MHT CET · Maths · Application of Derivatives
The function \(f(x)=\cot ^{-1} x+x\) is increasing in the interval.
- A \((-\infty, \infty)\)
- B \((0,3)\)
- C \((1, \infty)\)
- D \((-1, \infty)\)
Answer & Solution
Correct Answer
(A) \((-\infty, \infty)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(\mathrm{x})=\cot ^{-1} \mathrm{x}+\mathrm{x} \)
\( \therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\mathrm{x}^2}+1=\frac{-1+1+\mathrm{x}^2}{1+\mathrm{x}^2}=\frac{\mathrm{x}^2}{1+\mathrm{x}^2} \)
\( \text { Here } \mathrm{x}^2 \geq 0 \Rightarrow \frac{\mathrm{x}^2}{1+\mathrm{x}^2} \geq 0\)
Hence \(\mathrm{f}(\mathrm{x})\) is always increasing.
\( \therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{-1}{1+\mathrm{x}^2}+1=\frac{-1+1+\mathrm{x}^2}{1+\mathrm{x}^2}=\frac{\mathrm{x}^2}{1+\mathrm{x}^2} \)
\( \text { Here } \mathrm{x}^2 \geq 0 \Rightarrow \frac{\mathrm{x}^2}{1+\mathrm{x}^2} \geq 0\)
Hence \(\mathrm{f}(\mathrm{x})\) is always increasing.
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