TASK 1, parts 1-3

Let y=f(x)=x+ab, then x=y-ab. If x=g(y)=y-ab, so g(x)=x-ab.

But were given g(x)=cx-d, therefore cx-d≡x-ab. If we equate coefficients, c=1 and d=ab. Let a=2 and b=-3, then d=-6. f(x)=x-6, g(x)=x+6.

f(g(x))=x and g(f(x))=x if f and g are mutual inverses, so:

f(g(x))=f(x+6)=(x+6)-6=x and g(f(x))=g(x-6)=(x-6)+6=x, confirming mutual inverses.

Part 4

x f(x) g(x)

-6 -12 0

-3 -9 3

0 -6 6

3 -3 9

6 0 12

f(x) is red, g(x) is blue and y=x is green.

TASK 2

Part 1

2√(x+1)+7=5 is a radical equation with an extraneous solution.

Part 2

2√(x+1)+7=5, 2√(x+1)=-2, √(x+1)=-1, square both sides: x+1=1, x=0.

But 2√(x+1)+7=9 when x=0 and 9≠5. So x=0 is an extraneous solution.

2√(x+1)+5=7 is a radical equation with solution x=0.

Part 3

The extraneous solution is a result of squaring because (-1)²=1²=1. This creates an ambiguity. But √(x+1) specifically implies the positive root. This is the reason why the second equation works, because the positive root satisfies the equation.

TASK 3

Part 1

“Jasmine practises the piano for thirty minutes on Monday. Every day she increases her practice time by five minutes...”

f(n)=30+5(n-1) where n is the number of days after Monday, and f(n) is practice time. As an AP, a_n=30+5(n-1), or a₁=30 and a_(n+1)=a_n+5.

7th term (n=7), a₇=30+5×6=60 minutes.

Parts 2 and 3

“Anthony goes to the gym for fifty minutes on Monday. Every day he increases his gym time by eight percent...”

g(n)=50×1.08ⁿ⁻¹ where n is the number of days after Monday, and g(n) is gym time. As a GP, term T_n=50(1.08)ⁿ⁻¹, or T₁=50 and T_(n+1)=1.08T_n. 5th term (n=5), T₅=50(1.08)⁵=73.47 minutes (approx).

“If I go to the gym for 50 minutes on Monday and decide to increase my gym time by 8% for each subsequent day thereafter, by how many minutes will my gym time have increased by Saturday? (Accurate to 1 decimal place.)” Answer: 73.47-50=23.47, that is, 23.5 minutes increase.