Practice Questions

Q78.If A = [ 5a3 −b2 ] and A. adjA = A AT , then 5a + b is equal to (1) 4 (2) 13 (3) −1 (4) 5

Q78.Let A, be a 3 × 3 matrix, such that A2 −5A + 7I = O. Statement - I : A−1 = 71 (5I −A). Statement - II : The polynomial A3 −2A2 −3A + I ,can be reduced to 5(A −4I). Then : (1) Both the statements are true (2) Both the statements are false (3) Statement - I is true, but Statement - II is false (4) Statement - I is false, but Statement - II is true , then the determinant of the matrix (A2016 −2A2015 −A2014) is :

Q79.If A = [ −43 −11 ] JEE Main 2016 (10 Apr Online) JEE Main Previous Year Paper (1) −175 (2) 2014 (3) 2016 (4) −25

Q79. cos x sin x sin x The number of distinct real roots of the equation, sin x cos x sin x = 0 in the interval [−π4 , π4 ] is : sin x sin x cos x (1) 1 (2) 4 (3) 2 (4) 3

Q79.The system of linear equations x + λy −z = 0 λx −y −z = 0 x + y −λz = 0 has a non -trivial solution for (1) Exactly two values of λ (2) Exactly three values of λ (3) Infinitely many values of λ (4) Exactly one value of λ

Q80.For x ∈R, x ≠0, x ≠1, let f0(x) = 1−x1 and fn+1(x) = f0(fn(x)), n = 0, 1, 2, … . . Then the value of f100(3) + f1( 32 ) + f2( 32 ) is equal to : (1) 8 (2) 4 3 3 (3) 5 (4) 1 3 3 is differentiable at x = 1 , then ab is equal to

Q80.Let a, b ∈R, (a ≠0). If the function f , defined as , 0 ≤x < 1 ⎧ 2x2a f(x) = a, 1 ≤x < √2 ,is continuous in the interval [0, ∞), then an ordered pair (a, b) can be ⎨ 2b2−4b ⎩ x3 , √2 ≤x < 8 1 −1 + −√3) (2) (√2, √3) (1) (−√2, 1 1 + −√3) (4) (−√2, √3) (3) (√2,

Q80.If f(x) + 2f( x1 ) = 3x, x ≠0, and S = {x ∈R : f(x) = f(−x)}, then S (1) Contains exactly two elements (2) Contains more than two elements (3) Is an empty set (4) Contains exactly one element

Q81.If the function f(x) = { a + cos−1(x−x, + b), 1 ≤xx < 1≤2 (1) π+2 (2) π−2 2 2 (3) −π−2 (4) −1 −cos−1 (2) 2

Q81.Let C be a curve given by y(x) = 1 + √4x −3 , x > 43 . If P is a point on C, such that the tangent at P has slope 2 , then a point through which the normal at P passes, is : 3 (1) (1, 7) (2) (3, −4) (3) (4, −3) (4) (2, 3)

Q81.For x ∈R, f(x) = |log 2 −sin x| and g(x) = f(f(x)), then (1) g′(0) = −cos(log 2) (2) g is differentiable at x = 0 and g′(0) = −sin(log 2) (3) g is not differentiable at x = 0 (4) g′(0) = cos(log 2) x π π 1−sin x x ∈(0, 2 ). A normal to y = f(x) at x = 6 also passes through the

Q82.Let f(x) = sin4x + cos4x. Then, f is an increasing function in the interval: (1) ] 5π8 , 3π4 [ (2) ] π2 , 5π8 [ (3) ] π4 , π2 [ (4) ]0, π4 [

Q82.Consider f(x) = tan−1(√1+sin ), point (1) ( π6 , 0) (2) ( π4 , 0) (3) (0, 0) (4) (0, 2π3 )

Q82.The minimum distance of a point on the curve y = x2 −4 from the origin is (1) √15 units 2 units (2) √192 (4) √19 units units 2 (3) √152 JEE Main 2016 (09 Apr Online) JEE Main Previous Year Paper

Q83.A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then (1) x = 2r (2) 2x = r (3) 2x = (π + 4)r (4) (4 −π)x = πr

Q83.If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3 −1, t ∈R, meets the curve again at a point Q, then the coordinates of Q are : (1) (16t2 + 3, −64t3 −1) (2) (4t2 + 3, −8t3 −1) (3) (t2 + 3, t3 −1) (4) (t2 + 3, −t3 −1)

Q83.The integral ∫ dx is equal to (1+√x)√x−x2 (1) (2) + c + c −2√1+√x1−√x −√1−√x1+√x (3) (4) −2 + c + c √1−√x1+√x √1+√x1−√x

Q84.The integral ∫ 2x12+5x9 dx, is equal to (x5+x3+1)3 (1) x5 + c (2) −x10 + c 2(x5+x3+1)2 2(x5+x3+1)2 (3) −x5 + c (4) x10 + c (x5+x3+1)2 2(x5+x3+1)2

Q84.For x ∈R, x ≠0, if y(x) is a differentiable function such that x ∫x y(t)dt = (x + 1) ∫x ty(t)dt, then y(x) 1 1 equals (where C is a constant) (1) Cx3 e x1 (2) C e−1x x2 (3) C x (4) C e−1x x e−1 x3 dx, where [x] denotes the greatest integer less than or equal to x, is

Q84.If ∫ dx = (tan x)A + C(tan x)B + k, where k is a constant of integration, then A + B + C equals cos3 x √2 sin 2x (1) 16 (2) 27 5 10 (3) 7 (4) 21 10 5

Q85.The value of the integral ∫10 [x2−28x+196]+[x2][x2] 4 (1) 1 (2) 6 3 (3) 7 (4) 3

Q85.If 2 ∫1 tan−1 xdx = ∫1 cot−1(1 −x + x2)dx, then ∫1 tan−1(1 −x + x2)dx is equal to 0 0 0 (1) π 2 + ln 2 (2) ln 2 (3) π 2 −ln 4 (4) ln 4

Q85.The area (in sq. units) of the region {(x, y) : y2 ≥2x and x2 + y2 ≤4x, x ≥0, y ≥0} is (1) π −4√23 (2) π2 −2√23 (3) π −43 (4) π −83 JEE Main 2016 (03 Apr) JEE Main Previous Year Paper

Q86.The solution of the differential equation dx dy + 2y sec x = tan2y x , where 0 ≤x < π2 and y(0) = 1 , is given by (1) y2 = 1 + sec x+tanx x (2) y = 1 + sec x+tanx x (3) y = 1 − sec x+tanx x (4) y2 = 1 − sec x+tanx x y−2

Q86.The area (in sq. units) of the region described by A = {(x, y) y ≥x2 −5x + 4, x + y ≥1, y ≤0} is (1) 19 (2) 17 6 6 (3) 7 (4) 13 2 6