Limits, Continuity & Differentiability
Continuity of Piecewise Function
nta_pyq_2025_apr
Grade 12
Question:
If the function $f(x) = \begin{cases} \frac{2}{x}\{\sin(k_1+1)x + \sin(k_2-1)x\}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x}\log_e\!\left(\frac{2+k_1x}{2+k_2x}\right), & x > 0 \end{cases}$ is continuous at $x = 0$, then $k_1^2 + k_2^2$ is equal to:
Step-by-Step Solution
Key Concept: For continuity at $x=0$, LHL = RHL = 4. Use $\lim_{x\to0}\frac{\sin ax}{x}=a$ for LHL, and $\lim_{x\to0}\frac{1}{x}\ln(1+u)\approx u/x$ for RHL.
LHL: $\lim_{x\to0^-}\frac{2}{x}(\sin(k_1+1)x+\sin(k_2-1)x)=2(k_1+1)+2(k_2-1)=4 \Rightarrow k_1+k_2=2$. RHL: $\lim_{x\to0^+}\frac{2}{x}\ln\frac{2+k_1x}{2+k_2x}=\frac{k_1-k_2}{2}\cdot2=4\Rightarrow k_1-k_2=4$. Solving: $k_1=3,k_2=-1$. $k_1^2+k_2^2=9+1=10$.
Correct Answer: 10