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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2023 · Paper 1 · Mathematics · Application of Derivatives

The function f(x)= ^ -1 ( x+ x) is an increasing function in

The function f(x)=tan⁡−1(sin⁡x+cos⁡x)f(x)=\tan ^{-1}(\sin x+\cos x) is an increasing function in

Options

  1. A(π4,π2)\left(\frac{\pi}{4}, \frac{\pi}{2}\right)
  2. B(−π2,π4)\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)
  3. C(0,π2)\left(0, \frac{\pi}{2}\right)
  4. D(−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Correct answer

B. (−π2,π4)\left(-\frac{\pi}{2}, \frac{\pi}{4}\right)

Step-by-step solution

Since, f(x)=tan⁡−1(sin⁡x+cos⁡x)f(x)=\tan ^{-1}(\sin x+\cos x)
∴\mathrmf′(x)=11+(sin⁡x+cos⁡x)2(cos⁡x−sin⁡x)\therefore \mathrmf^{\prime}(\mathrm{x})=\frac{1}{1+(\sin \mathrm{x}+\cos \mathrm{x})^2}(\cos \mathrm{x}-\sin \mathrm{x})
=2cos⁡(x+π4)1+(sin⁡x+cos⁡x)2=\frac{\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)}{1+(\sin x+\cos x)^2}
f(x)f(x) is increasing if f′(x)>0⇒cos⁡(x+π4)>0f^{\prime}(x) > 0 \Rightarrow \cos \left(x+\frac{\pi}{4}\right) > 0
⇒−π2<x+π4<π2⇒−3π2<x+π4\Rightarrow \quad-\frac{\pi}{2} < x+\frac{\pi}{4} < \frac{\pi}{2} \Rightarrow-\frac{3 \pi}{2} < x+\frac{\pi}{4}
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