Practice Questions

Q74.Let S be the set of all the natural numbers, for which the line xa + yb = 2 is a tangent to the curve ( xa ) n + ( yb ) n = 2 at the point (a, b), ab ≠0. Then (1) S = ϕ (2) n(S) = 1 (3) S = {2k : k ∈N} (4) S = N

Q74. I = ∫ π 3 ( 8 sin x−sinx 2x )dx. Then 4 (1) π 2 < I < 3π4 (2) π5 < I < 5π12 (3) 5π 12 < I < √23 π (4) 3π4 < I < π

Q74.The area of the region given by 𝐴= 𝑥, 𝑦: 𝑥2 ≤𝑦≤min𝑥+ 2, 4 - 3𝑥 is (1) 31 (2) 17 8 6 19 27 (3) (4) 6 8 JEE Main 2022 (25 Jul Shift 1) JEE Main Previous Year Paper

Q74.The area of the region S = {(x, y) : y2 ≤8x, y ≥√2x, x ≥1} is (1) 5√2 (2) 19√2 6 6 (3) 13√2 (4) 11√2 6 6 pass + e x = x + + e x y ]x dxdy y ]y

Q74.If ∫1x √1−x1+x + π3 (1) loge( √3+1√3−1 ) + π3 (2) loge( √3+1√3−1 ) (3) loge( √3−1√3+1 ) −π3 (4) 13 loge( √3−1√3+1 ) −π6

Q74.Let f be a differentiable function in (0, π2 ). If ∫1cos x t2f(t)dt = sin3 x + cos x, then √31 f ′( √31 ) (1) 6 −9√2 (2) 6 + 9 √2 (3) 6 − 9 (4) 3 + √2 √2 dx, where [⋅] denotes the greatest integer function, is equal to

Q74.If f(α) = ∫α1 log101+t t dt, (1) 9 (2) 92 (3) 9 (4) 9 loge(10) 2 loge(10) is equal to

Q74.The value of the integral ∫ −π2 2 (1+ex)(sin6dxx+cos6 x) is equal to (1) 2π (2) 0 (3) π (4) π 2

Q74.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 (1) x2 + 81y2 = 2 (2) y29 −x2 = 8 (3) y = 4x2 + 5 (4) x3 −y2 = 2

Q74. lim 2n1 1 + 1 + 1 + … . + 1 is equal to n→∞ ( √1−12n √1−22n √1−32n √1−2n−12n ) (1) 1 (2) 1 2 (3) 2 (4) −2

Q74.If the line 𝑦= 4 + 𝑘𝑥, 𝑘> 0, is the tangent to the parabola 𝑦= 𝑥- 𝑥2 at the point 𝑃 and 𝑉 is the vertex of the parabola, then the slope of the line through 𝑃 and 𝑉 is (1) 3 (2) 26 2 9 5 23 (3) (4) 2 6

Q75.The area of the region bounded by y2 = 8x and y2 = 16(3 −x) is equal to (1) 32 (2) 40 3 3 (3) 16 (4) 9

Q75.If the angle made by the tangent at the point 𝑥0, 𝑦0 on the curve 𝑥= 12𝑡+ sin𝑡cos𝑡, 𝜋 𝜋 𝑦= 121 + sin𝑡2, 0 < 𝑡< 2, with the positive 𝑥-axis is 3, then 𝑦0 is equal to (1) 63 + 2√2 (2) 37 + 4√3 (3) 27 (4) 48 𝜋 𝑛∈ℕ, then

Q75.The area of the region enclosed by y ≤4x2, x2 ≤9y and y ≤4 , is equal to (1) 40 (2) 56 3 3 (3) 112 (4) 80 3 3

Q75.A wire of length 22m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is (1) 22 (2) 66 9+4√3 9+4√3 (3) 22 (4) 66 4+9√3 4+9√3 t, is equal toQ76. ∫50 cos(π(x −[ x2 ]))dx, where [t] denotes greatest integer less than or equal to (1) 0 (2) 2 (3) −3 (4) 4 JEE Main 2022 (29 Jun Shift 1) JEE Main Previous Year Paper

Q75.The sum of absolute maximum and absolute minimum values of the function f(x) = 2x2 + 3x −2 + sin x cos x in the interval [0, 1] is 1 sin(1) cos2( (1) 2 ) (2) 3 + 12 (1 + 2 cos(1)) sin(1) 3 + 2 (3) 5 + 12 (sin(1) + sin(2)) (4) 2 + sin( 21 ) cos( 12 )

Q75.The area of the bounded region enclosed by the curve y = 3 −x −12 −|x + 1| and the x-axis is (1) 9 (2) 45 4 16 (3) 278 (4) 1663 x x −4xe y2 = 0 such that x(1) = 0.

Q75.Let the solution curve y = y(x) of the differential equation, [ √x2−y2x [ √x2−y2x through the points (1, 0) and (2α, α), α > 0 . Then α is equal to (1) 2 1 exp( π6 + √e −1) (2) 12 exp( π3 + √e −1) (3) exp( π6 + √e + 1) (4) 2 exp( π3 + √e −1)

Q75.If the solution curve of the differential equation 𝑑𝑦 𝑥+ 𝑦- 2 passes through the point 2, 1 and 𝑘+ 1, 2, k > 0, 𝑑𝑥= 𝑥- 𝑦 then (1) 2tan-11 + 1 𝑘= loge𝑘2 + 1 (2) tan-11𝑘= loge𝑘2 1 𝑘2 + 1 (3) 2tan-1 = loge𝑘2 + 2𝑘+ 2 (4) 2tan-11 𝑘+ 1 𝑘= loge 𝑘2

Q75.If ∫20 (√2x −√2x −x2)dx + I , then I equal to + ∫21 (2 −y22 )dy ∫10 (1 −√1 −y2 −y22 )dy −y2 + + √1 −y2)dy (2) ∫10 ( y22 −√1 1)dy (1) ∫10 (1 + √1 −y2 + 1)dy (3) ∫10 (1 −√1 −y2)dy (4) ∫10 ( y22

Q75. n→∞(lim (n2+1)(n+1)n2 + (n2+4)(n+2)n2 + (n2+9)(n+3)n2 + … + (n2+n2)(n+n)n2 ) is equal to (1) π 8 + 14 ln 2 (2) π4 + 18 ln 2 (3) π 4 −18 ln 2 (4) π8 + ln √2

Q75.The value of ∫0 1 + cos2𝑥ecos𝑥+ e-cos𝑥d𝑥 is equal to (1) 𝜋2 (2) 𝜋 4 4 (3) 𝜋 (4) 𝜋2 6 2

Q75.Let y = y(x) be the solution curve of the differential equation dx 1 1 y = ( x−1x+1 ) 2 , x > 1 passing through x2−1 the point . Then √7y(8) is equal to 3 (2, √1 ) (1) 11 + 6 loge 3 (2) 19 (3) 12 −2 loge 3 (4) 19 −6 loge 3

Q75.The slope of the tangent to a curve 𝐶: 𝑦= 𝑦𝑥 at any point [𝑥, 𝑦) on it is 2e2x - 6e-x + 9 . If 𝐶 passes through the 2 + 9e-2x 1 𝜋 1 points 0, + and 𝛼, then 𝑒𝛼 is equal to 2 2√2 2e2𝛼 (1) 3 + √2 (2) 3 3 + √2 3 - √2 √2 3 - √2 (3) 1 √2 + 1 (4) √2 + 1 √2 √2 - 1 √2 - 1

Q75.Let the solution curve of the differential equation 𝑥𝑑𝑦= √𝑥2 + 𝑦2 + 𝑦𝑑𝑥, 𝑥> 0, intersect the line x = 1 at 𝑦= 0 and the line 𝑥= 2 at 𝑦= 𝛼. Then the value of 𝛼 is (1) 1 (2) 3 2 2 3 5 (3) - (4) 2 2