JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    Passage (Q. - 90) Let f(x) be a polynomial with positive leading coefficient satisfying \[f(0)=0\]and \[f(f(x))=\]\[x\int\limits_{0}^{x}{f(x)}dx\forall x\in R\] If tangents are drawn to the curve y = f(x) from any point on the line \[y+\sqrt{3}/4=0,\] then the chord of contacts will be concurrent at the point

    A) \[\left( 0,\frac{\sqrt{3}}{4} \right)\]                        

    B) \[\left( 0,-\frac{\sqrt{3}}{2} \right)\]

    C) \[\left( \frac{\sqrt{3}}{4},\frac{3}{2} \right)\]                     

    D) \[(3\sqrt{3},-3)\]

    Correct Answer: A

    Solution :

    Let f(x) has degree n so\[{{n}^{2}}=1+n+1\Rightarrow n=2\] \[\Rightarrow \]f(x) is quadratic with\[f(0)=0\] so let\[f(x)=a{{x}^{2}}+bx\] so\[a{{(a{{x}^{2}}+bx)}^{2}}+b(a{{x}^{2}}+bx)=x\left[ \frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2} \right]\forall x\in R\]\[\Rightarrow \]\[{{a}^{3}}-\frac{a}{3}=0;2{{a}^{2}}b-\frac{b}{2}=0;a{{b}^{2}}+ab=0\]and\[{{b}^{2}}=0\]\[\Rightarrow \]b = 0 and \[a=\pm \frac{1}{\sqrt{3}}\] \[\therefore \]given leading coefficient is positive \[\Rightarrow \]\[f(x)=\frac{{{x}^{2}}}{\sqrt{3}}\] So required curve is parabola Hence perpendicular tangents intersect on directrix. Chord of contact of any point on directrix always passes through focus.

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