Practice Questions

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

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

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. 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

Q85.The area (in sq. units) of the region A = {(x, y) : x2 ≤y ≤x + 2} is (1) 136 (2) 316 (3) 9 (4) 10 2 3 dy

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 solution of the differential equation, dy dx = (x −y)2 , when y(1) = 1, is: (1) loge 2−x2−y = x −y (2) −loge 1+x−y1−x+y = 2(x −1) (3) −loge 1−x+y1+x−y = x + y −2 (4) loge 2−x2−y = 2(y −1)

Q85.The area (in sq. units) of the region bounded by the curves 𝑦= 2𝑥 and 𝑦= 𝑥+ 1, in the first quadrant is 3 1 1 (1) - (2) 2 log𝑒⁡2 2 3 3 (3) log𝑒⁡2 + 2 (4) 2

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, 2 ≤x ≤y + 4} (1) 30 (2) 18 (3) 53 (4) 16 3

Q85.Let 𝑓𝑥= ∫ 𝑔𝑡𝑑𝑡, where 𝑔 is a non-zero even function. If 𝑓𝑥+ 5 = 𝑔𝑥, then ∫ 𝑓( 𝑡) 𝑑𝑡 equals 0 0 𝑥+ 5 5 (1) (2) ∫ 𝑔( 𝑡) 𝑑𝑡 ∫ 𝑔( 𝑡) 𝑑𝑡 5 𝑥+ 5 5 𝑥+ 5 (3) (4) 5 ∫ 𝑔( 𝑡) 𝑑𝑡 2 ∫ 𝑔( 𝑡) 𝑑𝑡 𝑥+ 5 5

Q85.A curve amongst the family of curves represented by the differential equation, (x2 −y2) dx + 2xy dy = 0 which passes through (1, 1), is (1) A circle with centre on the x− axis. (2) A circle with centre on the y− axis. (3) A hyperbola with transverse axis along the x− (4) An ellipse with major axis along the y− axis. axis. x f( x1 )

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 bounded by the parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is (1) 17 (2) 21 4 2 (3) 15 (4) 15 2 4

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 →

Q85.The region represented by |x −y| ≤2 and |x + y| ≤2 is bounded by a (1) rhombus of area 8√2 sq. units. (2) rhombus of side length 2 units. (3) square of area 16 sq. units. (4) square of side length 2√2 units. x ∈(−π2 , π2 ) , such that

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 ∫ dθ = 1 − , > , then the value of k is √2k sec θ √2 0 (k 0) (1) 21 (2) 1 (3) 2 (4) 4 JEE Main 2019 (09 Jan Shift 2) JEE Main Previous Year Paper