Question

Let $f(x)$ be continuous on $[0,4]$, differentiable on $(0,4), f(0)=4$ and $f(4)=-2 .$ If $g(x)=\frac{f(x)}{x+2}$, then the value of $g^{\prime}(c)$ for some Lagrange's constant $c \in(0,4)$ is

• A. $\frac{1}{2}$
• B. $\frac{5}{12}$
• C. $-\frac{5}{12}$
• D. $-\frac{7}{12}$

Answer 1

Answer: (D)

Let $f(x)$ be continuous on $[0,4]$, differentiable on $(0,4)$
$F(0)=4$ and $F(4)=-2\, g(x)=\frac{f(x)}{x+2}$
At $x=0, g(0)=\frac{f(0)}{0+2}=\frac{4}{2}=2$
At $x=4, g(4)=\frac{f(4)}{4+2}=\frac{-2}{6}=\frac{-1}{3}$
Now, $g'(c)=\frac{g(4)-g(0)}{4-0}=\frac{\frac{-1}{3}-2}{4}=\frac{-7}{3 \times 4} $
$ g' (c)=\frac{-7}{12}$