Practice Questions

Q81.If x = −1 and x = 2 are extreme points of f(x) = α log|x| + βx2 + x, then (1) α = 2, β = −12 (2) α = 2, β = 12 (3) α = −6, β = 12 (4) α = −6, β = −12

Q81.If f(x) = ( 35 )x + ( 45 )x −1, (1) No solution (2) More than two solutions (3) One solution (4) Two solutions

Q81.Let f(x) = x|x|, g(x) = sin x and h(x) = (g ∘f)(x). Then (1) h(x) is not differentiable at x = 0. (2) h(x) is differentiable at x = 0, but h′(x) is not continuous at x = 0 (3) h′(x) is continuous at x = 0 but it is not (4) h′(x) is differentiable at x = 0 differentiable at x = 0

Q81.If the volume of a spherical ball is increasing at the rate of 4π cc / sec then the rate of increase of its radius (in cm / sec), when the volume is 288π cc is (1) 1 (2) 1 9 6 (3) 1 (4) 1 24 36

Q82.If the Rolle's theorem holds for the function f(x) = 2x3 + ax2 + bx in the interval [−1, 1] for the point c = 21 , then the value of 2a + b is: (1) -1 (2) 2 (3) 1 (4) -2 Q83. ∫ sin8 x−cos8 x dx is equal to (1−2 sin2 x cos2 x) (1) −12 sin 2x + c (2) −sin2 x + c (3) −12 sin x + c (4) 21 sin 2x + c Q84. 21 dx equals The integral ∫ ln(1+2x)1+4x2 0 (1) π 4 ln 2 (2) 16π ln 2 (3) π 8 ln 2 (4) 32π ln 2

Q82.The slope of the line touching both the parabolas y2 = 4x and x2 = −32y is (1) 1 (2) 2 8 3 (3) 1 (4) 3 2 2 x dx, is equal to

Q82.For the curve y = 3 sin θ cos θ, x = eθ sin θ, 0 ≤θ ≤π, the tangent is parallel to x-axis when θ is: (1) 3π (2) π 4 2 (3) π (4) π 4 6

Q82.Let f and g be two differentiable functions on R such that f ′(x) > 0 and g′(x) < 0 for all x ∈R. Then for all x : (1) f(g(x)) > f(g(x −1)) (2) f(g(x)) > f(g(x + 1)) (3) g(f(x)) > g(f(x −1)) (4) g(f(x)) < g(f(x + 1)) JEE Main 2014 (12 Apr Online) JEE Main Previous Year Paper

Q82.If non-zero real numbers b and c are such that min f(x) > max g(x), where f(x) = x2 + 2bx + 2c2 and g(x) = −x2 −2cx + b2, (x ∈R); then cb lies in the interval , (1) (√2, ∞) (2) [ 12 1 ) √2 , √2] (3) (0, 12 ) (4) [ √21

Q83. sin2 x cos2 x The integral dx is equal to: ∫ 2 (sin3 x + cos3 x) 1 1 (1) + c (2) − + x (1+cot3 x 3(1+tan3 c) (3) sin3 x + c (4) − cos3 x + c (1+cos3 x 3(1+sin3 x

Q83.The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius = √3 is: (1) 3 4 √3π (2) 83 √3π (3) 4π (4) 2π > 0) is equal to:

Q83.The integral ∫(1 + x −1x )ex+ 1 JEE Main 2014 (06 Apr) JEE Main Previous Year Paper (1) (x + 1)ex+ x1 + c (2) −xex+ x1 + c (3) (x −1)ex+ x1 + c (4) xex+ x1 + c π x 2 −4 sin x2 dx equals

Q83.If m is a non-zero number and ∫x5m−1+2x4m−1 dx = f(x) + c, then f(x) is equal to (x2m+xm+1)3 (1) (x5m−x4m) (2) 1 x4m 2m(x2m+xm+1)2 2m (x2m+xm+1)2 (3) x5m (4) 2m(x5m+x4m) 2m(x2m+xm+1)2 (x2m+xm+1)2

Q84.The integral ∫ √1 + 4 sin2 0 (1) 4√3 −4 (2) 4√3 −4 −π3 (3) π −4 (4) 2π3 −4 −4√3

Q84.If [ ] denotes the greatest integer function, then the integral ∫π0 [cos xdx is equal to: (1) π (2) 0 2 (3) −1 (4) −π2

Q84.The integral ∫x cos−1 ( 1+x21−x2 )dx(x (1) −x + (1 + x2) tan−1 x + c (2) x −(1 + x2) cot−1 x + c (3) −x + (1 + x2) cot−1 x + c (4) x −(1 + x2) tan−1 x + c

Q84.Let, the function F be defined as F(x) = ∫x1 ett dt, x > 0, then the value of the integral ∫x1 t+aet dt, where a > 0, is (1) ea[F(x) −F(1 + a)] (2) e−a[F(x + a) −F(a)] (3) ea[F(x + a) −F(1 + a)] (4) e−a[F(x + a) −F(1 + a)]

Q85.If for n ≥1, Pn = ∫e1 (log xn)dx, then P10 −90P8 is equal to: (1) −9 (2) 10e (3) −9e (4) 10 Φ, is given by

Q85.The area (in sq. unit) of the region described by A = {(x, y) : x2 + y2 ≤1 and y2 ≤1 −x} is (1) π 2 −23 (2) π2 + 32 (3) π 2 + 34 (4) π2 −43

Q85.If for a continuous function f(x), ∫t−π(f(x) + xdx) = π2 −t2 , for all t ≥−π, then f (−π3 ) is equal to: (1) π (2) π2 (3) π (4) π 3 6

Q85.Let A = {(x, y) : y2 ≤4x, y −2x ≥−4}. The area of the region A in square units is (1) 10 (2) 8 (3) 9 (4) 11

Q85.The area of the region (in square units ) above the x-axis bounded by the curve y = tan x, 0 ≤x ≤π2 and the tangent to the curve at x = π4 is (1) 2 1 (log 2 −12 ) (2) 12 (1 + log 2) (3) 1 2 (1 −log 2) (4) 12 (log 2 + 12 )

Q86.If dxdy + ytan x=sin 2x and y(0) = 1, then y(π) is equal to (1) −1 (2) 5 (3) 1 (4) −5 → → →

Q86.The general solution of the differential equation, sin 2x −y = 0, is : dx ( dy −√tan x) (1) y√tan x = x + c (2) y√cot x = tan x + c (3) y√tan x = cot x + c (4) y√cot x = x + c

Q86.If the general solution of the differential equation y′ = xy + Φ ( xy ), for some function y ln |cx| = x, where c is an arbitrary constant, then Φ(2) is equal to: (1) 4 (2) 1 4 (3) −4 (4) −14 →