JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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If the function

f\left( x \right) = {x^3} - 6{x^2} + ax + b

satisfies Rolle’s theorem in the interval [1, 3] and

f'\left( {\frac{{2\sqrt 3 + 1}}{{\sqrt 3 }}} \right) = 0

, then

View answer and explanation

Correct option: D

a = 11

Option D (

a = 11

) is correct because it satisfies Rolle's theorem conditions, where the derivative at a point within the interval [1, 3] is zero, allowing us to solve for 'a'.

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
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