Question

The value of $\theta$ of mean value theorem for the function $f\left(x\right)={ax}^2+bx+c$ in $[1,2]$ is

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{3}{2}$

Value of $\theta$ mean value theorem

$f\left(x\right)=a{x}^2+bx+c$ in $[1,2]$

$\to f\left(x\right)=a{x}^2+bx+c$ in $[a,b]$

Where $a=7,b=2$

$\to$ Condition of mean value theorem

(i) $f\left(x\right)$ is continuous at $(a,b)$

(ii) $f\left(x\right)$ is derivable at $(a,b)$

If both condition satisfied then there exist some $\theta$ vallue in $(a,b)$ such that $f^{\prime }\left(\theta \right)=\frac{f\left(b\right)-\left(a\right)}{b-a}$

$\to$ condition : $1$

$f\left(x\right)=a{x}^2+bx+c$

$f\left(x\right)$ is polynomial & every polynomial $f^n$ is continuous

$f\left(x\right)$ is continuous at $x\in (1,7)$

$\to$ Condition $2$

$f\left(x\right)=a{x}^2+bx+c$

$f\left(x\right)$ is polynomial & every polynomial $f^n$ is differentiable

-so, $f\left(x\right)$ is differentiable at $x\in (1,2)$

Now $f\left(x\right)=a{x}^2+bx+c$

$f^{\prime }\left(x\right)=2ax+b$

$f^{\prime }\left(\theta \right)=2a\theta +b$

Also, $f\left(1\right)=a(1{)}^2+b(1)+c=a+b+c$

$f\left(2\right)=a(2{)}^2+b(a)+c=4a+2b+c$

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

$\therefore 24\theta +b=4a+2b+c-(a+b+c)$

$24\theta +b=3a+b$

$\therefore 24\theta =3a$

$\theta =\frac{3}{2}$