Question

If $f$ and $g$ are differentiable functions in $(0,1)$ satisfying $f(0)=2=g(1),g(0)=0$ and $f(1)=6$, then for some $c\in ]0,1[$

• A. $2f^{\prime }(c)=g^{\prime }(c)$
• B. $2f^{\prime }(c)=3g^{\prime }(c)$
• C. $f^{\prime }(c)=g^{\prime }(c)$
• D. $f^{\prime }(c)=2g^{\prime }(c)$

Answer 1

Answer: (D)

Given, $f(0)=2=g(1),g(0)=0$ and $f(1)=6$
$f$ and $g$ are differentiable in $(0,1)$.
Let $h(x)=f(x)-2g(x)....$ (i)
$h(0)=f(0)-2g(0)=2-0=0$
and $h(1)=f(1)-2g(1)=6-2(2)=2$
$h(0)=h(1)=2$
Hence, using Rolle's theorem.
$h^{\prime }(c)=0$, such that $c\in (0,1)$
Differentiating Eq. (i) at $c$, we get
$\Rightarrow f^{\prime }(c)-2g^{\prime }(c)=0$
$\Rightarrow f^{\prime }(c)=2g^{\prime }(c)$