JEE-Mains 2014 (12/30)

Calculus Level 2

If f f and g g are differentiable functions in ( 0 , 1 ) (0,1) satisfying f ( 0 ) = 2 = g ( 1 ) f(0)=2=g(1) , g ( 0 ) = 0 g(0)=0 and f ( 1 ) = 6 f(1)=6 , then for some c ∈ ( 0 , 1 ) c \in (0,1) :

f ′ ( c ) = g ′ ( c ) f'(c)=g'(c) 2 f ′ ( c ) = 3 g ′ ( c ) 2f'(c)=3g'(c) f ′ ( c ) = 2 g ′ ( c ) f'(c)=2g'(c) 2 f ′ ( c ) = g ′ ( c ) 2f'(c)=g'(c)

Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Since f f and g g are differentiable functions in ( 0 , 1 ) (0,1) means that it is continuous too. Hence for any c ∈ ( 0 , 1 ) c \in (0,1)

f ′ ( c ) = f ( 1 ) − f ( 0 ) 1 − 0 f'(c) = \dfrac{f(1) - f(0)}{1 - 0} by Lagrange's Mean Value Theorem and,

g ′ ( c ) = g ( 1 ) − g ( 0 ) 1 − 0 g'(c) = \dfrac{g(1) - g(0)}{1 - 0}

f ′ ( c ) = 4 , g ′ ( c ) = 2 ⇒ f ′ ( c ) = 2 g ′ ( c ) f'(c) = 4 , g'(c) = 2 \Rightarrow f'(c) = 2g'(c)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Apply cauchy

Qwerty Asdfghjkl - 3 years, 2 months ago

How can you say that both happen for same value of c c ?

From your solution, one can only say that f ′ ( x 1 ) = 2 g ′ ( x 2 ) f'(x_1)=2g' (x_2) where x 1 , x 2 ∈ ( 0 , 1 ) x_1, x_2 \in (0,1) . You are still left to prove x 1 = x 2 x_1=x_2 .

Akshat Sharda - 3 years ago

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