Consider the mapping f: A B is defined by f(x) = x - 1 such that f is a bijection. Based on the above information, answer the following questions. 1. Domain of f is: 1. R - 2 2. R 3. R - 1, 2 4. R - 0 2. Range of f is: 1. R 2. R - 2 3. R - 0 4. R - 1, 2 3. If g: R - 2 R - 1 is defined by g(x) = 2f(x) - 1, then g(x) in terms of x is: 1. x+2 x 2. x+1 x-2 3. x-2 x 4. x x-2 4. The function g defined above, is: 1. One-one 2. Many-one 3. into 4. None of these 5. A function f(x) is said to be one-one iff. 1. f(x_1) = f(x_2) -x_1= x_2 2. f(-x_1) = f(-x_2) -x_1 = x_2 3. f(x_1) = f(x_2) x_1 = x_2 4. None of these

1. (a) R - 2 Solution: For f(x) to be defined x - 2; ≠ 0 i.e., x; ≠ 2. Domain of f = R - 2 2. (b) R - 2 Solution: Let y = f(x), then y= x-1 x-2 xy - 2y = x - 1 xy - x = 2y - = 2y-1 y-1 Since, x R - 2 , therefore y ≠ 1 Hence, range of f = R - 1 3. (d) x x-2 Solution: We have, g(x) = 2f(x) - 1 =2 ( x-1 x-2 )-1= 2x-2-x+2 x-2 = x x-2 4. (a) One-one Solution: We have, g(x) = x x-2 Let g(x_1) = g(x_2) x_1 x_ 1 -2 = x_2 x_ 2 -2 x_1x_2 - 2x_1 = x_1x_2 - 2x_2 2x_1 = 2x_2 x_1 = x_2 Thus, g(x_1) = g(x_2) x_1= x_2 Hence, g(x) is one-one. 5. (c) f(x_1) = f(x_2) x_1 = x_2

Consider the mapping f: A B is defined by f(x) = x - 1 such that …

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Consider the mapping $f: A \rightarrow B$ is defined by $f(x) = x - $1 such that $f$ is a bijection.
Based on the above information, answer the following questions.

Domain of $f$ is:

$R - \{2\}$
$R$
$R -\{1, 2\}$
$R - \{0\}$

Range of $f$ is:

$R$
$R - \{2\}$
$R -\{0\}$
$R - \{1, 2\}$

If $g: R - \{2\} \rightarrow R - \{1\}$ is defined by $g(x) = 2f(x) - 1,$ then $g(x)$ in terms of $x$ is:

$\frac{\text{x}+2}{\text{x}}$
$\frac{\text{x}+1}{\text{x}-2}$
$\frac{\text{x}-2}{\text{x}}$
$\frac{\text{x}}{\text{x}-2}$

The function g defined above, is:

One-one
Many-one
into
None of these

A function $f(x)$ is said to be one-one iff.

$f(x_1) = f(x_2) \Rightarrow -x_1= x_2$
$f(-x_1) = f(-x_2) \Rightarrow -x_1 = x_2$
$f(x_1) = f(x_2) \Rightarrow x_1 = x_2$
None of these

✓

Answer

(a) $R - \{2\}$

Solution:

For $f(x)$ to be defined $x - 2; ≠ 0$ i.e., $x; ≠ 2.$

$\therefore$ Domain of $f = R - \{2\}$

(b) $R - \{2\}$

Solution:

Let $y = f(x)$, then $\text{y}=\frac{\text{x}-1}{\text{x}-2}$

$\Rightarrow xy - 2y = x - 1 \Rightarrow xy - x = 2y -$
$\Rightarrow\text{x}=\frac{2\text{y}-1}{\text{y}-1}$

Since, $x \in R - \{2\},$ therefore $y ≠ 1$

Hence, range of $f = R - \{1\}$

(d) $\frac{\text{x}}{\text{x}-2}$

Solution:

We have, $g(x) = 2f(x) - 1$

$=2\Big(\frac{\text{x}-1}{\text{x}-2}\Big)-1=\frac{2\text{x}-2-\text{x}+2}{\text{x}-2}=\frac{\text{x}}{\text{x}-2}$

(a) One-one

Solution:

We have, $g(x) =\frac{\text{x}}{\text{x}-2}$

Let $g(x_1) = g(x_2) \Rightarrow\frac{\text{x}_1}{\text{x}_{1}-2}=\frac{\text{x}_2}{\text{x}_{2}-2}$

$\Rightarrow x_1x_2 - 2x_1 = x_1x_2 - 2x_2 \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2$

Thus, $g(x_1) = g(x_2) \Rightarrow x_1= x_2$

Hence, $g(x)$ is one-one.

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Shreya got a rectangular parallelepiped shaped box and spherical ball inside it as return gift. Sides of the box are $x, 2x,$ and $\frac{\text{x}}{3},$ while radius of the ball is $r.$

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$\frac{2\text{x}^2}{3}+\frac{4}{3}\pi\text{r}^2$
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If sum of the surface areas of box and ball are given to be constant $k^2$ then $x$ is equal to.

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$\sqrt{\frac{\text{k}^2-4\pi}{6}}$
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When the doctor arrives late, what is the probability that he comes by metro?

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When the doctor arrives late, what is the probability that he comes by other means of transport?

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Find the probability that daughter is at one end, given that father and mother are in the middle.

$1$
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Find the probability that mother is at right end, given that son and daughter are together.

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$
$0$

Find the probability that father and mother are in the middle, given that son is at right end.

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{1}{3}$
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Find the probability that father and son are standing together, given that mother and daughter are standing together.

$0$
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Find the probability that father and mother are on either of the ends, given that son is at second position from the right end.

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