Piecewise models and continuity Problem set A Worked solutions

Find a so that \(f\) is continuous at x = 2: \(f(x)=\begin{cases}3x+a & x<2\\ x^2-5 & x\ge2\end{cases}\).

• Left at 2: \(3(2)+a=6+a\)
• Right at 2: \(2^2-5=-1\)
• \(6+a=-1\), so a = -7

Find a so that \(g\) is continuous at x = 3: \(g(x)=\begin{cases}ax^2-1 & x\le3\\ 2x+5 & x>3\end{cases}\).

• Left at 3: \(a(3)^2-1=9a-1\)
• Right at 3: \(2(3)+5=11\)
• \(9a-1=11\), so \(a=\tfrac{4}{3}\) = 1.33

Find a so that \(h\) is continuous at x = 3: \(h(x)=\begin{cases}x^a & 1\le x<3\\ 5x+5 & x\ge3\end{cases}\).

• Left at 3: \(3^a\). Right at 3: \(5(3)+5=20\)
• \(3^a=20\), so \(a\ln3=\ln20\)
• \(a=\ln20\div\ln3=2.73\)

Find a so that \(f\) is continuous at x = 4: \(f(x)=\begin{cases}(x+1)^a & x<4\\ 2x^2+3 & x\ge4\end{cases}\).

• Left at 4: \((4+1)^a=5^a\). Right at 4: \(2(4)^2+3=35\)
• \(5^a=35\), so \(a=\ln35\div\ln5=2.21\)

Find a so that \(h\) is continuous at x = 2: \(h(x)=\begin{cases}e^{ax} & x<2\\ x^2+4 & x\ge2\end{cases}\).

• Left at 2: \(e^{2a}\). Right at 2: \(2^2+4=8\)
• \(e^{2a}=8\), so \(2a=\ln8\)
• \(a=\ln8\div2=1.04\)

Find a so that \(g\) is continuous at x = 4: \(g(x)=\begin{cases}\ln(ax) & x<4\\ 3x-10 & x\ge4\end{cases}\).

• Left at 4: \(\ln(4a)\). Right at 4: \(3(4)-10=2\)
• \(\ln(4a)=2\), so \(4a=e^{2}\)
• \(a=e^{2}\div4=1.85\)

A sample of a radioactive substance has mass 80 g. Each day 6% of the mass decays, so after n days the mass is \(80\times0.94^n\) g. Find the least whole number of days after which the mass is less than 10 g.

• \(80\times0.94^n<10\)
• Solve \(80\times0.94^n=10\): n = 33.6. Dividing by ln 0.94 (negative) turns < into >: n > 33.6
• Round up: n = 34 days

A taxi fare in pounds for a journey of d km is \(F=3+1.6d\) for \(0\le d\le8\), and \(F=ad+7\) for \(d>8\). The fare has no sudden jump at 8 km. Find a.

• At 8 the first rule gives 3 + 1.6(8) = 15.8
• The second rule gives a(8) + 7
• 8 a + 7 = 15.8, so a = 1.1

The mass of a drug in a pond, in mg, is \(120e^{-0.07d}\) after d days. Find the least whole number of days after which less than 15 mg remains.

• \(120 e^{-0.07d}<15\)
• Solve \(120e^{-0.07d}=15\): d = 29.7
• The amount is below 15 from then on. Round up: 30 days

A country charges income tax £T on an income of £x as follows. No tax on the first £12 000. 20% of the income between £12 000 and £50 000. 40% of any income above £50 000.

1. (a) Find the tax on an income of £65 000.

   • First £50 000: 0.2 × 38 000 = 7600
   • Above £50 000: 0.4 × 15 000 = 6000
   • Total: £13600.00

2. (b) Find the income that pays £15 000 in tax.

   • Tax at £50 000 is £7600. £15 000 is more, so use the 40% branch: 7600 + 0.4(x − 50 000) = 15 000
   • Solve for x: x = 68 500

The function \(f\) is continuous at x = 2: \(f(x)=\begin{cases}e^{ax} & x<2\\ \ln(ax)+3 & x\ge2\end{cases}\). Find the two values of a.

• At x = 2: \(e^{2a}=\ln(2a)+3\)
• This cannot be rearranged. Solve on the calculator; graph \(y=e^{2x}-\ln(2x)-3\) to find both roots
• a = 0.0806 or a = 0.571
• Check a = 0.0806: \(e^{0.161}=1.17\) and \(\ln(0.161)+3=1.17\)

A radioactive sample has mass 640 g and a half-life of 11 days. Find the least whole number of days after which less than 4 g remains.

• \(k=\ln2\div11=0.0630\)
• \(640e^{-0.0630t}<4\): solve \(640e^{-0.0630t}=4\), t = 80.5
• Round up: 81 days