We have, f(x)=3sin‌12x+4cos10x Period of 3sin‌12x is π. Period of 4cos16x is π. ∴ Period of f(x)=3sin‌12x+4cos16x is π Maximum of f(x)= Maximum of f(x) in [0,π] f(x)=3sin‌12x+4cos16x On differentiating w.r.t. x, we get ‌f′(x)=36sin‌11x‌cos‌x+64cos15x(−sin‌x) ‌⇒f′(x)=4sin‌x‌cos‌x(9sin‌10x−16cos14x) ‌⇒f′(x)=2sin‌2x(9sin‌10x−16cos14x) On putting f′(x)=0, we get ‌2sin‌2x(9sin‌10x−16cos14x)=0 ‌⇒‌‌2sin‌2x=0⇒x=0,‌
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,π ‌‌ x ‌0r∕2r ‌f(x)43‌‌4 Clearly, the maximum value of f(x) is 4 .