Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Most Important Selected Qs for JEE AdvancedMathematicsApplication of Derivatives

Let, f(0, ) R be a strictly increasing function such that f(x) - 1 x x 0 and f(x) f (f(x)+ 1 x )=1 x 0 , then

Options

  1. Af(1)= 1- 5 2
  2. Bf(1)= 1+ 5 2
  3. Cf(x)= 1- 5 2 x
  4. Df(x)= (1+ 5 ) x 2

Correct answer

C. f(x)= 1- 5 2 x

Step-by-step solution

Let f(1)=t Put x=1 in f(x) f (f(x)+ 1 x )=1 (1) We get t f(t+1)=1 t 0 and f(t+1)= 1 t Put x=t+1 in equation (1), we get f ( 1 t + 1 t+1 )=t=f(1) 1 t + 1 t+1 =1 t= 1 5 2 But for t = 1+ 5 2 , t 1- f (1) f (1+ t )= 1 t 1 a contradiction

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 Most Important Selected Qs for JEE Advanced PYQs