Question

Let we have a function as \[f\left( x \right)=\cos 5x+A\cos 4x+B\cos 3x+C\cos 2x+D\cos x+E\] and \[T=f\left( 0 \right)-f\left( \dfrac{\pi }{5} \right)+f\left( \dfrac{2\pi }{5} \right)-f\left( \dfrac{3\pi }{5} \right)+.....+f\left( \dfrac{8\pi }{5} \right)-f\left( \dfrac{9\pi }{5} \right).\] Then the value of T is:
(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 and E.
(d) Is independent of A, B, C, D, E

Answer 1

Hint: Find the expression for \[f\left( \pi -x \right)\] and \[f\left( \pi +x \right)\] using the given expression for f(x). Then substitute the values of \[x=\dfrac{\pi }{5},\dfrac{2\pi }{5},\dfrac{3\pi }{5},\dfrac{4\pi }{5}\] in \[f\left( \pi -x \right)\] and \[f\left( \pi +x \right).\] Simplify the expression of T and check which of the terms in A, B, C, D, E cancels out. The expression will be independent of the terms which get cancelled.

Complete step-by-step solution
Here, we have been provided with the function:
\[f\left( x \right)=\cos 5x+A\cos 4x+B\cos 3x+C\cos 2x+D\cos x+E\]
Replacing x with \[\pi -x\] we get,
\[\Rightarrow f\left( \pi -x \right)=\cos \left( 5\pi -5x \right)+A\cos \left( 4\pi -4x \right)+B\cos \left( 3\pi -3x \right)+C\cos \left( 2\pi -2x \right)+D\cos \left( \pi -x \right)+E\]
\[\Rightarrow f\left( \pi -x \right)=-\cos 5x+A\cos 4x-B\cos 3x+C\cos 2x-D\cos x+E\]
Replacing x with \[\pi +x\] in f(x), we get,
\[\Rightarrow f\left( \pi +x \right)=\cos \left( 5\pi +5x \right)+A\cos \left( 4\pi +4x \right)+B\cos \left( 3\pi +3x \right)+C\cos \left( 2\pi +2x \right)+D\cos \left( \pi +x \right)+E\]
\[\Rightarrow f\left( \pi +x \right)=-\cos 5x+A\cos 4x-B\cos 3x+C\cos 2x-D\cos x+E\]
Clearly, we can see that,
\[f\left( \pi -x \right)=f\left( \pi +x \right)......\left( i \right)\]
So, substituting \[x=\dfrac{\pi }{5},\dfrac{2\pi }{5},\dfrac{3\pi }{5},\dfrac{4\pi }{5}\] in equation (i), we get the following four results.
\[\left( 1 \right)f\left( \dfrac{4\pi }{5} \right)=f\left( \dfrac{6\pi }{5} \right)\]
\[\left( 2 \right)f\left( \dfrac{3\pi }{5} \right)=f\left( \dfrac{7\pi }{5} \right)\]
\[\left( 3 \right)f\left( \dfrac{2\pi }{5} \right)=f\left( \dfrac{8\pi }{5} \right)\]
\[\left( 4 \right)f\left( \dfrac{\pi }{5} \right)=f\left( \dfrac{9\pi }{5} \right)\]
Now, we have been provided with the expression:
\[T=f\left( 0 \right)-f\left( \dfrac{\pi }{5} \right)+f\left( \dfrac{2\pi }{5} \right)-f\left( \dfrac{3\pi }{5} \right)+.....+f\left( \dfrac{8\pi }{5} \right)-f\left( \dfrac{9\pi }{5} \right)\]
Applying the above four obtained results, we get,
\[\Rightarrow T=f\left( 0 \right)-2f\left( \dfrac{\pi }{5} \right)+2f\left( \dfrac{2\pi }{5} \right)-2f\left( \dfrac{3\pi }{5} \right)+2f\left( \dfrac{4\pi }{5} \right)-f\left( \dfrac{5\pi }{5} \right)\]
\[\Rightarrow T=f\left( 0 \right)-f\left( \pi \right)+2\left[ -f\left( \dfrac{\pi }{5} \right)+f\left( \dfrac{2\pi }{5} \right)-f\left( \dfrac{3\pi }{5} \right)+f\left( \dfrac{4\pi }{5} \right) \right]\]
The above expression can be written as
\[\Rightarrow T=f\left( 0 \right)-f\left( \pi \right)+2\left[ \left\{ f\left( \dfrac{4\pi }{5} \right)-f\left( \dfrac{\pi }{5} \right) \right\}+f\left( \dfrac{2\pi }{5} \right)-f\left( \dfrac{3\pi }{5} \right) \right]......\left( ii \right)\]
Now subtracting \[f\left( \pi -x \right)\] from f(x), we get,
\[\Rightarrow f\left( x \right)-f\left( \pi -x \right)=2\cos 5x+2B\cos 3x+2D\cos x\]
Substituting x = 0, we get,
\[\Rightarrow f\left( 0 \right)-f\left( \pi \right)=2\cos 0+2B\cos 0+2D\cos 0\]
\[\Rightarrow f\left( 0 \right)-f\left( \pi \right)=2+2B+2D......\left( iii \right)\]
Substituting \[x=\dfrac{\pi }{5},\] we get,
\[f\left( \dfrac{\pi }{5} \right)-f\left( \dfrac{4\pi }{5} \right)=2\cos \dfrac{5\pi }{5}+2B\cos \dfrac{3\pi }{5}+2D\cos \dfrac{\pi }{5}\]
\[\Rightarrow f\left( \dfrac{\pi }{5} \right)-f\left( \dfrac{4\pi }{5} \right)=2\cos \pi +2B\cos \dfrac{3\pi }{5}+2D\cos \dfrac{\pi }{5}\]
\[\Rightarrow f\left( \dfrac{\pi }{5} \right)-f\left( \dfrac{4\pi }{5} \right)=-2+2B\cos \dfrac{3\pi }{5}+2D\cos \dfrac{\pi }{5}\]
\[\Rightarrow f\left( \dfrac{4\pi }{5} \right)-f\left( \dfrac{\pi }{5} \right)=2-2B\cos \dfrac{3\pi }{5}-2D\cos \dfrac{\pi }{5}......\left( iv \right)\]
Substituting \[x=\dfrac{2\pi }{5},\] we get,
\[\Rightarrow f\left( \dfrac{2\pi }{5} \right)-f\left( \dfrac{3\pi }{5} \right)=2\cos \dfrac{5\times 2\pi }{5}+2B\cos \dfrac{6\pi }{5}+2D\cos \dfrac{2\pi }{5}\]
\[\Rightarrow f\left( \dfrac{2\pi }{5} \right)-f\left( \dfrac{3\pi }{5} \right)=2\cos 2\pi +2B\cos \left( \pi +\dfrac{\pi }{5} \right)+2D\cos \dfrac{2\pi }{5}\]
\[\Rightarrow f\left( \dfrac{2\pi }{5} \right)-f\left( \dfrac{3\pi }{5} \right)=2-2B\cos \dfrac{\pi }{5}+2D\cos \dfrac{2\pi }{5}......\left( v \right)\]
Substituting the values of the expression (iii), (iv) and (v) in the equation (ii), we get,
\[\Rightarrow T=2+2B+2D+2\left[ \left\{ 2-2B\cos \dfrac{3\pi }{5}-2D\cos \dfrac{\pi }{5} \right\}+\left\{ 2-2B\cos \dfrac{\pi }{5}+2D\cos \dfrac{2\pi }{5} \right\} \right]\]
On simplifying the above expression, we get,
\[\Rightarrow T=10+2B+2D-4B\left( \cos \dfrac{3\pi }{5}+\cos \dfrac{\pi }{5} \right)+4D\left( \cos \dfrac{2\pi }{5}-\cos \dfrac{\pi }{5} \right)\]
Clearly, we can see that the value of T does not contain the terms of A, C and E but contains the terms of B and D. Therefore, the value of T depends on B, D but independent of A, C and E.
Hence, option (c) is the right answer.

Note: One must remember the sign of cosine function in all four quadrants. Also, remember that cosine function is negative in the second and third quadrant and positive in the first quadrant and fourth quadrant. Now, to solve the above question, we have to group the terms properly so that our calculation gets reduced and we do not get confused in so many terms.