Let a 1 and 0

Question

(D) Given the function $f(x) = \begin{cases} a^x, & -\infty  1$ and $0 < b < 1$.For $x < 0$,

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$(D)$ Neither one-one nor onto

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(D) Given the function $f(x) = \begin{cases} a^x, & -\infty  1$ and $0 For $x  1$,as $x 	o -\infty$,$f(x) 	o 0$,and as $x 	o 0^-$,$f(x) 	o 1$. Thus,the range for this part is $(0, 1)$.For $x \geq 0$,$f(x) = b^x$. Since $0 Combining these,the range of $f(x)$ is $(0, 1]$.$1$. One-one check: For any $y \in (0, 1)$,there exist two values of $x$,one negative and one positive,such that $f(x) = y$. For example,if $y = 0.5$,there is an $x_1  0$ such that $b^{x_2} = 0.5$. Thus,$f(x)$ is not one-one.$2$. Onto check: The codomain is given as $[0, 1]$. Since the range is $(0, 1]$,the value $0$ is not in the range (as $a^x > 0$ and $b^x > 0$ for all $x$). Therefore,$f(x)$ is not onto.Thus,$f(x)$ is neither one-one nor onto.

Let $a > 1$ and $0 < b < 1$. If $f: R \rightarrow [0, 1]$ is defined by $f(x) = \begin{cases} a^x, & -\infty < x < 0 \\ b^x, & 0 \leq x < \infty \end{cases}$,then $f(x)$ is

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