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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Evening)

  • question_answer
    Let a function \[f:\,\,[0,\,\,5]\to R\] be continuous,  \[f(1)=3\] and F be defined as: \[F(x)=\int\limits_{1}^{x}{{{t}^{2}}}g(t)\,dt,\] where \[g(t)=\int\limits_{1}^{t}{f(u)\,du.}\] Then for the function F, the point \[x=1\] is [JEE MAIN Held on 09-01-2020 Evening]

    A) a point of inflection.

    B) not a critical point.

    C) a point of local minima.

    D) a point of local maxima.

    Correct Answer: C

    Solution :

    \[f\left( x \right)=\int\limits_{1}^{x}{{{t}^{2}}g\left( t \right)dt,}\]   \[g\left( t \right)=\int\limits_{1}^{t}{f\left( u \right)\,du}\] \[f'(x)={{x}^{2}}g(x),\]  \[g'(t)=f(t)\] and \[g(1)=0\] \[\therefore \,\,\,\,\,\,\,f'(1)=g(1)=0\] Also,   \[f''(x)={{x}^{2}}g'(x)+2x\,\,g(x)\] \[f''(1)=g'(1)+2g(1)=3>0\] \[\therefore \]    Local Minima at \[x=1\]


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