If $f\left( x \right) = \sin x/{e^x}$ in $\left[ {0,\pi } \right]$, then $f\left( x \right)$

Satisfies Rolle’s Theorem and $c = \frac{\pi }{4}$, So that ${f'^{\left( {\frac{\pi }{4}} \right)}} = 4$

Does not satisfy Rolle’s Theorem but $f'\left( {\frac{\pi }{4}} \right) > 0$

Satisfies Lagrange’s Mean Value Theorem but $f'\left( {\frac{\pi }{4}} \right) \ne 0$

JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Analyzing +4 | -1

f\left( x \right) = \sin x/{e^x}

\left[ {0,\pi } \right]

, then

f\left( x \right)

View answer and explanation

Correct option: C

Satisfies Rolle’s Theorem but

f'\left( {\frac{\pi }{4}} \right) = 0

Option C is correct because the function satisfies Rolle's Theorem as f(0) = f(pi) = 0, and the derivative f'(pi/4) equals 0.

Practice All Chapter MCQs →

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)