Let $f(x) = \tan^{-1}x$. Then $f'(x) + f''(x)$ is = $0$, when $x$ is equal to
- $0$
- $+1$
- $i$
- $-i$
The Correct Option is B
Solution and Explanation
Given, \(f(x)=\tan ^{-1} x\)
\(\Rightarrow f^{\prime}(x)=\frac{1}{1+x^{2}}\)
\(\Rightarrow f^{\prime \prime}(x)=-\frac{1}{\left(1+x^{2}\right)^{2}}(2 x)=-\frac{2 x}{\left(1+x^{2}\right)^{2}}\)
\(\because f^{\prime}(x)+f^{\prime \prime}(x)=0\)
\(\therefore \frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}=0\)
\(\Rightarrow \frac{\left(1+x^{2}\right)-2 x}{\left(1+x^{2}\right)^{2}}=0\)
\(\Rightarrow (1-x)^{2}=0 \Rightarrow x=1\)
So, the correct answer is (B): +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.