• # 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) $(3\sqrt{3},-3)$

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.