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WAEC Further Mathematics 2017 Paper

Question : 39

Total: 40

If P =

\{n^{2} + 1: n = 0,2,3\}

\{n + 1: n = 2,3,5\}

\cap

Solution:

Concept:

Sets described by a rule are expanded by substituting the given values, then their common elements are found.

Explanation:

For

P

, substitute

n = 0, 2, 3

into

n^{2} + 1

n=0

0^{2}+1 = 1

n=2

2^{2}+1 = 5

n=3

3^{2}+1 = 10

So,

P = \{1, 5, 10\}

For

Q

, substitute

n = 2, 3, 5

into

n + 1

n=2

2+1 = 3

n=3

3+1 = 4

n=5

5+1 = 6

So,

Q = \{3, 4, 6\}

Now find the intersection

P \cap Q

There is no element common to both sets.

Therefore,

P \cap Q = \{ \}

Answer:

\{ \}

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