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