JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Let

f

be twice differentiable function satisfying

f(1)=1, f(2)=4, f(3)=9,

then

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Correct option: C

There exists at least one

x \in(1,3)

such that

f^{\prime \prime}(x)=2

Option C is correct because by defining a function g(x) = f(x) - x² and applying Rolle's theorem twice, we can show that there exists at least one x in (1,3) such that f''(x) = 2.

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f(x)

and

g(x)

are differentiable functions in

[0,1]

such that

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

2 , then there exists

0 < \mathrm{c} < 1

such that
Assertion (A): If Rolle's theorem be applied in

f(x)

, then L.M.V. theorem is also applicable for

f(x)

.
Reason (R): Both Rolle's theorem & L.M.V. theorem can not be applied in

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

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

The codes are given below, select the correct codes. Let

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

be real valued function, then

f'\left( x \right)

for

1 < x < 26

The Rolle’s theorem is applicable in the interval

- 1 \le x \le 1

for the function If

f(x)

satisfies the conditions for Rolle's theorem on

[3,5]

, then

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

equals The value of c, in the Lagrange’s Mean value theorem

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

for the function

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

in the interval [0, 1/2], is