Question

For the function $f(x)=x^2-4x-3,x\in (a,b)$, where $a=1$ and $b=1$, the value of $c$ for mean value theorem where $c\in (a,b)$ is

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

Answer 1

Answer: (D)

(i) $f(x)$ is continuous in $[a,b]$
(ii) $f(x)$ is differentiable in $(a,b)$.
Then, there will be atleast one value of $c\in (a,b)$ such that $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$.
Here, $f(x)=x^2-4x-3,x\in [1,4]$
which is a polynomial function, so it is continuous and derivable at all $x\in R$, therefore
(i) $f(x)$ is continuous on $[1,4]$.
(ii) $f(x)$ is derivable on $(1,4)$.
$\therefore$ Conditions of Lagrange's theorem are satisfied on $[1,4]$.
Hence, there is atleast one real number $c\in (1,4)$. Such that
$f^{\prime }(c)=\frac{f(4)-f(1)}{4-1}\left(\therefore f^{\prime }(c)=-\frac{f(b)-f(a)}{b-a}\right)$
$\Rightarrow 2c-4=\frac{\left(4^2-4\times 4-3\right)-\left(1^2-4\times 1-3\right)}{4-1}=1$
$\Rightarrow \left[\because f^{\prime }(x)=\frac{d}{dx}\left(x^2-4x-3\right)=2x-4\right]$
$2c-4=1\Rightarrow c=\frac{5}{2}\in (1,4)$
$\therefore$ MVT is verified for $f(x)$ in $[1,4]$.