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Differentiation question

2019 · 10 Apr · Shift 2 · Q23
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Differentiation question

2019 · 10 Apr · Shift 2 · Q23

JEE MainMathematicsDifferentiationMCQ+4 / −1
Let f(x) = loge(sin x), (0 < x < π\piπ) and g(x) = sin–1 (e–x ), (x ≥\ge≥ 0). If α\alphaα is a positive real number such that a = (fog)'(α\alphaα) and b = (fog)(α\alphaα), then :
  1. A
    a α\alphaα 2 + b α\alphaα - a = -2 α\alphaα 2
  2. B
    a α\alphaα 2 + b α\alphaα + a = 0
  3. C
    a α\alphaα 2 - b α\alphaα - a = 0
  4. D
    a α\alphaα 2 - b α\alphaα - a = 1
View written solutionFree

Correct answer: D

  1. Given functions f(x)=ln⁡(sin⁡x),0<x<πf(x)=\ln(\sin x), \quad 0<x<\pif(x)=ln(sinx),0<x<π g(x)=sin⁡−1(e−x),x≥0g(x)=\sin^{-1}(e^{-x}), \quad x\ge 0g(x)=sin−1(e−x),x≥0

    We need: a=(f∘g)′(α),b=(f∘g)(α)a=(f\circ g)'(\alpha), \qquad b=(f\circ g)(\alpha)a=(f∘g)′(α),b=(f∘g)(α)

  2. Find f(g(x))f(g(x))f(g(x))

    Since g(x)=sin⁡−1(e−x),g(x)=\sin^{-1}(e^{-x}),g(x)=sin−1(e−x), we have sin⁡(g(x))=e−x.\sin(g(x))=e^{-x}.sin(g(x))=e−x.

    Therefore, f(g(x))=ln⁡(sin⁡(g(x)))=ln⁡(e−x)=−x.f(g(x))=\ln(\sin(g(x)))=\ln(e^{-x})=-x.f(g(x))=ln(sin(g(x)))=ln(e−x)=−x.

    Hence, (f∘g)(x)=−x. (f\circ g)(x)=-x.(f∘g)(x)=−x.

  3. Compute aaa and bbb at x=αx=\alphax=α

    Since (f∘g)(x)=−x, (f\circ g)(x)=-x,(f∘g)(x)=−x, we get b=(f∘g)(α)=−α.b=(f\circ g)(\alpha)=-\alpha.b=(f∘g)(α)=−α.

    Differentiating, (f∘g)′(x)=−1, (f\circ g)'(x)=-1,(f∘g)′(x)=−1, so a=(f∘g)′(α)=−1.a=(f\circ g)'(\alpha)=-1.a=(f∘g)′(α)=−1.

  4. Check each option

    We substitute a=−1a=-1a=−1 and b=−αb=-\alphab=−α.

    Option A: aα2+bα−a=(−1)α2+(−α)(α)−(−1)=−α2−α2+1=−2α2+1a\alpha^2+b\alpha-a=(-1)\alpha^2+(-\alpha)(\alpha)-(-1)=-\alpha^2-\alpha^2+1=-2\alpha^2+1aα2+bα−a=(−1)α2+(−α)(α)−(−1)=−α2−α2+1=−2α2+1 This is not equal to −2α2-2\alpha^2−2α2.

    Option B: aα2+bα+a=−α2−α2−1=−2α2−1≠0a\alpha^2+b\alpha+a=-\alpha^2-\alpha^2-1=-2\alpha^2-1\ne 0aα2+bα+a=−α2−α2−1=−2α2−1=0 So B is false.

    Option C: aα2−bα−a=−α2−(−α)(α)−(−1)=−α2+α2+1=1≠0a\alpha^2-b\alpha-a=-\alpha^2-(-\alpha)(\alpha)-(-1)=-\alpha^2+\alpha^2+1=1\ne 0aα2−bα−a=−α2−(−α)(α)−(−1)=−α2+α2+1=1=0 So C is false.

    Option D: aα2−bα−a=−α2−(−α)(α)−(−1)=1a\alpha^2-b\alpha-a=-\alpha^2-(-\alpha)(\alpha)-(-1)=1aα2−bα−a=−α2−(−α)(α)−(−1)=1 This is true.

  5. Conclusion

    The correct option is: D\boxed{\text{D}}D​

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