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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Continuity and Differentiability

Let K be the set of all real values of x where the function f(x)=...

Let K\mathrm{K} be the set of all real values of xx where the function f(x)=sin⁡∣x∣−∣x∣+2(x−π)cos⁡∣x∣f(x)=\sin |x|-|x|+2(x-\pi) \cos |x| is not differentiable. Then the set KK is equal to :

Options

  1. Aϕ\phi (an empty set)
  2. B(π}(\pi\}
  3. C{0}\{0\}
  4. D{0,π}\{0, \pi\}

Correct answer

A. ϕ\phi (an empty set)

Step-by-step solution

f(x)=sin⁡∣x∣−∣x∣+2(x−π)cos⁡∣x∣f(x)=\sin |x|-|x|+2(x-\pi) \cos |x| There are two cases, Case (1),x>0(1), x>0 f(x)=sin⁡x−x+2(x−π)cos⁡xf(x)=\sin x-x+2(x-\pi) \cos x f′(x)=cos⁡x−1+2(1−0)cos⁡x−2sin⁡(x−π)f^{\prime}(x)=\cos x-1+2(1-0) \cos x-2 \sin (x-\pi) f′(x)=3cos⁡x−2(x−π)sin⁡x−1f^{\prime}(x)=3 \cos x-2(x-\pi) \sin x-1 Then, function f(x)f(x) is differentiable for all x>0x>0 Case (2)x<0(2) x < 0 f(x)=−sin⁡x+x+2(x−π)cos⁡xf(x)=-\sin x+x+2(x-\pi) \cos x f′(x)=−cos⁡x+1−2(x−π)sin⁡x+2cos⁡xf^{\prime}(x)=-\cos x+1-2(x-\pi) \sin x+2 \cos x f′(x)=cos⁡x+1−2(x−π)sin⁡xf^{\prime}(x)=\cos x+1-2(x-\pi) \sin x Then, function f(x)f(x) is differentiable for all x<0x < 0 Now check for x=0x=0 f′(0∗)R⋅HD=3−1=2f^{\prime}\left(0^{*}\right) \mathrm{R} \cdot \mathrm{HD}=3-1=2 f′(0)L.H.D.=1+1=2f^{\prime}(0) \mathrm{L} . \mathrm{H.D.}=1+1=2 L.\mathrm{L} .\mathrm{HD}=\mathrm{R}.H.D. \mathrm{H.D} Then, function f(x)f(x) is differentiable for x=0.x=0 . So it is differentiable everywhere Hence, k=ϕk=\phi
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