Given that $\mathrm{f}\left( x \right)=27{{x}^{3}}-18{{x}^{2}}+6nx+2n+3$, show that \(x=

Question

Given that $\mathrm{f}\left( x \right)=27{{x}^{3}}-18{{x}^{2}}+6nx+2n+3$, show that $x=-1$ is a solution to $\mathrm{f}\left( \frac{x}{3} \right)=0$ for all values of $n$. Show that for any real number $m$ the curve $y={{x}^{3}}+5{{x}^{2}}-m$ and the line $y=6-mx$ meet at the point where $x=1$. Find the range of values of $m$ such that the curve $y={{x}^{3}}+5{{x}^{2}}-m$ and the line $y=6-mx$ meet only at the point $x=1$.

Accepted Answer

(a) $x=-1$ (b)(i) $x=1$ (b)(ii) $m>3$

Quadratic Functions, Equations and Inequalities - Nature of Roots of Quadratic Equations

Quadratic Functions, Equations and Inequalities - Nature of Roots of Quadratic Equations

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