$(i)$  $f (x)$ is continuous and defined for all real numbers

$(ii)$ $f '(-5) = 0 \,; \,f '(2)$ is not defined and $f '(4)  = 0$

$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f (x)$

$(iv)$ $f ''(2)$ is undefined, but $f ''(x)$ is negative everywhere else.

$(v)$ the signs of  $f '(x)$ is given below

Possible graph of $y = f (x)$ is

Let $f :[2,4] \rightarrow R$ be a differentiable function such that $\left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1$, $x \in[2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Consider the following two statements:

$(A): f(x) \leq 1$, for all $x \in[2,4]$

$(B)$ : $f(x) \geq \frac{1}{8}$, for all $x \in[2,4]$

Then,

Verify Mean Value Theorem, if $f(x)=x^{3}-5 x^{2}-3 x$ in the interval $[a, b],$ where $a=1$ and $b=3 .$ Find all $c \in(1,3)$ for which $f^{\prime}(c)=0$

The function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfy the conditions of Rolle's theorem in $[1, 3]. $ The values of  $a $ and $ b $ are

If $f(x) = \cos x,0 \le x \le {\pi \over 2}$, then the real number $ ‘c’ $ of the mean value theorem is

If for $f(x) = 2x - {x^2}$, Lagrange’s theorem satisfies in $[0, 1]$, then the value of $c \in [0,\,1]$ is