if u r aware with sin(x) graph then u can directly say answer is b
now see rational proof:
f(x)=sinx , f'(x)=cos(x) , f"(x)=-sin(x)
put f'(x)=0
cos(x)=0
therefore x=n(pi)+pi/2, hence f"(x)=-sin(n*pi+pi/2)= -cos(n*pi)
therefore for all odd values of n, f"(x) will be positive hence local minima will occur here
now in given question if x=3pi/2 i.e x= pi+pi/2 therefore n= 1 (odd) option b is correct
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@Arjun Sir the problem is not with the ...