Question

Let $f(x)=\cos 5x+A\cos 4x+B\cos 3x+C\cos 2x+D\cos x+E$, and $T=f\left(0\right)-f\left(\frac{\pi }{5}\right)+f\left(\frac{2\pi }{5}\right)-f\left(\frac{3\pi }{5}\right)+\dots +f\left(\frac{8\pi }{5}\right)-f\left(\frac{9\pi }{5}\right)$ Then, $T$

• A. depends on $A,B,C,D,E$
• B. depends on $A,C,E$, but independent of $B$ and $D$
• C. depends on $B,D$ , but independent of $A,C,E$
• D. is independent of $A,B,C,D,E$

Answer 1

Answer: (C)

We have,
$f(x)=\cos 5x+A\cos 4x+B\cos 3x+C\cos 2x+D\cos x+E$
We know that,
$\cos x=\cos (2\pi -x)$
$\therefore f(x)=f(2\pi -x)$
$[\because f(x)$ contains only cosines terms]
$\because f\left(\frac{\pi }{5}\right)=f\left(2\pi -\frac{\pi }{5}\right)=f\left(\frac{9\pi }{5}\right)$
Similarly , $f\left(\frac{2\pi }{5}\right)=f\left(\frac{8\pi }{5}\right),f\left(\frac{3\pi }{5}\right)=f\left(\frac{7\pi }{5}\right)$
$f\left(\frac{4\pi }{5}\right)=f\left(\frac{6\pi }{5}\right)$
Let $T=f\left(0\right)-f\left(\frac{\pi }{5}\right)+f\left(\frac{2\pi }{5}\right)-f\left(\frac{3\pi }{5}\right)+\dots f\left(\frac{9\pi }{5}\right)$
$T=f\left(0\right)-2\left[f\left(\frac{\pi }{5}\right)+f\left(\frac{3\pi }{5}\right)\right]$
$+2\left[f\left(\frac{2\pi }{5}\right)+f\left(\frac{4\pi }{5}\right)\right]-f\left(\pi \right)$
Now, $f\left(0\right)=1+A+B+C+D+E$
$f\left(\pi \right)=-1+A-B+C-D+E$
$\because f\left(0\right)-f\left(\pi \right)=2\left(1+B+D\right)$
$f\left(\frac{\pi }{5}\right)+f\left(\frac{3\pi }{5}\right)$
$=2\left(1+B\cos \frac{3\pi }{5}+D\cos \frac{\pi }{5}\right)$
$f\left(\frac{2\pi }{5}\right)+f\left(\frac{4\pi }{5}\right)$
$=2\left(1+B\cos \frac{6\pi }{5}+D\cos \frac{2\pi }{5}\right)$
Clearly, $T$ contains only $B$ and $D$ terms