Practice Questions

Q83. sin5𝑥2 ∫ 𝑑𝑥, is equal to sin𝑥 2 (1) 𝑥+ 2sin𝑥+ sin2𝑥+ 𝑐(2) 2𝑥+ sin𝑥+ sin2𝑥+ 𝑐(3) 𝑥+ 2sin𝑥+ 2sin2𝑥+ 𝑐(4) 2𝑥+ sin𝑥+ 2sin2𝑥+ 𝑐 𝜋 Q84. 4 2 - 𝑥cos𝑥 If 𝑓𝑥= and 𝑔(𝑥) = log𝑒⁡𝑥, then the value of the integral ∫ 𝑔𝑓𝑥𝑑𝑥 is 2 + 𝑥cos𝑥 -𝜋 4 (1) log𝑒⁡𝑒 (2) log𝑒⁡2 (3) log𝑒⁡1 (4) log𝑒⁡3

Q83.Let, n ≥2 be a natural number and 0 < θ < 1 dθ, is equal to 2 . Then ∫(sinnθ−sinθ)sinn+1θn cosθ (1) n 1 n+1n (2) n 1 n+1n n2−1 (1 − sinn+1θ ) + c n2+1 (1 − sinn−1θ ) + c (3) n+1 n+1 + 1 ) n + c n2−1 n (1 − sinn−1θ1 ) n + c (4) n2−1n (1 sinn−1θ

Q83.The value of ∫ [x]+[sin x] + 4 , −π/2 (1) 20 3 (4π −3) (2) 103 (4π −3) (3) 12 1 (7π −5) (4) 121 (7π + 5) x 1 1 is

Q83.A value of α such that ∫ (x+α)(x+α+1) α (1) −12 (2) 21 (3) −2 (4) 2

Q83.The integral ∫ {( e )2x −( x )x}loge 1 JEE Main 2019 (12 Jan Shift 2) JEE Main Previous Year Paper (1) 3 2 −e − 2e21 (2) 12 −e − e21 (3) −12 + 1e − 2e21 (4) 32 −1e − 2e21

Q83.For, 𝑥2 ≠𝑛𝜋+ 1, 𝑛∈𝑁 (the set of natural numbers), the integral ∫𝑥√ 2sin𝑥2 - 1 - sin2𝑥2 - 1 is equal to 2sin𝑥2 - 1 + sin2𝑥2 - 1𝑑𝑥, (where 𝑐 is a constant of integration). 𝑥2 - 1 1 (1) (2) log𝑒 2sec2𝑥2 - 1 + 𝑐 logesec 4 + 𝑐 1 𝑥2 - 1 + 𝑐 (3) 2log𝑒sec𝑥2 - 1 + 𝑐 (4) log𝑒sec2 2

Q83.The integral ∫cos(lnx)dx, is equal to (1) x 2 (cos(lnx) −sin(ln x)) + C (2) x(cos(lnx) −sin(ln x)) + C (3) x(cos(lnx) + sin(ln x)) + C (4) x2 (cos(lnx) + sin(ln x)) + C

Q84.If ∫ f(t)dt = x2 + ∫ t2f(t)dt, then f ′( 2 ) 0 x JEE Main 2019 (10 Jan Shift 2) JEE Main Previous Year Paper (1) 18 (2) 24 25 25 (3) 4 (4) 6 5 25

Q84. lim + +. . . . . + n→∞( n4/3 n4/3 n4/3 ) is equal to (1) 3 4 (2)4/3 −34 (2) 34 (2)3/4 (3) 3 4 (2)4/3 (4) 34 (2)4/3 −43

Q84.If the area (in sq. units) bounded by the parabola y2 = 4λx and the line y = λx, λ > 0, is 91 , then λ is equal to (1) 4√3 (2) 2√6 (3) 48 (4) 24

Q84.Let I = ∫ba (x4 −2x2)dx. If I is minimum then the ordered pair (a, b) is (1) (0, √2) (2) (√2, −√2) √2, (3) (− 0) (4) (−√2, √2)

Q84.The integral ∫π sec 3𝑥· cosec 3𝑥𝑑𝑥 is equal to 6 7 5 (1) 3 6 - 3 6 (2) 3 43 - 3 13 5 2 (3) 3 6 - 3 3 (4) 3 53 - 3 13

Q84.If f(x) = ∫ (5x8+7x6) dx, (x ≥0), and f(0) = 0, then the value of f(1) is (x2+1+2x7)2 (1) −1 (2) 1 4 2 (3) 4 1 (4) −12 π/3 tan θ 1

Q84.If f : R →R is a differentiable function and f(2) = 6, then lim ∫f(x)6 (x−2)2tdt is: x→2 (1) 0 (2) 2f '(2) (3) 24f '(2) (4) 12f '(2) y2 is: y) :

Q84.The value of ∫ sinx+cosx 0 (1) π−1 (2) π−2 2 8 (3) π−1 (4) π−2 4 4

Q84.Let f and g be continuous functions on [0, a] such that f(x) = f(a −x) and g(x) + g(a −x) = 4, then ∫a0 f(x)g(x)dx is equal to (1) ∫a0 f(x)dx (2) −3 ∫a0 f(x)dx (3) 4 ∫a0 f(x)dx (4) 2 ∫a0 f(x)dx

Q84. n→∞(lim n2+12n + n2+22n + n2+32n +. . … . + 5n21 ) is equal to (1) π (2) tan−1(2) 4 (3) π (4) tan−1 (3) 2 ,

Q84.The area (in sq. units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2,5) and the coordinate axes is : (1) 8 (2) 37 3 24 (3) 187 (4) 14 24 3

Q84.The value of the integral ∫2−2 [ sin2 π ]+ 2 (1) 0 (2) sin 4 (3) 4 (4) 4 −sin 4

Q85.If a curve passes through the point (1, −2) and has slope of the tangent at any point (x, y) on it as x2−2yx then the curve also passes through the point (1) (√3, 0) (2) (−1, 2) (3) (−√2, 1) (4) (3, 0) → →

Q85.If the area (in sq. units) of the region 𝑥, 𝑦: 𝑦2 ≤4𝑥, 𝑥+ 𝑦≤1, 𝑥≥0, 𝑦≥0 is 𝑎√2 + 𝑏, then 𝑎- 𝑏 is equal to 10 (1) 6 (2) 3 (3) -2 (4) 8 3 3 1

Q85.The area (in sq. units) bounded by the parabola 𝑦= 𝑥2 - 1, the tangent at the point 2, 3 to it and the 𝑦-axis is 14 8 (1) (2) 3 3 32 56 (3) (4) 3 3

Q85.The area (in sq. units) of the region 𝐴= 𝑥, 𝑦∈𝑅× 𝑅0 ≤𝑥≤3, 0 ≤𝑦≤4, 𝑦≤𝑥2 + 3𝑥 is (1) 26 (2) 8 (3) 53 (4) 59 3 6 6

Q85.The area (in sq. units) of the region A = {(x, 2 ≤x ≤y + 4} (1) 30 (2) 18 (3) 53 (4) 16 3

Q85.The general solution of the differential equation (y2 −x3)dx −xydy = 0, (x ≠0) is (where c is a constant of integration) (1) y2 + 2x2 + cx3 = 0 (2) y2 −2x2 + cx3 = 0 (3) y2 −2x3 + cx2 = 0 (4) y2 + 2x3 + cx2 = 0 →