Continuity and Differentiability
Differentiability — Matching Value and Derivative
nta_pyq_2026_jan
Grade 12
Question:
Let $\alpha,\beta\in\mathbb{R}$ be such that the function $f(x)=\begin{cases}2\alpha(x^2-2)+2\beta x & x<1\\(\alpha+3)x+(\alpha-\beta) & x\geq1\end{cases}$ be differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to
Step-by-Step Solution
Key Concept: Continuity at $x=1$: $f(1^-)=2\alpha(-1)+2\beta=f(1)=2\alpha+3$. So $4\alpha-3\beta+3=0$ ...(1). Differentiability at $x=1$: $f'(1^-)=4\alpha x|_{x=1}+2\beta=4\alpha+2\beta$ and $f'(1^+)=\alpha+3$. So $4\alpha+2\beta=\alpha+3$ ...(2).
$\alpha=3/17$, $\beta=21/17$. $34(\alpha+\beta)=48$.
Correct Answer: 4