Question

If $x^3+x-5=0$ has atleast one real solution in $(a,b);$ where $a,b\in Z$, then $(a+b)=$

Answer 1

$f\left(x\right)=x^3+x-5$
Case $1:$ for $x<0,f\left(x\right)<0$ (no solution)

Case $2:f\left(0\right)<0$ and $f\left(1\right)<0$
$\Rightarrow f\left(x\right)=0$ no solution in $(0,1).$

Case $3:f\left(1\right)<0$ and $f\left(2\right)>0$
$\Rightarrow f\left(x\right)=0$ has atleast one solution in $(1,2).$

Case $4:$ for $x>2,f\left(x\right)>0$
So, no solution.

$\Rightarrow (a+b)=1+2=3$