A function $f$ is defined by $f\left( x \right) = {e^x}\sin x$ in $\left[ {0,\;\pi } \right]$. Which of the following is not correct?

$f$ f is continous in $\left[ {0,\;\pi } \right]$

$f$ is differentiable in $\left[ {0,\;\pi } \right]$

$f\left( 0 \right) = f\left( \pi \right)$

JEE MATHEMATICS MCQ: MEAN VALUES THEOREM

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Analyzing +4 | -1

A function

f

is defined by

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

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

. Which of the following is not correct?

View answer and explanation

Correct option: D

Rolle’s theorem is not true in

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

Option B (a decreasing function) is correct because the derivative of the function is negative for all

x > 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)