Continuity & Differentiability: Numerical Problems Set (5)

Question

Find the values of $a$ and $b$ such that $f(x)$ is continuous at $x = 1$ and differentiable there: $f(x) = \begin{cases} ax + b, & x < 1 \\ x^2, & x \ge 1 \end{cases}$

Solution — Step by Step

The values are $a = 2$ and $b = -1$.

Why This Works

Continuity at a point requires the function to “join up” — left limit = right limit = function value. Differentiability is stricter: the slopes from the left and right must also match. Together, two conditions pin down two unknowns.

This is the standard JEE/CBSE setup: a piecewise function with two parameters, joined at a single point. The trick is to write both conditions cleanly and solve the linear system.

Alternative Method

Compute the left and right derivatives from first principles. For $x < 1$: $f'(x) = a$, so $f'(1^-) = a$. For $x \ge 1$: $f'(x) = 2x$, so $f'(1^+) = 2$. Equating gives $a = 2$. Then continuity gives $b = -1$. Same answer.

Common Mistake

Imposing only continuity and trying to find both $a$ and $b$ — leads to infinitely many solutions. The fix: read the problem carefully. If “differentiable” is stated, both conditions apply. CBSE 2023 had this exact phrasing.