JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    The value of c in the Lagrange's mean value theorem for the function \[f(x)={{x}^{3}}4{{x}^{2}}+8x+11,\] when \[x\in [0,1]\]  is [JEE MAIN Held on 07-01-2020 Evening]

    A) \[\frac{4-\sqrt{7}}{2}\]          

    B) \[\frac{4-\sqrt{5}}{3}\]

    C) \[\frac{2}{3}\]

    D) \[\frac{\sqrt{7}-2}{3}\]

    Correct Answer: A

    Solution :

    [a] \[\because \,\,\,\,f'(c)=\frac{f(b)-f(a)}{b-a}\] \[\therefore \,\,\,\,\,\,\,\,\,\,3{{c}^{2}}-8c+8=\frac{f(1)-f(0)}{1-0}\] \[3{{c}^{2}}-8c+8=16-11\] \[3{{c}^{2}}-8c+3=0\] \[\therefore \,\,\,\,c=\frac{8\pm 2\sqrt{7}}{6}\] \[\therefore \,\,\,\,c=\frac{4-\sqrt{7}}{3}\]      as \[c\in (0,1)\]

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