axis.

Eccentricity, e=1-b2a2

=1-59 = 23

Focii of ellipse x29+y25=1 is ±ae,0 = ±2,0

Focii of ellipse x25+y29=1 is 0,±be = 0,±2

Quadrilateral formed is a square of side length =22 + 22 = 22

Area of square = (side)2 =222 = 8 sq. units

Topic: Mathematics, Ellipse, JEE Main
The area (in square units) of the region bounded by the curve $y=|\sin 2 x|$ and the $X$-axis in $[0,2 \pi]$ is

Options:
A $0$
B $3$
C $4$
D $1$

The correct answer is:
$4$

Solution:
$y=(\sin 2 x)$
$\begin{aligned} & x \in[0,2 \pi] \\ & \int_0^{2 \pi}|\sin 2 x| d x \\ & =4 \int_0^{\frac{\pi}{2}} \sin 2 x d x \\ & =4\left[\frac{-\cos 2 x}{2}\right]_0^{\frac{\pi}{2}} \\ & =\frac{4}{2}[\cos 0-\cos \pi]=\frac{4}{2} \times 2=4\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) of the region bounded by the curves y+2x2=0 and y+3x2=1, is equal to

Options:
A 34 sq. units
B 13 sq. units
C 35 sq. units
D 43 sq. units

The correct answer is:
43 sq. units

Solution:
To find point of intersection of two given curves

Put y=-2x2 in y+3x2=1

⇒ x2=1

⇒ x=±1

Let the given equations be y1=1-3x2 and y2=-2x2.

The desired area would be,

∫-11y1-y2dx =∫-111-3x2--2x2dx

=∫-111-x2dx

=x- x33-11= 1-13- -1+13

=23- -23

=43 sq. units .

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2015 (10 Apr Online)
The area (in square units) of the region bounded by the parabola y2=4(x-2) and the line y=2x-8.

Options:
A 8
B 9
C 6
D 7

The correct answer is:
9

Solution:
Given curve is y2=4x-2 and line y=2x-8

Finding the point of intersection of y=2x-4 and y2=4x.

⇒x=y+82=y2+84

⇒2y+16=y2+8

⇒y2-2y-8=0

⇒y-4y+2=0

⇒y=4,-2

⇒x=6,3

Area of the bounded region is given by, A=∫-24y+82-y24+2dy

⇒A=y24+2y-y312-24

⇒A=4+8-163-1+4-23

⇒A=9 sq. units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The area (in square units) of the region bounded by $x^2=8 y, x=4$ and $X$-axis, is

Options:
A $\frac{2}{3}$
B $\frac{4}{3}$
C $\frac{8}{3}$
D $\frac{10}{3}$

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

Solution:
Given equations are
$\begin{aligned} x^2 & =8 y \\ x & =4 \\ y & =0\end{aligned}$
and
On solving Eqs. (i) and (ii), we get $(4,2)$
The required area
$=\int_0^4 y d x=\int_0^4 \frac{x^2}{8} d x=\left[\frac{x^3}{24}\right]_0^4=\frac{64}{24}=\frac{8}{3}$
The required area $=\frac{8}{3}$ sq units

Topic: Mathematics, Area Under Curves, JEE Main
The area in square units of the region bounded by $y^{2}=9 x$ and $y=3 x$ is

Options:
A 2
B $\frac{1}{4}$
C $\frac{1}{2}$
D 1

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

Solution:
Intersecting points of $y^{2}=9 x$ and $y=3 x$ is $(0,0)$ and $(1,3)$.
$\begin{aligned} \therefore \text { Hequired area } &=\int_{0}^{1}(\sqrt{9 x}-3 x) d x \\ &=3 \int_{0}^{1}(\sqrt{x}-x) d x=3\left[\frac{x^{3 / 2}}{3 / 2}-\frac{x^{2}}{2}\right]_{0}^{1} \\ &=3\left(\frac{2}{3}-\frac{1}{2}\right)=3\left(\frac{1}{6}\right)=\frac{1}{2} \text { squnit } \end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) of the region enclosed between the parabola $y^2=2 x$ and the line $y=4 x-1$

Options:
A $\frac{9}{32}$
B $\frac{7}{23}$
C $\frac{16}{3}$
D $\frac{15}{4}$

The correct answer is:
$\frac{9}{32}$

Solution:
Given equation of parabola $y^2=2 x$ $\ldots(\mathrm{i})$
and equation of line $y=4 x-1$ $\ldots(\mathrm{ii})$
Intersecting points of Eqs. (i) and (ii)
$\operatorname{are}\left(\frac{1}{8}, \frac{-1}{2}\right)$ and $\left(\frac{1}{2}, 1\right)$
Required area $A O B$
$=\int_{-1 / 2}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y$
$=\frac{1}{4} \int_{-1 / 2}^1(y+1) d y-\frac{1}{2} \int_{-1 / 2}^1 y^2 d y$
$=\frac{1}{4}\left[\frac{y^2}{2}+y\right]_{-1 / 2}^1-\frac{1}{2}\left[\frac{y^3}{3}\right]_{-1 / 2}^1$
$=\frac{1}{4}\left[\left(\frac{1}{2}+1\right)-\left(\frac{1}{8}-\frac{1}{2}\right)\right]-\frac{1}{6}\left[1-\left(-\frac{1}{8}\right)\right]$
$=\frac{1}{4}\left[\frac{3}{2}+\frac{3}{8}\right]-\frac{1}{6}\left(\frac{9}{8}\right)=\frac{1}{4} \times \frac{15}{8}-\frac{3}{16}$
$=\frac{15-6}{32}=\frac{9}{32}$ sq units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) of the region enclosed by the ellipse $x^2+3 y^2=18$ in the first quadrant below the line $y=x$ is

Options:
A $\sqrt{3} \pi-\frac{3}{4}$
B $\sqrt{3} \pi+1$
C $\sqrt{3} \pi$
D $\sqrt{3} \pi+\frac{3}{4}$

The correct answer is:
$\sqrt{3} \pi$

Solution:
$\frac{x^2}{18}+\frac{y^2}{6}=1$
$\begin{aligned}
& \frac{x^2}{18}+\frac{3 x^2}{18}=1 \Rightarrow 4 x^2=18 \Rightarrow x^2=\frac{9}{2} \\
& \int_{\frac{3}{\sqrt{2}}}^{\sqrt[3]{2}} \frac{\sqrt{18-x^2}}{\sqrt{3}} d x \\
& =\frac{1}{\sqrt{3}}\left(\frac{x \sqrt{18-x^2}}{2}+\frac{18}{2} \sin ^{-1} \frac{x}{3 \sqrt{2}}\right)_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\
& =\frac{1}{\sqrt{3}}\left(9 \times \frac{\pi}{2}-\frac{3}{2 \sqrt{2}} \times \frac{3 \sqrt{3}}{\sqrt{2}}-9 \times \frac{\pi}{6}\right)
\end{aligned}$
Required Area
$\begin{aligned}
& =\frac{1}{2} \times \frac{9}{2}+\left(\frac{18 \pi}{6}-\frac{9 \sqrt{3}}{4}\right) \frac{1}{\sqrt{3}} \\
& =\sqrt{3} \pi
\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

Options:
A $\frac{2 \pi}{3}+\frac{\sqrt{3}}{2}$
B $\frac{\pi}{3}+\frac{\sqrt{3}}{2}$
C $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
D $\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}$

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

Solution:
Intersection point of two circles
$x^2+y^2=1$ $\ldots$ (i)
$(x-1)^2+y^2=1$ $\ldots$ (ii)
is given by
$(x-1)^2+\left(1-x^2\right)=1$
$\Rightarrow \quad x^2+1-2 x-x^2=0$
$\Rightarrow \quad x=\frac{1}{2}$
From Eq. (i), $\frac{1}{4}+y^2=1$
$y^2=1-\frac{1}{4} \Rightarrow y= \pm \frac{\sqrt{3}}{2}$
Point $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $C\left(\frac{1}{2}, \frac{-\sqrt{3}}{2}\right)$
So, Area of region $O A B C O$
$=2 \times$ Area of region $O A B D O$
Area of $O A B D O$
$=$ Area of $O A D O+$ Area of $A B D A$
$=\int_0^{1 / 2} \sqrt{1-(x-1)^2} d x+\int_{1 / 2}^1 \sqrt{1-x^2} d x$
$=\left[\frac{1}{2} \cdot(x-1) \sqrt{1-(x-1)^2}+\frac{1}{2} \sin ^{-1}\left(\frac{x-1}{1}\right)\right]_0^{1 / 2}$ $+\left[\frac{1}{2} x \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}\left(\frac{x}{1}\right)\right]_{1 / 2}^1$
$=\left[-\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \sin ^{-1}\left(\frac{-1}{2}\right)-\frac{1}{2} \cdot 0-\frac{1}{2} \sin ^{-1}(-1)\right]$ $+\left[\frac{1}{2} \cdot 0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{4} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)\right]$
$=\left(-\frac{\sqrt{3}}{8}\right)-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)+\frac{1}{2} \sin ^{-1}(1)$ $+\frac{1}{2} \sin ^{-1}(1)-\frac{\sqrt{3}}{8}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)$
$=\sin ^{-1}(1)-\sin ^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{4}$
$=\frac{\pi}{2}-\frac{\pi}{6}-\frac{\sqrt{3}}{4}$
$=\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)$
from Eq. (i),
Area of region $O A B C O=2 \times\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right)$
$=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in square units) of the triangle bounded by x=4 and the lines y2-x2+2x=1 is equal to

Options:
A 3
B 6
C 12
D 9

The correct answer is:
9

Solution:
The given lines are x=4, y=x-1, y=1-x
Hence, the bounded area =12322=9 sq. units

Topic: Mathematics, Straight Lines, JEE Main
The area in the first quadrant between $x^2+y^2=\pi^2$ and $y=\sin x$ is

Options:
A $\frac{\left(\pi^3-8\right)}{4}$
B $\frac{\pi^3}{4}$
C $\frac{\left(\pi^3-16\right)}{4}$
D $\frac{\left(\pi^3-8\right)}{2}$

The correct answer is:
$\frac{\left(\pi^3-8\right)}{4}$

Solution:
Area of the circle in first quadrant is $\frac{\pi\left(\pi^2\right)}{4}$ i.e., $\frac{\pi^3}{4}$. Also area bounded by curve $y=\sin x$ and $x$-axis is 2 sq. unit. Hence required area is $\frac{\pi^3}{4}-2=\frac{\pi^3-8}{4}$.

Topic: Mathematics, Area Under Curves, JEE Main
The area of a circle is $A_1$ and the area of a regular pentagon inscribed in the circle is $A_2$. Then $A_1: A_2$ is

Options:
A $\frac{\pi}{5} \cos \frac{\pi}{10}$
B $\frac{2 \pi}{5} \sec \frac{\pi}{10}$
C $\frac{2 \pi}{5} \operatorname{cosec} \frac{\pi}{10}$
D None of these

The correct answer is:
$\frac{2 \pi}{5} \sec \frac{\pi}{10}$

Solution:
If in the $\triangle O A B, O A=O B=r$ and $\angle A O B=\frac{360^{\circ}}{5}=72^{\circ}$
$\therefore \quad \operatorname{ar}(\triangle A O B)=\frac{1}{2} \cdot r \cdot r \sin 72^{\circ}=\frac{1}{2} r^2 \cos 18^{\circ}$
$\therefore \quad A_1: A_2=\frac{2 \pi r^2}{5 r^2 \cos 18^{\circ}}=\frac{2 \pi}{5}$ sec $\frac{\pi}{10}$

Topic: Mathematics, Properties of Triangles, JEE Main
The area of a parallelogram whose two adjacent sides are represented by the vector $3 \mathbf{i}-\mathbf{k}$ and $\mathbf{i}+2 \mathbf{j}$ is

Options:
A $\frac{1}{2} \sqrt{17}$
B $\frac{1}{2} \sqrt{14}$
C $\sqrt{41}$
D $\frac{1}{2} \sqrt{7}$

The correct answer is:
$\sqrt{41}$

Solution:
The area of parallelogram is given by $=|\overrightarrow{A B} \times \overrightarrow{A D}|=\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B D}|$ Here we are given adjacent sides and so
$\overrightarrow{A B} \times \overrightarrow{A D}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 0 & -1 \\
1 & 2 & 0
\end{array}\right|=2 \mathbf{i}-\mathbf{j}+6 \mathbf{k}$
Hence required area is $=\mid2 \mathbf{i}-\mathbf{j}+6 \mathbf{k} \mid=\sqrt{41}$.

Topic: Mathematics, Vector Algebra, JEE Main
The Area of a region bounded by \( y=\sin a x, y=0 \) from \( x=\frac{\pi}{3 a} \) to \( x=\frac{\pi}{a} \) is \( 3 \) square unit, find the integral value of
\( a(a>0) \)
Given that graph of \( y=\sin a x \) lies above the axis from \( x=\frac{\pi}{a} \) to \( \left.x=\frac{\pi}{3 a}\right\} \)

Options:
A \( 3 \)
B \( \frac{1}{3} \)
C \( \frac{1}{2} \)
D \( \frac{1}{1} \)

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

Solution:
We have, y=sinax

Given that the given curve is above the axis from x=πa to x=π3a.

Then, the area bounded by the curve from x=πa to x=π3a=∫π/aπ/3aydx

⇒A=∫π3aπasinaxdx
⇒A=1a(−cosax)π/aπ/3a
⇒A=1a32

∴ A=32a=3 Given

⇒a=12

Topic: Mathematics, Area Under Curves, JEE Main
The area of closed region bounded by the parabolas \( y=4 x^{2}, y=\frac{x^{2}}{9} \) and the line \( y=2 \) is

Options:
A \( \frac{20 \sqrt{2}}{3} \) sq. unit
B \( \frac{10 \sqrt{2}}{3} \) sq. unit
C \( \frac{40 \sqrt{2}}{3} \) sq. unit
D None of these

The correct answer is:
\( \frac{20 \sqrt{2}}{3} \) sq. unit

Solution:
Area=2â«029y-y4 dy=23.23y32-12.23y3202=253.y3202=103.22

Topic: Mathematics, Area Under Curves, JEE Main
The area of cross-section of a large tank is 0.5 m2. It has a narrow opening near the bottom having area of cross-section 1 cm2. A load of 25 kg is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is 40 cm above the bottom, will be _____ cm s-1.

[Take g=10 m s-2 ]

Options:

The correct answer is:
300

Solution:
Pressure at point 1 is P1=P0, Pressure at point 2 is P2=P0+mgA

Applying Bernoulli's equation at top and bottom

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2⇒P0+12ρv2+0=P0+mgA+0+ρgh

⇒2505+103×10×0.4=12×103×v2

v=3 m s-1=300 cm s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The area of cross-section of a railway track is 0.01 m2. The temperature variation is 10 °C. Coefficient of linear expansion of material of track is 10-5 C-1∘. The energy stored per meter in the track is J m-1. (Young's modulus of material of track is 1011 N m-2)

Options:

The correct answer is:
5

Solution:
Elastic energy =Y2(strain)2× Area × length

⇒ Elastic energy per unit length =Y2(strain)2× Area

(strain =Δℓℓ=αΔT=10-5×10=10-4)
=10112×10-42×10-2=5 J m-1

Topic: Physics, Mechanical Properties of Solids, JEE Main
The area of cross-section of a railway track is 0.01 m2. The temperature variation is 10 °C. Coefficient of linear expansion of material of track is 10-5 C-1∘. The energy stored per meter in the track is J m-1. (Young's modulus of material of track is 1011 N m-2)

Options:

The correct answer is:
5

Solution:
Elastic energy =Y2(strain)2× Area × length

⇒ Elastic energy per unit length =Y2(strain)2× Area

(strain =Δℓℓ=αΔT=10-5×10=10-4)
=10112×10-42×10-2=5 J m-1

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The area of region $\left\{(x, y): x^2+y^2 \leq 1 \leq x+y\right\}$ is

Options:
A $\frac{\pi^2}{5}$
B $\frac{\pi^2}{2}$
C $\frac{\pi^2}{3}$
D $\frac{\pi}{4}-\frac{1}{2}$

The correct answer is:
$\frac{\pi}{4}-\frac{1}{2}$

Solution:
$x^2+y^2=1, x+y=1$ meet when
$x^2+(1-x)^2=1 \Rightarrow x^2+1+x^2-2 x=1$
$\Rightarrow 2 x^2-2 x=0 \Rightarrow 2 x(x-1)=0$
$\Rightarrow x=0, x=1 \Rightarrow y=1, y=0$, i.e., $A(1,0) ; B(0,1)$
$\begin{aligned} \text { Required area } & =\int_0^1\left[\sqrt{1-x^2}-(1-x)\right] d x \\ & =\left[\frac{x \sqrt{1-x^2}}{2}+\frac{1}{2} \sin ^{-1} x-x+\frac{x^2}{2}\right]_0^1 \\ & =\frac{1}{2} \cdot \frac{\pi}{2}-1+\frac{1}{2}=\frac{\pi}{4}-\frac{1}{2} .\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of smaller part between the circle $x^2+y^2=4$ and the line $x=1$ is

Options:
A $\frac{4 \pi}{3}-\sqrt{3}$
B $\frac{8 \pi}{3}-\sqrt{3}$
C $\frac{4 \pi}{3}+\sqrt{3}$
D $\frac{5 \pi}{3}+\sqrt{3}$

The correct answer is:
$\frac{8 \pi}{3}-\sqrt{3}$

Solution:
Area of smaller part $=2 \int_1^2 \sqrt{4-x^2} d x$
$\begin{aligned}& =2\left[\frac{x}{2} \sqrt{4-x^2}+2 \sin ^{-1} \frac{x}{2}\right]_1^2=2\left[2 \cdot \frac{\pi}{2}-\left[\frac{\sqrt{3}}{2}-2 \cdot \frac{\pi}{6}\right]\right] \\& =2\left[\pi-\left[\frac{\sqrt{3}}{2}-\frac{\pi}{3}\right]\right]=\frac{8 \pi}{3}-\sqrt{3}\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the bounded region enclosed by the curve y=3-x-12-x+1 and the x-axis is

Options:
A 94
B 4516
C 278
D 6316

The correct answer is:
278

Solution:
Plotting the graph of the function y=3-x+1-x-12 and x-axis, we get

Now applying the formula of area of trapezium which is 12×height×sum of parallel sides we get,

Required area of TrapABCD=12×32×3+32=278

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (28 Jun Shift 2)
The area of the closed figure bounded by \( x=-1, x=2 \) and \( y=\left\{\begin{array}{cl}-x^{2}+2, & x \leq 1 \\ 2 x-1, & x>1\end{array}\right. \) and the \( x \)-axis is

Options:
A \( \frac{16}{3} \) sq. units
B \( \frac{10}{3} \) sq. units
C \( \frac{13}{3} \) sq. units
D \( \frac{7}{3} \) sq. units

The correct answer is:
\( \frac{16}{3} \) sq. units

Solution:
A=∫-11-x2+2dx+∫122x-1dx
=-x33+2x-11+x2-x2
=163 sq.units

Topic: Mathematics, Area Under Curves, JEE Main