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Application of Derivatives question

2007 · Shift 0 · Q53
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Application of Derivatives question

2007 · Shift 0 · Q53

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
The function f(x)=tan⁡−1(sin⁡x+cos⁡x)f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)f(x)=tan−1(sinx+cosx) is an incresing function in
  1. A
    (0,π2)\left( {0,{\pi \over 2}} \right)(0,2π​)
  2. B
    (−π2,π2)\left( { - {\pi \over 2},{\pi \over 2}} \right)(−2π​,2π​)
  3. C
    (π4,π2)\left( { {\pi \over 4},{\pi \over 2}} \right)(4π​,2π​)
  4. D
    (−π2,π4)\left( { - {\pi \over 2},{\pi \over 4}} \right)(−2π​,4π​)
View written solutionFree

Correct answer: D

  1. We need to find where f(x)=tan⁡−1(sin⁡x+cos⁡x)f(x)=\tan^{-1}(\sin x+\cos x)f(x)=tan−1(sinx+cosx) is increasing.

Since tan⁡−1(u)\tan^{-1}(u)tan−1(u) is an increasing function of uuu, we can also check the sign of f′(x)f'(x)f′(x).

  1. Differentiate: f′(x)=cos⁡x−sin⁡x1+(sin⁡x+cos⁡x)2.f'(x)=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}.f′(x)=1+(sinx+cosx)2cosx−sinx​.

Now simplify the denominator: 1+(sin⁡x+cos⁡x)2=1+sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=2+sin⁡2x.1+(\sin x+\cos x)^2=1+\sin^2x+\cos^2x+2\sin x\cos x=2+\sin 2x.1+(sinx+cosx)2=1+sin2x+cos2x+2sinxcosx=2+sin2x.

So, f′(x)=cos⁡x−sin⁡x2+sin⁡2x.f'(x)=\frac{\cos x-\sin x}{2+\sin 2x}.f′(x)=2+sin2xcosx−sinx​.

  1. Determine the sign of the denominator.

Since −1≤sin⁡2x≤1-1\le \sin 2x\le 1−1≤sin2x≤1, 1≤2+sin⁡2x≤3.1\le 2+\sin 2x\le 3.1≤2+sin2x≤3. Hence the denominator is always positive.

Therefore, the sign of f′(x)f'(x)f′(x) depends only on cos⁡x−sin⁡x.\cos x-\sin x.cosx−sinx.

  1. Solve for increasing condition: f′(x)>0  ⟺  cos⁡x−sin⁡x>0.f'(x)>0 \iff \cos x-\sin x>0.f′(x)>0⟺cosx−sinx>0.

Rewrite: cos⁡x−sin⁡x=2cos⁡(x+π4).\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right).cosx−sinx=2​cos(x+4π​).

Thus,

\iff \cos\left(x+\frac{\pi}{4}\right)>0.$$ This gives $$-\frac{\pi}{2}<x+\frac{\pi}{4}<\frac{\pi}{2}$$ for the principal interval, i.e. $$-\frac{3\pi}{4}<x<\frac{\pi}{4}.$$ So the function is increasing on intervals of the form $$\left(-\frac{3\pi}{4}+2n\pi,\;\frac{\pi}{4}+2n\pi\right),\quad n\in\mathbb Z.$$ 5. Check the given options: - **A:** $\left(0,\frac{\pi}{2}\right)$ Here for $x>\frac{\pi}{4}$, $f'(x)<0$. So not increasing on the whole interval. - **B:** $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ Again, for $x>\frac{\pi}{4}$, $f'(x)<0$. Not increasing on the whole interval. - **C:** $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$ In this interval, $\cos x-\sin x<0$, so $f$ is decreasing. - **D:** $\left(-\frac{\pi}{2},\frac{\pi}{4}\right)$ For all $x$ in this interval, $$\cos x-\sin x>0,$$ so $f'(x)>0$. Hence $f$ is increasing here. 6. Therefore, the correct option is $$\boxed{D}.$$
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