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JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

  • question_answer
    Let \[f(t)={{t}^{2}}\] for \[0\le t\le l\] and \[g(t)={{t}^{3}}\] for\[0\le t\le l\]. The value of c with \[0<c<l\] at which \[\frac{f(l)-f(0)}{g(l)-g(0)}=\frac{f'(c)}{g'(c)}\], is

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

    B)  \[\frac{1}{3}\]

    C)  \[\frac{1}{2}\]                                  

    D)  \[\frac{1}{6}\]

    Correct Answer: A

    Solution :

    We have \[\frac{f(1)\,-f(0)}{g(1)\,-g(0)}\,\,=\frac{'f(c)}{g'(c)}\]             \[\Rightarrow \,\,\frac{1-0}{1-0}\,=\frac{2c}{3{{c}^{2}}}\,\Rightarrow \,\frac{3c}{2}=1\,(As\,\,0<c<1)\]           \[\Rightarrow c=\frac{2}{3}\]


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