Question

# If $$f(x)=\cos x\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x$$, then $$f'\left (\dfrac {\pi}{4}\right )$$ is

A
2
B
12
C
1
D
none of these

Solution

## The correct option is A $$\sqrt 2$$Given that$$f\left( x \right) =\cos { x } \cos { 2x } \cos { 4x } \cos { 8x } \cos { 16x }$$Multiply & divide by $$2\sin { x }$$$$f\left( x \right) =\displaystyle \dfrac { 2\sin { x } \cos { x } \cos { 2x } \cos { 4x } \cos { 8x } \cos { 16x } }{ 2\sin { x } }$$Since $$\sin { 2x } =2\sin { x } \cos { x }$$$$f\left( x \right) =\displaystyle \dfrac { \sin { 2x } \cos { 2x } \cos { 4x } \cos { 8x } \cos { 16x } }{ 2\sin { x } }$$ Multiplying and dividing by $$2$$, we get$$f\left( x \right) =\displaystyle \dfrac { 2\sin { 2x } \cos { 2x } \cos { 4x } \cos { 8x } \cos { 16x } }{ { 2 }^{ 2 }\sin { x } }$$$$f\left( x \right) =\displaystyle \dfrac { \sin { 4x } \cos { 4x } \cos { 8x } \cos { 16x } }{ { 2 }^{ 2 }\sin { x } }$$Similarly continuing upto $$\cos { 16x }$$, we get$$f\left( x \right) =\displaystyle \dfrac { \sin { { 2 }^{ 5 }x } }{ { 2 }^{ 5 }\sin { x } } =\dfrac { \sin { 32x } }{ 32\sin { x } }$$Differentiating w.r. to $$x$$$$f^{ ' }\left( x \right) =\displaystyle \dfrac { 1 }{ 32 } \left[ \dfrac { 32\sin { x } \cos { 32x } -\cos { x } \sin { 32x } }{ { \left( \sin { x } \right) }^{ 2 } } \right]$$$$\therefore f^{ ' }\left( \displaystyle \dfrac { \pi }{ 4 } \right) =\displaystyle \dfrac { 1 }{ 32 } \left[ \displaystyle \dfrac { 32\sin {\displaystyle \dfrac { \pi }{ 4 } } \cos { 8\pi } -\cos {\displaystyle \dfrac { \pi }{ 4 } } \sin { 8\pi } }{ { \left( \sin {\displaystyle \dfrac { \pi }{ 4 } } \right) }^{ 2 } } \right]$$$$\therefore f^{ ' }\left(\displaystyle \dfrac { \pi }{ 4 } \right) =\displaystyle \dfrac { 2 }{ \sqrt { 2 } } =\sqrt { 2 }$$Mathematics

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