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Relations and Functions Ncert Solutions Maths class 11th Maths Class 11th Relations and Functions

25. The relation f is defined by $f (x) = {\begin{cases} x^{2}, 0 \leq x \leq 3 \\ 3 x, 3 \leq x \leq 10 \end{cases}$

The relation g is defined by $g (x) = {\begin{cases} x^{2}, 0 \leq x \leq 2 \\ 3, 2 \leq x \leq 10 \end{cases}$

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

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Payal Gupta | Contributor-Level 10

4 months ago

25. Given, f(x)= ${\begin{array}{l} x^{2}, & 0 \leq x \leq 3 \\ 3 x, & 3 \leq x \leq 10 \end{array}$

f(x)={(0,0),(1,1),(2,4),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the elements in domain of f has one and only one image.

? f(x) is a function.

Given, g(x)= ${\begin{array}{l} x^{2}, & 0 \leq x \leq 2 \\ 3 x, & 2 \leq x \leq 10 \end{array}$ .

g(x)={(0,0),(1,1),(2,4),(2,6),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the element 2 of the domain has more than one image i.e., 4 and 6.

? g(x) is not a function.

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Circle S? : x² + y² - 10x - 10y + 41 = 0.
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Kindly consider then following figure

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36. Given, A={9,10,11,12,13}.

f(x)=the highest prime factor of n.

and f: A → N.

Then, f(9)=3 [? prime factor of 9=3]

f (10)=5 [? prime factor of 10=2,5]

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25. The relation f is defined by $f (x) = {\begin{cases} x^{2}, 0 \leq x \leq 3 \\ 3 x, 3 \leq x \leq 10 \end{cases}$

The relation g is defined by $g (x) = {\begin{cases} x^{2}, 0 \leq x \leq 2 \\ 3, 2 \leq x \leq 10 \end{cases}$

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

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25. The relation f is defined by f ( x ) = { x 2 , 0 ≤ x ≤ 3 3 x , 3 ≤ x ≤ 1 0 The relation g is defined by g ( x ) = { x 2 , 0 ≤ x ≤ 2 3 , 2 ≤ x ≤ 1 0 Show that f is a function and g is not a function.

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25. The relation f is defined by $f (x) = {\begin{cases} x^{2}, 0 \leq x \leq 3 \\ 3 x, 3 \leq x \leq 10 \end{cases}$

The relation g is defined by $g (x) = {\begin{cases} x^{2}, 0 \leq x \leq 2 \\ 3, 2 \leq x \leq 10 \end{cases}$

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