Question

Let $f$ be a monic polynomial function of degree $n ( n \geq 1)$ such that $\int\limits_0^{\frac{\pi}{2}} \sin ^4 xdx < \int\limits_0^{\frac{\pi}{2}}(\sin x )^{ n } dx <1$ If $f(3)=f^{\prime \prime}(3)=f(4)-4=0$, then find the value of $\frac{1}{2}\left(\int\limits_3^5 f(x) d x\right)$.

Answer 1

$\int\limits_0^{\frac{\pi}{2}} \sin ^4 x d x<\int \limits_0^{\frac{\pi}{2}}(\sin x)^n d x<1 \Rightarrow n$ must be either 2 or 3 $f (3)= f "(3)=0$ and $f (4)=4 \Rightarrow n$ must be 3 .
$\left\{\Theta\right.$ for $n=2, f^{\prime \prime}(x)$ is non-zero constant $\}$
Now,
$f(x)=(x-3)\left(x^2+b x+c\right) $
$f^{\prime}(x)=(x-3)(2 x+b)+\left(x^2+b x+c\right)$
$f^{\prime \prime}(x)=(x-3) \cdot 2+(2 x+b)+(2 x+b) $
$f^{\prime \prime}(3)=2(6+b)=0 \Rightarrow b=-6$
$\therefore f ( x ) =( x -3)\left( x ^2-6 x + 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 ^2-6 x +12\right) $
$ =( x -3)\left(( x -3)^2+3\right)$
$ =( x -3)^3+3( x -3)$
$I =\int\limits_3^5\left(( x -3)^3+3( x -3)\right) dx$
Putting $x -3= t \Rightarrow dx = dt$
$ I =\int\limits_0^2\left( t ^3+3 t \right) dt =\left(\frac{ t ^4}{4}+\frac{3 t ^2}{2}\right)_0^2=\frac{16}{4}+\frac{3 \cdot 4}{2}=10$
$\therefore \frac{ I }{2} =5 $