Tutorials 256 Check whether the following functions are onto or one to one Show

Tutorials 256 check whether the following functions

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Tutorials 2.5.6. Check whether the following functions are onto or one-to-one. Show your steps clearly. (i) f : R R defined by f ( x ) = - 3 x + 2 (ii) f : R R defined by f ( x ) = - x 2 - 5 (iii) f : R + R + defined by f ( x ) = x 2 (iv) f : R [ - 1 , 1] defined by f ( x ) = cos x 2.5.2 Inequalities and absolute value Example 2.5.7. Solve for x : 3 x 2 + 10 x 5 x + 2 . 3 x 2 + 10 x 5 x + 2 3 x 2 + 10 x - 5 x - 2 0 3 x 2 + 5 x - 2 0 (3 x - 1)( x + 2) 0 Critical values are: x = 1 3 and x = - 2 . We can use the table or number line method to find our solution: x < - 2 x = - 2 - 2 < x < 1 3 x = 1 3 x > 1 3 3 x - 1 - - - 0 + x + 2 - 0 + + + (3 x - 1)( x + 2) + 0 - 0 + From this table it is clear that - 2 x 1 3 . 64
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Example 2.5.8. Solve for x in 3 x - 1 x +2 2 . 3 x - 1 x + 2 2 3 x - 1 x + 2 - 2 0 3 x - 1 - 2( x + 2) x + 2 0 3 x - 1 - 2 x - 4 x + 2 0 x - 5 x + 2 0 ( x 6 = - 2) Critical values are: x = 5 and x = - 2 . We can use the table or number line method to find our solution: x < - 2 x = - 2 - 2 < x < 5 x = 5 x > 5 x - 5 - - - 0 + x + 2 - 0 + + + 3 x - 1 x +2 + - 0 + From the table we have solution: x < - 2 or x 5 , i.e. ( -∞ , - 2) [5 , ) . Example 2.5.9. Solve for x : | 3 x - 4 | > x + 1 . 3 x - 4 > x + 1 or - (3 x - 4) > x + 1 3 x - x > 1 + 4 or - 3 x + 4 - x > 1 2 x > 5 or - 4 x > - 3 x > 5 2 or x < 3 4 So from this we conclude ( -∞ , 3 4 ) ( 5 2 , ) . Example 2.5.10. Solve for x in | x |- 3 x - 2 < 2 . 65
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Case 1: If x > 2 . | x | - 3 x - 2 < 2 ⇒ | x | - 3 < 2( x - 2) ⇒ | x | < 2 x - 4 + 3 ⇒ | x | < 2 x - 1 ⇒ - (2 x - 1) < x < 2 x - 1 ⇒ - 2 x + 1 < x < 2 x - 1 ⇒ - 2 x + 1 < x and x < 2 x - 1 ⇒ - 3 x < - 1 and - x < - 1 x < 1 3 and x > 1 x > 1 . Case 2: If x < 2 . | x | - 3 x - 2 > 2 ⇒ | x | - 3 > 2( x - 2) ⇒ | x | > 2 x - 4 + 3 ⇒ | x | > 2 x - 1 x > 2 x - 1 OR - x > 2 x - 1 ⇒ - x > - 1 OR - 3 x > - 1 x < 1 OR x < 1 3 So from this we conclude ( -∞ , 1 3 ) (1 , ) . Tutorials 2.5.11. In each of the following, solve for x . 1. ( x + 2) 2 < 4 66
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2. 4 x +1 x - 1 1 3. | 5 x - 1 | ≤ 2 x - 3 4. x +3 2 x - 1 < 1 5. | x | +5 x - 3 < 2 “God does arithmetic”. Carl Friedrich Gauss 67
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Chapter 3 MATHEMATICAL INDUCTION, BINOMIAL THEOREM AND COMBINATORICS ”God used beautiful mathematics in creating the world”. Paul Dirac 3.1 Mathematical Induction Definition 3.1.1. Natural numbers are the numbers used in counting. Thus we denote the set of natural numbers as: N = { 1 , 2 , 3 , 4 , . . . } . The Principle of Mathematical Induction is a method of proving results about the natural numbers or positive integers. This is based on the fact that the natural numbers can be arranged in a sequence, with a definite starting point 1, and in such case, each natural number has a unique successor. 68
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The Principle of Mathematical Induction: Let P ( n ) be a predicate on the set of natural numbers N satisfying: (i) P (1) is a true statement; (ii) If P ( k ) is true P ( k + 1) is also true for some k N , then P ( n ) is true for all natural numbers. To see how this works, let’s go through some examples. Example 3.1.2. Use the Principle of Mathematical Induction to prove that: 1 3 + 2 3 + 3 3 + . . . + n 3 = n 2 ( n + 1) 2 4 n N . Solution: Let P ( n ) be the statement 1 3 + 2 3 + 3 3 + . . . + n 3 = n 2 ( n +1) 2 4 n N . P (1) : 1 3 = 1 = 1 2 (1+1) 2 4 = 4 4 , hence P (1) is true. Assume P ( k ) is true i.e. 1 3 + 2 3 + 3 3 + . . . + k 3 = k 2 ( k +1) 2 4 for some k N . Then we need to deduce that P ( k + 1) is also true, i.e. 1 3 + 2 3 + 3 3 + . . . + k 3 + ( k + 1) 3 = ( k + 1) 2 [( k + 1) + 1] 2 4 = ( k + 1) 2 ( k + 2) 2 4 Now LHS = 1 3 + 2 3 + 3 3 + . . . + k 3 + ( k + 1) 3 = k 2 ( k + 1) 2 4 + ( k + 1) 3 (by assumption) , = ( k + 1) 2 [ k 2 4 + ( k + 1)] (taking common factor of ( k + 1) 2 ) , = ( k + 1) 2 [ k 2 + 4 k + 4 4 ] = ( k + 1) 2 [ ( k + 2)( k + 2) 4 ] = ( k + 1) 2 ( k + 2) 2 4 = RHS.
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