Linear growth and decay

Given \(F(2)=F(9)-63\), we get \(F(9)-F(2)=63\). For the linear function \(F(t)=45+rt\), \(F(9)-F(2)=(45+9r)-(45+2r)=7r\). Hence \(7r=63\), so \(r=9\). Option 7 is the time interval \(9-2\), not the rate \(r\). Exam tip: taking the difference of two function values cancels the constant term.

Given \(G(4)=G(12)+72\), we have \(G(4)-G(12)=72\). Now, \(G(4)=180-4r\) and \(G(12)=180-12r\), so their difference is \(8r\). Thus, \(8r=72\), which gives \(r=9\). If \(r=8\), the difference would be only 64, so it is not correct. Exam tip: Taking the difference of a linear function at two times cancels the constant term \(180\).

Given \(P(n)=a+18n\), substitute \(P(5)=170\): \(a+18\times5=170\), so \(a=80\). Hence, \(P(11)=80+18\times11=80+198=278\). Therefore, 278 is the correct answer. The value 260 can result from counting the increase incorrectly; from 5 to 11 there are 6 steps, each increasing by 18. Exam tip: first find the constant \(a\) from the given value, then substitute the required \(n\).

Given \(V(y)=c-2200y\). Substituting \(V(4)=31200\) gives \(c-2200(4)=31200\), so \(c=40000\). Therefore, \(V(9)=40000-2200(9)=40000-19800=20200\). Hence, 20200 is correct. The value 22400 would result from using 8 instead of 9. Exam tip: first find the constant \(c\) from the given value, then substitute the required value of \(y\).

Given \(y=24+5x\) and the condition \(y=9x\), we get \(24+5x=9x\). Subtracting \(5x\) from both sides gives \(24=4x\), so \(x=6\). Hence, option C is correct. For instance, if \(x=5\), then \(y=49\) but \(9x=45\), so option B is not correct. Exam tip: when one variable has two given expressions, equate them and solve for the required variable.

Given \(y=126-7x\), the required condition is \(y=2x\). Therefore, \(126-7x=2x\), so \(126=9x\) and \(x=14\). If \(x=12\), then \(y=42\), whereas \(2x=24\), so it does not satisfy the condition. Exam tip: In such questions, equate the given expression for \(y\) to the required condition and then solve for \(x\).

Given \(L(t)=4t+19\), we get \(L(5t)=4(5t)+19=20t+19\) and \(L(2t)=4(2t)+19=8t+19\). Therefore, \(L(5t)-L(2t)=(20t+19)-(8t+19)=12t\). Note that \(20t\) is only the variable part of \(L(5t)\), not the required difference. Exam tip: evaluate the function at each input separately before subtracting.

Given \(K(t)=150-6t\). Substituting \(4t\) for \(t\), we get \(K(4t)=150-6(4t)=150-24t\). Therefore, \(K(t)-K(4t)=(150-6t)-(150-24t)=150-6t-150+24t=18t\). Hence, \(18t\) is correct. \(24t\) is only the coefficient term in \(K(4t)\), not the value of the difference. Exam tip: when removing brackets after a minus sign, change the sign of every term inside the bracket.

Given \(Y(x)=a+7x\), we get \(Y(2)=a+14\) and \(Y(8)=a+56\). Hence, \(Y(2)+Y(8)=2a+70=150\). Therefore, \(2a=80\) and \(a=40\). If 45 were used, the sum would be \(160\), so it is not correct. Exam tip: Substitute each given \(x\)-value into the function separately before adding.

Given \(Z(x)=b-9x\), we get \(Z(3)=b-27\) and \(Z(5)=b-45\). Hence, \(Z(3)+Z(5)=2b-72\). Using \(2b-72=114\), we obtain \(2b=186\), so \(b=93\). Therefore, 93 is the correct option. If \(b=90\), the sum would be 108, not 114. Exam tip: evaluate the function at each given input before substituting into the stated sum.

For the two polynomials to be equal, set \(p(x)=q(x)\): \(27+6x=9x-6\). Subtracting \(6x\) from both sides gives \(27=3x-6\). Adding 6 to both sides gives \(33=3x\), so \(x=11\). At \(x=10\), the values of the two polynomials are not equal. Exam tip: To find where two expressions are equal, equate them first and solve the resulting linear equation.

Since \(p(x)=q(x)\), set \(180-10x=72+2x\). Subtracting 72 from both sides gives \(108-10x=2x\). Hence, \(108=12x\), so \(x=9\). If \(x=8\) is substituted, the two expressions do not have equal values, so it is a close but incorrect option. Exam tip: equate the two polynomials first, then collect all \(x\)-terms on one side.

In the linear function \(y=40+(k-6)x\), the coefficient of \(x\), or slope, is \(k-6\). For linear growth, the slope must be positive. Thus, \(k-6>0\), which gives \(k>6\). If \(k=6\), the slope is zero and \(y=40\) is constant, not growing. Exam tip: For linear growth, check that the coefficient of \(x\) is positive.

In a linear equation, the coefficient of \(x\) is the slope. Here the slope is \(-(k+2)\). For linear decay, the slope must be negative, so \(-(k+2)<0\). Multiplying by \(-1\) reverses the inequality, giving \(k+2>0\), or \(k>-2\). At \(k=-2\), the slope is zero, so the value remains constant rather than decaying. Exam tip: check the sign of the coefficient of \(x\) to identify growth or decay.

Here (R(0)=82) is directly the initial value. The difference gives (m=12), but the initial value remains the same.

\(S(2)=s-16\) and \(S(9)=s-72\). Hence, \(S(2)-S(9)=(s-16)-(s-72)=56\). The terms involving \(s\) cancel, so the given condition is true for every real value of \(s\). Values such as \(s=56\) or \(s=8\) are merely particular values, not required ones. Exam tip: In differences of function values, first check whether the parameter terms cancel.

Find the difference: \(B(t)-A(t)=(55+9t)-(34+5t)=21+4t\). The coefficient of \(t\) is 4, so the difference increases by 4 whenever \(t\) increases by 1. Here, 21 is the initial difference, not the rate of increase; therefore, option D is incorrect. Exam tip: The rate of change of the difference of two linear expressions is the difference of their \(t\)-coefficients.

Find the difference: \(C(t)-D(t)=(220-6t)-(150-13t)=70+7t\). Since the coefficient of \(t\) is 7, the difference increases by 7 whenever \(t\) increases by 1. Option B has the correct magnitude of rate but the wrong direction. Exam tip: when subtracting one linear expression from another, change the signs of all terms in the second expression.

In \(y=100-5x\), every increase of 1 in \(x\) decreases \(y\) by exactly 5, so the decay rate is constant. A relation containing \(x^2\) is not linear. Exam tip: look for a negative constant coefficient of \(x\).

In \(y=12-4x\), the coefficient of x is \(-4\), so y falls by 4 units for every 1-unit rise in x. Option B represents growth, while C and D are not linear. Exam tip: linear decay has a constant negative coefficient of x.