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JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Application +4 | -1
Let f(x)=ex,x∈[0,1],f(x)=e^x, x \in[0,1], then a number c of the Lagrange's mean value theorem is
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Correct option: A
log⁑e(eβˆ’1)\log _e(e-1)
The correct answer is A (log⁑e(eβˆ’1)\log _e(e-1)) because applying Lagrange's Mean Value Theorem to the function f(x)=exf(x) = e^x leads to c=log⁑e(eβˆ’1)c = \log_e(e-1).

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If f(x)f(x) and g(x)g(x) are differentiable functions in [0,1][0,1] such that f(0)=2,f(1)=6,g(0)=0,g(1)=f(0)=2, f(1)=6, g(0)=0, g(1)= 2 , then there exists 0<c<10 < \mathrm{c} < 1 such that
Assertion (A): If Rolle's theorem be applied in f(x)f(x), then L.M.V. theorem is also applicable for f(x)f(x).
Reason (R): Both Rolle's theorem & L.M.V. theorem can not be applied in f(x)=∣sin⁑∣x∣∣f(x)=|\sin | x || in [βˆ’Ο€3,Ο€3]\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]
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Let f(x)=xβˆ’1+x+24βˆ’10xβˆ’1,β€…β€Š1≀x≀26f\left( x \right) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} } ,\;1 \le x \le 26 be real valued function, then fβ€²(x)f'\left( x \right) for 1<x<261 < x < 26 The Rolle’s theorem is applicable in the interval βˆ’1≀x≀1- 1 \le x \le 1 for the function If f(x)f(x) satisfies the conditions for Rolle's theorem on [3,5][3,5], then ∫35f(x) dx\int\limits_3^5 {f\left( x \right)} \,dx equals The value of c, in the Lagrange&rsquo;s Mean value theorem f(b)βˆ’f(a)bβˆ’a=fβ€²(c)\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( c \right) for the function f(x)=x(xβˆ’1)(xβˆ’2)f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right) in the interval [0, 1/2], is
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