Suppose that $f (0) = – 3$ and $f ' (x) \le 5$ for all values of $x$. Then the largest value which $f (2)$ can attain is

The function $f(x) = {(x – 3)^2}$ satisfies all the conditions of mean value theorem in $[3, 4].$ A point on $y = {(x – 3)^2}$, where the tangent is parallel to the chord joining $ (3, 0)$  and $(4, 1)$  is

If the equation

${a_n}{x^{n – 1}} + \,{a_{n – 1}}{x^{n – 1}} + \,……\, + \,{a_1}x = 0,\,{a_1} \ne 0,n\, \geqslant \,2,$

has a positive root $x= \alpha ,$ then the equation 

$n{a_n}{x^{n – 1}} + \,(n – 1){a_{n – 1}}{x^{n – 1}} + \,……\, + \,{a_1} = 0$

has a positive root which is

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 ?

Consider the function $f(x) = {e^{ – 2x}}$ $sin\, 2x$ over the interval $\left( {0,{\pi \over 2}} \right)$. A real number $c \in \left( {0,{\pi \over 2}} \right)\,,$ as guaranteed by Rolle’s theorem, such that $f'\,(c) = 0$ is