Let f(x) be a quadratic polynomial with leadi | vedclass

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(A) Let $f(x) = x^2 + bx + c$. Since the leading coefficient is $1$ and $f(0) = p$,we have $c = p$. Thus,$f(x) = x^2 + bx + p$.Given $f(1) = 1 + b + p

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Let $f(x) = x^2 + bx + c$. Since the leading coefficient is $1$ and $f(0) = p$,we have $c = p$. Thus,$f(x) = x^2 + bx + p$.Given $f(1) = 1 + b + p = \frac{1}{3}$,so $b = \frac{1}{3} - 1 - p = -\frac{2}{3} - p$.Let $\alpha$ be the common root of $f(x) = 0$ and $f(f(f(f(x)))) = 0$. Since $f(\alpha) = 0$,we have $f(f(f(f(\alpha)))) = f(f(f(0))) = f(f(p)) = 0$.This implies that $f(p)$ must be a root of $f(x) = 0$,so $f(f(p)) = 0$ means $f(p) = \alpha$ or $f(p) = \beta$,where $\alpha, \beta$ are roots of $f(x) = 0$.Since $f(x) = (x-\alpha)(x-\beta)$,we have $f(0) = \alpha \beta = p$.Also,$f(1) = (1-\alpha)(1-\beta) = \frac{1}{3}$.If $f(p) = \alpha$,then $(p-\alpha)(p-\beta) = \alpha$. Substituting $p = \alpha \beta$,we get $(\alpha \beta - \alpha)(\alpha \beta - \beta) = \alpha$.$\alpha(\beta-1) \beta(\alpha-1) = \alpha$. Since $\alpha 
eq 0$,$(\beta-1)(\alpha-1)\beta = 1$.We know $(1-\alpha)(1-\beta) = \frac{1}{3}$,so $(\alpha-1)(\beta-1) = \frac{1}{3}$.Substituting this into the equation,we get $\frac{1}{3} \beta = 1$,which implies $\beta = 3$.Then $(1-\alpha)(1-3) = \frac{1}{3} \Rightarrow -2(1-\alpha) = \frac{1}{3} \Rightarrow 1-\alpha = -\frac{1}{6} \Rightarrow \alpha = \frac{7}{6}$.Thus,$f(x) = (x-\frac{7}{6})(x-3)$.Finally,$f(-3) = (-3 - \frac{7}{6})(-3 - 3) = (-\frac{25}{6})(-6) = 25$.

Let $f(x)$ be a quadratic polynomial with leading coefficient $1$ such that $f(0)=p, p \neq 0$ and $f(1)=\frac{1}{3}$. If the equations $f(x)=0$ and $f(f(f(f(x))))=0$ have a common real root,then $f(-3)$ is equal to $........$

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