Question

If the function
\(f(x) =   \begin{cases}    \frac {log_e(1−x+x^2)+log_e(1+x+x_2)}{sec⁡\ x−cos⁡\ x},      & \quad x∈(−\frac \pi2,\frac \pi2)−{0}\\     k,  & \quad x=0   \end{cases}\)
is continuous at \(x = 0\), then k is equal to

• A. 1
• B. -1
• C. e
• D. 0

Answer 1

The Correct Option is A

\(f(x) =   \begin{cases}    \frac {log_e(1−x+x^2)+log_e(1+x+x_2)}{sec⁡\ x−cos⁡\ x},      & \quad x∈(−\frac \pi2,\frac \pi2)−{0}\\     k,  & \quad x=0   \end{cases}\)
for continuity at \(x = 0\)
\(\lim\limits_{x→0}f(x)=k\)

∴ \(k=\lim\limits_{x→0}\) \(\frac {log_e(1−x+x^2)+log_e(1+x+x_2)}{sec⁡\ x−cos⁡\ x}\)   (\(\frac 00\) form)

=\(\lim\limits_{x→0}\) \(\frac {cos\ ⁡x log_e(x^4+x^2+1)}{sin^2⁡x}\)

=\(\lim\limits_{x→0}\) \(log_e \frac {(x_4+x_2+1)}{x_2}\)

=\(\lim\limits_{x→0}\) \(\frac {ln(1+x^2+x^4)}{x^2+x4} ⋅ \frac {x^2+x^4}{x^2}\)
=\(1\)

So, the correct option is (A): \(1\)