Question

If the function $f(x)=2 x^{2}+3 x+5$ satisfies LMVT at $x=2$ on the closed interval $[1, a]$ then the value of '$a$' is equal to_____.

Answer 1

$f'(x)=4x+3$
Now $f'(2)=11=\frac{f(a)-f(1)}{a-1}=\frac{\left(2 a^{2}+3 a+5\right)-10}{a-1}$
$11a-11=2 a^{2}+3 a-5$
$2 a^{2}-8 a+6=0$
$\Rightarrow a^{2}-4 a+3=0$
$(a-3)(a-1)=0$
$ \Rightarrow a=3$