Question

The number of real solution of the equation $(\frac{9}{10}{)}^x=-3+x-x^2$ is-

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Let $f\left(x\right)={\left(\frac{9}{10}\right)}^x$

$g\left(x\right)=-3+x-x^2$

For $x<0$

$f\left(x\right)>1$

$g\left(x\right)-1=-x^2+x-4$

$△=1-4(-1)(-4)$

$=-15<0$

and co efficient of $x^2=-1<0$

$\therefore g\left(x\right)-1<0$

$\therefore g\left(x\right)<1$

$\therefore f\left(x\right)-g\left(x\right)$ has no solution if $x<0$

For $x>0$

$f\left(x\right)<1$

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

$-x^2+x-3-f\left(x\right)=0$

$△=1-4(-1)(-3-f\left(x\right))$

$=1-12+4f\left(x\right)$

$△=-11+4f\left(x\right)$

$f\left(x\right)<0$

$4f\left(x\right)-11<-11$

$\therefore △<0$

$f\left(x\right)=g\left(x\right)$ has no real solution for $x>0$

At $x=0$

$f\left(x\right)=1,g\left(x\right)=-3$

$\therefore f\left(x\right)=g\left(x\right)$ has no real solution $\forall xϵR$