Question

Let $f$ be a monic polynomial function of degree $n(n\ge 1)$ such that $\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{4}xdx<\underset{0}{\overset{\frac{\pi }{2}}{\int }}(\sin x{)}^{n}dx<1$ If $f(3)=f^{\prime \prime }(3)=f(4)-4=0$, then find the value of $\frac{1}{2}\left(\underset{3}{\overset{5}{\int }}f(x)dx\right)$.

Answer 1

Answer: 5

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{4}xdx<\underset{0}{\overset{\frac{\pi }{2}}{\int }}(\sin x{)}^{n}dx<1\Rightarrow n$ must be either 2 or 3 $f(3)=f"(3)=0$ and $f(4)=4\Rightarrow n$ must be 3 .
$\{\Theta$ for $n=2,f^{\prime \prime }(x)$ is non-zero constant $\}$
Now,
$f(x)=(x-3)\left(x^2+bx+c\right)$
$f^{\prime }(x)=(x-3)(2x+b)+\left(x^2+bx+c\right)$
$f^{\prime \prime }(x)=(x-3)\cdot 2+(2x+b)+(2x+b)$
$f^{\prime \prime }(3)=2(6+b)=0\Rightarrow b=-6$
$\therefore f(x)=(x-3)\left(x^2-6x+c\right)=(x-3)\left((x-3{)}^2+c-9\right)$
$f(4)=4\Rightarrow 1+c-9=4\Rightarrow c=12$
$f(x)=(x-3)\left(x^2-6x+12\right)$
$=(x-3)\left((x-3{)}^2+3\right)$
$=(x-3{)}^3+3(x-3)$
$I=\underset{3}{\overset{5}{\int }}\left((x-3{)}^3+3(x-3)\right)dx$
Putting $x-3=t\Rightarrow dx=dt$
$I=\underset{0}{\overset{2}{\int }}\left(t^3+3t\right)dt={\left(\frac{t^4}{4}+\frac{3t^2}{2}\right)}_0^2=\frac{16}{4}+\frac{3\cdot 4}{2}=10$
$\therefore \frac{I}{2}=5$