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Differentiation question

2020 · 8 Jan · Shift 1 · Q36
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  5. /2020 · 8 Jan · Shift 1 · Q36

Differentiation question

2020 · 8 Jan · Shift 1 · Q36

JEE MainMathematicsDifferentiationMCQ+4 / −1
Let ƒ(x) = (sin(tan–1x) + sin(cot–1x))2 – 1, |x| > 1. If dydx=12ddx(sin⁡−1(f(x))){{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)dxdy​=21​dxd​(sin−1(f(x))) and y(3)=π6y\left( {\sqrt 3 } \right) = {\pi \over 6}y(3​)=6π​, then y(−3{ - \sqrt 3 }−3​) is equal to :
  1. A
    5π6{{5\pi } \over 6}65π​
  2. B
    −π6- {\pi \over 6}−6π​
  3. C
    π3{\pi \over 3}3π​
  4. D
    2π3{{2\pi } \over 3}32π​
View written solutionFree

Correct answer: B

  1. Simplify f(x)f(x)f(x)

Given

f(x)=(sin⁡(tan⁡−1x)+sin⁡(cot⁡−1x))2−1,∣x∣>1.f(x)=\left(\sin(\tan^{-1}x)+\sin(\cot^{-1}x)\right)^2-1, \qquad |x|>1.f(x)=(sin(tan−1x)+sin(cot−1x))2−1,∣x∣>1.

Use standard triangle identities:

  • If θ=tan⁡−1x\theta=\tan^{-1}xθ=tan−1x, then sin⁡(tan⁡−1x)=x1+x2.\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}.sin(tan−1x)=1+x2​x​.
  • If ϕ=cot⁡−1x\phi=\cot^{-1}xϕ=cot−1x, then tan⁡ϕ=1x\tan\phi=\frac1xtanϕ=x1​, so sin⁡(cot⁡−1x)=11+x2.\sin(\cot^{-1}x)=\frac{1}{\sqrt{1+x^2}}.sin(cot−1x)=1+x2​1​.

Hence

sin⁡(tan⁡−1x)+sin⁡(cot⁡−1x)=x+11+x2.\sin(\tan^{-1}x)+\sin(\cot^{-1}x)=\frac{x+1}{\sqrt{1+x^2}}.sin(tan−1x)+sin(cot−1x)=1+x2​x+1​.

Therefore,

f(x)=(x+1)21+x2−1=x2+2x+1−(1+x2)1+x2=2x1+x2.f(x)=\frac{(x+1)^2}{1+x^2}-1 =\frac{x^2+2x+1-(1+x^2)}{1+x^2} =\frac{2x}{1+x^2}.f(x)=1+x2(x+1)2​−1=1+x2x2+2x+1−(1+x2)​=1+x22x​.

So,

f(x)=2x1+x2.f(x)=\frac{2x}{1+x^2}.f(x)=1+x22x​.
  1. Use the given differential relation

We are given

dydx=12ddx(sin⁡−1(f(x))).\frac{dy}{dx}=\frac12\frac{d}{dx}\left(\sin^{-1}(f(x))\right).dxdy​=21​dxd​(sin−1(f(x))).

Substitute f(x)f(x)f(x):

dydx=12ddx(sin⁡−1(2x1+x2)).\frac{dy}{dx}=\frac12\frac{d}{dx}\left(\sin^{-1}\left(\frac{2x}{1+x^2}\right)\right).dxdy​=21​dxd​(sin−1(1+x22x​)).

Integrating both sides with respect to xxx,

y=12sin⁡−1(2x1+x2)+C.y=\frac12\sin^{-1}\left(\frac{2x}{1+x^2}\right)+C.y=21​sin−1(1+x22x​)+C.
  1. Simplify sin⁡−1(2x1+x2)\sin^{-1}\left(\frac{2x}{1+x^2}\right)sin−1(1+x22x​) using x=tan⁡θx=\tan\thetax=tanθ

Let x=tan⁡θx=\tan\thetax=tanθ. Then

2x1+x2=2tan⁡θ1+tan⁡2θ=sin⁡2θ.\frac{2x}{1+x^2}=\frac{2\tan\theta}{1+\tan^2\theta}=\sin 2\theta.1+x22x​=1+tan2θ2tanθ​=sin2θ.

So,

sin⁡−1(2x1+x2)=sin⁡−1(sin⁡2θ).\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\sin^{-1}(\sin 2\theta).sin−1(1+x22x​)=sin−1(sin2θ).

Because ∣x∣>1|x|>1∣x∣>1, we handle the relevant points directly.

  1. Use the condition y(3)=π6y(\sqrt3)=\frac\pi6y(3​)=6π​

At x=3x=\sqrt3x=3​,

f(3)=231+3=32.f(\sqrt3)=\frac{2\sqrt3}{1+3}=\frac{\sqrt3}{2}.f(3​)=1+323​​=23​​.

Thus

sin⁡−1(f(3))=sin⁡−1(32)=π3.\sin^{-1}(f(\sqrt3))=\sin^{-1}\left(\frac{\sqrt3}{2}\right)=\frac\pi3.sin−1(f(3​))=sin−1(23​​)=3π​.

Now,

y(3)=12⋅π3+C=π6+C.y(\sqrt3)=\frac12\cdot\frac\pi3+C=\frac\pi6+C.y(3​)=21​⋅3π​+C=6π​+C.

Given y(3)=π6y(\sqrt3)=\frac\pi6y(3​)=6π​, we get

C=0.C=0.C=0.

Hence

y=12sin⁡−1(2x1+x2).y=\frac12\sin^{-1}\left(\frac{2x}{1+x^2}\right).y=21​sin−1(1+x22x​).
  1. Find y(−3)y(-\sqrt3)y(−3​)

At x=−3x=-\sqrt3x=−3​,

f(−3)=2(−3)1+3=−32.f(-\sqrt3)=\frac{2(-\sqrt3)}{1+3}=-\frac{\sqrt3}{2}.f(−3​)=1+32(−3​)​=−23​​.

Therefore,

sin⁡−1(f(−3))=sin⁡−1(−32)=−π3.\sin^{-1}(f(-\sqrt3))=\sin^{-1}\left(-\frac{\sqrt3}{2}\right)=-\frac\pi3.sin−1(f(−3​))=sin−1(−23​​)=−3π​.

So,

y(−3)=12(−π3)=−π6.y(-\sqrt3)=\frac12\left(-\frac\pi3\right)=-\frac\pi6.y(−3​)=21​(−3π​)=−6π​.
  1. Check options

The value is

−π6,-\frac\pi6,−6π​,

which matches Option B.

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