Question

The number of elements in the range of the function $f:(-\infty, 1) \rightarrow \mathbb{R}$ defined by $f(x)=[9^x-3^x+1]$ (where [.] is the greatest integer function) is not less than

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (D)

7

Answer 2

To find the number of elements in the range of the function $f(x)=[9^x-3^x+1]$, we first analyze the expression inside the greatest integer function.

Let $y = 3^x$. Given the domain of $f$ is $x \in (-\infty, 1)$. As $x \rightarrow -\infty$, $3^x \rightarrow 0$. As $x \rightarrow 1$, $3^x \rightarrow 3^1 = 3$. So, the range of $y=3^x$ is $y \in (0, 3)$.

Now, substitute $y$ into the expression: Let $g(y) = 9^x - 3^x + 1 = (3^x)^2 - 3^x + 1 = y^2 - y + 1$. We need to find the range of $g(y)$ for $y \in (0, 3)$. The function $g(y)$ is a quadratic in $y$, representing a parabola opening upwards. The vertex of the parabola $g(y) = y^2 - y + 1$ occurs at $y = -\frac{(-1)}{2(1)} = \frac{1}{2}$. Since $y=\frac{1}{2}$ is within the interval $(0, 3)$, this is where the minimum value of $g(y)$ occurs. The minimum value is $g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1-2+4}{4} = \frac{3}{4}$.

Now, let's evaluate $g(y)$ at the boundaries of the interval $(0, 3)$: As $y \rightarrow 0^+$, $g(y) \rightarrow (0)^2 - (0) + 1 = 1$. As $y \rightarrow 3^-$, $g(y) \rightarrow (3)^2 - (3) + 1 = 9 - 3 + 1 = 7$.

Since the function $g(y)$ is continuous on $(0, 3)$, and its minimum value is $3/4$ (at $y=1/2$), and it approaches $7$ as $y \rightarrow 3^-$, the range of $g(y)$ for $y \in (0, 3)$ is $[3/4, 7)$.

The function $f(x)$ is defined as $f(x) = [g(y)]$, where $[.]$ denotes the greatest integer function. We need to find the integers that $f(x)$ can take based on the range of $g(y)$, which is $[3/4, 7)$. The greatest integer function $[Z]$ gives the largest integer less than or equal to $Z$. For $Z \in [3/4, 7)$:

• The smallest value $Z$ can take is $3/4$. So, $[3/4] = 0$. This value is achieved when $y=1/2$, i.e., $3^x=1/2$, which means $x=\log_3(1/2) = -\log_3 2$. This $x$ is in $(-\infty, 1)$.
• As $Z$ increases from $3/4$ up to $7$ (but not including $7$), the possible integer values for $[Z]$ are:
  • If $0.75 \le Z < 1$, then $[Z]=0$.
  • If $1 \le Z < 2$, then $[Z]=1$.
  • If $2 \le Z < 3$, then $[Z]=2$.
  • If $3 \le Z < 4$, then $[Z]=3$.
  • If $4 \le Z < 5$, then $[Z]=4$.
  • If $5 \le Z < 6$, then $[Z]=5$.
  • If $6 \le Z < 7$, then $[Z]=6$.

Since $g(y)$ is a continuous function on $(0,3)$ and its range covers the interval $[3/4, 7)$, it takes on all values in this interval. Thus, it will take on values that allow its floor to be $0, 1, 2, 3, 4, 5, 6$.

The integers in the range of $f(x)$ are $\{0, 1, 2, 3, 4, 5, 6\}$. The number of elements in this set is 7.

The question asks for the number of elements in the range of the function is "not less than" a certain value. This means "greater than or equal to". Since the number of elements is exactly 7, it is not less than 7.