Limits, Continuity & Differentiability
Continuity And Differentiability
nta_abhyas_2025
Grade 12
Question:
If $h(x) = \begin{cases} \frac{\lambda\sqrt{2x+3}}{0 \leq x \leq 3} \\ \mu x + 12, & 3 < x \leq 9 \end{cases}$ is differentiable at $x = 3$, then the value of $\lambda + \mu$ is equal to
Step-by-Step Solution
Key Concept: For differentiability at a point, both continuity and equality of left and right derivatives are required.
Since $f(x)$ is continuous at $x = 3$, we require $\text{LHL} = \text{RHL} = f(3)$. Computing: $\sqrt{2(3)} + 3 = \mu + \lambda \implies \mu + \lambda = 12$. For differentiability at $x = 3$, the left and right derivatives must be equal. From the left, $f'(3^-) = 2$ (from the derivative of the first piece). From the right, $f'(3^+) = \mu$. Therefore $\mu = 2$. Substituting into the continuity equation: $2 + \lambda = 12 \implies \lambda = 6$.
Correct Answer: μ = 2, λ = 6