Practice Questions

Q81.Let k and K be the minimum and the maximum values of the function f(x) = (1+x)0.6 in [0, 1], respectively, 1+x0.6 then the ordered pair (k, K) is equal to: (1) (2−0.4, 1) (2) (2−0.6, 1) (3) (2−0.4, 20.6) (4) (1,20.6) JEE Main 2015 (11 Apr Online) JEE Main Previous Year Paper 1

Q81.The normal to the curve x2 + 2xy −3y2 = 0 , at (1, 1) (1) Meets the curve again in the fourth quadrant (2) Does not meet the curve again (3) Meets the curve again in the second quadrant (4) Meets the curve again in the third quadrant

Q82.If ∫ log(t+√1+t2) dt = 2 (g(t))2 + c, where c is a constant, then g(2), is equal to √1+t2 (1) 2 + + √5) (2) log(2 √5) 1 log(2 √5 + log + (3) log(2 √5) (4) 12 (2 √5)

Q82.The distance from the origin, of the normal to the curve, x = 2 cos t + 2t sin t, y = 2 sin t −2t cos t at t = π4 , is : (1) √2 (2) 2√2 (3) 4 (4) 2

Q82.Let f(x) be a polynomial of degree four and having its extreme values at x = 1 and x = 2. If f(x) lim + = 3, then f(2) is equal to [1 x2 ] x→0 (1) 4 (2) −8 (3) −4 (4) 0 JEE Main 2015 (04 Apr) JEE Main Previous Year Paper

Q83.The integral ∫ dx 3 equals to x2(x4+1) 4 4 (1) x4+1 1 1 4 (2) x4+1 + c + c −( x4 ) ( x4 ) (3) 14 (4) 41 (x4 + 1) + c −(x4 + 1) + c logx2 dx is equal to

Q83.The integral ∫ 3dx 5 , is equal to (x+1) 4 (x−2) 4 (1) 1 1 4 + c 4( x+1x−2 ) 4 + c (2) −43 ( x+1x−2 ) (3) 1 1 4 + c 4( x−2x+1 ) 4 + c (4) −43 ( x−2x+1 )

Q83.Let f : R →R be a function such that f(2 −x) = f(2 + x) and f(4 −x) = f(4 + x), for all x ∈R and 2 50 ∫ f(x)dx = 5. Then the value of ∫ f(x)dx is 0 10 (1) 100 (2) 125 (3) 80 (4) 200

Q84.For x > 0, let f(x) = ∫x1 log1+tt dt. Then f(x) + f( x1 ) is equal to (1) 1 (log x)2 (2) log x 2 (3) 1 4 log x2 (4) 14 (log x)2

Q84.Let f : (−1, 1) →R be a continuous function. If ∫sin0 x f(t) dt = √32 x, then f( √32 ) is equal to: (1) √3 (2) √3 2 (3) 1 (4) 2 √32

Q84.The integral ∫4 logx2+log(6−x)2 2 (1) 6 (2) 2 (3) 4 (4) 1

Q85.The area (in sq. units) of the region described by [(x, y) : y2 ≤2x and y ≥4x −1] is (1) 32 9 sq. units (2) 327 sq. units (3) 64 5 sq. units (4) 6415 sq. units

Q85.The solution of the differential equation ydx −(x + 2y2)dy = 0 is x = f(y). If f(−1) = 1, then f(1) is equal to (1) 2 (2) 3 (3) 4 (4) 1 −−−−−

Q85.The area (in square units) of the region bounded by the curves y + 2x2 = 0 and y + 3x2 = 1 , is equal to (1) 4 3 sq. units (2) 13 sq. units (3) 5 3 sq. units (4) 43 sq. units

Q86.Let y (x) be the solution of the differential equation (x log x) dxdy + y = 2x log x, (x ≥1). Then y (e) is equal to (1) 2e (2) e (3) 0 (4) 2 →

Q86.If y (x) is the solution of the differential equation (x + 2) dxdy = x2 + 4x −9, x ≠ −2 and y(0) = 0, then y(−4) is equal to (1) −1 (2) 1 (3) 0 (4) 2 × , then 2→c is equal to:

Q86.In a parallelogram ABCD, AB→ = a, AD→ = b & AC→ = c. DB→ ⋅AB→ has the value: (1) 1 2 (a2 + b2 + c2) (2) 14 (a2 + b2 −c2) (3) 3 1 (b2 + c2 −a2) (4) 12 (a2 −b2 + c2)

Q87.Let →a, b and →c be three non - zero vectors such that no two of them are collinear and × →c→a. If θ is the angle between vectors b and →c, then a value of sin θ is = 13 b (→a → → → b) ×→c (1) −2√3 (2) 2√2 3 3 (3) −√2 (4) 2 3 3

Q87.Let →aand→b be two unit vectors such that →a+→b = √3. If →c=→a+ 2→b + (→a →b) (1) √51 (2) √37 (3) √43 (4) √55

Q87.A plane containing the point (3, 2, 0) and the line x−11 = y−25 = z−34 also contains the point (1) (0, 7, −10) (2) (0, 7, 10) (3) (0, 3, 1) (4) (0, −3, 1)

Q88.The distance of the point (1, 0, 2) from the point of intersection of the line x−23 = y+14 = z−212 and the plane x −y + z =16, is (1) 13 (2) 2√14 (3) 8 (4) 3√21

Q88.If the points (1, 1, λ) & (−3, 0, 1), are equidistant from the plane, 3x + 4y −12z + 13 = 0, then λ satisfies the equation: (1) 3x2 + 10x + 7 = 0 (2) 3x2 + 10x −13 = 0 (3) 3x2 −10x + 7 = 0 (4) 3x2 −10x + 21 = 0 JEE Main 2015 (10 Apr Online) JEE Main Previous Year Paper

Q88.The shortest distance between the z - axis and the line x + y + 2z −3 = 0 = 2x + 3y + 4z −4, is (1) 1 (2) 2 (3) 3 (4) 4

Q89.If the shortest distance between the line x−1α = y+1−1 = 1z , (α ≠−1) , and x + y + z + 1 = 0 = 2x −y + z + 3 is 1 ,then value of α is : √3 (1) −1916 (2) 3219 (3) −1619 (4) 1932

Q89.If the mean and the variance of a binomial variate X are 2 & 1 respectively, then the probability that X takes a value greater than or equal to one is: (1) 1 (2) 9 16 16 (3) 3 (4) 15 4 16