JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Application +4 | -1

Suppose

f

and

g

are differentiable functions on

[0,1]

and

f(0)=2, g(0)=0, f(1)=6, g(1)=2

. Then, there exists

c \in(0,1)

such that

f^{\prime}(c)=k g^{\prime}(c)

where the value of

k

View answer and explanation

Correct option: A

Option A (2) is correct because by constructing the function

H(x) = f(x) - 2g(x)

and applying Rolle's theorem, we find that

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

for some

c

(0,1)

More Questions from MEAN VALUES THEOREM

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f(x)

g(x)

[0,1]

f(0)=2, f(1)=6, g(0)=0, g(1)=

0 < \mathrm{c} < 1

f(x)

f(x)

f(x)=|\sin | x ||

\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]

f\left( x \right) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} } ,\;1 \le x \le 26

f'\left( x \right)

1 < x < 26

- 1 \le x \le 1

f(x)

[3,5]

\int\limits_3^5 {f\left( x \right)} \,dx

\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( c \right)

f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)