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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2015

  • question_answer
    If \[({{x}^{2}}-1)y'-xy=0\]and \[({{x}^{2}}-1)y''+(x-1)y'=0\], then \[2\{{{(x-a)}^{2}}+{{(y-a)}^{2}}\}={{(x+y)}^{2}}\]is equal to

    A)  \[2\sqrt{2}a\]

    B) \[\sqrt{2}a\]

    C)  \[\text{si}{{\text{n}}^{\text{2}}}\beta =\text{ 3si}{{\text{n}}^{\text{2}}}\theta \]

    D)  \[\text{co}{{\text{s}}^{\text{2}}}\theta \]

    Correct Answer: A

    Solution :

    Given, \[y=\sin \left[ \log \left( \frac{2x+3}{3-2x} \right) \right]\] On differentiating both sides w.r.t. x, we get \[\frac{dy}{dx}=\cos \left[ \log \left( \frac{2x+3}{3-2x} \right) \right].\frac{3-2x}{2x+3}\] \[\frac{(3-2x)2-(2x+3)(-2)}{{{(3-2x)}^{2}}}\]                 \[\Rightarrow \] \[\frac{dy}{dx}=\cos \left[ \log \left( \frac{2x+3}{3-2x} \right) \right].\frac{3-2x}{2x+3}.\frac{12}{{{(3-2x)}^{2}}}\]                                 \[=\frac{12}{9-4{{x}^{2}}}.\cos \left[ \log \left( \frac{2x+3}{3-2x} \right) \right]\]


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