Practice Questions

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.The value of the integral ∫10 xcot−1(1 −x2 + x4)dx is (1) π 4 −12 loge2 (2) π4 −loge2 (3) π 2 −loge2 (4) π2 −12 loge2

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 integral ∫π/4π/6 sin 2x(tan5dxx+cot5 x) equals: (1) 20 1 tan−1 ( 9√31 ) (2) 101 ( π4 −tan−1 ( 9√31 )) (3) π (4) 1 40 5 ( π4 −tan−1 ( 3√31 ))

Q83.Given that the slope of the tangent to a curve 𝑦= 𝑦( 𝑥) at any point 𝑥, 𝑦 is 2𝑦𝑥2. If the curve passes through the centre of the circle 𝑥2 + 𝑦2 - 2𝑥- 2𝑦= 0, then its equation is (1) 𝑥2log𝑒⁡|𝑦| = - 2(𝑥- 1) (2) 𝑥log𝑒⁡|𝑦| = 2(𝑥- 1) (3) 𝑥log𝑒⁡|𝑦| = - 2(𝑥- 1) (4) 𝑥log𝑒⁡|𝑦| = 𝑥- 1 1

Q83.The value of ∫2π [sin 2x(1 + cos 3x)]dx , where [t] denotes the greatest integer function is 0 (1) π (2) 2π (3) −π (4) −2π (n+1)1/3 (n+2)1/3 (2n)1/3

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

Q83.If ∫𝑥5𝑒-𝑥2𝑑𝑥= 𝑔𝑥𝑒-𝑥2 + 𝑐, where 𝑐 is a constant of integration, then 𝑔-1 is equal to 5 (1) - (2) -1 2 (3) 1 (4) -1 2 π 2 4 3

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.If x = 3 tant and y = 3 sect, then the value of dx2d2y π at t = 4 , is: (1) 1 (2) 1 6 6√2 (3) 1 (4) 3 3√2 2√2

Q83.If ∫ √1−x2x4 dx = A(x)(√1 −x2) m constant of integration, then (A(x))m equals : (1) −1 (2) −1 27x9 3x3 (3) 1 (4) 1 27x6 9x4 x dx (where [x] denotes the greatest integer less than or equal to x) is x 1

Q83.A value of α such that ∫ (x+α)(x+α+1) α (1) −12 (2) 21 (3) −2 (4) 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.The value of 𝜋cos𝑥3𝑑𝑥 is ∫0 2 (1) (2) 0 3 (3) 4 (4) -4 3 3

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

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

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. 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.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.If ∫ 𝑑𝑥 2 = 𝑥𝑓𝑥1 + 𝑥6 3 + 𝐶, where 𝐶 is a constant of integration, then the function 𝑓𝑥 is equal to 𝑥31 + 𝑥6 3 (1) 3 (2) - 1 𝑥2 2𝑥3 1 1 (3) - (4) - 6𝑥3 2𝑥2 𝑥 𝑥

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