Question

The solution set of ${\left(\frac{3}{5}\right)}^x=x-x^2-9$ is

• A. $\left\{0\right\}$
• B. $\left\{1\right\}$
• C. $\mathrm{\varphi }$
• D. None of these

Answer 1

The correct option is C

$\mathrm{\varphi }$

Explanation for the correct option:

The given equation: ${\left(\frac{3}{5}\right)}^x=x-x^2-9$.

As an exponential expression always has a positive value.

Thus, ${\left(\frac{3}{5}\right)}^x>0$.

So, $\left(x-x^2-9\right)$ should also be positive.

Now, $\left(x-x^2-9\right)=-\left(x^2-x+9\right)$.

$$\begin{gathered} \Rightarrow \left(x-x^2-9\right)=-\left[x^2-2·\left(x\right)·\left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^2+\frac{35}{4}\right] \\ \Rightarrow \left(x-x^2-9\right)=-\left[{\left(x-\frac{1}{2}\right)}^2+\frac{35}{4}\right] \end{gathered}$$

It is known that, ${\left(x-\frac{1}{2}\right)}^2\ge 0$ (Since squaring a number gives only positive values)

$$\begin{gathered} \Rightarrow {\left(x-\frac{1}{2}\right)}^2+\frac{35}{4}\ge 0 \\ \Rightarrow -\left[{\left(x-\frac{1}{2}\right)}^2+\frac{35}{4}\right]<0 \\ \Rightarrow x-x^2-9<0 \end{gathered}$$

Thus, $\left(x-x^2-9\right)$ is always negative.

Hence, the given equation has no real roots.

Therefore, the solution set of ${\left(\frac{3}{5}\right)}^x=x-x^2-9$ is $\mathrm{\varphi }$.

Hence, option C is correct.