Question

The solution set of ${(\dfrac{3}{5})^x} = x - {x^2} - 9$ is
A. $0$
B. $1$
C. $\phi $
D. None of these

Answer 1

Hint: We have an algebraic equation in this question, and we have to find its solution. By solution, we mean the value of $x$ at which the left-hand side of the given equation is equal to the right-hand side of the given equation. For finding the solution of the given equation, we will simplify the right-hand side using algebraic identities.

Complete step-by-step answer:
We are given ${(\dfrac{3}{5})^x} = x - {x^2} - 9$
We have to find its solution set. Simplifying the right-hand side of the above equation, we get:
$x - {x^2} - 9 = - ({x^2} - x + 9)$
Adding and subtracting $\dfrac{1}{4}$ from the above equation, we get:
$
  x - {x^2} - 9 = - ({x^2} + \dfrac{1}{4} - x + 9 - \dfrac{1}{4}) \\
   = - [{(x)^2} + {(\dfrac{1}{2})^2} - 2 \times \dfrac{1}{2} \times x + \dfrac{{36 - 1}}{4}] \\
 $
We know the algebraic identity ${(a - b)^2} = {a^2} + {b^2} - 2ab$ , using this identity we get:
${x^2} + {(\dfrac{1}{2})^2} - 2 \times \dfrac{1}{2} \times x = {(x - \dfrac{1}{2})^2}$
Now, we put this value in the above simplified equation of $x - {x^2} - 9$
$x - {x^2} - 9 = - [{(x - \dfrac{1}{2})^2} + \dfrac{{35}}{4}]$
We know that $\dfrac{{35}}{4} > 0$ as it is a positive number, we also know that square of any number is always positive, so ${(x - \dfrac{1}{2})^2} > 0$
$ \Rightarrow {(x - \dfrac{1}{2})^2} + \dfrac{{35}}{4} > 0$
If we put a negative sign before a positive number, then it represents a negative number, so
$ - [{(x - \dfrac{1}{2})^2} + \dfrac{{35}}{4}] < 0$
$ \Rightarrow x - {x^2} - 9 < 0$
We know $x - {x^2} - 9 = {(\dfrac{3}{5})^x}$
So, ${(\dfrac{3}{5})^x} < 0$
But, we also know that when a positive number is raised to any power, then the outcome is always positive.
So, ${(\dfrac{3}{5})^x} > 0$
Which is contradicting, so no solution exists for the equation ${(\dfrac{3}{5})^x} = x - {x^2} - 9$ .
Thus the solution set is $\phi $ .

The correct option is option C.

Note: The solution of the given equation came out to be $\phi $ , which means that there is no value of $x$ at which the left-hand side of the given equation is equal to its right-hand side. So, the given equality doesn’t hold true for any value of $x$ .