question_answer

If \[x=2\sin \theta -\sin 2\theta \] and \[y=2\cos \theta -\cos 2\theta ,\] \[\theta \in [0,\,\,2\pi ],\] then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\] at \[\theta =\pi \] is  [JEE MAIN Held on 09-01-2020 Evening]

A) \[\frac{3}{4}\]

B) \[\frac{3}{2}\]

C) \[-\frac{3}{8}\]

D) \[-\frac{3}{4}\]

Correct Answer: A , B , C , D

Solution :

(Bonus)  \[x=2\sin \theta -\sin 2\theta ,\]  \[y=2\cos \theta -\cos 2\theta \] \[\frac{dx}{d\theta }=2\cos \theta -2\cos 2\theta ,\] \[\frac{dy}{d\theta }=-2\sin \theta +2\sin 2\theta \] \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{\sin 2\theta -\sin \theta }{\cos \theta -\cos 2\theta }=\frac{\cos \frac{3\theta }{2}\sin \frac{\theta }{2}}{\sin \frac{3\theta }{2}\sin \frac{\theta }{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\frac{dy}{dx}=\cot \frac{3\theta }{2},\] again diff w.r.t x \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\,\,\left( \cot \frac{3\theta }{2} \right)\] \[=\frac{-3}{2}\text{cose}{{\text{c}}^{2}}\frac{3\theta }{2}\cdot \frac{d\theta }{dx}\] \[=\frac{-3}{2}\text{cose}{{\text{c}}^{2}}\frac{3\theta }{2}\cdot \frac{1}{2(\cos \theta -\cos 2\theta )}\] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=\bar{x}}}=\frac{-3}{2}\cdot 1\cdot \frac{1}{2(-1-1)}=\frac{3}{8}\]

Solution :

(Bonus)  \[x=2\sin \theta -\sin 2\theta ,\]  \[y=2\cos \theta -\cos 2\theta \] \[\frac{dx}{d\theta }=2\cos \theta -2\cos 2\theta ,\] \[\frac{dy}{d\theta }=-2\sin \theta +2\sin 2\theta \] \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{\sin 2\theta -\sin \theta }{\cos \theta -\cos 2\theta }=\frac{\cos \frac{3\theta }{2}\sin \frac{\theta }{2}}{\sin \frac{3\theta }{2}\sin \frac{\theta }{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\frac{dy}{dx}=\cot \frac{3\theta }{2},\] again diff w.r.t x \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\,\,\left( \cot \frac{3\theta }{2} \right)\] \[=\frac{-3}{2}\text{cose}{{\text{c}}^{2}}\frac{3\theta }{2}\cdot \frac{d\theta }{dx}\] \[=\frac{-3}{2}\text{cose}{{\text{c}}^{2}}\frac{3\theta }{2}\cdot \frac{1}{2(\cos \theta -\cos 2\theta )}\] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=\bar{x}}}=\frac{-3}{2}\cdot 1\cdot \frac{1}{2(-1-1)}=\frac{3}{8}\]

Solution :

(Bonus)  \[x=2\sin \theta -\sin 2\theta ,\]  \[y=2\cos \theta -\cos 2\theta \] \[\frac{dx}{d\theta }=2\cos \theta -2\cos 2\theta ,\] \[\frac{dy}{d\theta }=-2\sin \theta +2\sin 2\theta \] \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{\sin 2\theta -\sin \theta }{\cos \theta -\cos 2\theta }=\frac{\cos \frac{3\theta }{2}\sin \frac{\theta }{2}}{\sin \frac{3\theta }{2}\sin \frac{\theta }{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\frac{dy}{dx}=\cot \frac{3\theta }{2},\] again diff w.r.t x \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\,\,\left( \cot \frac{3\theta }{2} \right)\] \[=\frac{-3}{2}\text{cose}{{\text{c}}^{2}}\frac{3\theta }{2}\cdot \frac{d\theta }{dx}\] \[=\frac{-3}{2}\text{cose}{{\text{c}}^{2}}\frac{3\theta }{2}\cdot \frac{1}{2(\cos \theta -\cos 2\theta )}\] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=\bar{x}}}=\frac{-3}{2}\cdot 1\cdot \frac{1}{2(-1-1)}=\frac{3}{8}\]

Solution :

(Bonus)  \[x=2\sin \theta -\sin 2\theta ,\]  \[y=2\cos \theta -\cos 2\theta \] \[\frac{dx}{d\theta }=2\cos \theta -2\cos 2\theta ,\] \[\frac{dy}{d\theta }=-2\sin \theta +2\sin 2\theta \] \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{\sin 2\theta -\sin \theta }{\cos \theta -\cos 2\theta }=\frac{\cos \frac{3\theta }{2}\sin \frac{\theta }{2}}{\sin \frac{3\theta }{2}\sin \frac{\theta }{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\frac{dy}{dx}=\cot \frac{3\theta }{2},\] again diff w.r.t x \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\,\,\left( \cot \frac{3\theta }{2} \right)\] \[=\frac{-3}{2}\text{cose}{{\text{c}}^{2}}\frac{3\theta }{2}\cdot \frac{d\theta }{dx}\] \[=\frac{-3}{2}\text{cose}{{\text{c}}^{2}}\frac{3\theta }{2}\cdot \frac{1}{2(\cos \theta -\cos 2\theta )}\] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=\bar{x}}}=\frac{-3}{2}\cdot 1\cdot \frac{1}{2(-1-1)}=\frac{3}{8}\]