Consider f(x)=ax2+bx+c f(2)=4a+2b+c=2⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) f(3)=9a+3b+c=5⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) f(4)=16a+4b+c=10⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(III) Subtract equation (I) from equation (II), 5a+b=3⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(IV) Subtract equation (II) from equation (III), 7a+b=5⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(V) Subtract equation (IV) from equation (V), 2a=2 a=1 Substitute the value in equation (IV), 5×1+b=3 b=−2 Now, f′(x)=2x−2 f′′(x)=2 since, f′′(x)>1 for some x∈(2,4)