JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer The function \[f(x)={{(x-3)}^{2}}\] satisfies all the conditions of mean value theorem in \[\{3,\,\,4\}\]. A point on \[y={{(x-3)}^{2}}\], where the tangent is parallel to the chord joining \[(3,\,\,0)\] and \[(4,\,\,1)\] is

    A) \[\left( \frac{7}{2},\,\,\frac{1}{2} \right)\]                  

    B) \[\left( \frac{7}{2},\,\,\frac{1}{4} \right)\]

    C) \[(1,\,\,4)\]                   

    D) \[(4,\,\,1)\]

    Correct Answer: B

    Solution :

     Let the point be\[({{x}_{1}},\,\,{{y}_{1}})\]. Therefore\[={{({{x}_{1}}-3)}^{2}}\]                                ... (i) \[\therefore \]Now slope of the tangent at \[({{x}_{1}},\,\,{{y}_{1}})\] is \[2({{x}_{1}}-3),\] but it is equal to \[1\]. Therefore,\[2({{x}_{1}}-3)=1\Rightarrow {{x}_{1}}=\frac{7}{2}\]             \[{{y}_{1}}={{\left( \frac{7}{2}-3 \right)}^{2}}=\frac{1}{4}\]. Hence the point is\[\left( \frac{7}{2},\,\,\frac{1}{4} \right)\].


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