PYQs Guides Notes Mocks Board Questions JEE Mains Maths Application of Integrals 117 JEE Mains Maths previous year questions (PYQs) from 74 papers, 2021 to 2026, with answers and worked solutions.
About this chapter Most-asked subtopics: Circles, Ellipses and Other Conics (22), Curves and Lines: Vertical Strips (20), Trigonometric, Exponential and Reciprocal Curves (18). Updated 25 July 2026. #1 · Curves and Lines: Vertical Strips
The area of the region { ( x , y ) : x 2 − 8 x ≤ y ≤ − x } \left\{ (x,y):x^{2}- 8x \leq y \leq - x \right\} { ( x , y ) : x 2 − 8 x ≤ y ≤ − x } is: #2 · Curves and Lines: Vertical Strips
If the area of the region { ( x , y ) : 1 − 2 x ≤ y ≤ 4 − x 2 \left\{ (x,y):1 - 2x \leq y \leq 4 -x^{2} \right. { ( x , y ) : 1 − 2 x ≤ y ≤ 4 − x 2 , x ≥ 0 , y ≥ 0 } x \geq 0,y \geq 0\} x ≥ 0 , y ≥ 0 } is α β , α , β , ∈ N , g c d ( α , β ) = 1 \frac{\alpha}{\beta},\alpha,\beta, \in N,gcd(\alpha,\beta) = 1 β α , α , β , ∈ N , g c d ( α , β ) = 1 , then the value of ( α + β ) (\alpha + \beta) ( α + β ) is : #3 · Curves and Lines: Vertical Strips
Let A 1 A_{1} A 1 be the bounded area enclosed by the curves y = x 2 + 2 , x + y = 8 y =x^{2}+ 2,x + y = 8 y = x 2 + 2 , x + y = 8 and y y y -axis that lies in the first quadrant. Let A 2 A_{2} A 2 be the bounded area enclosed by the curves y = x 2 + 2 , y 2 = x , x = 2 y =x^{2}+ 2,y^{2}= x,x = 2 y = x 2 + 2 , y 2 = x , x = 2 , and y y y -axis that lies in the first quadrant. Then A 1 − A 2 A_{1}-A_{2} A 1 − A 2 is equal to A 2 3 ( 2 2 + 1 ) \frac{2}{3}(2\sqrt{2} + 1) 3 2 ( 2 2 + 1 ) B 2 3 ( 4 2 + 1 ) \frac{2}{3}(4\sqrt{2} + 1) 3 2 ( 4 2 + 1 ) C 2 3 ( 2 + 1 ) \frac{2}{3}(\sqrt{2} + 1) 3 2 ( 2 + 1 ) D 2 3 ( 3 2 + 1 ) \frac{2}{3}(3\sqrt{2} + 1) 3 2 ( 3 2 + 1 ) JEE Mains · 2024 · 31 January 2024 · Q161 Moderate Add #4 · Curves and Lines: Vertical Strips
The area of the region enclosed by the parabola y = 4 x − x 2 y = 4x -x^{2} y = 4 x − x 2 and 3 y = ( x − 4 ) 2 3y = (x - 4)^{2} 3 y = ( x − 4 ) 2 is equal to JEE Mains · 2023 · 11 April 2023 · Q180 Moderate Add #5 · Curves and Lines: Vertical Strips
If A A A is the area in the first quadrant enclosed by the curve C : 2 x 2 − y + 1 = 0 C:2x^{2}- y + 1 = 0 C : 2 x 2 − y + 1 = 0 , the tangent to C C C at the point ( 1 , 3 ) (1,3) ( 1 , 3 ) and the line x + y = 1 x + y = 1 x + y = 1 , then the value of 60 A 60\text{ }A 60 A is Done a few? Here’s your next move
JEE Mains · 2024 · 6 April 2024 · Q79 Moderate Add #6 · Curves and Lines: Vertical Strips
Let the area of the region enclosed by the curves y = 3 x , 2 y = 27 − 3 x y = 3x,2y = 27 - 3x y = 3 x , 2 y = 27 − 3 x and y = 3 x − x x y = 3x - x\sqrt{x} y = 3 x − x x be A. Then 10 A 10\text{ }A 10 A is equal to JEE Mains · 2022 · 28 June 2022 · Q69 Moderate Add #7 · Curves and Lines: Vertical Strips
The area of the region S = { ( x , y ) : y 2 ≤ 8 x , y ≥ 2 x , x ≥ 1 } S =\left\{ (x,y):y^{2}\leq 8x,y \geq\sqrt{2}x,x \geq 1 \right\} S = { ( x , y ) : y 2 ≤ 8 x , y ≥ 2 x , x ≥ 1 } is A 13 2 6 \frac{13\sqrt{2}}{6} 6 13 2 B 11 2 6 \frac{11\sqrt{2}}{6} 6 11 2 C 5 2 6 \frac{5\sqrt{2}}{6} 6 5 2 D 19 2 6 \frac{19\sqrt{2}}{6} 6 19 2 JEE Mains · 2022 · 29 June 2022 · Q66 Moderate Add #8 · Curves and Lines: Vertical Strips
The area enclosed by y 2 = 8 x y^{2}= 8x y 2 = 8 x and y = 2 x y=\sqrt{2}x y = 2 x that lies outside the triangle formed by y = 2 x , x = y=\sqrt{2}x,x= y = 2 x , x = 1 , y = 2 2 1,y= 2\sqrt{2} 1 , y = 2 2 , is equal to: A 16 2 6 \frac{16\sqrt{2}}{6} 6 16 2 B 11 2 6 \frac{11\sqrt{2}}{6} 6 11 2 C 13 2 6 \frac{13\sqrt{2}}{6} 6 13 2 D 5 2 6 \frac{5\sqrt{2}}{6} 6 5 2 JEE Mains · 2023 · 30 January 2023 · Q83 Moderate Add #9 · Curves and Lines: Vertical Strips
Let α \alpha α be the area of the larger region bounded by the curve y 2 = 8 x y^{2}= 8x y 2 = 8 x and the lines y = x y = x y = x and x = 2 x = 2 x = 2 , which lies in the first quadrant. Then the value of 3 α 3\alpha 3 α is equal to JEE Mains · 2023 · 30 January 2023 · Q174 Moderate Add #10 · Curves and Lines: Vertical Strips
Let A A A be the area of the region { ( x , y ) : y ≥ x 2 , y ≥ ( 1 − x ) 2 , y ≤ 2 x ( 1 − x ) } \left\{ (x,y):y \geq x^{2},y \geq (1 - x)^{2},y \leq 2x(1 - x) \right\} { ( x , y ) : y ≥ x 2 , y ≥ ( 1 − x ) 2 , y ≤ 2 x ( 1 − x ) } . Then 540 A 540\text{ }A 540 A is equal to JEE Mains · 2022 · 14 June 2022 · Q89 Moderate Add #11 · Curves and Lines: Vertical Strips
Let S S S be the region bounded by the curves y = x 3 y =x^{3} y = x 3 and y 2 = x y^{2}= x y 2 = x . The curve y = 2 ∣ x ∣ y = 2|x| y = 2∣ x ∣ divides S S S into two regions of areas R 1 R_{1} R 1 and R 2 R_{2} R 2 . If max { R 1 , R 2 } = R 2 \max\left\{ R_{1},R_{2} \right\}=R_{2} max { R 1 , R 2 } = R 2 , then R 2 R 1 \frac{R_{2}}{R_{1}} R 1 R 2 is equal to JEE Mains · 2022 · 25 July 2022 · Q73 Moderate Add #12 · Curves and Lines: Vertical Strips
Let the locus of the centre ( α , β ) , β > 0 (\alpha,\beta),\beta > 0 ( α , β ) , β > 0 , of the circle which touches the circle x 2 + ( y − 1 ) 2 = 1 x^{2}+ (y - 1)^{2}= 1 x 2 + ( y − 1 ) 2 = 1 externally and also touches the x x x -axis be L L L . Then the area bounded by L L L and the line y = 4 y = 4 y = 4 is: A 32 2 3 \frac{32\sqrt{2}}{3} 3 32 2 B 40 2 3 \frac{40\sqrt{2}}{3} 3 40 2 C D JEE Mains · 2025 · 7 Apr 2025 · Q66 Moderate Add #13 · Curves and Lines: Vertical Strips
If the area of the region bounded by the curves y = 4 − x 2 4 y= 4 -\frac{x^{2}}{4} y = 4 − 4 x 2 and y = x − 4 2 y=\frac{x- 4}{2} y = 2 x − 4 is equal to α \alpha α , then 6 α 6\alpha 6 α equals. JEE Mains · 2023 · 12 Apr 2023 · Q73 Moderate Add #14 · Curves and Lines: Vertical Strips
The area of the region enclosed by the curve y = x 3 y =x^{3} y = x 3 and its tangent at the point ( − 1 , − 1 ) ( - 1, - 1) ( − 1 , − 1 ) is JEE Mains · 2023 · 8 Apr 2023 · Q77 Moderate Add #15 · Curves and Lines: Vertical Strips
The area of the region { ( x , y ) : x 2 ≤ y ≤ 8 − x 2 , y ≤ 7 } \left\{ (x,y):x^{2}\leq y \leq 8 -x^{2},y \leq 7 \right\} { ( x , y ) : x 2 ≤ y ≤ 8 − x 2 , y ≤ 7 } is JEE Mains · 2023 · 25 Jan 2023 · Q161 Moderate Add #16 · Curves and Lines: Vertical Strips
Let T T T and C C C respectively be the transverse and conjugate axes of the hyperbola 16 x 2 − y 2 + 64 x + 4 y + 44 = 0 16x^{2}-y^{2}+ 64x + 4y + 44 = 0 16 x 2 − y 2 + 64 x + 4 y + 44 = 0 . Then the area of the region above the parabola x 2 = y + 4 x^{2}= y + 4 x 2 = y + 4 , below the transverse axis T T T and on the right of the conjugate axis C C C is: A 4 6 + 28 3 4\sqrt{6} + \frac{28}{3} 4 6 + 3 28 B 4 6 − 44 3 4\sqrt{6} - \frac{44}{3} 4 6 − 3 44 C 4 6 + 44 3 4\sqrt{6} + \frac{44}{3} 4 6 + 3 44 D 4 6 − 28 3 4\sqrt{6} - \frac{28}{3} 4 6 − 3 28 JEE Mains · 2024 · 5 Apr 2024 · Q83 Moderate Add #17 · Curves and Lines: Vertical Strips
The area of the region enclosed by the parabolas y = x 2 − 5 x y=x^{2}- 5x y = x 2 − 5 x and y = 7 x − x 2 y= 7x-x^{2} y = 7 x − x 2 is JEE Mains · 2021 · Paper 18 · Q86 Moderate Add #18 · Curves and Lines: Vertical Strips
The area of the region S = { ( x , y ) : 3 x 2 ≤ 4 y ≤ 6 x + 24 } S=\left\{ (x,y):3x^{2}\leq 4y\leq 6x+ 24 \right\} S = { ( x , y ) : 3 x 2 ≤ 4 y ≤ 6 x + 24 } is JEE Mains · 2021 · Paper 17 · Q69 Moderate Add #19 · Curves and Lines: Vertical Strips
The area of the region bounded by y − x = 2 y-x= 2 y − x = 2 and x 2 = y x^{2}= y x 2 = y is equal to :- JEE Mains · 2021 · Paper 2 · Q79 Moderate Add #20 · Curves and Lines: Vertical Strips
The area of the region: R = { ( x , y ) : 5 x 2 ≤ y ≤ 2 x 2 + 9 } R=\left\{ (x,y):5x^{2}\leq y\leq 2x^{2}+ 9 \right\} R = { ( x , y ) : 5 x 2 ≤ y ≤ 2 x 2 + 9 } is: A 9 3 9\sqrt{3} 9 3 square units B 12 3 12\sqrt{3} 12 3 square units C 11 3 11\sqrt{3} 11 3 square units D 6 3 6\sqrt{3} 6 3 square units #21 · Horizontal Strips and Sideways Parabolas
The area of the region { ( x , y ) : 0 ≤ y ≤ 6 − x , y 2 ≥ 4 x − 3 , x ≥ 0 } \{(x,y):0 \leq y \leq 6 - x,\ y^{2}\geq 4x - 3,\ x \geq 0\} {( x , y ) : 0 ≤ y ≤ 6 − x , y 2 ≥ 4 x − 3 , x ≥ 0 } is : JEE Mains · 2026 · 4 Apr 2026 Shift 2 · Q68 Moderate Add #22 · Horizontal Strips and Sideways Parabolas
The area of the region bounded by the curves x + 3 y 2 = 0 x+ 3y^{2}= 0 x + 3 y 2 = 0 and x + 4 y 2 = 1 x+ 4y^{2}= 1 x + 4 y 2 = 1 is equal to : JEE Mains · 2025 · 4 April 2025 · Q134 Moderate Add #23 · Horizontal Strips and Sideways Parabolas
A line passing through the point A ( − 2 , 0 ) A\left( - 2,0 \right) A ( − 2 , 0 ) , touches the parabola P : y 2 = P:y^{2}= P : y 2 = x − 2 x- 2 x − 2 at the point B B B in the first quadrant. The area, of the region bounded by the line AB , parabola P and the x -axis, is :- JEE Mains · 2024 · 4 April 2024 · Q159 Moderate Add #24 · Horizontal Strips and Sideways Parabolas
The area (in sq. units) of the region described by { ( x , y ) : y 2 ≤ 2 x \left\{ (x,y):y^{2}\leq 2x \right. { ( x , y ) : y 2 ≤ 2 x , and y ≥ 4 x − 1 } \left. \ y \geq 4x - 1 \right\} y ≥ 4 x − 1 } is JEE Mains · 2022 · 27 July 2022 · Q158 Moderate Add #25 · Horizontal Strips and Sideways Parabolas
The area of the region enclosed by y ≤ 4 x 2 , x 2 ≤ 9 y y \leq 4x^{2},x^{2}\leq 9y y ≤ 4 x 2 , x 2 ≤ 9 y and y ≤ 4 y \leq 4 y ≤ 4 , is equal to: JEE Mains · 2022 · 29 June 2022 · Q158 Moderate Add #26 · Horizontal Strips and Sideways Parabolas
If ∫ 0 2 ( 2 x − 2 x − x 2 ) d x = \int_{0}^{2} \left( \sqrt{2x}-\sqrt{2x -x^{2}} \right)dx = ∫ 0 2 ( 2 x − 2 x − x 2 ) d x = ∫ 0 1 ( 1 − 1 − y 2 − y 2 2 ) d y + ∫ 1 2 ( 2 − y 2 2 ) d y + I \int_{0}^{1} \left( 1 -\sqrt{1 -y^{2}}-\frac{y^{2}}{2} \right)dy +\int_{1}^{2} \left( 2 -\frac{y^{2}}{2} \right)dy + I ∫ 0 1 ( 1 − 1 − y 2 − 2 y 2 ) d y + ∫ 1 2 ( 2 − 2 y 2 ) d y + I then I I I is equal to A ∫ 0 1 ( 1 + 1 − y 2 ) d y \int_{0}^{1} \left( 1 + \sqrt{1 - y^{2}} \right)dy ∫ 0 1 ( 1 + 1 − y 2 ) d y B ∫ 0 1 ( y 2 2 − 1 − y 2 + 1 ) d y \int_{0}^{1} \left( \frac{y^{2}}{2} - \sqrt{1 - y^{2}} + 1 \right)dy ∫ 0 1 ( 2 y 2 − 1 − y 2 + 1 ) d y C ∫ 0 1 ( 1 − 1 − y 2 ) d y \int_{0}^{1} \left( 1 - \sqrt{1 - y^{2}} \right)dy ∫ 0 1 ( 1 − 1 − y 2 ) d y D ∫ 0 1 ( y 2 2 + 1 − y 2 + 1 ) d y \int_{0}^{1} \left( \frac{y^{2}}{2} + \sqrt{1 - y^{2}} + 1 \right)dy ∫ 0 1 ( 2 y 2 + 1 − y 2 + 1 ) d y JEE Mains · 2022 · 25 June 2022 · Q156 Moderate Add #27 · Horizontal Strips and Sideways Parabolas
The area of the region enclosed between the parabolas y 2 = 2 x − 1 y^{2}= 2x - 1 y 2 = 2 x − 1 and y 2 = 4 x − 3 y^{2}= 4x - 3 y 2 = 4 x − 3 is JEE Mains · 2022 · 14 June 2022 · Q176 Moderate Add #28 · Horizontal Strips and Sideways Parabolas
The area (in sq. units) of the region enclosed between the parabola y 2 = 2 x y^{2}= 2x y 2 = 2 x and the line x + y = 4 x+y= 4 x + y = 4 is JEE Mains · 2022 · 26 June 2022 · Q157 Moderate Add #29 · Horizontal Strips and Sideways Parabolas
The area of the region bounded by y 2 = 8 x y^{2}= 8x y 2 = 8 x and y 2 = y^{2}= y 2 = 16 ( 3 − x ) 16(3 - x) 16 ( 3 − x ) is equal to:- JEE Mains · 2025 · 28 Jan 2025 · Q134 Moderate Add #30 · Horizontal Strips and Sideways Parabolas
The area of the region bounded by the curves x ( 1 + y 2 ) = 1 x\left( 1 +y^{2} \right)= 1 x ( 1 + y 2 ) = 1 and y 2 = 2 x y^{2}= 2x y 2 = 2 x is : A 2 ( π 2 − 1 3 ) 2\left( \frac{\pi}{2} - \frac{1}{3} \right) 2 ( 2 π − 3 1 ) B π 4 − 1 3 \frac{\pi}{4} - \frac{1}{3} 4 π − 3 1 C π 2 − 1 3 \frac{\pi}{2} - \frac{1}{3} 2 π − 3 1 D 1 2 ( π 2 − 1 3 ) \frac{1}{2}\left( \frac{\pi}{2} - \frac{1}{3} \right) 2 1 ( 2 π − 3 1 ) #31 · Horizontal Strips and Sideways Parabolas
The area of the region enclosed by the curves y = x 2 − 4 x + 4 y =x^{2}- 4x + 4 y = x 2 − 4 x + 4 and y 2 = 16 − 8 x y^{2}= 16 - 8x y 2 = 16 − 8 x is : JEE Mains · 2023 · 15 Apr 2023 · Q88 Moderate Add #32 · Horizontal Strips and Sideways Parabolas
If the area bounded by the curve 2 y 2 = 3 x 2y^{2}= 3x 2 y 2 = 3 x , lines x + y = 3 , y = 0 x+y= 3,y= 0 x + y = 3 , y = 0 and outside the circle ( x − 3 ) 2 + y 2 = 2 (x- 3)^{2}+y^{2}= 2 ( x − 3 ) 2 + y 2 = 2 is A A A , then 4 ( π + 4 A ) 4(\pi+ 4A) 4 ( π + 4 A ) is equal to JEE Mains · 2023 · 24 Jan 2023 · Q64 Moderate Add #33 · Horizontal Strips and Sideways Parabolas
The area enclosed by the curves y 2 + 4 x = 4 y^{2}+ 4x = 4 y 2 + 4 x = 4 and y − 2 x = 2 y - 2x = 2 y − 2 x = 2 is: JEE Mains · 2023 · 24 Jan 2023 · Q179 Moderate Add #34 · Horizontal Strips and Sideways Parabolas
If the area of the region bounded by the curves y 2 − 2 y = − x , x + y = 0 y^{2}- 2y = - x,x + y = 0 y 2 − 2 y = − x , x + y = 0 is A A A , then 8 A 8\text{ }A 8 A is equal to JEE Mains · 2024 · 30 Jan 2024 · Q68 Moderate Add #35 · Horizontal Strips and Sideways Parabolas
The area (in square units) of the region bounded by the parabola y 2 = 4 ( x − 2 ) y^{2}= 4(x- 2) y 2 = 4 ( x − 2 ) and the line y = 2 x − 8 y= 2x- 8 y = 2 x − 8 is JEE Mains · 2024 · 30 Jan 2024 · Q179 Moderate Add #36 · Horizontal Strips and Sideways Parabolas
The area of the region enclosed by the parabola ( y − 2 ) 2 = x − 1 (y- 2)^{2}=x- 1 ( y − 2 ) 2 = x − 1 , the line x − 2 y + 4 = 0 x- 2y+ 4 = 0 x − 2 y + 4 = 0 and the positive coordinate axes is JEE Mains · 2024 · 27 Jan 2024 · Q84 Moderate Add #37 · Horizontal Strips and Sideways Parabolas
Let the area of the region { ( x , y ) : x − 2 y + 4 ≥ 0 \{(x,y):x- 2y+ 4 \geq 0 {( x , y ) : x − 2 y + 4 ≥ 0 , x + 2 y 2 ≥ 0 , x + 4 y 2 ≤ 8 , y ≥ 0 } \left. \ x+ 2y^{2}\geq 0,x+ 4y^{2}\leq 8,y\geq 0 \right\} x + 2 y 2 ≥ 0 , x + 4 y 2 ≤ 8 , y ≥ 0 } be m n \frac{m}{n} n m , where m m m and n n n are coprime numbers. Then m + n m + n m + n is equal to JEE Mains · 2021 · Paper 21 · Q76 Moderate Add #38 · Horizontal Strips and Sideways Parabolas
The area of the region bounded by the parabola ( y − 2 ) 2 = ( x − 1 ) (y- 2)^{2}= (x- 1) ( y − 2 ) 2 = ( x − 1 ) , the tangent to it at the point whose ordinate is 3 and the x x x -axis is: JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q74 Moderate Add #39 · Circles, Ellipses and Other Conics
If the area of the region bounded by 16 x 2 − 9 y 2 = 16x^{2} - 9y^{2} = 16 x 2 − 9 y 2 = 144 and 8 x − 3 y = 24 8x - 3y = 24 8 x − 3 y = 24 is A , then 3 ( A + 6 log e ( 3 ) ) 3\left( \text{ }A + 6\log_{e}(3) \right) 3 ( A + 6 log e ( 3 ) ) is equal to _ _ _ _ \_\_\_\_ ____ . #40 · Circles, Ellipses and Other Conics
The area of the region, inside the ellipse x 2 + 4 y 2 = 4 x^{2}+ 4y^{2}= 4 x 2 + 4 y 2 = 4 and outside the region bounded by the curves y = ∣ x ∣ − 1 y = |x| - 1 y = ∣ x ∣ − 1 and y = 1 − ∣ x ∣ y = 1 - |x| y = 1 − ∣ x ∣ , is : A B 2 π − 1 2 2\pi - \frac{1}{2} 2 π − 2 1 C D #41 · Circles, Ellipses and Other Conics
The area of the region enclosed between the circles x 2 + y 2 = 4 x^{2}+y^{2}= 4 x 2 + y 2 = 4 and x 2 + ( y − 2 ) 2 = 4 x^{2}+ (y - 2)^{2}= 4 x 2 + ( y − 2 ) 2 = 4 is : A 2 3 ( 2 π − 3 3 ) \frac{2}{3}(2\pi - 3\sqrt{3}) 3 2 ( 2 π − 3 3 ) B 4 3 ( 2 π − 3 3 ) \frac{4}{3}(2\pi - 3\sqrt{3}) 3 4 ( 2 π − 3 3 ) C 4 3 ( 2 π − 3 ) \frac{4}{3}(2\pi - \sqrt{3}) 3 4 ( 2 π − 3 ) D 2 3 ( 4 π − 3 3 ) \frac{2}{3}(4\pi - 3\sqrt{3}) 3 2 ( 4 π − 3 3 ) JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q62 Moderate Add #42 · Circles, Ellipses and Other Conics
The area of the region A = { ( x , y ) : 4 x 2 + y 2 ≤ 8 A =\left\{ (x,y):4x^{2}+y^{2}\leq 8 \right. A = { ( x , y ) : 4 x 2 + y 2 ≤ 8 and y 2 ≤ 4 x } \left. \ y^{2}\leq 4x \right\} y 2 ≤ 4 x } is : A π 2 + 2 \frac{\pi}{2} + 2 2 π + 2 B π + 2 3 \pi + \frac{2}{3} π + 3 2 C D π 2 + 1 3 \frac{\pi}{2} + \frac{1}{3} 2 π + 3 1 JEE Mains · 2023 · 11 April 2023 · Q73 Moderate Add #43 · Circles, Ellipses and Other Conics
Area of the region { ( x , y ) : x 2 + ( y − 2 ) 2 ≤ 4 \left\{ (x,y):x^{2}+ (y - 2)^{2}\leq 4 \right. { ( x , y ) : x 2 + ( y − 2 ) 2 ≤ 4 , x 2 ≥ 2 y } \left. \ x^{2}\geq 2y \right\} x 2 ≥ 2 y } is A 2 π − 16 3 2\pi - \frac{16}{3} 2 π − 3 16 B π − 8 3 \pi - \frac{8}{3} π − 3 8 C π + 8 3 \pi + \frac{8}{3} π + 3 8 D 2 π + 16 3 2\pi + \frac{16}{3} 2 π + 3 16 JEE Mains · 2024 · 8 April 2024 · Q161 Moderate Add #44 · Circles, Ellipses and Other Conics
The area of the region in the first quadrant inside the circle x 2 + y 2 = 8 x^{2}+y^{2}= 8 x 2 + y 2 = 8 and outside the parabola y 2 = 2 x y^{2}= 2x y 2 = 2 x is equal to : A π 2 − 1 3 \frac{\pi}{2} - \frac{1}{3} 2 π − 3 1 B π − 2 3 \pi - \frac{2}{3} π − 3 2 C π 2 − 2 3 \frac{\pi}{2} - \frac{2}{3} 2 π − 3 2 D π − 1 3 \pi - \frac{1}{3} π − 3 1 JEE Mains · 2024 · 9 April 2024 · Q62 Moderate Add #45 · Circles, Ellipses and Other Conics
The parabola y 2 = 4 x y^{2}= 4x y 2 = 4 x divides the area of the circle x 2 + y 2 = 5 x^{2}+y^{2}= 5 x 2 + y 2 = 5 in two parts. The area of the smaller part is equal to : A 2 3 + 5 sin − 1 ( 2 5 ) \frac{2}{3} + 5\sin^{- 1}\left( \frac{2}{\sqrt{5}} \right) 3 2 + 5 sin − 1 ( 5 2 ) B 1 3 + 5 sin − 1 ( 2 5 ) \frac{1}{3} + 5\sin^{- 1}\left( \frac{2}{\sqrt{5}} \right) 3 1 + 5 sin − 1 ( 5 2 ) C 1 3 + 5 sin − 1 ( 2 5 ) \frac{1}{3} + \sqrt{5}\sin^{- 1}\left( \frac{2}{\sqrt{5}} \right) 3 1 + 5 sin − 1 ( 5 2 ) D 2 3 + 5 sin − 1 ( 2 5 ) \frac{2}{3} + \sqrt{5}\sin^{- 1}\left( \frac{2}{\sqrt{5}} \right) 3 2 + 5 sin − 1 ( 5 2 ) JEE Mains · 2024 · 9 April 2024 · Q155 Moderate Add #46 · Circles, Ellipses and Other Conics
The area (in square units) of the region enclosed by the ellipse x 2 + 3 y 2 = 18 x^{2}+ 3y^{2}= 18 x 2 + 3 y 2 = 18 in the first quadrant below the line y = x y = x y = x is : A 3 π + 3 4 \sqrt{3}\pi + \frac{3}{4} 3 π + 4 3 B C 3 π − 3 4 \sqrt{3}\pi - \frac{3}{4} 3 π − 4 3 D JEE Mains · 2024 · 29 January 2024 · Q88 Moderate Add #47 · Circles, Ellipses and Other Conics
The area (in sq. units) of the part of circle x 2 + y 2 = 169 x^{2}+y^{2}= 169 x 2 + y 2 = 169 which is below the line 5 x − y = 13 5x-y= 13 5 x − y = 13 is π α 2 β − 65 2 + α β sin − 1 ( 12 13 ) \frac{\pi\alpha}{2\beta}-\frac{65}{2}+\frac{\alpha}{\beta}\sin^{- 1}\left( \frac{12}{13} \right) 2 β π α − 2 65 + β α sin − 1 ( 13 12 ) where α , β \alpha,\beta α , β are coprime numbers. Then α + β \alpha+\beta α + β is equal to JEE Mains · 2023 · 6 April 2023 · Q84 Moderate Add #48 · Circles, Ellipses and Other Conics
If the area of the region S = { ( x , y ) : 2 y − y 2 ≤ x 2 ≤ 2 y , x ≥ y } S =\left\{ (x,y):2y -y^{2}\leq x^{2}\leq 2y,x \geq y \right\} S = { ( x , y ) : 2 y − y 2 ≤ x 2 ≤ 2 y , x ≥ y } is equal to n + 2 n + 1 − π n − 1 \frac{n + 2}{n + 1}-\frac{\pi}{n - 1} n + 1 n + 2 − n − 1 π , then the natural number n n n is equal to JEE Mains · 2022 · 29 July 2022 · Q76 Moderate Add #49 · Circles, Ellipses and Other Conics
The area of the region { ( x , y ) : ∣ x − 1 ∣ ≤ y ≤ 5 − x 2 } \left\{ (x,y):|x - 1| \leq y \leq\sqrt{5 -x^{2}} \right\} { ( x , y ) : ∣ x − 1∣ ≤ y ≤ 5 − x 2 } is equal to : A 5 2 sin − 1 ( 3 5 ) − 1 2 \frac{5}{2}\sin^{- 1}\left( \frac{3}{5} \right) - \frac{1}{2} 2 5 sin − 1 ( 5 3 ) − 2 1 B 5 π 4 − 3 2 \frac{5\pi}{4} - \frac{3}{2} 4 5 π − 2 3 C 3 π 4 + 3 2 \frac{3\pi}{4} + \frac{3}{2} 4 3 π + 2 3 D 5 π 4 − 1 2 \frac{5\pi}{4} - \frac{1}{2} 4 5 π − 2 1 JEE Mains · 2022 · 27 July 2022 · Q69 Moderate Add #50 · Circles, Ellipses and Other Conics
The area of the smaller region enclosed by the curves y 2 = 8 x + 4 y^{2}= 8x + 4 y 2 = 8 x + 4 and x 2 + y 2 + 4 3 x − 4 = 0 x^{2}+y^{2}+ 4\sqrt{3}x - 4 = 0 x 2 + y 2 + 4 3 x − 4 = 0 is equal to A 1 3 ( 2 − 12 3 + 8 π ) \frac{1}{3}(2 - 12\sqrt{3} + 8\pi) 3 1 ( 2 − 12 3 + 8 π ) B 1 3 ( 2 − 12 3 + 6 π ) \frac{1}{3}(2 - 12\sqrt{3} + 6\pi) 3 1 ( 2 − 12 3 + 6 π ) C 1 3 ( 4 − 12 3 + 8 π ) \frac{1}{3}(4 - 12\sqrt{3} + 8\pi) 3 1 ( 4 − 12 3 + 8 π ) D 1 3 ( 4 − 12 3 + 6 π ) \frac{1}{3}(4 - 12\sqrt{3} + 6\pi) 3 1 ( 4 − 12 3 + 6 π ) Showing the first 50 of 117.
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