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

  • question_answer
    Let f(x) be a polynomial of degree 5 such that \[x=\pm 1\] are its critical points. If \[\underset{x\to 0}{\mathop{\lim }}\,\left( 2+\frac{f(x)}{{{x}^{3}}} \right)=4\], then which one of the following is not true? [JEE MAIN Held on 07-01-2020 Evening]

    A) f is an odd function

    B) x =1 is a point of minima and \[x=1\] is a point of maxima of f.

    C) \[f\left( 1 \right)4f\left( 1 \right)=4\]

    D) x =1 is a point of maxima and \[x=1\] is a point of minimum of f.

    Correct Answer: B

    Solution :

    [b] \[\because \] f(x) is a five degree polynomial such that
    \[\underset{x\to 0}{\mathop{\lim }}\,\left( 2+\frac{f(x)}{{{x}^{3}}} \right)=4\]
    let \[f(x)=a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}\]
    \[\underset{x\to 0}{\mathop{\lim }}\,\left( 2+\frac{a\,{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}}{{{x}^{3}}} \right)=4\]
    \[\Rightarrow \,\,2+c=4\Rightarrow \,c=2\]
    Now, \[f'(x)=5a{{x}^{4}}+4b{{x}^{3}}+3c{{x}^{2}}\]
    \[\because \,\,f'(1)\,=0\Rightarrow \,5a+4b+6=0\]
    and \[f'(-1)=0\Rightarrow \,5a-4b+6=0\]
    \[\therefore \,\,b=0,\,a=-\frac{6}{5}\]
    \[\therefore \,f(x)=-\frac{6}{5}{{x}^{5}}+2{{x}^{3}}\]
    It is clear that maxima at x = 1 and minima at\[x=1\].
    and \[f\left( 1 \right)4f\left( 1 \right)=4\]

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