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Paper 1 Answers

Time2 hours 30 minutes

Marks90

SetSet 1

PaperLevel 1 - Calculus Paper 1

Information

Section A: Limits and Continuity
Section B: Differential Calculus
Section C: Applications of Derivatives
Section D: Integral Calculus
Section E: Applications of Integrals and Series
Section F: Multivariable Calculus and Differential Equations

Marking Guidance

These worked answers show the expected method and final result. Equivalent correct reasoning should receive credit.

Section A: Limits and Continuity

Answer 1. Limits, continuity, and graph behaviour

[15 marks]

This question concerns limiting behaviour, continuity, and asymptotes of single-variable functions.

a) Evaluate \(\displaystyle \lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}\).

[4 marks]

Direct substitution gives \(0/0\), so simplify before taking the limit. Rationalise the numerator: \[ \frac{\sqrt{x+5}-3}{x-4}\cdot\frac{\sqrt{x+5}+3}{\sqrt{x+5}+3}=\frac{x+5-9}{(x-4)(\sqrt{x+5}+3)}. \] Since \(x+5-9=x-4\), for \(x\ne4\) the expression becomes \[ \frac{1}{\sqrt{x+5}+3}. \] Now substitute \(x=4\): \[ \lim_{x\to4}\frac{\sqrt{x+5}-3}{x-4}=\frac{1}{3+3}=\frac16. \] Therefore the limit is \(\frac16\).

Answer 1. Limits, continuity, and graph behaviour

[15 marks]

b) Let \(f(x)=ax+2\) for \(x<1\), and \(f(x)=x^2+b\) for \(x\ge1\). Find the relationship between the constants \(a\) and \(b\) required for \(f\) to be continuous at \(x=1\).

[4 marks]

Continuity at \(x=1\) requires the left-hand limit, right-hand value, and function value to agree. From the left, \[ \lim_{x\to1^-}(ax+2)=a+2. \] From the right branch and from the definition at \(x=1\), \[ f(1)=1^2+b=b+1. \] Thus continuity gives \[ a+2=b+1, \] so \[ b=a+1. \] Therefore all parameter pairs satisfying \(b=a+1\) make \(f\) continuous at \(x=1\).

Answer 1. Limits, continuity, and graph behaviour

[15 marks]

c) Find the vertical and horizontal asymptotes of \(g(x)=\frac{2x+1}{x-2}\).

[3 marks]

The denominator is zero when \(x-2=0\), so \(x=2\). The numerator at \(x=2\) is \(2(2)+1=5\ne0\), so \(x=2\) is a vertical asymptote. For the horizontal asymptote, compare leading terms: \[ \frac{2x+1}{x-2}=\frac{2+1/x}{1-2/x}. \] As \(x\to\pm\infty\), the fractions \(1/x\) and \(2/x\) approach \(0\), so the limit is \(2\). Therefore the horizontal asymptote is \(y=2\). The asymptotes are \(x=2\) and \(y=2\).

Answer 1. Limits, continuity, and graph behaviour

[15 marks]

d) Using first principles, find the derivative of \(f(x)=x^2-3x\).

[4 marks]

Use the derivative definition: \[ f'(x)=\lim_{h\to0}\frac{(x+h)^2-3(x+h)-(x^2-3x)}{h}. \] Expand the numerator: \[ x^2+2xh+h^2-3x-3h-x^2+3x=2xh+h^2-3h. \] Factor \(h\): \[ 2xh+h^2-3h=h(2x+h-3). \] For \(h\ne0\), the difference quotient is \(2x+h-3\). Taking \(h\to0\) gives \[ f'(x)=2x-3. \] Therefore the derivative from first principles is \(2x-3\).

Section B: Differential Calculus

Answer 2. Derivative methods

[15 marks]

This question uses differentiation rules, implicit differentiation, and local approximation.

a) Differentiate \(f(x)=(x^2+1)e^{-x}\).

[4 marks]

Use the product rule with \(u=x^2+1\) and \(v=e^{-x}\). Then \(u'=2x\) and \(v'=-e^{-x}\). Hence \[ f'(x)=2xe^{-x}+(x^2+1)(-e^{-x}). \] Factor \(e^{-x}\): \[ f'(x)=e^{-x}(2x-x^2-1). \] Therefore \(f'(x)=e^{-x}(2x-x^2-1)\).

Answer 2. Derivative methods

[15 marks]

b) For the implicit curve \(xy^2+x^2=5\), find \(\frac{dy}{dx}\).

[4 marks]

Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\). For \(xy^2\), use the product rule: \[ \frac{d}{dx}(xy^2)=y^2+x\cdot2y\frac{dy}{dx}. \] The derivative of \(x^2\) is \(2x\), and the derivative of \(5\) is \(0\). Thus \[ y^2+2xy\frac{dy}{dx}+2x=0. \] Solve for \(dy/dx\): \[ 2xy\frac{dy}{dx}=-(y^2+2x), \] so \[ \frac{dy}{dx}=-\frac{y^2+2x}{2xy}, \] where \(xy\ne0\).

Answer 2. Derivative methods

[15 marks]

c) For \(f(x)=x^4-6x^2+3\), find \(f''(2)\).

[3 marks]

First differentiate: \[ f'(x)=4x^3-12x. \] Differentiate again: \[ f''(x)=12x^2-12. \] Evaluate at \(x=2\): \[ f''(2)=12(2)^2-12=48-12=36. \] Therefore \(f''(2)=36\).

Answer 2. Derivative methods

[15 marks]

d) Use linear approximation about \(x=9\) to estimate \(\sqrt{9.3}\).

[4 marks]

Let \(f(x)=\sqrt{x}\) and choose the nearby base point \(a=9\). Then \[ f(9)=3,\qquad f'(x)=\frac{1}{2\sqrt{x}},\qquad f'(9)=\frac16. \] The linear approximation is \[ L(x)=f(9)+f'(9)(x-9)=3+\frac16(x-9). \] At \(x=9.3\), \[ L(9.3)=3+\frac{0.3}{6}=3+0.05=3.05. \] Therefore \(\sqrt{9.3}\approx3.05\).

Section C: Applications of Derivatives

Answer 3. Using derivatives to answer applied questions

[15 marks]

This question concerns optimisation, related rates, curve information, and indeterminate limits.

a) A rectangle has side lengths \(x\) and \(y\), with \(x+y=12\). Find the side lengths that maximise its area.

[4 marks]

The area is \[ A(x)=x(12-x)=12x-x^2, \] where \(0<x<12\). Differentiate: \[ A'(x)=12-2x. \] Set \(A'(x)=0\): \[ 12-2x=0,\qquad x=6. \] Then \(y=12-6=6\). Since \(A(x)\) is a downward-opening quadratic, this critical point gives the maximum. Therefore the maximum area is obtained by a \(6\) by \(6\) rectangle.

Answer 3. Using derivatives to answer applied questions

[15 marks]

b) A sphere has radius \(r\) increasing at \(0.5\) units per second. How fast is the volume increasing when \(r=2\)?

[4 marks]

The volume of a sphere is \[ V=\frac43\pi r^3. \] Differentiate with respect to time: \[ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}. \] Substitute \(r=2\) and \(dr/dt=0.5\): \[ \frac{dV}{dt}=4\pi(2)^2(0.5)=8\pi. \] Therefore the volume is increasing at \(8\pi\) cubic units per second.

Answer 3. Using derivatives to answer applied questions

[15 marks]

c) For \(f(x)=x^3-6x^2+1\), find and classify the local stationary points.

[4 marks]

Differentiate: \[ f'(x)=3x^2-12x=3x(x-4). \] Critical points occur at \(x=0\) and \(x=4\). The second derivative is \[ f''(x)=6x-12. \] At \(x=0\), \(f''(0)=-12<0\), so there is a local maximum. At \(x=4\), \(f''(4)=12>0\), so there is a local minimum. The point values are \(f(0)=1\) and \(f(4)=64-96+1=-31\). Therefore the local maximum is \((0,1)\), and the local minimum is \((4,-31)\).

Answer 3. Using derivatives to answer applied questions

[15 marks]

d) Use L'Hopital's rule to evaluate \(\displaystyle \lim_{x\to0}\frac{e^{2x}-1}{x}\).

[3 marks]

Substitution gives \((e^0-1)/0=0/0\), so L'Hopital's rule applies. Differentiate numerator and denominator: \[ \lim_{x\to0}\frac{e^{2x}-1}{x}=\lim_{x\to0}\frac{2e^{2x}}{1}. \] Now substitute \(x=0\): \[ 2e^0=2. \] Therefore the limit is \(2\).

Section D: Integral Calculus

Answer 4. Integral methods

[15 marks]

This question uses antiderivatives, substitution, integration by parts, partial fractions, and the fundamental theorem of calculus.

a) Evaluate \(\displaystyle \int 2x(x^2+4)^5\,dx\).

[3 marks]

Let \(u=x^2+4\). Then \[ du=2x\,dx. \] The integral becomes \[ \int 2x(x^2+4)^5\,dx=\int u^5\,du. \] Integrate: \[ \int u^5\,du=\frac{u^6}{6}+C. \] Substitute back \(u=x^2+4\): \[ \int 2x(x^2+4)^5\,dx=\frac{(x^2+4)^6}{6}+C. \]

Answer 4. Integral methods

[15 marks]

b) Evaluate \(\displaystyle \int x\cos x\,dx\).

[4 marks]

Choose \(u=x\), so \(du=dx\), and choose \(dv=\cos x\,dx\), so \(v=\sin x\). By integration by parts, \[ \int x\cos x\,dx=x\sin x-\int \sin x\,dx. \] Since \(\int\sin x\,dx=-\cos x\), \[ \int x\cos x\,dx=x\sin x+\cos x+C. \] Therefore the integral is \(x\sin x+\cos x+C\).

Answer 4. Integral methods

[15 marks]

c) Evaluate \(\displaystyle \int\frac{5x+6}{x^2-4}\,dx\), for \(x\ne\pm2\).

[4 marks]

Factor the denominator: \[ x^2-4=(x-2)(x+2). \] Write \[ \frac{5x+6}{x^2-4}=\frac{A}{x-2}+\frac{B}{x+2}. \] Multiplying by \((x-2)(x+2)\) gives \[ 5x+6=A(x+2)+B(x-2). \] Set \(x=2\): \(16=4A\), so \(A=4\). Set \(x=-2\): \(-4=-4B\), so \(B=1\). Therefore \[ \int\frac{5x+6}{x^2-4}\,dx=4\ln|x-2|+\ln|x+2|+C. \]

Answer 4. Integral methods

[15 marks]

d) Differentiate \(\displaystyle A(x)=\int_1^{x^2}\ln(1+t^2)\,dt\).

[4 marks]

Let \[ A(x)=\int_1^{x^2}\ln(1+t^2)\,dt. \] By the fundamental theorem of calculus, the derivative with respect to the upper limit \(u\) is \(\ln(1+u^2)\). Here \(u=x^2\), so use the chain rule: \[ A'(x)=\ln(1+(x^2)^2)\cdot2x. \] Thus \[ A'(x)=2x\ln(1+x^4). \] Therefore the derivative is \(2x\ln(1+x^4)\).

Section E: Applications of Integrals and Series

Answer 5. Integral applications and infinite processes

[15 marks]

This question combines accumulated area, volumes, improper integrals, and series convergence.

a) Find the area enclosed by \(y=x\) and \(y=x^2\).

[4 marks]

Find intersections first: \[ x=x^2 \quad\Longrightarrow\quad x^2-x=0 \quad\Longrightarrow\quad x(x-1)=0. \] Thus the curves meet at \(x=0\) and \(x=1\). On \([0,1]\), \(x\ge x^2\), so the area is \[ A=\int_0^1(x-x^2)\,dx. \] Integrate: \[ A=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac12-\frac13=\frac16. \] Therefore the enclosed area is \(\frac16\).

Answer 5. Integral applications and infinite processes

[15 marks]

b) Find the volume formed by rotating the region under \(y=\sqrt{x}\) from \(x=0\) to \(x=4\) about the \(x\)-axis.

[4 marks]

Rotating about the \(x\)-axis gives disks of radius \(R(x)=\sqrt{x}\). The cross-sectional area is \[ A(x)=\pi R(x)^2=\pi x. \] Thus the volume is \[ V=\pi\int_0^4 x\,dx. \] Evaluate: \[ V=\pi\left[\frac{x^2}{2}\right]_0^4=\pi\cdot\frac{16}{2}=8\pi. \] Therefore the volume is \(8\pi\).

Answer 5. Integral applications and infinite processes

[15 marks]

c) Determine whether \(\displaystyle \int_2^\infty \frac{1}{(x-1)^2}\,dx\) converges, and evaluate it if it does.

[4 marks]

Write the improper integral as a limit: \[ \int_2^\infty \frac{1}{(x-1)^2}\,dx=\lim_{b\to\infty}\int_2^b (x-1)^{-2}\,dx. \] An antiderivative is \[ -(x-1)^{-1}=-\frac{1}{x-1}. \] Evaluate from \(2\) to \(b\): \[ -\frac{1}{b-1}-\left(-\frac{1}{1}\right)=1-\frac{1}{b-1}. \] Taking \(b\to\infty\) gives \(1\). Therefore the improper integral converges and equals \(1\).

Answer 5. Integral applications and infinite processes

[15 marks]

d) Use the ratio test to determine whether \(\displaystyle \sum_{n=1}^{\infty}\frac{3^n}{n^2}\) converges.

[3 marks]

Use the ratio test with \[ a_n=\frac{3^n}{n^2}. \] Then \[ \left|\frac{a_{n+1}}{a_n}\right|=\frac{3^{n+1}}{(n+1)^2}\cdot\frac{n^2}{3^n}=3\left(\frac{n}{n+1}\right)^2. \] Taking the limit gives \[ \lim_{n\to\infty}3\left(\frac{n}{n+1}\right)^2=3. \] Since \(3>1\), the ratio test shows that the series diverges.

Section F: Multivariable Calculus and Differential Equations

Answer 6. Several variables and differential equations

[15 marks]

This question concerns functions of several variables and elementary differential equations.

a) For \(f(x,y)=x^2y+3xy^2\), find \(\nabla f(1,2)\).

[4 marks]

Compute the first partial derivatives by holding the other variable constant. With respect to \(x\), \[ f_x=2xy+3y^2. \] With respect to \(y\), \[ f_y=x^2+6xy. \] At \((1,2)\), \[ f_x(1,2)=2(1)(2)+3(2)^2=4+12=16, \] and \[ f_y(1,2)=1^2+6(1)(2)=1+12=13. \] Therefore \(\nabla f(1,2)=(16,13)\).

Answer 6. Several variables and differential equations

[15 marks]

b) Using the function from part (a), find the directional derivative at \((1,2)\) in the unit direction \(\mathbf u=(3/5,4/5)\).

[3 marks]

The directional derivative in a unit direction \(\mathbf u\) is \[ D_{\mathbf u}f=\nabla f\cdot\mathbf u. \] Using \(\nabla f(1,2)=(16,13)\) and \(\mathbf u=(3/5,4/5)\), \[ D_{\mathbf u}f=16\cdot\frac35+13\cdot\frac45=\frac{48}{5}+\frac{52}{5}=\frac{100}{5}=20. \] Therefore the directional derivative is \(20\).

Answer 6. Several variables and differential equations

[15 marks]

c) Let \(z=f(x,y)=x^2y+3xy^2\), where \(x=t\) and \(y=t^2\). Find \(\frac{dz}{dt}\) at \(t=1\).

[4 marks]

For \(z=f(x,y)\), the chain rule gives \[ \frac{dz}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}. \] Here \(f_x=2xy+3y^2\) and \(f_y=x^2+6xy\). Since \(x=t\), \(dx/dt=1\). Since \(y=t^2\), \(dy/dt=2t\). At \(t=1\), \(x=1\) and \(y=1\), so \[ f_x(1,1)=2(1)(1)+3(1)^2=5, \] and \[ f_y(1,1)=1^2+6(1)(1)=7. \] Thus \[ \frac{dz}{dt}\bigg|_{t=1}=5(1)+7(2)=19. \] Therefore \(dz/dt=19\) at \(t=1\).

Answer 6. Several variables and differential equations

[15 marks]

d) Solve \(\frac{dy}{dx}=2xy\) with initial condition \(y(0)=3\).

[4 marks]

Separate variables: \[ \frac{1}{y}\,dy=2x\,dx, \] assuming \(y\ne0\). Integrate both sides: \[ \ln|y|=x^2+C. \] Exponentiating gives \[ y=Ae^{x^2}. \] Use \(y(0)=3\): \[ 3=Ae^0=A. \] Therefore the solution is \[ y=3e^{x^2}. \]