Question

If the function $f(x)= \begin{cases} \frac{\log _e\left(1-x+x^2\right)+\log_e\left(1+x+x^2\right)}{\sec x-\cos x}, x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\} \\k, \,\,\,\,\, x=0\end{cases}$ 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

 To find the value of \( k \) that makes the function \( f(x) \) continuous at \( x = 0 \), we need to ensure that the left-hand limit, right-hand limit, and value at \( x = 0 \) of \( f(x) \) are equal.

Let's first compute the limit of the function as \( x \) approaches 0:

The given function is \(f(x) = \begin{cases} \frac{\log_e(1-x+x^2) + \log_e(1+x+x^2)}{\sec x - \cos x}, & x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)-\{0\} \\ k, & x = 0 \end{cases}\).

We focus on the expression \(\frac{\log_e(1-x+x^2) + \log_e(1+x+x^2)}{\sec x - \cos x}\)as \( x \to 0 \).

Step 1: Simplify the Numerator

Using the logarithmic identity \(\log_e a + \log_e b = \log_e (ab)\), we have:

\(\log_e \left((1-x+x^2)(1+x+x^2)\right) = \log_e \left(1 - x^2 + x^4 + 2x^2\right) = \log_e \left(1 + x^2(1+x^2)\right).\)

For small \( x \), \( \log_e(1 + y) \approx y \). Thus, \(\log_e(1 + x^2 + x^4) \approx x^2 + x^4\)is negligible, so \(\log_e(1 + x^2) \approx x^2\).

Step 2: Simplify the Denominator

The expression involves \(\sec x - \cos x\). Using the Taylor series expansions:

• \(\sec x = 1 + \frac{x^2}{2} + \text{higher order terms}\)and
• \(\cos x = 1 - \frac{x^2}{2} + \text{higher order terms}\).

Therefore, \(\sec x - \cos x \approx \left(1 + \frac{x^2}{2}\right) - \left(1 - \frac{x^2}{2}\right) = x^2\).

 

Step 3: Compute the Limit

Substituting the approximations back into the function, we get:

\(\lim_{x \to 0} \frac{x^2}{x^2} = \lim_{x \to 0} 1 = 1.\)

The limit of \( f(x) \) as \( x \to 0 \) is 1. For continuity at \( x = 0 \), \( f(0) \) should also be equal to this limit.

Conclusion:

Therefore, for \( f(x) \) to be continuous at \( x = 0 \), we must have \(k = 1\).

The correct answer is 1.