Question

Let $f\left(x\right)$ and $g\left(x\right)$ be differentiable for $0\le x\le 1$, such that $f\left(0\right)=0$, $g\left(0\right)=0$, $f\left(1\right)=6$. Let there exists a real number $c$ in $(0,1)$ such that $f^{\prime }\left(c\right)=2g^{\prime }\left(c\right)$. Then the value of $g\left(1\right)$ must be

• A. $1$
• B. $3$
• C. $-2$
• D. $-1$

Answer 1

Answer: (B)

$3$

From lagrange's theorem

$\frac{f\left(b\right)-f\left(a\right)}{b-a}=2\times \frac{g\left(b\right)-g\left(a\right)}{b-a}$

$\frac{f\left(1\right)-f\left(0\right)}{1-0}=2\times \frac{g\left(1\right)-g\left(0\right)}{1-0}$
$\frac{(6-0)}{1-0}=2\times \frac{\left(g\right(1)-0)}{1-0}$

$6=2g\left(1\right)$
$g\left(1\right)=3$