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sin ax +cos ax and IsinxI +Icos xI are periodic functions of same fundamental period, then a equals (a) 0 (b) 1 (C) 2 (d) 4
Why?
Differentiate the equ. sin(ax)+cos(ax)=>acos(ax)-asin(ax)
To find the maxima putting the above equation equal to 0.Then acos(ax)=asin(ax)=>sin(ax)=cos(ax)
This is possible only when ax=(pie)/4 or 5(pie)/4 or 9(pie)/4
=(pie)/4a or 5(pie)/4a or 9(pie)/4a
One of the solution must be Maxima and other must be Minima, 9(pie)/4 wil again be the Maxima.So the period of the equation must be 9(pie)/4a-(pie)/4a=2(pie)/a. Since the period of the equation mod(sin(x))+mod(cos(x))=(pie)
So 2(pie)/a should be equal to pie so a=2
Hi Aditi,
|sinx| + |cosx| has fundamental period of pi/2
Since f(x+pi/2) = f(x)
So, for sinax+cosax to have fundamental period as pi/2, "a" must be 4.
Hope that helps.
Best Regards,
Ashwin (IIT Madras).
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