Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$RHS=-3+x-x^2$

$=-(x^2-x+\frac{1}{4}-\frac{1}{4}+3)$

$=-\left[\right(x-\frac{1}{2}{)}^2+\frac{11}{4}]$ ......... Always negative

$LHS=(\frac{9}{10}{)}^x=(0.9{)}^x$ ....... Always positive

$\therefore$ No solution