JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumLimit, Continuity and Differentiability420× since 2002If the function f(x)={2x{sin(k1+1)x+sin(k2−1)x},x<04,x=02xloge(2+k1x2+k2x),x>0f(x) = \begin{cases} \frac{2}{x} \left\{ \sin (k_1 + 1) x + \sin (k_2 - 1) x \right\}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x > 0 \end{cases}f(x)=⎩⎨⎧x2{sin(k1+1)x+sin(k2−1)x},4,x2loge(2+k2x2+k1x),x<0x=0x>0 is continuous at x=0x = 0x=0, then k12+k22k_1^2 + k_2^2k12+k22 isAOption A. 2BOption B. 5COption C. 8DOption D. 10CheckSkip · show solution