Question

The real values of $k$ for which the equation $4x^3+3x^2-6x+k=0$ has two distinct real roots in $[0,1]$ lie in the inteval ?

Answer 1

$f\left(x\right)=4x^3+3x^2-6x+k$

$f^{\prime }\left(x\right)=12x^2+6x-6=0$

$\Rightarrow 6(2x^2+x-1)=0$

$\Rightarrow 2x^2+x-1=0$

$\Rightarrow 2x^2+2x-x-1=0$

$\Rightarrow 2x(x+1)-1(x+1)=0$

$\Rightarrow (x+1)(2x-1)=0$

$\Rightarrow x=-1,\frac{1}{2}$

Thus,$x=\frac{1}{2}$ lies in the interval $[0,1]$

Put $x=\frac{1}{2}$ in $4x^3+3x^2-6x+k=0$

$\Rightarrow 4\times {\left(\frac{1}{2}\right)}^3+3\times {\left(\frac{1}{2}\right)}^2-6\times \frac{1}{2}+k=0$

$\Rightarrow 4\times \frac{1}{8}+3\times \frac{1}{4}-6\times \frac{1}{2}+k=0$

$\Rightarrow \frac{1}{2}+\frac{3}{4}-3+k=0$

$\Rightarrow \frac{2}{4}+\frac{3}{4}-\frac{12}{4}+k=0$

$\Rightarrow \frac{2+3-12}{4}+k=0$

$\Rightarrow \frac{-7}{4}+k=0$

$\therefore k=\frac{7}{4}$