If $f\left(x\right) = \cos^{-1} \left[\frac{1-\left(\log x\right)^{2}}{1+\left(\log x\right)^{2}}\right] $ then the value of $f'(e)$ is equal to
- 1
- $\frac{1}{e}$
- $\frac{2}{e}$
- $\frac{2}{e^2}$
The Correct Option is B
Solution and Explanation
Rewriting the equation as \(\cos [ f ( x )]=\frac{1-(\log x )^{2}}{1+(\log x)^{2}}\).
Now taking derivatives we get,
\(-\sin [ f ( x )] f '( x )=\frac{\left[1+(\log x)^{2}\right][-2(\log x ) x ]-\left[1-(\log x )^{2}\right][2(\log x ) x ]}{\left[1+(\log x)^{2}\right]^{2}}\)
Evaluating at \(x = e\) we get,
\(-f'(x)=\frac{(2)(-2 / e)-(0)(2 / 2)}{2^{2}}\)
This implies that \(f '( x )=1 / e\).
Therefore, the correct option is (B): \(\frac 1e\)
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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.