Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main201912 Jan 2019Evening ShiftMathematicsDifferentiationActual

Let f be a differentiable function such that f 1 = 2 and f ' x = f x for all x ∈ R . If h x = f f x , then h ' 1 is equal to :

Options

  1. A4 e 2
  2. B2 e
  3. C  4 e
  4. D2 e 2

Correct answer

C.   4 e

Step-by-step solution

Given h x = f f x ⇒ h ' x = f ' f x . f ' x ⇒ h ' x = f f x . f x     . . . 1 (as f ' x = f x ) Now f ' x = f x ⇒ f ' x f x = 1 Integrating both sides with respect to x , we get ln f x = x + c ⇒ f x = k . e x ⇒ f x = 2 e .   e x           . . . 2     ∵   f 1 = 2 Putting x = 1 in equation 1 , we get h ' 1 = f f 1 . f 1 = f 2 . 2 = 2 . 2 e e 2 = 4 e (using equation 2 )

Practice Differentiation on Quantrex Academy →

More from Differentiation

Let R denote the set of all real numbers. Consider the polynomial function f: R R defined by f(x) = d¹⁰ dx¹⁰ ((x^2 - 1)¹⁰ ) , for all x R . Here d¹⁰ dx¹⁰ ((x^2 - 1)¹⁰ ) is the 10th 2026Let f(x) and g(x) be twice differentiable functions satisfying f''(x) = g''(x) for all x R , f'(1) = 2g'(1) = 4 and g(2) = 3f(2) = 9 . Then f(25) - g(25) is equal to : 2026Let f be a real polynomial of degree n such that f(x) = f'(x) f''(x) , for all x R . If f(0) = 0 , then 36 (f'(2) + f''(2) + ₀^2 f(x) ,dx ) is equal to: 2026Let f(x)=x³+x² f^ (1)+2 x f^ (2)+f^ (3), x R . Then the value of f^ (5) is: 2026Let R denote the set of all real numbers. Let f: R R and g: R (0,4) be functions defined by f(x)= _e (x^2+2 x+4 ) , and g(x)= 4 1+e^ -2 x Define the composite function f g⁻¹ by (f 2025Let f: R R be a twice differentiable function such that ( x y)(f(2 x+2 y)-f(2 x-2 y))=( x y )(f(2 x +2 y )+f(2 x -2 y )) , for all x , y R . If f^ (0)= 1 2 , then the value of 24 f 2025If _e y=3 ⁻¹ x , then (1-x^2 ) y^ -x y^ at x= 1 2 is equal to 2024Let f(x)=a x^3+b x^2+c x+41 be such that f(1)=40, f^ (1)=2 and f^ (1)=4 . Then a ^2+ b ^2+ c ^2 is equal to: 2024 Full Differentiation list All JEE Main PYQs