Question

If $f\left(x\right)$ is diffrentiable in $\left[a,b\right]$ such that $f\left(a\right)=2,f\left(b\right)=6,$ then there exists at least one c, $a<c<b,$ such that $\left(b^3-a^3\right)f^{\prime }\left(c=\right)$

• A. $c^2$
• B. $2c^2$
• C. $-3c^2$
• D. $12c^2$

Answer 1

The correct option is D $12c^2$

Let $h\left(x\right)=f\left(x\right)-f\left(a\right)+A(x^3-a^3)$

where $A$ is obtained from the relation $h\left(b\right)=0$

$\Rightarrow h\left(b\right)=4+A(b^3-a^3)=0$

$\Rightarrow A=-\frac{4}{b^3-a^3}$

Also,

$\Rightarrow h\left(a\right)=0$

$\Rightarrow h\left(a\right)=h\left(b\right)=0$

Clearly, $h\left(x\right)$ is continuous on $[a,b]$ and differentiable in $(a,b)$.

Hence, Rolle's theorem holds for $h\left(x\right)$ in $[a,b]$, so there exists $c\in (a,b)$ such that $h^{\prime }\left(c\right)=0$

$\Rightarrow f^{\prime }\left(c\right)+A\left(3c^2\right)=0$

$\Rightarrow f^{\prime }\left(c\right)=-A\left(3c^2\right)$

$\Rightarrow f^{\prime }\left(c\right)=3c^2\left(\frac{4}{b_3-a^3}\right)$

$\therefore (b^3-a^3)f^{\prime }\left(c\right)=12c^2$