JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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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.

View answer and explanation

Correct option: B

Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).

Option B (Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).) is correct because while both the assertion and the reason are true statements, the reason does not directly explain the assertion.

More Questions from MEAN VALUES THEOREM

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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
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 If the functions

f(x)

and

g(x)

are continuous on

[a, b]

and differentiable on

(a, b),

then in the interval

(a, b),

the equation

\left|\begin{array}{ll}f^{\prime}(x) & f(a) \\ g^{\prime}(x) & g(a)\end{array}\right|=\frac{1}{a-b}\left|\begin{array}{ll}f(a) & f(b) \\ g(a) & g(b)\end{array}\right|