Practice Questions

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 S be the set of all values of x for which the tangent to the curve y = f(x) = x3 −x2 −2x at (x, y) is parallel to the line segment joining the points (1, f(1)) and (−1, f(−1)), then S is equal to (1) {−13 , −1} (2) {−13 , 1} (3) { 31 , 1} (4) { 13 , −1} 3 xdx is equal to Q83. ∫sec2x ⋅cot 4 3 x + C (1) 3tan−13 x + C (2) −34 tan−4 (3) −3tan−13 x + C (4) −3cot−13 x + C π/2 sin3x dx is:

Q82.If ∫x5e−4x3dx = 481 e−4x3f(x) + C , where C is a constant of integration, then f(x) is equal to (1) −4x3 −1 (2) −2x3 + 1 (3) −2x3 −1 (4) 4x3 + 1 π/2 dx where [t] denotes the greatest integer less than or equal to t, is

Q82.If ∫esecx(secx tan xf(x) + (secx tan x + sec2x))dx = esecxf(x) + C, then a possible choice of f(x) is: (1) secx −tanx −12 (2) secx + tanx + 12 (3) xsecx + tanx + 12 (4) secx + xtanx −12

Q82.The maximum area (in sq. units) of a rectangle having its base on the x− axis and its other two vertices on the parabola, y = 12 −x2 such that the rectangle lies inside the parabola, is : (1) 20√2 (2) 32 (3) 36 (4) 18√3

Q82.If 𝜃 denotes the acute angle between the curves, 𝑦= 10 - 𝑥2 and 𝑦= 2 + 𝑥2 at a point of their intersection, then tan⁡𝜃 is equal to: (1) 4 (2) 8 9 17 7 8 (3) (4) 17 15

Q82.If ∫ x+1 dx = f(x)√2x −1 + C, where C is a constant of integration, then f(x) is equal to: √2x−1 (1) 3 1 (x + 1) (2) 32 (x + 2) (3) 3 2 (x −4) (4) 31 (x + 4)

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

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. 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 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