Question 24 of 33advanced🔍 AnalyzeLong Answer4 marks

The relation ff is defined by

f(x)={x2,0≤x≤33x,3≤x≤10f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\ 3x, & 3 \le x \le 10 \end{cases}

The relation gg is defined by

g(x)={x2,0≤x≤23x,2≤x≤10g(x) = \begin{cases} x^2, & 0 \le x \le 2 \\ 3x, & 2 \le x \le 10 \end{cases}

Show that ff is a function and gg is not a function.

Correct Answer

For a relation to be a function, each element in the domain must have exactly one image in the codomain.

For relation ff:

The domain of ff is [0,10][0, 10]. At the point x=3x = 3, which is common to both intervals, we check the values:

  • From first case: f(3)=32=9f(3) = 3^2 = 9
  • From second case: f(3)=3×3=9f(3) = 3 \times 3 = 9

Since both cases give the same value f(3)=9f(3) = 9, there is no conflict. Every element in the domain [0,10][0, 10] has exactly one image. Hence, ff is a function.

For relation gg:

The domain of gg is [0,10][0, 10]. At the point x=2x = 2, which is common to both intervals, we check the values:

  • From first case: g(2)=22=4g(2) = 2^2 = 4
  • From second case: g(2)=3×2=6g(2) = 3 \times 2 = 6

Since g(2)=4g(2) = 4 and g(2)=6g(2) = 6 are two different values, the element x=2x = 2 has two images. This violates the definition of a function. Hence, gg is not a function.

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 1 | (Page 17)
For More Understanding

Explanation

This question tests the fundamental definition of a function from Chapter 2. A function requires that each element in the domain maps to exactly one element in the codomain.

For piecewise-defined relations, we must check the boundary points where intervals overlap. In relation ff, the boundary point x=3x = 3 yields the same value (99) from both cases, satisfying the function definition.

In relation gg, the boundary point x=2x = 2 yields different values (44 and 66), violating the one-to-one mapping requirement, making gg not a function.

Solution Steps

  1. Step 1: Identify the domain for both relations ff and gg as [0,10][0, 10].

  2. Step 2: For ff, check the common point x=3x = 3: f(3)=32=9f(3) = 3^2 = 9 and f(3)=3×3=9f(3) = 3 \times 3 = 9. Both values are equal.

  3. Step 3: Conclude ff is a function since every element has exactly one image.

  4. Step 4: For gg, check the common point x=2x = 2: g(2)=22=4g(2) = 2^2 = 4 and g(2)=3×2=6g(2) = 3 \times 2 = 6. The values are different.

  5. Step 5: Conclude gg is not a function since x=2x = 2 has two different images.

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