Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 393 Question 393 of 401 Medium MCQ
If the function
\[
f(x)=\left\{\[\begin{gathered} \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}, \quad x0 \end{gathered}\]\right.
\]
is continuous at x = 0 x=0 x = 0 , then k 1 2 + k 2 2 k_1^2+k_2^2 k 1 2 + k 2 2 is equal to :
Solutions · 1 Post your solution Official solution
\[ lim x → 0 − 2 x { sin ( k 1 + 1 ) x + sin ( k 2 − 1 ) x } = 4 ⇒ 2 ( k 1 + 1 ) + 2 ( k 2 − 1 ) = 4 ⇒ k 1 + k 2 = 2 ⇒ lim x → 0 + 2 x ln ( 2 + k 1 x 2 + k 2 x ) = 4 ⇒ lim x → 0 + 1 x ln ( 1 + ( k 1 − k 2 ) x 2 + k 2 x ) = 2 ⇒ k 1 − k 2 2 = 2 ⇒ k 1 − k 2 = 4 ∴ k 1 = 3 , k 2 = − 1 k 1 2 + k 2 2 = 9 + 1 = 10 \begin{aligned}
& \lim _{x \rightarrow 0^{-}} \frac{2}{\mathrm{x}}\left\{\sin \left(\mathrm{k}_1+1\right) \mathrm{x}+\sin \left(\mathrm{k}_2-1\right) \mathrm{x}\right\}=4 \\
& \Rightarrow 2\left(\mathrm{k}_1+1\right)+2\left(\mathrm{k}_2-1\right)=4 \\
& \Rightarrow \mathrm{k}_1+\mathrm{k}_2=2 \\
& \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{2}{\mathrm{x}} \ln \left(\frac{2+\mathrm{k}_1 \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=4 \\
& \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{1}{\mathrm{x}} \ln \left(1+\frac{\left(\mathrm{k}_1-\mathrm{k}_2\right) \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=2 \\
& \Rightarrow \frac{\mathrm{k}_1-\mathrm{k}_2}{2}=2 \\
& \Rightarrow \mathrm{k}_1-\mathrm{k}_2=4 \\
& \therefore \mathrm{k}_1=3, \mathrm{k}_2=-1 \\
& \mathrm{k}_1^2+\mathrm{k}_2^2=9+1=10
\end{aligned} x → 0 − lim x 2 { sin ( k 1 + 1 ) x + sin ( k 2 − 1 ) x } = 4 ⇒ 2 ( k 1 + 1 ) + 2 ( k 2 − 1 ) = 4 ⇒ k 1 + k 2 = 2 ⇒ x → 0 + lim x 2 ln ( 2 + k 2 x 2 + k 1 x ) = 4 ⇒ x → 0 + lim x 1 ln ( 1 + 2 + k 2 x ( k 1 − k 2 ) x ) = 2 ⇒ 2 k 1 − k 2 = 2 ⇒ k 1 − k 2 = 4 ∴ k 1 = 3 , k 2 = − 1 k 1 2 + k 2 2 = 9 + 1 = 10 \] Answer the question to unlock