Linear growth and decay

Given \(f(x)=10x+35\), we get \(f(4x)=10(4x)+35=40x+35\). Therefore, \(f(4x)-f(x)=(40x+35)-(10x+35)=30x\). Setting \(30x=150\) gives \(x=5\), so option B is correct. For instance, if \(x=4\), the difference is only \(120\), not \(150\). Exam tip: while finding \(f(4x)\), replace every \(x\) in the function by \(4x\).

Given \(g(x)=300-15x\), we get \(g(5x)=300-15(5x)=300-75x\). Hence, \(g(x)-g(5x)=(300-15x)-(300-75x)=60x\). Setting \(60x=180\) gives \(x=3\). Therefore, option B is correct. For example, \(x=2\) gives a difference of only \(120\), not \(180\). Exam tip: while finding \(g(5x)\), replace every \(x\) in the rule by \(5x\).

This is a linear function, so the sum of its values at two inputs equals twice its value at the midpoint of those inputs. The midpoint of 6 and 18 is \(\frac{6+18}{2}=12\). Hence, \(P(6)+P(18)=2P(12)\). Checking directly, \(P(6)=111\) and \(P(18)=183\), so their sum is 294, while \(2P(12)=2\times147=294\). Exam tip: for a linear function, use \(P(a)+P(b)=2P\left(\frac{a+b}{2}\right)\).

The function \(Q(t)=320-8t\) is linear. The midpoint of \(5\) and \(17\) is \(\frac{5+17}{2}=11\), so for a linear function, \(Q(5)+Q(17)=2Q(11)\). Hence, \(Q(11)\) is correct. \(Q(10)\) is a close distractor, but 10 is not the midpoint of 5 and 17. Exam tip: In such questions, find the average of the two inputs to determine \(k\).

In \(y=500-12x\), the coefficient of x is the constant negative number \(-12\), so y decreases at a constant rate. The \(x^2\) relation is not linear. Exam tip: linear decay has a negative coefficient of x.

The change in x is 15 − 7 = 8. In y = 360 − bx, y decreases by b units for every 1-unit increase in x. Hence, the total decrease is 8b = 136. Therefore, b = 136/8 = 17. If b were 16, the decrease would be 8 × 16 = 128, not 136. Exam tip: For a linear expression, multiply the change in x by its coefficient to find the total change in y.

The condition is \(A(t)=B(t)+6\). So, \(50+7t=92+4t+6\), which gives \(3t=48\) and hence \(t=16\). Therefore, the correct option is \(t=16\). At \(t=14\), the difference is only \(50+98-(92+56)=0\), not 6. Exam tip: translate “6 more than” as \(A=B+6\).

The condition is (310-12t=70+6t-6). This gives (246=18t), so (t=\frac{41}{3}), and none of the options is correct.

The sum is ((a+16)+(a+48)+(a+80)=3a+144). From (3a+144=252), (a=36), which is not in the options.

Given \(g(t)=b-7t\), we get \(g(4)=b-28\), \(g(8)=b-56\), and \(g(12)=b-84\). Their sum is \(3b-(28+56+84)=3b-168\). Hence, \(3b-168=258\), so \(3b=426\) and \(b=142\). If 138 were used, the sum would be 246, not 258. Exam tip: First write the function value for each given input, then combine like terms carefully.

In \(P(t)=80-5t\), the power of \(t\) is 1, so it is linear. The coefficient \(-5\) means the quantity falls by 5 per time unit. \(80+5t\) represents growth, not decay. Exam tip: for linear decay, check for a negative coefficient of \(t\).

In linear decay, the quantity decreases by an equal amount in equal time intervals, so its graph has a constant negative slope. Option B shows exponential decay because the quantity is halved each time. Exam tip: a constant negative slope indicates linear decay.

For the models to be equal, set \(64+9t=184-6t\). Moving \(6t\) to the left gives \(15t=120\), so \(t=8\). Although \(t=7\) is a close distractor, the two model values are not equal at that time. Exam tip: when one linear model increases and the other decreases, equate their expressions and solve for \(t\).

For equality, set \(75+8t=219-10t\). Moving \(10t\) to the left gives \(18t=144\), so \(t=8\). Substituting \(t=6\) does not give equal values, so it is incorrect despite being a close distractor. Exam tip: When equating linear expressions, collect all terms containing \(t\) on one side first.

Given \(f(x)=a+15x\) and \(f(4)=112\), we get \(a+15(4)=112\). Thus, \(a+60=112\), so \(a=52\). Now \(f(11)=52+15(11)=52+165=217\). Hence, option B is correct. The value 232 can result from incorrectly adding the growth term. Exam tip: first find the unknown constant using the given function value, then substitute the required value of \(x\).

Given \(g(x)=b-13x\), use \(g(5)=116\): \(b-13(5)=116\), so \(b-65=116\) and hence \(b=181\). Now \(g(12)=181-13(12)=181-156=25\). Therefore, 25 is correct. A value such as 38 results from an incorrect subtraction. Exam tip: first find the unknown constant \(b\) using the given function value, then substitute the required value of \(x\).

In \(y=18-3x\), the coefficient of \(x\) is \(-3\), so \(y\) decreases by 3 for every increase of 1 in \(x\). This is linear decay at a constant rate. \(18-3x^2\) is not linear because it contains \(x^2\). Exam tip: in \(y=a+bx\), \(b<0\) indicates decay.

For \(P(0)=240\), the constant term must be 240. Uniform decay requires a negative slope, so \(-12t\) is correct. \(240-12t^2\) is not linear. Exam tip: in a linear model, the variable has power 1.

In a linear polynomial \(p(x)=ax+b\), increasing x by 1 changes p(x) by \(a\). Here \(a=-3\), so the value falls by 3 each step. \(-3x^2+11\) is not linear. Exam tip: a negative coefficient of x indicates linear decay.

In \(12-2x\), the coefficient of x is \(-2\), so the quantity decreases by 2 for every increase of 1 in x. In \(12-2x^2\), the decrease is not constant. Exam tip: linear decay has degree 1 and a negative coefficient of x.