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Differential Equations question

2017 · Shift 2 · Q31
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  5. /2017 · Shift 2 · Q31

Differential Equations question

2017 · Shift 2 · Q31

JEE AdvancedMathematicsDifferential EquationsMultiple correct+4 / −2
If g(x)=∫sin⁡xsin⁡(2x)sin⁡−1(t) dtg(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dtg(x)=∫sinxsin(2x)​sin−1(t)dt, then
  1. A
    g′(−π2)=0g'\left( { - {\pi \over 2}} \right) = 0g′(−2π​)=0
  2. B
    g′(−π2)=−2πg'\left( { - {\pi \over 2}} \right) = - 2\pig′(−2π​)=−2π
  3. C
    g′(π2)=2πg'\left( {{\pi \over 2}} \right) = 2\pig′(2π​)=2π
  4. D
    g′(π2)=0g'\left( {{\pi \over 2}} \right) = 0g′(2π​)=0
View written solutionFree

Correct answer: A, D

  1. Given function

g(x)=∫sin⁡xsin⁡(2x)sin⁡−1(t) dtg(x)=\int_{\sin x}^{\sin(2x)} \sin^{-1}(t)\,dtg(x)=∫sinxsin(2x)​sin−1(t)dt

Here, sin⁡−1(t)\sin^{-1}(t)sin−1(t) means arcsin⁡(t)\arcsin(t)arcsin(t).

  1. Differentiate using Leibniz rule

If F(x)=∫u(x)v(x)f(t) dt,F(x)=\int_{u(x)}^{v(x)} f(t)\,dt,F(x)=∫u(x)v(x)​f(t)dt, then F′(x)=f(v(x)) v′(x)−f(u(x)) u′(x).F'(x)=f(v(x))\,v'(x)-f(u(x))\,u'(x).F′(x)=f(v(x))v′(x)−f(u(x))u′(x).

So, g′(x)=sin⁡−1(sin⁡2x)⋅ddx(sin⁡2x)−sin⁡−1(sin⁡x)⋅ddx(sin⁡x).g'(x)=\sin^{-1}(\sin 2x)\cdot \frac{d}{dx}(\sin 2x)-\sin^{-1}(\sin x)\cdot \frac{d}{dx}(\sin x).g′(x)=sin−1(sin2x)⋅dxd​(sin2x)−sin−1(sinx)⋅dxd​(sinx).

Since ddx(sin⁡2x)=2cos⁡2x,ddx(sin⁡x)=cos⁡x,\frac{d}{dx}(\sin 2x)=2\cos 2x, \qquad \frac{d}{dx}(\sin x)=\cos x,dxd​(sin2x)=2cos2x,dxd​(sinx)=cosx, we get g′(x)=2cos⁡2x sin⁡−1(sin⁡2x)−cos⁡x sin⁡−1(sin⁡x).g'(x)=2\cos 2x\,\sin^{-1}(\sin 2x)-\cos x\,\sin^{-1}(\sin x).g′(x)=2cos2xsin−1(sin2x)−cosxsin−1(sinx).

  1. Evaluate at x=−π2x=-\dfrac{\pi}{2}x=−2π​

First compute the trigonometric values: sin⁡(2x)=sin⁡(−π)=0,cos⁡(2x)=cos⁡(−π)=−1,\sin(2x)=\sin(-\pi)=0, \qquad \cos(2x)=\cos(-\pi)=-1,sin(2x)=sin(−π)=0,cos(2x)=cos(−π)=−1, sin⁡x=sin⁡(−π2)=−1,cos⁡x=cos⁡(−π2)=0.\sin x=\sin\left(-\frac{\pi}{2}\right)=-1, \qquad \cos x=\cos\left(-\frac{\pi}{2}\right)=0.sinx=sin(−2π​)=−1,cosx=cos(−2π​)=0.

Also, sin⁡−1(0)=0,sin⁡−1(−1)=−π2.\sin^{-1}(0)=0, \qquad \sin^{-1}(-1)=-\frac{\pi}{2}.sin−1(0)=0,sin−1(−1)=−2π​.

Thus, g′(−π2)=2(−1)(0)−0(−π2)=0.g'\left(-\frac{\pi}{2}\right)=2(-1)(0)-0\left(-\frac{\pi}{2}\right)=0.g′(−2π​)=2(−1)(0)−0(−2π​)=0.

So A is correct and B is false.

  1. Evaluate at x=π2x=\dfrac{\pi}{2}x=2π​

Now, sin⁡(2x)=sin⁡(π)=0,cos⁡(2x)=cos⁡(π)=−1,\sin(2x)=\sin(\pi)=0, \qquad \cos(2x)=\cos(\pi)=-1,sin(2x)=sin(π)=0,cos(2x)=cos(π)=−1, sin⁡x=sin⁡(π2)=1,cos⁡x=cos⁡(π2)=0.\sin x=\sin\left(\frac{\pi}{2}\right)=1, \qquad \cos x=\cos\left(\frac{\pi}{2}\right)=0.sinx=sin(2π​)=1,cosx=cos(2π​)=0.

Also, sin⁡−1(0)=0,sin⁡−1(1)=π2.\sin^{-1}(0)=0, \qquad \sin^{-1}(1)=\frac{\pi}{2}.sin−1(0)=0,sin−1(1)=2π​.

Hence, g′(π2)=2(−1)(0)−0(π2)=0.g'\left(\frac{\pi}{2}\right)=2(-1)(0)-0\left(\frac{\pi}{2}\right)=0.g′(2π​)=2(−1)(0)−0(2π​)=0.

So D is correct and C is false.

  1. Final conclusion

Correct options are: A,D\boxed{A, D}A,D​

This matches the stored correct answer.

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