Consider  $f (x) = | 1 – x | \,;\,1 \le x \le 2 $   and $g (x) = f (x) + b sin\,\frac{\pi }{2}\,x$, $1 \le x \le 2$  then which of the following is correct ?

If the function $f(x) = {x^3} – 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is

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

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,

Let $f(x) = 8x^3 – 6x^2 – 2x + 1,$ then