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dear meghana, we see that f(x)= g(x) + h(x) where g(x)= |x – pi/2|^3 and h(x)= sin^2x 
so, f”(x)= g”(x) + h”(x)
now, h’(x)= 2sinxcosx= sin2x so h”(x)= 2cos2x so that h”(pi/2)= 2cospi= – 2
now, to find g”(x) at pi/2, we first find RHL and then LHL: 
RHL= Lt x tends to pi/2+ [g’(x) – g’(pi/2)]/(x – pi/2) and LHL= Lt x tends to pi/2- [g’(x) – g’(pi/2)]/(x – pi/2) 
now, g’(pi/2+)= Lt x tends to pi/2+ [g(x) – g(pi/2)]/(x – pi/2)= Lt x tends to pi/2+ (x – pi/2)^3/(x – pi/2)
= Lt x tends to pi/2+ (x – pi/2)^2= 0
similarly g’(pi/2-)= Lt x tends to pi/2– [g(x) – g(pi/2)]/(x – pi/2)= Lt x tends to pi/2- (pi/2 – x)^3/(x – pi/2)
= Lt x tends to pi/2- – (x – pi/2)^2= 0
hence, g’(pi/2)= 0 since g’(pi/2+)= g’(pi/2-)= 0
so, RHL=  Lt x tends to pi/2+ [g’(x) – 0]/(x – pi/2)=  Lt x tends to pi/2+ g’(x)/(x – pi/2)
but for x greater than pi/2, g’(x)=  Lt h tends to 0 [g(x+h) – g(x)]/h
= Lt h tends to 0 [(x+h – pi/2)^3 – (x – pi/2)^3]/h= 3(x – pi/2)^2
so RHL= Lt x tends to pi/2+ g’(x)/(x – pi/2)= Lt x tends to pi/2+ 3(x – pi/2)^2/(x – pi/2)= 0
similarly LHL= Lt x tends to pi/2- g’(x)/(x – pi/2)
for x less than pi/2, g’(x)= Lt h tends to 0 [g(x+h) – g(x)]/h= Lt h tends to 0 [(pi/2 – x – h)^3 – (pi/2 – x)^3]/h= 3(pi/2 – x)^2*( – 1)= – 3(x – pi/2)^2
so, LHL= Lt x tends to pi/2- g’(x)/(x – pi/2)= LHL= Lt x tends to pi/2- – 3(x – pi/2)^2/(x – pi/2)= 0
since LHL= RHL= 0, we have g”(pi/2)= 0
hence, f”(pi/2)= g”(pi/2) + h”(pi/2)
= 0 + ( – 2)
= – 2
KINDLY APPROVE :))