Linear growth and decay

Given \(F(2)=F(5)-18\), we get \(F(5)-F(2)=18\). For the linear function \(F(t)=20+rt\), \(F(5)-F(2)=(20+5r)-(20+2r)=3r\). Hence \(3r=18\), so \(r=6\). If \(r=5\), the difference would be only 15, so it is not correct. Exam tip: when subtracting two values of a linear function, the constant term 20 cancels out.

Given \(G(2)=G(7)+35\), we have \(G(2)-G(7)=35\). Substituting \(G(2)=100-2r\) and \(G(7)=100-7r\) gives \((100-2r)-(100-7r)=5r\). Thus, \(5r=35\), so \(r=7\). If \(r=5\), the difference would be only \(25\), so it is not correct. Exam tip: subtracting function values at two times cancels the constant term \(100\).

Given \(P(n)=a+12n\). Putting \(n=3\), we get \(a+12(3)=86\), so \(a+36=86\) and \(a=50\). Therefore, \(P(8)=50+12(8)=50+96=146\). The option 158 results from using an incorrect increase. Exam tip: first find the constant term \(a\) from the given value, then substitute the required \(n\).

Given \(V(y)=c-1500y\). Substituting \(V(2)=27000\) gives \(c-1500(2)=27000\), so \(c-3000=27000\) and hence \(c=30000\). Now, \(V(7)=30000-1500(7)=30000-10500=19500\). Therefore, 19500 is the correct answer. 18000 would result from an incorrect depreciation calculation. Exam tip: first find \(c\) from the given value, then substitute the required value of \(y\).

Given \(y=15+3x\), the required condition is \(y=6x\). Therefore, \(15+3x=6x\). Subtracting \(3x\) from both sides gives \(15=3x\), so \(x=5\). For \(x=6\), \(y=33\), whereas \(6x=36\), so it is not correct. Exam tip: In such questions, equate the two expressions for \(y\) and then solve for \(x\).

Given \(y=80-5x\) and the condition \(y=3x\), set the two expressions equal: \(80-5x=3x\). Adding \(5x\) to both sides gives \(80=8x\), so \(x=10\). If \(x=8\), then \(y=40\) while \(3x=24\), so it does not satisfy the condition. Exam tip: When two expressions represent the same variable, equate them and solve the resulting linear equation.

Given \(L(t)=2t+11\), we get \(L(3t)=2(3t)+11=6t+11\). Therefore, \(L(3t)-L(t)=(6t+11)-(2t+11)=4t\). \(6t\) is part of the value of \(L(3t)\), not the difference. Exam tip: substitute \(3t\) into the function first, then subtract the two expressions using brackets.

Given \(K(t)=90-4t\). Substituting \(2t\) for \(t\) gives \(K(2t)=90-4(2t)=90-8t\). Therefore, \(K(t)-K(2t)=(90-4t)-(90-8t)=90-4t-90+8t=4t\). Hence, \(4t\) is correct. The distractor \(-4t\) may result from not changing every sign while subtracting the second bracket. Exam tip: when finding a function at \(2t\), replace every \(t\) by \(2t\), then distribute the minus sign carefully.

Given \(Y(x)=a+5x\), we get \(Y(2)=a+10\) and \(Y(6)=a+30\). Therefore, \(Y(2)+Y(6)=2a+40=86\). Hence \(2a=46\), so \(a=23\). For example, choosing 28 would give a sum of 96, not 86. Exam tip: Substitute each given \(x\)-value into the function separately before adding.

Given \(Z(x)=b-6x\), we get \(Z(1)=b-6\) and \(Z(4)=b-24\). Hence, \(Z(1)+Z(4)=(b-6)+(b-24)=2b-30\). Using \(2b-30=90\), we obtain \(2b=120\), so \(b=60\). If 63 were used, the sum would be 96, not 90. Exam tip: Substitute each given value of \(x\) separately before adding the function values.

For equality, set \(p(x)=q(x)\): \(10+3x=4x-2\). Subtracting \(3x\) from both sides gives \(10=x-2\), so \(x=12\). On checking, \(p(12)=46\) and \(q(12)=46\). For example, at \(x=10\), their values are 40 and 38, so it is not correct. Exam tip: To find when two polynomials are equal, equate their expressions first and solve for \(x\).

Since \(p(x)=q(x)\), set \(120-7x=60+5x\). Subtracting 60 from both sides gives \(60-7x=5x\), so \(60=12x\). Hence, \(x=5\). If \(x=6\) is substituted, the two expressions do not have equal values. Exam tip: equate the two linear expressions and collect all \(x\)-terms on one side.

In \(y=18+(k-2)x\), the coefficient of \(x\), or slope, is \(k-2\). For linear growth, the slope must be positive, so \(k-2>0\). Hence, \(k>2\). When \(k=2\), \(y=18\) is constant, not increasing. Exam tip: in \(y=mx+c\), check \(m>0\) for linear growth.

In \(y=100-(k+3)x\), the coefficient of \(x\), or slope, is \(-(k+3)\). For linear decay, the slope must be negative, so \(-(k+3)<0\). This gives \(k+3>0\), hence \(k>-3\). When \(k=-3\), the slope is zero and \(y=100\) is constant, not decay. Exam tip: check the sign of the coefficient of \(x\) to identify linear growth or decay.

For a linear rule R(x) = 50 + mx, the constant term 50 is the initial value and equals R(0), because R(0) = 50 + m(0) = 50. The given difference can be used as a check: R(8) - R(3) = (50 + 8m) - (50 + 3m) = 5m = 45, so m = 9. However, this slope does not alter the value at x = 0. Therefore R(0) = 50, making option C correct. Option A is the slope, while the other numbers confuse the difference or combine values incorrectly.

\(S(2)=s-4(2)=s-8\) and \(S(9)=s-4(9)=s-36\). Therefore, \(S(2)-S(9)=(s-8)-(s-36)=28\). The terms containing \(s\) cancel, so the condition is true for every real \(s\). Neither \(s=28\) nor \(s=4\) is a specially required value. Exam tip: In differences of function values, write both values first and check whether the parameter terms cancel.

Find the difference: (B(t)-A(t))=(30+5t)-(18+2t)=12+3t. The coefficient of t is 3, so the difference increases by 3 for every increase of 1 in t. Therefore, option A is correct. The value 12 is the initial difference, not the rate of increase, so option D is incorrect. Exam tip: when subtracting linear expressions, subtract the constant terms and the coefficients of t separately.

Subtracting gives \(C(t)-D(t)=(150-3t)-(120-8t)=30+5t\). Thus, for every increase of 1 in \(t\), the difference increases by 5. Option B is incorrect because \(D(t)\) decreases 5 units faster than \(C(t)\), so the gap between them increases. Exam tip: When subtracting linear expressions, change the signs of every term in the second bracket carefully.

In \(y=80-5x\), every increase of 1 in x decreases y by a constant 5, so it shows linear decay. Option B has a positive slope and represents growth. Exam tip: linear decay always has a negative coefficient of x.

In \(P(x)=240-8x\), the constant term 240 is the initial value and the coefficient \(-8\) shows a decrease as x increases. \(240+8x\) represents growth. Exam tip: a negative slope indicates decay.