Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20206 Sep 2020Evening ShiftMathematicsApplication of DerivativesActual

For all twice differentiable functions f : R → R , with f 0 = f 1 = f ' 0 = 0 ,

Options

  1. Af " x ≠ 0 at every point x ε 0 , 1
  2. Bf " x = 0 , for some x   ε   0 , 1
  3. Cf " 0 = 0
  4. Df " x = 0 , at every point x   ε 0 , 1

Correct answer

D. f " x = 0 , at every point x   ε 0 , 1

Step-by-step solution

Applying Rolle's theorem in [0,1] for function f ( x ) f ' c = 0 ,   c   ε 0 , 1 again applying Rolle's theorem in [0, c] for function f ' ( x ) , f ' ' x = 0 ,   c ∈ 0 , 1

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? 2026Let 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 2026Let 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: 2026Let 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: 2026The 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 : 2026Let 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 2026Consider 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