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J & K CETJ & K CET 2007Application of Derivatives
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Solution:
Let y=sin3x+cos3x dxdy=3sin2xcosx−3cos2xsinx =3sinxcosx(sinx−cosx) Put dxdy=0 ⇒3sinxcosx(sinx−cosx)=0 ⇒x=0,2π or 4π
Now, y has its maximum value at x=0 or 2π,
and ymax=1