Continuity & Differentiability: Speed-Solving Techniques (2)

Question

For what values of $a$ and $b$ is the function $f(x) = \begin{cases} ax + b & x \leq 1 \\ x^2 + 3 & x > 1 \end{cases}$ both continuous and differentiable at $x = 1$? JEE Main 2023 pattern, CBSE 2024 boards.

Solution — Step by Step

Why This Works

A function is differentiable at a point only if it’s continuous there — but the converse is false. So we always check continuity first, then differentiability. Continuity gives us one equation, differentiability gives a second; with two unknowns ($a$ and $b$), we can solve.

Geometrically: the line $ax + b$ must meet the parabola $x^2 + 3$ smoothly at $x = 1$. “Smoothly” means same height (continuous) and same slope (differentiable). Two conditions, two unknowns.

Alternative Method

The “smooth join” interpretation gives the answer almost instantly. The parabola at $x = 1$ has $y = 4$ and slope $y' = 2 \cdot 1 = 2$. So the line $y = ax + b$ at $x = 1$ must have $y = 4$ (giving $a + b = 4$) and slope $a = 2$. Solve: $a = 2, b = 2$.