Question

Let $f\left(x\right)$ and $g\left(x\right)$ be differentiable for $0\le x\le 1$, such that $f\left(0\right)=2,g\left(0\right)=0,f\left(1\right)=6$. Let these exist a real number c in $[0,1]$ such that $f^{\prime }\left(c\right)=2g^{\prime }\left(c\right)$ then $g\left(1\right)=$

• A. $1$
• B. $2$
• C. $-2$
• D. $-1$

Answer 1

Answer: (B)

$2$

Given that $f\left(0\right)=2,f\left(1\right)=6,g\left(0\right)=0$

Given $f\left(x\right),f^{{}^{\prime }}\left(x\right)$ are differentiable on $(0,1)$.

Therefore, Lagrange's mean value theorem can be applied.

Lagrange's mean value theorem states that if $f\left(x\right)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists some $c$ between $a$ and $b$ such that $f^{{}^{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Given that $f^{{}^{\prime }}\left(c\right)=2g^{{}^{\prime }}\left(c\right)$

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

$⟹6-2=2\left(g\right(1)-0)$

$⟹2g\left(1\right)=4$

$⟹g\left(1\right)=2$