Practice Questions

Q81.If 𝑚 is the minimum value of 𝑘 for which the function 𝑓𝑥= 𝑥√𝑘𝑥- 𝑥2 is increasing in the interval [0, 3] and 𝑀 is the maximum value of 𝑓 in [0, 3] when 𝑘= 𝑚, then the ordered pair ( 𝑚, 𝑀) is equal to: (1) 4, 3√3 (2) 5, 3√6 (3) 3, 3√3 (4) 4, 3√2

Q81.If the tangent to the curve 𝑦= 𝑥2 - 3, 𝑥∈𝑅, 𝑥≠± √3, at a point 𝛼, 𝛽≠0, 0 on it is parallel to the line 2𝑥+ 6𝑦- 11 = 0, then: (1) 2𝛼+ 6𝛽= 19 (2) 2𝛼+ 6𝛽= 11 (3) 6𝛼+ 2𝛽= 19 (4) 6𝛼+ 2𝛽= 9

Q81.The tangent to the curve, y = xex2 passing through the point (1, e) also passes through the point: (1) ( 34 , 2e) (2) (2, 3e) (3) ( 53 , 2e) (4) (3, 6e)

Q81.If 𝑆1 and 𝑆2 are respectively the sets of local minimum and local maximum points of the function, 𝑓𝑥= 9𝑥4 + 12𝑥3 - 36𝑥2 + 25, 𝑥∈𝑅, then (1) 𝑆1 = -2; 𝑆2 = {0,1} (2) 𝑆1 = -1; 𝑆2 = 0,2 (3) 𝑆1 = -2,0; 𝑆2 = {1} (4) 𝑆1 = -2,1; 𝑆2 = {0}

Q81.If the function f given by f(x) = x3 −3(a −2)x2 + 3ax + 7, for some a ∈R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, f(x)−14 = 0, (x ≠1) is : (x−1)2 (1) 7 (2) −7 (3) 6 (4) 5

Q81.If the tangent to the curve, y = x3 + ax–b at the point (1, –5) is perpendicular to the line, –x + y + 4 = 0, then which one of the following points lies on the curve? (1) (2, –2) (2) (2, –1) (3) (–2, 1) (4) (–2, 2) JEE Main 2019 (09 Apr Shift 1) JEE Main Previous Year Paper

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.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.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.The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is: 2 (1) √3 (2) 3√3 (3) √6 (4) 2 √3

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.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.If ∫ dx = A(tan−1( x−13 ) + x2−2x+10f(x) ) (x2−2x+10)2 (1) A = 271 and f(x) = 9(x −1) (2) A = 811 and f(x) = 3(x −1) (3) A = 541 and f(x) = 9(x −1)2 (4) A = 541 and f(x) = 3(x −1) JEE Main 2019 (10 Apr Shift 1) JEE Main Previous Year Paper

Q82.Let f be a differentiable function from R to R such that |f(x) −f(y)| ≤2|x −y|3/2, for all x, y ∈R. If 1 f(0) = 1 then ∫ f 2(x)dx is equal to 0 (1) 0 (2) 1 (3) 2 (4) 21

Q82.The integral ∫ 3x13+2x11 dx, is equal to (2x4+3x2+1)4 (1) x4 + C (2) x4 + C 6(2x4+3x2+1)3 (2x4+3x2+1)3 (3) x12 + C (4) x12 + C (2x4+3x2+1)3 6(2x4+3x2+1)3 e x e x dx is equal to

Q82.The shortest distance between the point ( 23 , 0) and the curve y = √x, (x > 0) , is (1) √3 (2) 5 2 4 (3) 3 (4) √5 2 2 π

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.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.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.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)

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.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.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.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