A) \[\frac{3}{\pi \sqrt{{{\pi }^{2}}-3}}\]
B) \[\frac{2}{\pi \sqrt{{{\pi }^{2}}-2}}\]
C) \[\frac{-3}{\pi \sqrt{{{\pi }^{2}}+3}}\]
D) None of these
Correct Answer: A
Solution :
[a] \[\because {{(siny)}^{sin\frac{\pi }{2}}}+\frac{\sqrt{3}}{2}.se{{c}^{-1}}(2x)\] \[+{{2}^{x}}.\tan (\log (x+1))=0\] Differentiating w.r.t x, on both sides/ we have \[{{(siny)}^{sin\frac{\pi }{2}}}\log (\sin y).\cos \frac{\pi x}{2}.\frac{\pi }{2}\] \[+\sin \frac{\pi x}{2}.\sin {{y}^{{{\left( \sin \frac{\pi x}{2} \right)}^{-1}}}}\times \cos y.\frac{dy}{dx}\] \[+\frac{\sqrt{3}}{2}.\frac{2}{2\left| x \right|\sqrt{4{{x}^{2}}-2}}\] \[+\frac{2x.se{{c}^{2}}\left( log\left( x+2 \right) \right)}{x+2}\] \[+{{2}^{x}}.log22.tan\text{ }(log(x+2))=0\] Putting \[x=-1\And y=\frac{\sqrt{3}}{2}\], we have \[{{\frac{dy}{dx}}_{at\left( -1,\frac{\sqrt{-3}}{\pi } \right)}}=\frac{{{\left( -\frac{\sqrt{3}}{\pi } \right)}^{2}}}{\sqrt{1-{{\left( -\frac{\sqrt{3}}{\pi } \right)}^{2}}}}=\frac{3}{\pi \sqrt{{{\pi }^{2}}-3}}\] Hence, option [a]is correct.
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