Question

The value of $c$ for which the conclusion of Lagrange's Theorem holds for the function $f\left(x\right)=\sqrt{a^2-x^2},a>1$ on the interval $[1,a]$

• A. $\frac{a(a+1)}{2}$
• B. $\frac{a+1}{2}$
• C. $\sqrt{\frac{a(a+1)}{2}}$
• D. $\frac{a(a-1)}{2}$

Answer 1

The correct option is C $\sqrt{\frac{a(a+1)}{2}}$

$\frac{f\left(a\right)-f\left(1\right)}{a-1}=f^{\prime }\left(c\right)=\frac{-c}{\sqrt{a^2-c^2}}$
$\Rightarrow \frac{\sqrt{a^2-a^2}-\sqrt{a^2-1}}{a-1}=\frac{-c}{\sqrt{a^2-c^2}}$
$\Rightarrow -\sqrt{\frac{a+1}{a-1}}=-\frac{c}{\sqrt{a^2-c^2}}$
$\Rightarrow \frac{a+1}{a-1}=\frac{c^2}{a^2-c^2}$
$\Rightarrow (a+1)(a^2-c^2)=c^2(a-1)$
$\Rightarrow c^2(-a-1+1-a)=-(a+1)a^2$
$\Rightarrow c^2\left(2a\right)=(a+1)a^2$
$\Rightarrow c^2=\frac{(a+1)a}{2}$
$\{\because c\in (1,a)\text{so}c=\sqrt{\frac{(a+1)a}{2}}\}$