Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Advanced2014MathematicsApplication of DerivativesActual

Let f : 0 , ∞ → R be given by f x = ∫ 1 x x e - t + 1 t d t t , then

Options

  1. Af x is monotonically increasing on 1 , ∞
  2. Bf x is monotonically decreasing on (0, 1)
  3. Cf x + f 1 x = 0 , for all x ∈ 0 , ∞
  4. Df 2 x is an odd function of x on R

Correct answer

A. f x is monotonically increasing on 1 , ∞

Step-by-step solution

f ′ x = 2 e - x + 1 x x Which is increasing in 1 , ∞ Also, f x + f 1 x = 0 g x = f 2 x = ∫ 2 - x 2 x e - t + 1 t t d t g - x = ∫ 2 x 2 - x e - t + 1 t t d t = - g x Hence, an odd function

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 Advanced PYQs