Question

Let f(x) be a polynomial with positive leading coefficient satisfying f(0)=0 & f(f(x))=x∫x0f(t)dt ∀x∈R.

Two perpendicular tangents to the curve y=f(x) will intersect on the curve/line

Answer 1

y = -√(3)/4

Answer 2

Let the given equation be $f(f(x)) = x \int_0^x f(t) dt$ for all $x \in R$. $f(x)$ is a polynomial with positive leading coefficient and $f(0)=0$. Let the degree of the polynomial $f(x)$ be $n$. Since $f(0)=0$, $n \ge 1$. The degree of the left side $f(f(x))$ is $n \cdot n = n^2$. For the right side, let's find the degree of $\int_0^x f(t) dt$. If $f(x)$ has degree $n$, then $\int_0^x f(t) dt$ has degree $n+1$ (since $f(0)=0$, the constant of integration is zero). Thus, $x \int_0^x f(t) dt$ has degree $1 + (n+1) = n+2$. Equating the degrees of both sides, we get $n^2 = n+2$, which implies $n^2 - n - 2 = 0$. Factoring the quadratic equation, we get $(n-2)(n+1) = 0$. Since $n \ge 1$, the only valid degree is $n=2$. So, $f(x)$ is a polynomial of degree 2 with $f(0)=0$. Let $f(x) = ax^2 + bx + c$. Since $f(0)=0$, $c=0$. So $f(x) = ax^2 + bx$. The leading coefficient is $a$, and it is positive, so $a>0$.

Substitute $f(x) = ax^2 + bx$ into the given equation: $f(f(x)) = f(ax^2+bx) = a(ax^2+bx)^2 + b(ax^2+bx) = a(a^2x^4 + 2abx^3 + b^2x^2) + abx^2 + b^2x$ $f(f(x)) = a^3x^4 + 2a^2bx^3 + (ab^2+ab)x^2 + b^2x$.

$\int_0^x f(t) dt = \int_0^x (at^2+bt) dt = \left[ a\frac{t^3}{3} + b\frac{t^2}{2} \right]_0^x = a\frac{x^3}{3} + b\frac{x^2}{2}$. $x \int_0^x f(t) dt = x \left( a\frac{x^3}{3} + b\frac{x^2}{2} \right) = \frac{a}{3}x^4 + \frac{b}{2}x^3$.

Equating the coefficients of the powers of $x$ in $f(f(x))$ and $x \int_0^x f(t) dt$: Coefficient of $x^4$: $a^3 = \frac{a}{3}$. Since $a>0$, we can divide by $a$. $a^2 = \frac{1}{3}$. Since $a>0$, $a = \frac{1}{\sqrt{3}}$. Coefficient of $x^3$: $2a^2b = \frac{b}{2}$. Substituting $a^2 = \frac{1}{3}$, we get $2(\frac{1}{3})b = \frac{b}{2}$, which is $\frac{2}{3}b = \frac{b}{2}$. This implies $(\frac{2}{3} - \frac{1}{2})b = 0$, so $\frac{1}{6}b = 0$, which means $b=0$. Coefficient of $x^2$: $ab^2+ab = 0$. With $b=0$, this becomes $a(0)^2 + a(0) = 0$, which is $0=0$. This is consistent. Coefficient of $x$: $b^2 = 0$. With $b=0$, this becomes $0^2=0$, which is $0=0$. This is consistent.

So, the polynomial is $f(x) = \frac{1}{\sqrt{3}} x^2$. The curve is $y = \frac{1}{\sqrt{3}} x^2$. This is a parabola. We can rewrite the equation as $x^2 = \sqrt{3} y$. This is a parabola with vertex at $(0,0)$ and axis along the positive y-axis. The standard form of such a parabola is $x^2 = 4Ay$, where $4A = \sqrt{3}$, so $A = \frac{\sqrt{3}}{4}$. The focus of this parabola is at $(0, A) = (0, \frac{\sqrt{3}}{4})$. The directrix of this parabola is the line $y = -A = -\frac{\sqrt{3}}{4}$.

The question asks for the locus of the intersection point of two perpendicular tangents to the curve $y = f(x)$. For any parabola, the locus of the intersection point of two perpendicular tangents is the directrix of the parabola. For the parabola $y = \frac{1}{\sqrt{3}} x^2$ or $x^2 = \sqrt{3} y$, the directrix is the line $y = -\frac{\sqrt{3}}{4}$.

Thus, two perpendicular tangents to the curve $y = f(x)$ will intersect on the line $y = -\frac{\sqrt{3}}{4}$.