Question

If the function f(x) = x³ − 9kx² + 27x + 30 is increasing on R, then
(a) −1 ≤ k < 1
(b) k < −1 or k > 1
(c) 0 < k < 1
(d) −1 < k < 0

Answer 1

(a)

$$\begin{gathered} f\left(x\right)=x^3-9k{x}^2+27x+30 \\ f\text{'}\left(x\right)=3x^2-18kx+27 \\ =3\left(x^2-6kx+9\right) \\ \mathrm{Given}:f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{on}R. \\ \Rightarrow f\text{'}\left(x\right)>0\mathrm{for}\mathrm{all}x\in R \\ \Rightarrow 3\left(x^2-6kx+9\right)>0\mathrm{for}\mathrm{all}x\in R \\ \Rightarrow x^2-6kx+9>0\mathrm{for}\mathrm{all}x\in R \\ \Rightarrow {\left(-6k\right)}^2-4\left(1\right)\left(9\right)<0\left[\because a{x}^2+bx+c>0\mathrm{for}\mathrm{all}x\in R\Rightarrow a>0\mathrm{and}\mathrm{Disc}<0\right] \\ \Rightarrow 36k^2-36<0 \\ \Rightarrow k^2-1<0 \\ \Rightarrow \left(k+1\right)\left(k-1\right)<0 \\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{possible}\mathrm{when}\left(k+1\right)<0\mathrm{and}\left(k-1\right)>0. \\ \Rightarrow k<-1\mathrm{and}k>1(\mathrm{Not}\mathrm{possible}) \\ \mathrm{or}\left(k+1\right)>0\mathrm{and}\left(k-1\right)<0 \\ \Rightarrow k>-1\mathrm{and}k<1 \\ \Rightarrow -1<k<1 \\ \mathrm{Disclaimer}:\left(a\right)\mathrm{part}\mathrm{should}\mathrm{be}-1<k<1\mathrm{instead}\mathrm{of}-1\le k<1. \end{gathered}$$