Question

In $y = a^x$ if $a>0$, if $x$ belongs to $(-infinity to 0)$ then $y$ will always be $(0,1)$

• A. True
• B. False

Answer 1

Answer: (B)

False

Answer 2

The statement claims that for any $a>0$ and any $x$ in the interval $(-\infty, 0)$, the value of $y = a^x$ will always lie within the interval $(0,1)$. We need to determine if this statement is true or false.

To evaluate the truthfulness of the statement, we examine the behavior of the function $y = a^x$ for $x < 0$ under different conditions for $a>0$.

• Case 1: $a > 1$. If $a > 1$, the function $y = a^x$ is an increasing function. As $x$ approaches $-\infty$, $y = a^x$ approaches $0$. As $x$ approaches $0$ from the left (i.e., $x \to 0^-$), $y = a^x$ approaches $a^0 = 1$. Thus, for $a > 1$ and $x \in (-\infty, 0)$, the range of $y$ is $(0,1)$. The statement holds for this case.

• Case 2: $a = 1$. If $a = 1$, the function becomes $y = 1^x$. For any real number $x$, $y = 1^x = 1$. Therefore, for $x \in (-\infty, 0)$, $y$ is always $1$. The interval $(0,1)$ does not include the value $1$. Hence, $y$ is not always in $(0,1)$ when $a=1$. The statement fails for this case.

• Case 3: $0 < a < 1$. If $0 < a < 1$, the function $y = a^x$ is a decreasing function. As $x$ approaches $-\infty$, $y = a^x$ approaches $+\infty$. As $x$ approaches $0$ from the left (i.e., $x \to 0^-$), $y = a^x$ approaches $a^0 = 1$. Thus, for $0 < a < 1$ and $x \in (-\infty, 0)$, the range of $y$ is $(1, \infty)$. The interval $(1, \infty)$ has no overlap with $(0,1)$. Hence, $y$ is not always in $(0,1)$ when $0 < a < 1$. The statement fails for this case.

Since the statement claims that $y$ will always be in $(0,1)$ for all $a>0$, and we have found cases ($a=1$ and $0<a<1$) where this is not true, the overall statement is false.

Counterexample 1: If $a=1$, then $y = 1^x = 1$ for all $x$. Since $1 \notin (0,1)$, the statement is false. Counterexample 2: If $0 < a < 1$, let $a=0.5$ and $x=-2$. Then $y = (0.5)^{-2} = (1/2)^{-2} = 2^2 = 4$. Since $4 \notin (0,1)$, the statement is false. The statement is only true for $a>1$.