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Differentiation question

2020 · 9 Jan · Shift 2 · Q33
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  5. /2020 · 9 Jan · Shift 2 · Q33

Differentiation question

2020 · 9 Jan · Shift 2 · Q33

JEE MainMathematicsDifferentiationMCQ+4 / −1
If x=2sin⁡θ−sin⁡2θx = 2\sin \theta - \sin 2\thetax=2sinθ−sin2θ and y=2cos⁡θ−cos⁡2θy = 2\cos \theta - \cos 2\thetay=2cosθ−cos2θ, θ∈[0,2π]\theta \in \left[ {0,2\pi } \right]θ∈[0,2π], then d2ydx2{{{d^2}y} \over {d{x^2}}}dx2d2y​ at θ\thetaθ=π\piπ is :
  1. A
    38{3 \over 8}83​
  2. B
    32{3 \over 2}23​
  3. C
    34{3 \over 4}43​
  4. D
    -34{3 \over 4}43​
View written solutionFree

Correct answer: A

  1. Given parametric equations

x=2sin⁡θ−sin⁡2θ,y=2cos⁡θ−cos⁡2θx=2\sin\theta-\sin2\theta,\qquad y=2\cos\theta-\cos2\thetax=2sinθ−sin2θ,y=2cosθ−cos2θ

We need to find

d2ydx2\frac{d^2y}{dx^2}dx2d2y​

at θ=π\theta=\piθ=π.

  1. First derivatives with respect to θ\thetaθ

Differentiate both xxx and yyy with respect to θ\thetaθ:

dxdθ=2cos⁡θ−2cos⁡2θ\frac{dx}{d\theta}=2\cos\theta-2\cos2\thetadθdx​=2cosθ−2cos2θ

dydθ=−2sin⁡θ+2sin⁡2θ\frac{dy}{d\theta}=-2\sin\theta+2\sin2\thetadθdy​=−2sinθ+2sin2θ

So,

=\frac{-2\sin\theta+2\sin2\theta}{2\cos\theta-2\cos2\theta}$$ Factor $2$: $$\frac{dy}{dx}=\frac{-\sin\theta+\sin2\theta}{\cos\theta-\cos2\theta}$$ Using $\sin2\theta=2\sin\theta\cos\theta$, $$-\sin\theta+\sin2\theta=-\sin\theta+2\sin\theta\cos\theta=\sin\theta(2\cos\theta-1)$$ Thus, $$\frac{dy}{dx}=\frac{\sin\theta(2\cos\theta-1)}{\cos\theta-\cos2\theta}$$ Now use $$\cos2\theta=2\cos^2\theta-1$$ Then $$\cos\theta-\cos2\theta=\cos\theta-(2\cos^2\theta-1)=1+\cos\theta-2\cos^2\theta$$ Factor: $$1+\cos\theta-2\cos^2\theta=(1-\cos\theta)(2\cos\theta+1)$$ Hence, $$\frac{dy}{dx}=\frac{\sin\theta(2\cos\theta-1)}{(1-\cos\theta)(2\cos\theta+1)}$$ But it is easier to use the direct expression for second derivative: $$\frac{d^2y}{dx^2}=\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\Big/\frac{dx}{d\theta}$$ --- 3. **Simplify $\dfrac{dy}{dx}$ first** From $$\frac{dy}{dx}=\frac{-2\sin\theta+2\sin2\theta}{2\cos\theta-2\cos2\theta}$$ use identities: $$-2\sin\theta+2\sin2\theta=2\sin\theta(2\cos\theta-1)$$ and $$2\cos\theta-2\cos2\theta=2(\cos\theta-\cos2\theta)$$ Now, $$\cos\theta-\cos2\theta=-2\sin\frac{3\theta}{2}\sin\left(-\frac{\theta}{2}\right)=2\sin\frac{3\theta}{2}\sin\frac{\theta}{2}$$ But a cleaner route is to convert using half-angle: $$\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2},\qquad 1-\cos\theta=2\sin^2\frac{\theta}{2}$$ After simplification, $$\frac{dy}{dx}=\frac{2\cos\frac{\theta}{2}(2\cos\theta-1)}{2\sin\frac{\theta}{2}(2\cos\theta+1)} =\cot\frac{\theta}{2}\cdot \frac{2\cos\theta-1}{2\cos\theta+1}$$ This is valid, but we only need the value near $\theta=\pi$, so we proceed by differentiation directly. --- 4. **Find $\dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)$ at $\theta=\pi$** Let $$u=\frac{dy}{d\theta}=-2\sin\theta+2\sin2\theta$$ $$v=\frac{dx}{d\theta}=2\cos\theta-2\cos2\theta$$ So, $$\frac{dy}{dx}=\frac{u}{v}$$ Then $$\frac{d}{d\theta}\left(\frac{dy}{dx}\right)=\frac{u'v-uv'}{v^2}$$ Now compute derivatives: $$u'=-2\cos\theta+4\cos2\theta$$ $$v'=-2\sin\theta+4\sin2\theta$$ At $\theta=\pi$: $$\sin\pi=0,\quad \sin2\pi=0,\quad \cos\pi=-1,\quad \cos2\pi=1$$ Therefore, $$u(\pi)=-2\sin\pi+2\sin2\pi=0$$ $$v(\pi)=2\cos\pi-2\cos2\pi=2(-1)-2(1)=-4$$ $$u'(\pi)=-2\cos\pi+4\cos2\pi=-2(-1)+4(1)=2+4=6$$ $$v'(\pi)=-2\sin\pi+4\sin2\pi=0$$ Hence, $$\frac{d}{d\theta}\left(\frac{dy}{dx}\right)_{\theta=\pi} =\frac{u'v-uv'}{v^2} =\frac{6(-4)-0}{(-4)^2} =\frac{-24}{16}=-\frac{3}{2}$$ --- 5. **Second derivative** $$\frac{d^2y}{dx^2}=\frac{\dfrac{d}{d\theta}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{d\theta}}$$ At $\theta=\pi$, $$\frac{d^2y}{dx^2}=\frac{-\frac{3}{2}}{-4}=\frac{3}{8}$$ --- 6. **Match with options** $$\frac{d^2y}{dx^2}=\frac{3}{8}$$ So the correct option is: **A. $\dfrac{3}{8}$** --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.
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