Limits, Continuity & Differentiability
Continuity — Finding Parameters
nta_pyq_2024_apr
Grade 12
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ be a function given by $f(x)=\begin{cases}\dfrac{1-\cos2x}{x^2}, & x<0\\ \alpha, & x=0\\ \dfrac{\beta\sqrt{1-\cos x}}{x}, & x>0\end{cases}$ where $\alpha,\beta\in\mathbb{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to:
Step-by-Step Solution
Key Concept: $f(0^-)=\lim_{x\to0^-}\frac{1-\cos2x}{x^2}=\frac{2\sin^2x}{x^2}\to2=\alpha$. $f(0^+)=\lim_{x\to0^+}\frac{\beta\sqrt{1-\cos x}}{x}=\beta\cdot\frac{\sqrt{2}|\sin(x/2)|}{x}\to\frac{\beta}{\sqrt{2}}=2\Rightarrow\beta=2\sqrt{2}$.
$\alpha=2$, $\beta=2\sqrt{2}$. $\alpha^2+\beta^2=12$.
Correct Answer: 2