Thank you for reporting, we will resolve it shortly
Q.
Suppose that $f(0)=-3$ and $f^{\prime}(x) \leq 5$ for all values of $x$ then the largest value which $f(2)$ can attain is
Application of Derivatives
Solution:
using LMVT in $[0,2]$
$f^{\prime}(c)=\frac{f(2)-f(0)}{2-0} \text { where } c \in(0,2)$
$f^{\prime}(c)=\frac{f(2)+3}{2}$
But $f ^{\prime}( x ) \leq 5$
$\frac{ f (2)+3}{2} \leq 5 \Rightarrow f (2)+3 \leq 10 \Rightarrow f (2) \leq 7$