Interleaved — Phase 4

Maths interleaved practice

printable — key stays hidden on paper

Instructions: Work each problem on its own. The topics are deliberately shuffled — before solving, decide which tool the problem demands (limit law, one-sided analysis, squeeze, first principles, a specific differentiation rule, IVT, etc.). Show all reasoning. Total: 50 marks.

1. (5 marks) Evaluate $\lim_{x\to 0}\frac{\sin(3x)}{5x} + \lim_{x\to 0}\frac{1-\cos x}{x^2}.$

2. (6 marks) Using the difference-quotient definition (first principles), find $f'(x)$ for $f(x)=\sqrt{2x+1}$ .

3. (5 marks) For $g(x)=\begin{cases}\dfrac{x^2-4}{x-2}, & x<2\\[2mm] 3x-2, & x\ge 2\end{cases}$ compute the left- and right-hand limits at $x=2$ and classify any discontinuity.

4. (6 marks) Find $\dfrac{dy}{dx}$ by implicit differentiation for $x^2y + \sin(xy) = 4$ .

5. (4 marks) Show that $\displaystyle\lim_{x\to 0} x^2\cos\!\left(\frac{1}{x}\right)=0$ , stating the theorem you use.

6. (5 marks) Differentiate $h(x)=\dfrac{e^{2x}}{\ln x}$ , naming each rule as you apply it.

7. (5 marks) Show that $p(x)=x^3 - 6x + 2$ has a root in $(0,1)$ . Name the theorem and verify its hypotheses.

8. (6 marks) A curve is given parametrically by $x=t^2$ , $y=t^3-3t$ . Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ at $t=1$ .

9. (4 marks) Evaluate $\displaystyle\lim_{x\to\infty}\frac{3x^2-5x+1}{2x^2+7}$ and state the equation of the horizontal asymptote.

10. (4 marks) Use the $\varepsilon$ – $\delta$ definition to prove $\displaystyle\lim_{x\to 3}(4x-1)=11$ .

Answer keyMark scheme & solutions

1. (Tests 4.1.6 Important limits + 4.1.2 limit laws)

First term: $\frac{\sin 3x}{5x}=\frac{3}{5}\cdot\frac{\sin 3x}{3x}\to \frac{3}{5}\cdot 1 = \frac35$ .
Second term: $\frac{1-\cos x}{x^2}\to \frac12$ (standard limit).
Sum $=\frac35+\frac12=\boxed{\frac{11}{10}}$ . Why: recognise both as disguised standard limits, not L'Hôpital fodder; adjust the argument to match $\sin u/u$ .

2. (Tests 4.1.10 first principles) $f'(x)=\lim_{h\to0}\frac{\sqrt{2(x+h)+1}-\sqrt{2x+1}}{h}.$ Rationalise: $=\lim_{h\to0}\frac{2h}{h\big(\sqrt{2x+2h+1}+\sqrt{2x+1}\big)}=\frac{2}{2\sqrt{2x+1}}=\boxed{\frac{1}{\sqrt{2x+1}}}.$ Why: problem explicitly demands the definition — no shortcut rules; multiply by the conjugate to clear the $0/0$ .

3. (Tests 4.1.3 one-sided limits + 4.1.7 continuity/discontinuity type)

Left: $\frac{x^2-4}{x-2}=x+2\to 4$ as $x\to2^-$ .
Right: $3x-2\to 4$ as $x\to2^+$ .
Both equal $4$ ; $g(2)=3(2)-2=4$ . So limit $=4=g(2)$ ⇒ continuous at $x=2$ (no discontinuity). Why: must test each side separately because the definition changes at $x=2$ ; the algebraic factor removes the apparent $0/0$ .

4. (Tests 4.1.22 implicit + 4.1.14 product + 4.1.16 chain) Differentiate: $2xy + x^2 y' + \cos(xy)\,(y + xy')=0.$ Group $y'$ : $y'(x^2 + x\cos(xy)) = -2xy - y\cos(xy),$ $\boxed{y' = -\frac{y\big(2x+\cos(xy)\big)}{x\big(x+\cos(xy)\big)}}.$ Why: $y$ is a function of $x$ , so every $y$ -term needs chain rule; product rule on $x^2y$ and on $xy$ inside the sine.

5. (Tests 4.1.5 squeeze theorem) Since $-1\le\cos(1/x)\le1$ , for $x\ne0$ : $-x^2\le x^2\cos(1/x)\le x^2$ . As $x\to0$ , both bounds $\to0$ . By the Squeeze Theorem, the limit $=\boxed{0}$ . Why: $\cos(1/x)$ has no limit at $0$ , so limit laws fail; bound the oscillation between vanishing envelopes.

6. (Tests 4.1.15 quotient + 4.1.19 $e^x$ + 4.1.20 $\ln x$ + chain) $h'(x)=\frac{(2e^{2x})(\ln x) - e^{2x}\cdot\frac1x}{(\ln x)^2}=\boxed{\frac{e^{2x}\big(2\ln x - \tfrac1x\big)}{(\ln x)^2}}.$ Why: quotient structure ⇒ quotient rule; numerator needs chain rule ( $\frac{d}{dx}e^{2x}=2e^{2x}$ ), denominator gives $1/x$ .

7. (Tests 4.1.8 IVT) $p$ is polynomial ⇒ continuous on $[0,1]$ . $p(0)=2>0$ , $p(1)=1-6+2=-3<0$ . Sign change ⇒ by the Intermediate Value Theorem, $\exists\,c\in(0,1)$ with $p(c)=0$ . $\checkmark$ Why: existence-of-root question with a sign change is the IVT signature; verify continuity + opposite signs.

8. (Tests 4.1.23 parametric derivatives) $\dot x = 2t$ , $\dot y = 3t^2-3$ . $\frac{dy}{dx}=\frac{3t^2-3}{2t}. \quad\text{At }t=1:\ \frac{0}{2}=\boxed{0}.$ Second derivative: $\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\dot x}.$ $\frac{d}{dt}\frac{3t^2-3}{2t}=\frac{d}{dt}\left(\frac{3t}{2}-\frac{3}{2t}\right)=\frac32+\frac{3}{2t^2}$ . At $t=1$ : $3$ . Divide by $\dot x=2$ : $\boxed{\tfrac32}$ . Why: must divide by $\dot x$ (not differentiate w.r.t. $x$ directly); the second derivative differentiates $dy/dx$ w.r.t. $t$ then divides again by $\dot x$ — a classic trap.

9. (Tests 4.1.4 limits at infinity / horizontal asymptote) Divide by $x^2$ : $\frac{3-5/x+1/x^2}{2+7/x^2}\to\frac{3}{2}$ . Horizontal asymptote $\boxed{y=\tfrac32}$ . Why: compare leading degrees; ratio of leading coefficients gives the asymptote.

10. (Tests 4.1.9 $\varepsilon$ – $\delta$ ) Given $\varepsilon>0$ , choose $\delta=\varepsilon/4$ . If $0<|x-3|<\delta$ then $|(4x-1)-11|=|4x-12|=4|x-3|<4\delta=\varepsilon.\ \blacksquare$ Why: formal proof — reverse-engineer $\delta$ from $|f(x)-L|=4|x-3|$ .

[
  {"claim":"Problem 1 sum equals 11/10",
   "code":"import sympy as sp\nx=sp.symbols('x')\na=sp.limit(sp.sin(3*x)/(5*x),x,0)\nb=sp.limit((1-sp.cos(x))/x**2,x,0)\nresult = sp.simplify(a+b - sp.Rational(11,10))==0"},
  {"claim":"Problem 2 derivative of sqrt(2x+1) is 1/sqrt(2x+1)",
   "code":"import sympy as sp\nx=sp.symbols('x')\nf=sp.sqrt(2*x+1)\nresult = sp.simplify(sp.diff(f,x)-1/sp.sqrt(2*x+1))==0"},
  {"claim":"Problem 8 d2y/dx2 at t=1 equals 3/2",
   "code":"import sympy as sp\nt=sp.symbols('t')\nx=t**2\ny=t**3-3*t\ndy=sp.diff(y,t)/sp.diff(x,t)\nd2=sp.diff(dy,t)/sp.diff(x,t)\nresult = sp.simplify(d2.subs(t,1)-sp.Rational(3,2))==0"}
]