Question

If the function $f\left(x\right)=2x^2+3x+5$ satisfies Lagrange's mean value theorem at $x=2$ on the closed interval $[1,a],$ then the value of $a$ is equal to

• A. $3$
• B. $4$
• C. $6$
• D. $1$

Answer 1

The correct option is A $3$

$f\left(x\right)=2x^2+3x+5$
Differentiating w.r.t. $x,$ we get
$f^{\prime }\left(x\right)=4x+3$
$\Rightarrow f^{\prime }\left(2\right)=11$
By LMVT, we have
$f^{\prime }\left(2\right)=\frac{f\left(a\right)-f\left(1\right)}{a-1}$
$\Rightarrow 11=\frac{(2a^2+3a+5)-10}{a-1}$
$\Rightarrow 11a-11=2a^2+3a-5$
$\Rightarrow a^2-4a+3=0$
$\Rightarrow (a-3)(a-1)=0$
$\Rightarrow a=3(\because a>1)$