CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
12
loader
C
1
loader
D
none of these
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image