WBJEE · Maths · Differentiation
Let \(\mathrm{f}(\mathrm{x})=\tan ^{-1} \mathrm{x}\). Then \(\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime \prime}(\mathrm{x})\) is \(=0\), when \(\mathrm{x}\) is equal to
- A 0
- B 1
- C 1
- D \(-i\)
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Hints: \(f(x)=\tan ^{-1} x\) \(\begin{aligned} & f^{\prime}(x)=\frac{1}{1+x^2} \\ & f^{\prime \prime}(x)=\frac{-1}{\left(1+x^2\right)} \cdot 2 x, \frac{1}{1+x^2}=\frac{2 x}{\left(1+x^2\right)^2} \\ & 1+x^2=2 x,(x-1)^2=0 \\ & x=1 \end{aligned}\)
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