Limits, Continuity & Differentiability
Continuity And Differentiability
nta_abhyas_2025
Grade 12
Question:
If $f(x) = \begin{cases} px + q & x \leq 2 \\ x^2 - 5x + 6 & 2 < x < 3 \\ a x^2 + bx + 1 & x \geq 3 \end{cases}$ is differentiable everywhere, then $|p| + |q| + \left|\frac{1}{3}\right| + |1|$ is equal to
Step-by-Step Solution
Key Concept: Continuity at multiple points creates a system of equations; differentiability adds derivative matching conditions.
From continuity at $x = 2$: $p(2) + q = 2^2 - 5(2) + 6 \implies q = -2p$. From continuity at $x = 3$: $p(3) + q(3) + 1 = 0 \implies 3p + 3q + 1 = 0$. Substituting $q = -2p$: $3p - 6p + 1 = 0 \implies p = \frac{1}{3}$, so $q = -\frac{2}{3}$. From differentiability at $x = 2$: $p = -1$ and from $x = 3$: $2a - 6 + b = -1$. Solving yields $p = -1, q = \frac{1}{3}, a = \frac{5}{2}, b = \frac{5}{2}$. Therefore $|p| + |q| + |\frac{1}{2}| + |\frac{1}{2}| = 1 + \frac{1}{3} + \frac{1}{2} + \frac{1}{2} = 3$.
Correct Answer: 3