Question

If the function $f\left(x\right)=2x^2+3x+5$ satisfies LMVT 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

Answer: (A)

$3$

Applying Lagrange's mean value theorem, we get

$\frac{f\left(a\right)-f\left(1\right)}{a-1}=f^{\prime }\left(2\right)$

$\frac{2a^2+3a+5-10}{a-1}=4x+3{|}_{x=2}$

$\frac{2a^2+3a-5}{a-1}=11$

$2a^2+3a-5=11a-11$

$2a^2-8a+6=0$

$a^2-4a+3=0$

$(a-3)(a-1)=0$

$a=3$ and $a=1$

However

$aϵ[1,a]$

Hence

$a=3$.