Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is D $0$

Given $\frac{(}{9}10{)}^x=-(x^2-x+3)$

Here $x^2-x+3>0$ because descriminant <0 [$\left(\sqrt{b^2-4ac}\right)=\sqrt{-8}]$

Therefore it has imaginary roots and graph of parabola lies entirely above x axis

$\Rightarrow$equation is always positive

But we have negative of that equation

$(\frac{9}{10}{)}^x<0$ which is impossible as it is always positive for all values of x

$\therefore$number of real roots of the equation is $0$