Good question

f'(1)=1/√3
f'(2)=√3
f'(3)=1
Now separating the integration(I) into two parts-I1 and I2
I1=∫f'(x).f"(x).dx (with limits from 2 to 3)
I2=∫f"(x).dx (with limits from 1 to 3)
Now applying IBP in I1 we get the value as 1-3/2..i.e -1
And I2=f'(3)-f'(1)=1-1/√3
Now required integration(I)= I1+I2 =-1/√3
Hence B is correct......:)

Limits are 2 to 3 and 1 to 3.

(B)