If the function f(x)={(2/x{sin(k1+1)x+sin(k2-1)x} , x<0), (4 , x=0), (2/xloge(2+k1x/2+k2x)) , x>0

Question

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 \(\mathrm{x}=0,\) then \(\mathrm{k}_{1}{ }^{2}+\mathrm{k}_{2}{ }^{2}\) is equal to

(1) 8

(2) 20

(3) 5

(4) 10

Answer 1

Correct option is (4) 10 

f(x) is continuous at x = 0,

\(\therefore \mathrm{f}\left(0^{-}\right)=\mathrm{f}(0) \)

\(\lim _{\mathrm{x} \rightarrow 0^{-}} \frac{2}{\mathrm{x}}\left[\sin \left(\mathrm{k}_1+1\right) \mathrm{x}+\sin \left(\mathrm{k}_2-1\right) \mathrm{x}\right] \)

\(\lim _{\mathrm{x} \rightarrow 0^{-}}\left[\frac{\sin \left(\mathrm{k}_1+1\right)}{\left(\mathrm{k}_1+1\right) \mathrm{x}} \times\left(\mathrm{k}_1+1\right)+\frac{\sin \left(\mathrm{k}_2-1\right)}{\left(\mathrm{k}_2-1\right) \mathrm{x}} \times\left(\mathrm{k}_2-1\right)\right]=2 \)

\({\left[\mathrm{k}_1+1+\mathrm{k}_2-1\right]=2} \)

\(\mathrm{k}_1+\mathrm{k}_2=2 \)     ....(1)

\( \text { again, } \mathrm{F}\left(0^{+}\right)=\mathrm{F}(0) \)

\(\frac{2}{\mathrm{x}}\left[In 2\left(1+\frac{\mathrm{k}_1 \mathrm{x}}{2}\right)-In 2\left(1+\frac{\mathrm{k}_2 \mathrm{x}}{2}\right)\right]=4 \)

\(\frac{2}{\mathrm{x}}\left[In 2+In \left(1+\frac{\mathrm{k}_1 \mathrm{x}}{2}\right)-In 2-In \left(1+\frac{\mathrm{k}_2 \mathrm{x}}{2}\right)\right]=4 \) 

\(\left[\left\{\frac{\ln \left(1+\frac{k_1 \mathrm{x}}{2}\right)}{\left(\frac{\mathrm{k}_1 \mathrm{x}}{2}\right)}\right\}, \frac{\mathrm{k}_1}{2}-\left(\frac{\ln \left(1+\frac{\mathrm{k}_2 \mathrm{x}}{2}\right)}{\left(\frac{\mathrm{k}_2 \mathrm{x}}{2}\right)}\right) \frac{\mathrm{k}_2}{2}\right]\)

\(\mathrm{k}_1-\mathrm{k}_2=4 \)  ...(2)

Adding (1) + (2)

\(2 \mathrm{k}_1=6 \)

\(k_1=3 \Rightarrow k_2=-1\)

\(\therefore \mathrm{k}_1^2+\mathrm{k}_2^2=9+1=10\)