Question

Let f be a nonnegative function defined on the interval $[0,\pi /2]$. If
${\int }_0^x\left(f^{\prime }\right(t)-\sin 2t)dt={\int }_x^{0}f\left(t\right)\tan tdt$ and $f\left(0\right)=1$, then

• A. $f\left(0\right)=1$ is the maximum value of f
• B. $f\left(0\right)=1$ is the minimum value of f
• C. $f\left(\pi /4\right)=\frac{3\sqrt{2}}{2}-1$
• D. $f\left(\pi /3\right)=1$

Answer 1

Answer: (A), (C), (D)

$f\left(\pi /3\right)=1$
$f\left(\pi /4\right)=\frac{3\sqrt{2}}{2}-1$
$f\left(0\right)=1$ is the maximum value of f

${\int }_0^x\left(f^{\prime }\right(t)-\sin 2t)dt={\int }_x^{0}f\left(t\right)\tan tdt$
Differentiating both sides, we have $f^{\prime }\left(x\right)-\sin 2x=-f\left(x\right)\tan x$
For $y=f\left(x\right)$, we have $y^{\prime }+y\tan x=\sin 2x$.
This is a linear equation whose integrating factor is $e^{\int \tan x.dx}=\sec x$.
The general solution is
$y\left(x\right)=\cos x[\int \sin xdx+C]=C\cos x-2{\cos }^{2}x$
Since $y\left(0\right)=f\left(0\right)=1$ so $C=3$. Thus
$f\left(x\right)=y=3\cos x-2{\cos }^{2}x=-2(\cos x-3/4{)}^2+\frac{9}{8}$
Thus $0\le f\left(x\right)\le 1$. The maximum value is $f\left(0\right)=1$.
$f\left(\pi /4\right)=\frac{3}{\sqrt{2}}-1$ and $f\left(\pi /3\right)=\frac{3}{2}-\frac{1}{2}=1$