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If f(x)=cosx.cos2x.cos4x.cos8x.cos16x ,...

If `f(x)=cosx.cos2x.cos4x.cos8x.cos16x` , then `f'(pi/4)` is :

Text Solution

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differentiate with respect to x
`f'(x)=-sinx*cos2x*cos4x*cos8x*cos16x-cosx*2sin2x*cos4x*cos16-4cosxcos2xsin4x*cos8x*cos16x-8cosxcos2xcos4xsin8xcos16x-16cosxcos2xcos4xcos8xsin16x`
`f'(pi/4)=-cos(pi/4)*2*sin(pi/2)cospicos2picos4pi`
`=-1/sqrt2*2*1*(-1)*1*1`
`=2/sqrt2=sqrt2`.
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Knowledge Check

  • if f(x) = cos x cos 2x cos 4x cos 8 x cos 16 x , then f' (pi //4) is

    A
    `1`
    B
    ` sqrt(2)`
    C
    `1 // sqrt(2)`
    D
    None

  • If f(x) = cos x cos 2x cos 4x cos 8x cos 16 x then f^1((pi)/(4)) equals

    A
    `sqrt(2)`
    B
    `0`
    C
    `1/(sqrt(2))`
    D
    `"Cosec" pi/4`

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