Consider the piecewise function, f(x)=⎩⎨⎧xx−∣x∣,5x2+a,b(x2−3x+2x2−1),−14, when x<0 when 0≤x≤1 when 1<x<3 when x≥3. The function is continuous at x=0 and x=3 LHL at x=0 is equal to RHL at x=0x→0−limf(x)=x→0+limf(x)x→0−limxx−∣x∣=x→0+lim5x2+ax→0−limxx−(−x)=x→0+lim5x2+ax→0−limx2x=x→0+lim5x2+a Solve further 2=aa=2 The function is continuous at x=3 LHL at x=3 is equal to RHL at x=3x→3−limf(x)=x→3+limf(x)x→3−limb(x2−3x+2x2−1)=x→3+lim−14x→3−lim((x−1)(x−2)(x+1)(x−1))=x→3+lim−14x→3−limb(x−2x+1)=x→3+lim−14 Solve further, b3−23+1=−144b=−14b=4−14b=−27 Therefore (a,b)=(2,−27)