Question

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