JEE Main & Advanced Mathematics Differentiation Question Bank Self Evaluation Test - Limits and Derivatives

  • question_answer
    Let \[f(x)=\sqrt{x-1}+\sqrt{x+24-10\sqrt{x-1}};\] \[1<x<26\] Be real valued function. Then \[f'(x)\]for \[1<x<26\] is

    A) 0

    B) \[\frac{1}{\sqrt{x-1}}\]

    C) \[2\sqrt{x-1}-5\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[f(x)=\sqrt{x-1}+\sqrt{25+(x-1)-10\sqrt{x-1}}\] \[=\sqrt{x-1}+\sqrt{{{(5-\sqrt{x-1})}^{2}}}\] \[=\sqrt{x-1}+\left| 5-\sqrt{x-1} \right|=5\] \[[\because \sqrt{x-1}<5for1<x<26]\] \[\therefore f'(x)=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner