The rela tion f is defined by The relation g is defined by Show that f is a function and g is not a function.

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Solution

The relation f is defined as,

f( x )={ x 2 , 0≤x≤3 3x, 3≤x≤10

It is observed that for

0≤x≤3, f( x )= x 2 3≤x≤10, f( x )=3x

For x=3 ,

f( x )= 3 2 =9

Or,

f( x )=3×3 =9

Therefore, for 0≤x≤10 , the images of f( x ) are unique.

Hence, the given relation is a function.

g( x )={ x 2 , 0≤x≤2 3x, 2≤x≤10

It is observed that for x=2 ,

g( x )= x 2 = 2 2 =4

And,

g( x )=3x =3×2 6

Hence, the element 2 of the domain of the relation g corresponds to two different images that are 4 and 6 . Hence, this relation is not a function.

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Similar questions

Q. The function f is defined by

f (x) = \{\begin{matrix} x^{2}, & 0 \leq x \leq 3 \\ 3 x, & 3 \leq x \leq 10 \end{matrix}

The relation g is defined by

g (x) = \{\begin{cases} x^{2}, & 0 \leq x \leq 2 \\ 3 x, & 2 \leq x \leq 10 \end{cases}

Show that f is a function and g is not a function.

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$g (x) = {\begin{matrix} x^{2}, & 0 \leq x \leq 3 3 x, & 2 \leq x \leq 10 \end{matrix}$

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If the relation f is defined by $f (x) = {\begin{matrix} x^{2}, & 0 \leq x \leq 3 3 x, & 3 \leq x \leq 10 \end{matrix}$
and the relation g is defined by $g (x) = {\begin{matrix} x^{2}, & 0 \leq x \leq 2 3 x, & 2 \leq x \leq 10 \end{matrix}$
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