Mathematics - Quadratic Equation Question with Solution | TestHub

Given $f(x)=x^3+x^2-5x-1$ $\therefore f^{\prime }(x)=3x^2+2x-5$.

The roots of $f^{\prime }(x)=0$ are $-\frac{5}{3}$ and 1

Writing the sign scheme for $f^{\prime }(x)$,

Also, $f(-\infty )=-\infty <0,f(\infty )=\infty >0$ $f(1)=-4,f\left(-\frac{5}{3}\right)=\frac{148}{27}$ Now, graph of $y=f(x)$ is as follows

$\begin{array}{rlrlrlr}& f(-3)=-27+9+15-1=-4<0& & & & & \end{array}$

$f(-2)=-8+4+10-1>0$

$f(-1)=4>0,f(0)=-1<0$

$f(2)=1>0$

$\therefore -3<\alpha <-2,-1<\beta <0,1<\gamma <2$

$|[\alpha ]+[\beta ]+[\gamma ]|=|-3-1+1|=3$