JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If the function
\(f(x)=\left\{\begin{array}{l}\frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}, \quad x \lt 0 \\ 4, \quad x=0 \\ \frac{2}{x} \log _e\left(\frac{2+k_1 x}{2+k_2 x}\right), \quad x\gt0\end{array}\right.\)
is continuous at \(\mathrm{x}=0\), then \(\mathrm{k}_1^2+\mathrm{k}_2^2\) is equal to
- A \(20\)
- B \(5\)
- C \(8\)
- D \(10\)
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{2}{\mathrm{x}}\left\{\sin \left(\mathrm{k}_1+1\right) \mathrm{x}+\sin \left(\mathrm{k}_2-1\right) \mathrm{x}\right\}=4 \\ & \Rightarrow 2\left(\mathrm{k}_1+1\right)+2\left(\mathrm{k}_2-1\right)=4 \\ & \Rightarrow…
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