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

The correct option is B $3$

Let $F\left(x\right)=f\left(x\right)-2g\left(x\right)$,which is continuous and differentiable in given interval.
$F\left(0\right)=f\left(0\right)-2g\left(0\right)=0$,
$F\left(1\right)=f\left(1\right)-2g\left(1\right)=6-2g\left(1\right)$

$F^{\prime }\left(x\right)=f^{\prime }\left(x\right)-2g^{\prime }\left(x\right)$
For Rolle's theorem to be applicable,
$F\left(0\right)=F\left(1\right)=0$
$\Rightarrow 6-2g\left(1\right)=0$
$\therefore g\left(1\right)=3$