JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    If \[y=\frac{(a-x)\sqrt{a-x}-(b-x)\sqrt{x-b}}{\sqrt{a-x}+\sqrt{x-b}}\], then \[\frac{dy}{dx}\] wherever it is defined is

    A) \[\frac{x+(a+b)}{\sqrt{(a-x)(x-b)}}\]

    B) \[\frac{2x-a-b}{2\sqrt{a-x}\sqrt{x-b}}\]

    C) \[-\frac{(a+b)}{2\sqrt{(a-x)(x-b)}}\]

    D) \[\frac{2x+(a+b)}{2\sqrt{(a-x)(x-b)}}\]

    Correct Answer: B

    Solution :

    [b] \[y=\frac{{{(a-x)}^{3/2}}+{{(x-b)}^{3/2}}}{\sqrt{a-x}+\sqrt{x-b}}\] \[=\frac{(\sqrt{a-x}+\sqrt{x-b})(a-x-\sqrt{a-x}\sqrt{a-b}+x-b)}{\sqrt{a-x}+\sqrt{x-b}}\]\[=a-b-\sqrt{a-x}\sqrt{x-b}\] or \[\frac{dy}{dx}=\frac{1}{2\sqrt{a-x}}\sqrt{x-b}-\frac{1}{2\sqrt{x-b}}\sqrt{a-x}\] \[=\frac{2x-a-b}{2\sqrt{a-x}\sqrt{x-b}}\]


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