If the function f(x)=2xsin⁡k1+1x+sin⁡k2−1x,x 0 is continuous at x=0, then k12 + k22 is equal to

limx→0− 2xsin⁡k1+1x+sin⁡k2−1x=4⇒2k1+1+2k2−1=4⇒k1+k2=2⇒limx→0+ 2xln⁡2+k1x2+k2x=4⇒limx→0+ 1xln⁡1+k1−k2x2+k2x=2⇒k1−k22=2⇒k1−k2=4∴k1=3,k2=−1k12+k22=9+1=10

If the function f(x)=2xsin⁡k1+1x+sin⁡k2−1x,x 0 is continuous at x=0, then k12 + k22 is equal to

limx→0− 2xsin⁡k1+1x+sin⁡k2−1x=4⇒2k1+1+2k2−1=4⇒k1+k2=2⇒limx→0+ 2xln⁡2+k1x2+k2x=4⇒limx→0+ 1xln⁡1+k1−k2x2+k2x=2⇒k1−k22=2⇒k1−k2=4∴k1=3,k2=−1k12+k22=9+1=10

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If the function $f (x) = \{\begin{array}{cc} \frac{2}{x} \{\sin (k_{1} + 1) x + \sin (k_{2} - 1) x\} & , x < 0 \\ 4 & , x = 0 \\ \frac{2}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}) & , x > 0 \end{array}$ is continuous at x=0, then k₁²+ k₂² is equal to

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Detailed Solution

$\begin{array}{l} lim_{x \to 0^{-}} \frac{2}{x} \{\sin (k_{1} + 1) x + \sin (k_{2} - 1) x\} = 4 \\ \Rightarrow 2 (k_{1} + 1) + 2 (k_{2} - 1) = 4 \\ \Rightarrow k_{1} + k_{2} = 2 \\ \Rightarrow lim_{x \to 0^{+}} \frac{2}{x} \ln (\frac{2 + k_{1} x}{2 + k_{2} x}) = 4 \\ \Rightarrow lim_{x \to 0^{+}} \frac{1}{x} \ln (1 + \frac{(k_{1} - k_{2}) x}{2 + k_{2} x}) = 2 \\ \Rightarrow \frac{k_{1} - k_{2}}{2} = 2 \\ \Rightarrow k_{1} - k_{2} = 4 \\ ∴ k_{1} = 3, k_{2} = - 1 \\ k_{1}^{2} + k_{2}^{2} = 9 + 1 = 10 \end{array}$