Question

Given a function $g$ continuous on $R$ such that $\underset{0}{\overset{1}{\int }}g\left(t\right)dt=2$ and $g\left(1\right)=5.$ If $f\left(x\right)=\frac{1}{2}\underset{0}{\overset{x}{\int }}(x-t{)}^{2}g(t)dt,$ then the value of $\left(f^{\prime \prime \prime }\right(1)-f^{\prime \prime }(1\left)\right)$ is equal to

• A. $0$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is B $3$

$f\left(x\right)=\frac{1}{2}\underset{0}{\overset{x}{\int }}(x-t{)}^{2}g(t)dt$
$\Rightarrow f\left(x\right)=\frac{1}{2}\left[x^2\underset{0}{\overset{x}{\int }}g\left(t\right)dt-2x\underset{0}{\overset{x}{\int }}tg\left(t\right)dt+\underset{0}{\overset{x}{\int }}{t}^{2}g\left(t\right)dt\right]$
Differentiating w.r.t. $x$ on both sides, we get
$f^{\prime }\left(x\right)=\frac{1}{2}\left[2x\underset{0}{\overset{x}{\int }}g\left(t\right)dt+x^2\left(g\right(x)\cdot 1-0)-2\underset{0}{\overset{x}{\int }}tg\left(t\right)dt-2x\left(xg\right(x)\cdot 1-0)+\left(x^{2}g\right(x)\cdot 1-0)\right]$
$\Rightarrow f^{\prime }\left(x\right)=x\underset{0}{\overset{x}{\int }}g\left(t\right)dt-\underset{0}{\overset{x}{\int }}tg\left(t\right)dt$
$f^{\prime \prime }\left(x\right)=\underset{0}{\overset{x}{\int }}g\left(t\right)dt+x\left(g\right(x)\cdot 1-0)-\left(xg\right(x)\cdot 1-0)$
$\Rightarrow f^{\prime \prime }\left(x\right)=\underset{0}{\overset{x}{\int }}g\left(t\right)dt$
$f^{\prime \prime \prime }\left(x\right)=g\left(x\right)$
$\therefore f^{\prime \prime }\left(1\right)=\underset{0}{\overset{1}{\int }}g\left(t\right)dt=2$ and $f^{\prime \prime \prime }\left(1\right)=g\left(1\right)=5$
Hence, the value of $f^{\prime \prime \prime }\left(1\right)-f^{\prime \prime }\left(1\right)$ is $3.$