+cos(2x2)] =−cos(x2)+2cos(x2)cos(2x2) =−cos(x2)+cos(3x2)+cos(x2) ⇒f(x)=cos(3x2) . . . (i) On differentiating w.r.t. x, we get f′(x)=−sin(3x2)(6x) For extremum, put f′(x)=0 ⇒−sin(3x2)(6x)=0 ⇒x=0,π Put x=0,π, in equation (i), we get f(0)=cos(0)=1 f(π)=cos(π)=−1