Home / IB DP Maths Topic 1.6 :Simple deductive proof, numerical and algebraic HL Paper 1

# IB DP Maths Topic 1.6 :Simple deductive proof, numerical and algebraic HL Paper 1

### Question:

Consider any three consecutive integers, n – 1, n and n + 1.
(a) Prove that the sum of these three integers is always divisible by 3.

Ans:  (n – 1) + n + (n + 1)

= 3n

which is always divisible by 3

(b) Prove that the sum of the squares of these three integers is never divisible by 3.

Ans: (n -1)2 +n2 + (n+1)2     (=n2 – 2n + 1 + n2 + n2 + 2n + 1)
attempts to expand either  (n -1)2 or  (n+1)2  ( do not accept n2 – 1  or n2 + 1)
= 3n2 + 2

demonstrating recognition that 2 is not divisible by 3 or $$\frac{2}{3}$$   seen after correct expression divided by 3
3n2  is divisible by 3 and so 3n2 + 2 is never divisible by 3  OR the first term is divisible by 3, the second is not

OR  $$3\left ( n^{2}+\frac{2}{3} \right ) OR \frac{3n^{2} + 2}{3} = n^{2} + \frac{2}{3}$$ hence the sum of the squares is never divisible by 3

### Question:

[Maximum mark: 8] [without GDC]

(a) Prove the identity  $$2x^{3}+7x^{2}-14x+5\equiv (x+5)(2x^{2}-3x+1)$$                              [1]

Ans: RHS to LHS (easy)

(b) Given that $$3x^{3}+13x^{2}-3x+35\equiv (x+5)(ax^{2}+bx+c)$$ find the values of $$a,b, c$$ .                   [2]

Ans: RHS = $$ax^{3}+bx^{2}+cx+5ax^{2}+5bx+5c=ax^{3}+(b+5a)x^{2}+(c+5b)x+5c$$

Compare with LHS:  a = 3, b + 5a =13⇔b = −2 , and 5c = 35⇔c = 7

(c) Given that $$ax^{3}+bx^{2}-23x+c\equiv (x+5)(2x^{2}+dx+2)$$ find the values of $$a,b, c, d$$ .                            [2]

Ans: RHS = $$2x^{3}+dx^{2}+2x+10x^{2}+5dx+10=2x^{3}+(d+10)x^{2}+(2+5d)x+10$$

Compare with LHS: a = 2 , b = d +10 , 2 + 5d = −23 and c =10

Thus a = 2 , b = 5 , c =10 and d = −5

### Question

[Maximum mark: 4] [without GDC]
Prove the following statement:
For all integers $$n$$ , if  n2+ 2n+5 is odd then $$n$$ is even.

Ans.

Suppose that the result is not true. That is $$n = 2k + 1$$ is odd. Then

$$n^{2}+2n+5=(2k+1)^{2}+2(2k+1)+5=4k^{2}+4k+1+4k+2+7=4k^{2}+8k+10$$

### Question:

Consider two consecutive positive integers, n and n + 1 .
Show that the difference of their squares is equal to the sum of the two integers.

Ans:

diff. of squares =$$\left(n+1\right)^2-n^2$$

=$$\left(n+1+n\right)\left(n+1-n\right)$$

= $$\left(n+1+n\right)\left(n+1-n\right)$$

=2n+1

=n+1+n

=sum of 2 integers

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