MCQ
If the function
$
f(x)=\left\{\begin{array}{cc}
\frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) 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{array}\right.
$
is continuous at $x =0$, then $k _1{ }^2+ k _2{ }^2$ is equal to
  • A
    8
  • B
    20
  • C
    5
  • 10

Answer

Correct option: D.
10
(D)
$
\begin{array}{l}
\lim _{x \rightarrow 0^{-}} \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}=4 \\
\Rightarrow 2\left(k_1+1\right)+2\left(k_2-1\right)=4 \\
\Rightarrow k_1+k_2=2 \\
\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{2}{x} \ln \left(\frac{2+k_1 x}{2+k_2 x}\right)=4 \\
\Rightarrow \lim _{x \rightarrow 0^{+}} \frac{1}{x} \ln \left(1+\frac{\left(k_1-k_2\right) x}{2+k_2 x}\right)=2 \\
\Rightarrow \frac{k_1-k_2}{2}=2 \\
\Rightarrow k_1-k_2=4 \\
\therefore k_1=3, k_2=-1 \\
k_1^2+k_2^2=9+1=10
\end{array}
$

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