Limits, Continuity & Differentiability
Differentiability of Piecewise Function — Finding Constants
nta_pyq_2024_jan
Grade 12
Question:
If the function $f(x)=\begin{cases}\dfrac{1}{|x|} & ,\; |x|\ge 2\\ax^2+2b & ,\; |x|<2\end{cases}$ is differentiable on $\mathbb{R}$, then $48(a+b)$ is equal to ______.
Step-by-Step Solution
Key Concept: For differentiability on $\mathbb{R}$, impose continuity and equal derivatives at $x=2$ (and by symmetry at $x=-2$). Continuity at $x=2$: $\frac{1}{2}=4a+2b$. Differentiability at $x=2$: $-\frac{1}{4}=2a(2)=4a$.
Continuity at $x=2$: $4a+2b=\frac{1}{2}$.
Differentiability at $x=2$: derivative of $ax^2+2b$ is $2ax$; at $x=2$: $4a=-\frac{1}{4}\Rightarrow a=-\frac{1}{16}$.
From continuity: $2b=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\Rightarrow b=\frac{3}{8}$.
$48(a+b)=48\left(-\frac{1}{16}+\frac{3}{8}\right)=48\cdot\frac{5}{16}=15$.
Correct Answer: 15