Linear growth and decay

Given \(f(x)=5x+20\), we get \(f(2x)=5(2x)+20=10x+20\). Therefore, \(f(2x)-f(x)=(10x+20)-(5x+20)=5x\). Setting \(5x=30\) gives \(x=6\). Hence, option C is correct. For instance, when \(x=5\), the difference is only \(25\), not \(30\). Exam tip: substitute \(2x\) for every \(x\) in \(f(x)\) before simplifying the subtraction.

Given \(g(x)=96-6x\), we get \(g(2x)=96-6(2x)=96-12x\). Therefore, \(g(x)-g(2x)=(96-6x)-(96-12x)=6x\). Setting \(6x=36\) gives \(x=6\), so option C is correct. For example, at \(x=5\), the difference would be only \(30\). Exam tip: while finding \(g(2x)\), replace every \(x\) in the function by \(2x\).

In a linear model, (P(2)+P(8)=2P(5)) because (5) is the midpoint. Checking gives (110) on both sides.

This is a linear function. The midpoint of 1 and 9 is \(\frac{1+9}{2}=5\), so \(Q(1)+Q(9)=2Q(5)\). Checking directly, \(Q(1)=96\) and \(Q(9)=64\), giving a sum of \(160\). Also, \(Q(5)=80\), hence \(2Q(5)=160\). \(Q(4)\) is not at the midpoint, so it is not correct. Exam tip: for a linear function, the sum of the values at two inputs equals twice the value at their midpoint.

The change in x is 9−4=5. In the linear expression y=70+ax, the change in y is a×5. Hence, 5a=35, so a=7. Option 5 represents the change in x, not the value of a. Exam tip: In y=mx+c, identify m as the rate of change.

The change in \(x\) is \(8-3=5\). In \(y=160-bx\), \(y\) decreases by \(b\) units for every 1-unit increase in \(x\). Hence, the total decrease is \(5b\). Given \(5b=45\), we get \(b=45/5=9\). Therefore, \(9\) is correct. If \(b=15\), the decrease would be \(5\times15=75\), not 45. Exam tip: For a linear relation, total change equals rate per unit times the change in the variable.

“\(A(t)\) is 4 more than \(B(t)\)” means \(A(t)=B(t)+4\). Thus, \(12+3t=18+2t+4\), which gives \(t=10\). Checking: \(A(10)=42\) and \(B(10)=38\), so the difference is 4. At \(t=9\), the difference is only 3. Exam tip: For “more than” questions, first write the difference as \(A(t)-B(t)\).

“\(C(t)\) is 12 less than \(D(t)\)” means \(C(t)=D(t)-12\). Thus, \(130-5t=70+3t-12\). Simplifying gives \(72=8t\), so \(t=9\). Option \(t=8\) is close, but the difference between the two values is not 12 there. Exam tip: For “less than” comparisons, subtract the stated amount from the larger expression before forming the equation.

Given f(t)=a+4t, we get f(1)=a+4, f(3)=a+12, and f(5)=a+20. Their sum is 3a+36. Hence, 3a+36=81, so 3a=45 and a=15. If a were 17, the sum would be 87, so it is not correct. Exam tip: substitute each given value of t into the function before adding the results.

Given \(g(t)=b-3t\), we get \(g(2)=b-6\), \(g(4)=b-12\), and \(g(6)=b-18\). Their sum is \((b-6)+(b-12)+(b-18)=3b-36\). Hence \(3b-36=144\), so \(3b=180\) and \(b=60\). If 54 is used instead, the sum becomes \(126\), not 144. Exam tip: In such questions, first evaluate the function at every given input and then form the sum equation.

In linear decay, the change over equal time intervals is constant. Here the change is always \(-7\), so the slope is fixed and the model has the form \(P=a-7t\). A 10% decrease depends on the current value, so it is not linear. Exam tip: check for equal first differences.

In \(Q(t)=Q_0-rt\), the quantity decreases by the same amount \(r\) in every unit of time, so it is a first-degree linear decay model. \(Q_0+rt\) shows growth, while \(e^{-rt}\) shows exponential decay. Exam tip: a constant rate of change indicates a linear relation.

For the models to be equal, set \(32+7t=86-2t\). Moving \(2t\) to the left and 32 to the right gives \(9t=54\). Therefore, \(t=6\). At \(t=5\), the two model values are not equal, so it is a plausible but incorrect option. Exam tip: when equating linear models, collect all terms containing \(t\) on one side first.

For equality, write \(55+5t=135-3t\). Moving \(3t\) to the left and 55 to the right gives \(8t=80\), so \(t=10\). Substituting \(t=8\) does not make the two values equal, so it is a close but incorrect distractor. Exam tip: collect all terms containing \(t\) on one side and constant terms on the other.

The coefficient of x is 9, so f(x) increases by 9 whenever x increases by 1. From 3 to 8, x increases by 5; therefore, f(x) increases by \(5\times 9=45\). Hence, \(f(8)=f(3)+45=64+45=109\). Option 100 represents an increase of only 36, not the correct increase over 5 steps. Exam tip: for a linear function, multiply the slope by \(\Delta x\) to find the change in the function value.

Given (g(x)=b-8x), first use (g(4)=91): (b-8(4)=91), so (b-32=91). Hence, (b=123). Now substitute (x=10): (g(10)=123-8(10)=123-80=43). Therefore, 43 is correct. A choice such as 51 can result from an arithmetic error in subtraction. Exam tip: find the unknown constant first, then substitute the required value of x.

In \(y=18-2x\), the coefficient of x is \(-2\), so y falls by 2 for every increase of 1 in x. Option B represents growth. Exam tip: linear decay has a negative x-coefficient and degree 1.

In \(P(t)=120-5t\), the constant negative coefficient \(-5\) means the quantity falls by 5 for each unit of time. \(120-5t^2\) is not linear because the variable has power 2. Exam tip: check that the variable’s power is 1.

In \(18-3x\), the coefficient of x is \(-3\), so the quantity falls by 3 units for every 1-unit rise in x. \(18+3x\) shows growth instead. Exam tip: a linear polynomial has x to the power 1.

In linear decay, the quantity decreases by an equal amount for each unit increase, so its graph is a straight line with negative slope. For example, \(p(x)=a-bx\), where \(b>0\). Exam tip: a negative coefficient of \(x\) indicates decay.