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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    If \[x=\sec \theta -\cos \theta \]and \[y={{\sec }^{n}}\theta -{{\cos }^{n}}\theta \], then [IIT 1989]

    A) \[({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{n}^{2}}({{y}^{2}}+4)\]

    B) \[({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{x}^{2}}({{y}^{2}}+4)\]

    C) \[({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}=({{y}^{2}}+4)\]     

    D) None of these

    Correct Answer: A

    Solution :

    • \[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }\]\[=\frac{n{{\sec }^{n}}\theta \tan \theta +n{{\cos }^{n-1}}\theta \sin \theta }{\sec \theta \tan \theta +\sin \theta }\]                       
    • \[=\frac{n({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}{\sec \theta +\cos \theta }\](Dividing \[{{N}^{r}}\] and \[{{D}^{r}}\]by \[\tan \theta \])                   
    • Þ  \[{{\left( \frac{dy}{dx} \right)}^{2}}=\frac{{{n}^{2}}{{({{\sec }^{n}}\theta +{{\cos }^{n}}\theta )}^{2}}}{{{(\sec \theta +\cos \theta )}^{2}}}\]                    \[=\frac{{{n}^{2}}[{{({{\sec }^{n}}\theta -{{\cos }^{n}}\theta )}^{2}}+4{{\sec }^{n}}\theta {{\cos }^{n}}\theta ]}{{{(\sec \theta -\cos \theta )}^{2}}+4\sec \theta .\cos \theta }=\frac{{{n}^{2}}({{y}^{2}}+4)}{{{x}^{2}}+4}\]           
    • Þ  \[({{x}^{2}}+4)\text{ }{{\left( \frac{dy}{dx} \right)}^{2}}={{n}^{2}}({{y}^{2}}+4)\].


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