Linear growth and decay

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.

For the function \(R(x)=82+mx\), the value at zero is determined by the constant term. Substituting \(x=0\) removes the variable part, since \(m\times0=0\). Therefore \(R(0)\) can be found without first solving for \(m\). The difference condition gives useful information about the slope, but it does not alter the constant value at zero.

Substitution gives \(R(0)=82+m(0)=82\), so option C is correct. The supplied condition is consistent: \(R(10)-R(4)=(82+10m)-(82+4m)=6m=72\), hence \(m=12\). This confirms the rate of change. However, 12 is the slope and 72 is the stated difference; neither is the value of \(R(0)\).

\(S(2)=s-16\) 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.

Given \(f(x)=8x+22\), we get \(f(3x)=8(3x)+22=24x+22\). Hence, \(f(3x)-f(x)=(24x+22)-(8x+22)=16x\). Setting \(16x=80\) gives \(x=5\). Therefore, \(x=5\) is correct. For \(x=4\), the difference would be only \(64\), not 80. Exam tip: while finding \(f(3x)\), replace every \(x\) in the function with \(3x\).

Given \(g(x)=180-9x\), we get \(g(3x)=180-9(3x)=180-27x\). Hence, \(g(x)-g(3x)=(180-9x)-(180-27x)=18x\). Setting \(18x=90\) gives \(x=5\). Therefore, option B is correct. For \(x=4\), the difference would be only \(72\), not \(90\). Exam tip: while finding \(g(3x)\), substitute the complete expression \(3x\) in place of \(x\).

Given \(P(t)=50+4t\), we get \(P(4)=66\) and \(P(12)=98\). Thus, \(P(4)+P(12)=164\), and half of this is \(82\). Since \(P(8)=50+4\times8=82\), we have \(P(4)+P(12)=2P(8)\). Although \(P(7)=78\) is close, twice it is \(156\), so it is not correct. Exam tip: for a linear function, the function value at the midpoint of two inputs equals the average of the two function values.

This is a linear function. The midpoint of 3 and 15 is \(\frac{3+15}{2}=9\), so for a linear function, \(Q(3)+Q(15)=2Q(9)\). Checking directly, \(Q(3)=189\) and \(Q(15)=105\), giving a sum of \(294\), while \(2Q(9)=2\times147=294\). Hence, Q(9) is correct. Q(10) is not the midpoint, so twice its value is not equal to the given sum. Exam tip: For a linear function, the sum of outputs at two equally spaced inputs equals twice the output at their midpoint.

In \(y=120+ax\), \(a\) is the rate of change of \(y\) with respect to \(x\). The change in \(x\) is \(14-6=8\), so the change in \(y\) is \(8a\). Given \(8a=88\), we get \(a=11\). If \(a=10\), the increase would be \(8\times10=80\), not 88. Exam tip: while finding a change in a linear expression, the constant term \(120\) cancels out.

The change in x is 13 - 5 = 8. In y = 260 - bx, y decreases by b for every increase of 1 in x. Hence, the total decrease over 8 units is 8b. So, 8b = 104, giving b = 104/8 = 13. Option 12 is close, but it would give a total decrease of 8 × 12 = 96, not 104. Exam tip: For a linear relation, use total change = rate per unit × change in x.

The condition is \(A(t)=B(t)+6\). Thus, \(30+5t=54+3t+6\). Simplifying gives \(2t=30\), so \(t=15\). At \(t=14\), \(A(t)-B(t)=4\), not 6. Exam tip: Translate “6 more than” as \(A=B+6\).

The condition is (200-8t=70+4t-10). It gives (140=12t), so (t=\frac{35}{3}), hence no option is correct.

Given \(f(t)=a+6t\), we have \(f(1)=a+6\), \(f(5)=a+30\), and \(f(9)=a+54\). Their sum is \(3a+90\). Hence, \(3a+90=180\), so \(3a=90\) and \(a=30\). If 36 were used, the sum would be 198, so it is not correct. Exam tip: Evaluate the function at each specified input first, then form the equation using their sum.

Given \(g(t)=b-5t\), we get \(g(2)=b-10\), \(g(6)=b-30\), and \(g(10)=b-50\). Their sum is \((b-10)+(b-30)+(b-50)=3b-90\). Thus, \(3b-90=210\), so \(3b=300\) and \(b=100\). If 105 were used, the sum would be \(225\), not 210. Exam tip: Write each function value separately before combining like terms.