CONCEPT: Second Derivative Test: Let f be a function defined on an interval I. - Calculate f′(x) - Solve f′(x)=0 and find the roots of f′(x)=0. Suppose x=c is the root of f′(x)=0. - Calculatef" (x) and put x=c to get the value of f "(c). - If f′′(c)<0 then x=c is a point of local maxima. - If f " (c)>0 then x=c is a point of local minima. - If f"(c)=0 then we need to use the first derivative test. Note: Maxima and Minima on a closed Interval: Let f(x) be a given function defined on [a,b]. Let the local minimum value of f(x) be m and let the local maximum value of f(x) be M. Then Minimum value of f(x) on [a,b] is the smallest of m,f(a) and f(b) Maximum value of f(x) on [a,b] is the greatest of M,f(a) and f(b) CALCULATION: Let f(x)=sin‌2‌x⋅cos‌2‌x ⇒f′(x)=2cos22x−2sin22x ⇒f′(x)=2⋅(cos22x−sin22x) As we know that, cos‌2‌x=cos2x−sin2x ⇒f′(x)=2‌cos‌4‌x If f′(x)=0 then 2‌cos‌4‌x=0⇒x=π∕8 ⇒f′(x)=−8‌sin‌4‌x ⇒f′′(x)=−8<0 So, x=π∕8 is the point of maxima. So, the maximum value of f(x)=sin‌2‌x⋅cos‌2‌x is given by f(π∕8)=sin(π∕4)⋅cos(π∕4)=1∕2 Hence, correct option is 1 .