Continuity at a point – 23 January 2025 (Shift 1)

Solution:

For continuity at $x=0$, LHL = RHL = $f(0) = 4$.
**LHL:**
$$\lim_{x \to 0^-} \frac{2}{x} \left(\sin(k_1+1)x + \sin(k_2-1)x\right) = 4$$
$$2(k_1+1) + 2(k_2-1) = 4 \Rightarrow 2(k_1+k_2) = 4 \Rightarrow k_1+k_2 = 2 \quad \dots(1)$$
**RHL:**
$$\lim_{x \to 0^+} \frac{2}{x} \ln\left(\frac{2+k_1x}{2+k_2x}\right) = 4$$
$$\lim_{x \to 0^+} \frac{2}{x} \ln\left(1 + \frac{(k_1-k_2)x}{2+k_2x}\right) = 4$$
$$2 \left( \frac{k_1-k_2}{2} \right) = 4 \Rightarrow k_1 – k_2 = 4 \quad \dots(2)$$
Solving (1) and (2):
$2k_1 = 6 \Rightarrow k_1 = 3$
$k_2 = -1$
$$k_1^2 + k_2^2 = (3)^2 + (-1)^2 = 9 + 1 = 10$$
Ans. (4)