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
\(f(x) = \begin{cases} \frac {log_e(1−x+x^2)+log_e(1+x+x_2)}{sec\ x−cos\ x}, & \quad x∈(−\frac \pi2,\frac \pi2)−{0}\\ k, & \quad x=0 \end{cases}\)
is continuous at \(x = 0\), then k is equal to
- 1
- -1
- e
- 0
The Correct Option is A
Solution and Explanation
\(f(x) = \begin{cases} \frac {log_e(1−x+x^2)+log_e(1+x+x_2)}{sec\ x−cos\ x}, & \quad x∈(−\frac \pi2,\frac \pi2)−{0}\\ k, & \quad x=0 \end{cases}\)
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)
=\(\lim\limits_{x→0}\) \(\frac {cos\ x log_e(x^4+x^2+1)}{sin^2x}\)
=\(\lim\limits_{x→0}\) \(log_e \frac {(x_4+x_2+1)}{x_2}\)
=\(\lim\limits_{x→0}\) \(\frac {ln(1+x^2+x^4)}{x^2+x4} ⋅ \frac {x^2+x^4}{x^2}\)
=\(1\)
So, the correct option is (A): \(1\)
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.