JEE Main202124 Feb 2021Morning ShiftMathematicsApplication of DerivativesActual

The function f x = 4 x 3 - 3 x 2 6 - 2 sin x + 2 x - 1 cos x :

Options

Aincreases in 1 2 , ∞
Bdecreases in - ∞ , 1 2
Cdecreases in 1 2 , ∞
Dincreases in - ∞ , 1 2

Correct answer

A. increases in 1 2 , ∞

Step-by-step solution

f x = 4 x 3 - 3 x 2 6 - 2 sin x + 2 x - 1 cos x f ' x = 2 x 2 - x - 2 cos x + 2 cos x - sin x 2 x - 1 . = 2 x - 1 x - sin x for x > 0 , x - sin x > 0 x < 0 , x - sin x < 0 for x ∈ - ∞ , 0 ∪ 1 2 , ∞ , f ' x ≥ 0 for x ∈ 0 , 1 2 , f ' x ≤ 0 ⇒   f x increases in 1 2 , ∞

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the function f: (0, ) (- , ) given by f(x) = x , _e(x) - x + 1 . Then which one of the following statements is TRUE? 2026 Let P be the point on the parabola y = x^2 such that the slope of the tangent to the parabola at the point P is 4 . Let Q be the point in the first quadrant lying on the circle x^2 2026 Let f: R R be a differentiable function such that f ( x+y 3 ) = f(x) + f(y) 3 for all x, y R , and f'(0) = 3 . Then the minimum value of the function g(x) = 3 + e^x f(x) , is: 2026 _ 0 x (16 ( x 2 ) ^3 ( x 2 ) ) is equal to: 2026 Let f(x) be a polynomial of degree 5 , and have extrema at x = 1 and x = -1 . If _ x 0 ( f(x) x^3 ) = -5 , then f(2) - f(-2) is equal to: 2026 The number of critical points of the function f(x) = cases | x x |, & x 0 1, & x = 0 cases in the interval (-2 , 2 ) is equal to : 2026 Let f be a differentiable function satisfying f(x)=1-2 x+ ₀^ x e ^ (x-t) f(t) dt , x R and let g (x)= ₀^ x (f( t )+2)¹⁵( t -4)⁶( t +12)¹⁷ dt , x R . If p and q are respectively the 2026 Consider the following three statements for the function f:(0, ) R defined by f(x)= | _ e x |-|x-1| : (I) f is differentiable at all x>0 . (II) f is increasing in (0,1) . (III) f i 2026 Full Application of Derivatives list All JEE Main PYQs