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

Answer: (A)

1

Answer 2

$f(x) = \begin{cases} \frac {log_e(1−x+x^2)+log_e(1+x+x_2)}{sec⁡\ x−cos⁡\ x}, & \quad x∈(−\frac \pi2,\frac \pi2)−{0}\\\ k, & \quad x=0 \end{cases}$
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): $1$