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  • The relation f is defined by Relation and Function Q-1

1. The relation f is defined by
f (x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 3 \\ 3x &3 \leq x \leq 10 \end{matrix}\right.
The relation g is defined by
g(x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 2 \\ 3x &2 \leq x \leq 10 \end{matrix}\right.
Show that f is a function and g is not a function.

Answers (1)

best_answer

It is given that
f (x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 3 \\ 3x &3 \leq x \leq 10 \end{matrix}\right.
 Now,
f(x) = x^2 \ for \ 0\leq x\leq 3
And
f(x) = 3x \ for \ 3\leq x\leq 10

At x = 3,f(x) = x^2 = 3^2 = 9

Also, at x = 3,f(x) = 3x = 3\times 3 = 9

We can see that for 0\leq x\leq 10, f(x) has unique images.

Therefore, By definition of a function, the given relation is function.

Now,
It is given that
g(x) = \left\{\begin{matrix} x^2 & 0 \leq x\leq 2 \\ 3x &2 \leq x \leq 10 \end{matrix}\right.
Now,
g(x) = x^2 \ for \ 0\leq x\leq 2
And
g(x) = 3x \ for \ 2\leq x\leq 10

At x = 2, g(x) = x^2 = 2^2 = 4
Also, at x = 2, g(x) = 3x = 3\times2 = 6
We can clearly see that element 2 of the domain of relation g(x) corresponds to two different images i.e. 4 and 6. Thus,  f(x) does not have unique images
Therefore, by definition of a function, the given relation is not a function

Hence proved 

 

Posted by

Gautam harsolia

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