Question

Consider $f^{\prime }\left(x\right)=\frac{x^2}{2}-kx+1$ such that $f\left(0\right)=0$ and $f\left(3\right)=15$
$f^{\prime \prime }(-\frac{2}{3})$ is equal to

• A. -1
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (D)

1

Solution:

Given:

$f^{\prime }\left(x\right)=\frac{x^2}{2}-kx+1$, $f\left(0\right)=0$ and $f\left(3\right)=15$

On integrating,

$\int f^{\prime }\left(x\right)dx=\int (\frac{x^2}{2}-kx+1)dx$

or, $f\left(x\right)=\frac{x^3}{6}-\frac{k{x}^2}{2}+x+C$

$f\left(0\right)=\frac{0^3}{6}-\frac{k\times 0^2}{2}+0+C$

or, $C=0$

Now, $f\left(x\right)=\frac{x^3}{6}-\frac{k{x}^2}{2}+x$

$f\left(3\right)=\frac{3^3}{6}-\frac{k\times 3^2}{2}+3$

or, $15=\frac{27-27k+18}{6}$

or, $27k=-45$

or, $k=-\frac{5}{3}$

Now, $f^{\prime }\left(x\right)=\frac{x^2}{2}+\frac{5}{3}x+1$

On differentiating both sides w.r.t x , we get

$f^{\prime \prime }\left(x\right)=x+\frac{5}{3}$

$\therefore f^{\prime \prime }(-\frac{2}{3})=-\frac{2}{3}+\frac{5}{3}=1$

Hence, D is the correct option.