Let f(x) = ax(x-1) for x < 1; x - 1 for 1 ≤ x ≤ 3; px² + qx + 2 for x > 3. Given that f(x) is continuous for all x but not differentiable at x = 1. Further f'(x) is continuous at x = 3. What is the value of p?

Continuity at x = 3: 3 - 1 = 9p + 3q + 2, so 9p + 3q = 0, i.e., q = -3p. Differentiability requires f'(3⁻) = f'(3⁺). f'(x) for 1≤x≤3 is 1; for x > 3 is 2px + q. At x = 3: 1 = 6p + q = 6p - 3p = 3p. So p = 1/3. Hmm, checking: actually re-examining, we need f' continuous at x=3 means 1 = 6p + q. Combined with 9p+3q=0 giving q=-3p, then 1 = 6p - 3p = 3p, so p = 1/3. But the marked answer should be (b) -1/3 based on options. Let me reconsider: At x=3, value from middle piece is 2, value from right piece is 9p+3q+2, equating gives 9p+3q=0. Derivative from middle is 1, from right at x=3 is 6p+q. So 6p+q=1 and q=-3p gives 3p=1, p=1/3. Answer is (c) 1/3.