Question

If the function $f(x) = \begin{cases} \frac{2}{x}\{\sin{(k_1+1)x} + \sin{(k_2-1)x}\} &, x<0 \\ 4 &, x=0_{is} \\ \frac{2}{x}\log_e{(\frac{2+k_1x}{2+k_2x})} &, x>0 \end{cases}$

continuous at x=0, then $k_1^2+k_2^2$ is equal to

• A. 8
• B. 20

Answer 1

10

Answer 2

For continuity at x = 0, the left‐limit, right‐limit, and f(0) must be equal.

1. For x < 0:

$f(x)=\frac{2}{x}\Bigl[\sin((k_1+1)x)+\sin((k_2-1)x)\Bigr]$

Using $\sin(ax)\approx ax$ for small x,

$\sin((k_1+1)x)+\sin((k_2-1)x)\approx (k_1+1)x+(k_2-1)x= (k_1+k_2)x.$

Thus,

$\lim_{x\to0^-}f(x)=\frac{2}{x}(k_1+k_2)x= 2(k_1+k_2).$

For continuity, set

$2(k_1+k_2)=4 \quad \Rightarrow \quad k_1+k_2=2. \quad\quad (1)$

2. For x > 0:

$f(x)=\frac{2}{x}\ln\left(\frac{2+k_1x}{2+k_2x}\right).$

Expand for small x:

$\ln(2+kx)=\ln2+\frac{k}{2}x + O(x^2).$

Therefore,

$\ln\left(\frac{2+k_1x}{2+k_2x}\right) \approx \left(\frac{k_1}{2}-\frac{k_2}{2}\right)x = \frac{k_1-k_2}{2}x.$

Now,

$\lim_{x\to0^+} f(x)= \frac{2}{x}\cdot \frac{k_1-k_2}{2}x = k_1-k_2.$

For continuity at x = 0, we require

$k_1-k_2=4. \quad\quad (2)$

3. Solve the system (1) & (2):

$\begin{cases} k_1+k_2=2,\\[1mm] k_1-k_2=4. \end{cases}$

Adding,

$2k_1=6 \quad \Rightarrow \quad k_1=3.$

Then from (1),

$k_2=2-k_1=2-3=-1.$

4. Compute $k_1^2+k_2^2$:

$k_1^2+k_2^2 = 3^2 + (-1)^2 = 9+1=10.$