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The Function F is Defined by F ( X ) = { X 2 , 0 ≤ X ≤ 3 3 X , 3 ≤ X ≤ 10 the Relation G is Defined by G ( X ) = { X 2 , 0 ≤ X ≤ 2 3 X , 2 ≤ X ≤ 10 Show that F is a Function and G is Not a Function. - Mathematics

The function f is defined by \[f\left( x \right) = \begin{cases}x^2 , & 0 \leq x \leq 3 \\ 3x, & 3 \leq x \leq 10\end{cases}\] The relation g is defined by \[g\left( x \right) = \begin{cases}x^2 , & 0 \leq x \leq 2 \\ 3x, & 2 \leq x \leq 10\end{cases}\]

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

 
 
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Solution

The function f is defined by

\[f\left( x \right) = \begin{cases}x^2 & 0 \leqslant x \leqslant 3 \\ 3x & 3 \leqslant x \leqslant 10\end{cases}\] 
It is observed that for 0 ≤ x < 3, (x) = x2 .
3 <  x ≤ 10, f (x) = 3x
Also, at x = 3, f(x) = 32 = 9. And
(x) = 3 × 3 = 9.
That is, at x = 3, f (x) = 9.
Therefore, for 0 ≤ x ≤ 10, the images of f (x) are unique.
Thus, the given relation is a function.
Again,
the relation g is defined as
\[g\left( x \right) = \begin{cases}x^2 , & 0 \leqslant x \leqslant 2 \\ 3x, & 2 \leqslant x \leqslant 10\end{cases}\]
It can be observed that for x = 2, g(x) = 22 = 4 and also, 
g(x) = 3 × 2 = 6.
Hence, 2 in the domain of the relation g corresponds to two different images, i.e. 4 and 6.
Hence, this relation is not a function.

Hence proved.
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 3 Functions
Exercise 3.1 | Q 16 | Page 8
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