CBSE (Science) Class 11CBSE
Share

Books Shortlist
Your shortlist is empty

# 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. - CBSE (Science) Class 11 - Mathematics

#### Question

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.

#### 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?

#### APPEARS IN

RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 3: Functions
Ex.3.10 | Q: 16 | Page no. 8
Solution 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. Concept: Functions.
S