Practice Questions

Q73.Let f : R →R be a differentiable function such that f( π4 ) = √2, f( π2 ) = 0 and f ′( π2 ) = 1 and let π lim g(x) = ∫ x4 (f ′(t) sec t + tan t sec tf(t))dt for x ∈[ π4 , π2 ). Then π x→( 2 )−g(x) is equal to (1) 2 (2) 3 (3) 4 (4) −3

Q73.Considering only the principal values of the inverse trigonometric functions, the domain of the function 𝑥2 - 4𝑥+ 2 𝑓𝑥= cos-1 is 𝑥2 + 3 1 1 (1) - ∞, (2) - ∞ 4 4, (3) -1 ∞ (4) - ∞, 1 3, 3

Q73.If m and n respectively are the number of local maximum and local minimum points of the function dt, then the ordered pair (m, n) is equal to f(x) = ∫x20 t2−5t+42+et (1) (2, 3) (2) (3, 2) (3) (2, 2) (4) (3, 4) is equal to

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.If 𝑡 denotes the greatest integer ≤t, then the value of ∫0 2𝑥- 3𝑥2 - 5𝑥+ 2 + 1𝑑𝑥 is JEE Main 2022 (29 Jul Shift 2) JEE Main Previous Year Paper (1) √37 + √13 - 4 (2) √37 - √13 - 4 6 6 (3) -√37 - √13 + 4 (4) -√37 + √13 + 4 6 6

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.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.The minimum value of the twice differentiable function 𝑓𝑥= 𝑥𝑒𝑥- 𝑡𝑓'𝑡𝑑𝑡- 𝑥2 - 𝑥+ 1𝑒𝑥, 𝑥∈𝑅, is ∫0 2 (1) - (2) -2√𝑒 √𝑒 2 (3) -√𝑒 (4) √𝑒

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

Q74.The area enclosed by the curves y = loge(x + e2), x = loge( 2y ) and (1) 2 + e −loge 2 (2) 1 + e −loge 2 (3) e −loge 2 (4) 1 + loge 2 dy +

Q74.Let f : R →R be continuous function satisfying f(x) + f(x + k) = n, for all x ∈R where k > 0 and n is a positive integer. If I1 = ∫4nk0 f(x)dx and I2 = ∫3k−k f(x)dx, then (1) I1 + 2I2 = 4nk (2) I1 + 2I2 = 2nk (3) I1 + nI2 = 4n2 K (4) I1 + nI2 = 6n2k

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 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. 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.Let 𝑔: 0, ∞→𝑅 be a differentiable function such that ∫ + d𝑥= + 𝐶, for all 𝑥> 0 e𝑥+ 1 e𝑥+ 12 e𝑥+ 1 , where 𝐶 is an arbitrary constant. Then 𝜋 𝜋 (1) 𝑔 is decreasing in 0, (2) 𝑔- 𝑔' is increasing in 0, 4 2 (3) 𝑔' is increasing in 0, 𝜋 (4) 𝑔+ 𝑔' is increasing in 0, 𝜋 4 2 𝜋 ecos𝑥sin𝑥

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 ∫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.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.Let f be a real valued continuous function on [0, 1] and f(x) = x + ∫10 (x −t)f(t)dt. Then which of the following points (x, y) lies on the curve y = f(x)? (1) (2, 4) (2) (1, 2) (3) (4, 17) (4) (6, 8) JEE Main 2022 (29 Jun Shift 2) JEE Main Previous Year Paper =

Q74.The value of the integral ∫2−2 (ex|x|+1)x3+x (1) 5e2 (2) 3e−2 (3) 4 (4) 6 dy ax−by+a

Q74.If a = n→∞∑n (1) 2√2f( a2 ) = f ′( a2 ) (2) f( a2 )f ′( a2 ) = √2 (3) √2f( a2 ) = f ′( a2 ) (4) f( a2 ) = √2f ′( a2 )

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

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.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 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