5 Marks Questions

Question 15 Marks

Let f = {(2, 4), (5, 6), (8, – 1), (10, – 3)}
g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}
be two real functions. Then Match the following:

Column I		Column II
(a)	$\text{f}-\text{g}$	(i)	$\Big\{\Big(2, \frac{4}{5}\Big), \Big(8, \frac{-1}{4}\Big), \Big(10, \frac{-3}{13}\Big)\Big\}$
(b)	$\text{f}+\text{g}$	(ii)	$\{(2, 20), (8, -4), (10, -39)\}$
(c)	$\text{f}\times\text{g}$	(iii)	$\{(2, 1), (8, -5), (10, -16)\}$
(d)	$\frac{\text{f}}{\text{g}}$	(iv)	$\{(2, 9), (8, 3), (10, 10)\}$

Answer

Domain of f(x) is {2, 5, 8, 10}.
Domain of g(x) is {2, 7, 8, 10, 11}.
Thus, domain of $\text{f}\pm\text{g}, \text{f}\times\text{g}$ and $\frac{\text{f}}{\text{g}}$ is {2, 8, 10}.
For function y = f(x), We have f(2) = 4, f(8) = -1 and f(10) = -3
For function y = g(x), We have g(2) = 5, g(8) = 4 and g(10) = 13

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Question 25 Marks

Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, $\text{g(x)} = \alpha\text{x} + \beta,$ then what values should be assigned to $\alpha$ and $\beta?$

Answer

Given that: g = {(1, 1), (2, 3), (3, 5), (4, 7)}
Since every element of the domain in this relation has unique image, so g is a function.
Now $\text{g(x)} = \alpha\text{x} + \beta,$
For (1, 1) g(1) $= \alpha(1) + \beta = 1 ⇒ \alpha + \beta = 1\ \dots(1)$
For (2, 3) g(2) $= \alpha(2) + \beta = 3 ⇒ 2\alpha+ \beta = 3 \ \dots(2)$
Solving eqn. (i) and (ii) we have
$\alpha= 2$ and $\beta = -1.$ [Note: We can take any other two ordered pairs]
Hence, the value of $\alpha= 2$ and $\beta = -1.$

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