1.If the functionf(x)={sin((k1+1)x)+sin((k2−1)x)x,x<04,x=02−loge(2+x)x,x>0f(x) = \begin{cases} \frac{\sin((k_1+1)x) + \sin((k_2-1)x)}{x}, & x < 0 \\ 4, & x = 0 \\ \frac{2 - \log_e(2+x)}{x}, & x > 0 \end{cases}f(x)=⎩⎨⎧xsin((k1+1)x)+sin((k2−1)x),4,x2−loge(2+x),x<0x=0x>0is continuous at x=0x = 0x=0, then k12+k22k_1^2 + k_2^2k12+k22 is equal toa.20b.5c.8d.10Login to continueOnly logged in users canattempt or see the solution.
