Practice Questions

Q82.Themaximum value of the finction f(x) = 3x3 −18x2 + 27x −40 on the set S = {x ∈R : x2 + 30 ≤11x} is : (1) -122 (2) -222 (3) 122 (4) 222 JEE Main 2019 (11 Jan Shift 1) JEE Main Previous Year Paper + C, for a suitable chosen integer m and a function A(x), where C is a

Q82.A spherical iron ball of radius 10 𝑐𝑚 is coated with a layer of ice of uniform thickness that melts at a rate of 50 𝑐𝑚3 / 𝑚𝑖𝑛. When the thickness of the ice is 5 𝑐𝑚, then the rate at which the thickness ( in 𝑐𝑚/ 𝑚𝑖𝑛) of the ice decreases, is : 1 1 (1) (2) 9π 36π (3) 1 (4) 5 18π 6π

Q82.Let 𝑓: 0, 2 →𝑅 be a twice differentiable function such that 𝑓''𝑥> 0, for all 𝑥∈0, 2 . If 𝜙𝑥= 𝑓𝑥+ 𝑓2 – 𝑥, then 𝜙 is (1) decreasing on 0,2 (2) increasing on 0,2 (3) increasing on ( 0,1 ) (4) decreasing on 0,1 and and decreasing on 1,2 increasing on ( 1,2 )

Q82.Let α ∈(0, π2 ) , be constant.If the integral ∫ tanx−tantanx+tanαα dx = A(x)cos2α + B(x)sin2α + C , where C is a constant of integration, then the functions A(x) and B(x) are respectively (1) x −α and loge|sin(x −α)| (2) x + α and loge|cos(x −α)| (3) x + α and loge|sin(x + α)| (4) x −α and loge|cos(x −α)| JEE Main 2019 (12 Apr Shift 2) JEE Main Previous Year Paper α+1 dx 9 = loge( 8 ) is

Q82.The integral ∫2𝑥3 - 1 is equal to 𝑥4 + 𝑥𝑑𝑥, (1) 2 (2) |𝑥3 + 1| 1 (𝑥3 + 1) + 𝐶 + 𝐶 log𝑒 𝑥2 2log𝑒 |𝑥3| (3) 𝑥3 + 1 (4) 1 |𝑥3 + 1| log𝑒 𝑥 + 𝐶 2log𝑒 𝑥2 + 𝐶

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. 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.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.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.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.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.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.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 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.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.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.Let 𝑓: 𝑅→𝑅 be a continuous and differentiable function such that 𝑓2 = 6 and 𝑓'2 = 48.1 If 𝑓( 𝑥) ∫6 4𝑡3𝑑𝑡= 𝑥- 2𝑔𝑥, then 𝑥→2𝑔𝑥lim is equal to (1) 24 (2) 18 (3) 12 (4) 36 π Q84. 2 cot𝑥 If ∫ 𝑚(π + 𝑛), then 𝑚𝑛 is equal to cot𝑥+ cosec𝑥𝑑𝑥= 0 (1) 1 (2) 1 2 1 (3) -1 (4) - 2

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

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(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 𝜋cos𝑥3𝑑𝑥 is ∫0 2 (1) (2) 0 3 (3) 4 (4) -4 3 3

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