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Answer the following.
Consider the mapping f: A → B is defined by f(x) = (x-1)/(x-2) such that f(x) is one-one onto…
Consider the mapping $f: A \to B$ is defined by $f(x) = \frac{x-1}{x-2}$ such that $f(x)$ is one-one onto.
(i)[2]
Domain of $f(x)$ is:
- (a)$\mathbb{R} - \{2\}$
- (b)$\mathbb{R}$
- (c)$\mathbb{R} - \{1, 2\}$
- (d)$\mathbb{R} - \{0\}$
(ii)[2]
Range of $f(x)$ is:
- (a)$\mathbb{R} - \{2\}$
- (b)$\mathbb{R}$
- (c)$\mathbb{R} - \{1\}$
- (d)$\mathbb{R} - \{0\}$
(iii)[2]
If $g(x) = 2f(x) - 1$, then $g(x)$ in terms of $x$ is:
- (a)$\frac{x+2}{x}$
- (b)$\frac{x+1}{x-2}$
- (c)$\frac{x-2}{x}$
- (d)$\frac{x}{x-2}$
(iv)[2]
A function $f(x)$ is said to be one-one if:
- (a)$f(x_1) = f(x_2) \implies x_1 = x_2$
- (b)$f(-x_1) = f(-x_2) \implies -x_1 = x_2$
- (c)$f(x_1) = f(x_2) \implies -x_1 = x_2$
- (d)$-f(x_1) = f(x_2) \implies x_1 = x_2$
Answer
Answer (i)
Correct option: (a)
Answer: (a) $\mathbb{R} - \{2\}$
$f(x)=\frac{x-1}{x-2}$ is undefined only where $x-2=0$.
Answer (ii)
Correct option: (c)
Answer: (c) $\mathbb{R} - \{1\}$
$y=\frac{x-1}{x-2}$ gives $x=\frac{2y-1}{y-1}$, which exists for all $y\ne1$.
Hence the range is $\mathbb{R}-\{1\}$.
Answer (iii)
Correct option: (d)
Answer: (d) $\frac{x}{x-2}$
$g(x)=2\cdot\frac{x-1}{x-2}-1=\frac{2x-2-x+2}{x-2}=\frac{x}{x-2}$.
Answer (iv)
Correct option: (a)
Answer: (a) $f(x_1) = f(x_2) \implies x_1 = x_2$
This is the definition of a one-one function.