The value of $\left[ \frac{\log (x/e)}{x - e} \right]$ for all $x > e$ is equal to (where $[.]$ denotes the greatest integer function).
- A$1$
- B$0$
- C$2$
- Ddoes not take a unique value
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Consider the function $f(x) = |x - 2| + |x - 5|, x \in R$.
Statement-$1$: $f'(4) = 0$.
Statement-$2$: $f$ is continuous in $[2, 5]$,differentiable in $(2, 5)$ and $f(2) = f(5)$.
Statement-$1$: $f'(4) = 0$.
Statement-$2$: $f$ is continuous in $[2, 5]$,differentiable in $(2, 5)$ and $f(2) = f(5)$.
Difficult
View SolutionLet $f: D \rightarrow R$ where $D=[0,1] \cup [2,4]$ be defined by $f(x)=\begin{cases} x, & \text{if } x \in [0,1] \\ 4-x, & \text{if } x \in [2,4] \end{cases}$. Then,
In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$,and $f(x) = \sqrt{x}$,then the value of $c$ is:
Easy
View SolutionIf $f(x)$ is differentiable on the interval $[2, 5]$ such that $f(2) = 1/5$ and $f(5) = 1/2$,then there exists a number $c$ such that $2 < c < 5$ and $f'(c) = \dots$
Difficult
View SolutionIn the Mean Value Theorem,$f'(c) = \frac{f(b) - f(a)}{b - a}$. If $a = 0$,$b = \frac{1}{2}$,and $f(x) = x(x - 1)(x - 2)$,then the value of $c$ is:
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