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ZIMSEC A Level · N2021

Pure Mathematics Paper 1 November 2021

Questions
78
Total marks
120

Sit this paper online

Questions
78
Pass mark
47
Sit this paper

Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]Exponential equations
Solve the equation 2e2x−7ex+6=02e^{2x} - 7e^x + 6 = 0, giving your answers in exact form.
  1. Ax=ln⁡2x = \ln 2 or x=ln⁡3x = \ln 3
  2. Bx=ln⁡12x = \ln\tfrac{1}{2} or x=ln⁡23x = \ln\tfrac{2}{3}
  3. Cx=ln⁡2x = \ln 2 or x=ln⁡32x = \ln\tfrac{3}{2}
  4. Dx=2x = 2 or x=32x = \tfrac{3}{2}

Question 102

[2 marks]Exponential equations
Let u=exu=e^x. Factorise 2u2−7u+62u^2-7u+6.

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Question 201

[1 marks]Trigonometric identities
Simplify sin⁡2θ1+cos⁡2θ\dfrac{\sin 2\theta}{1 + \cos 2\theta}.
  1. A2tan⁡θ2\tan\theta
  2. Bcot⁡θ\cot\theta
  3. Ctan⁡θ\tan\theta
  4. Dtan⁡2θ\tan 2\theta

Question 202

[2 marks]Trigonometric identities
Using double angle formulas, rewrite the numerator sin⁡2θ\sin2\theta and denominator 1+cos⁡2θ1+\cos2\theta of sin⁡2θ1+cos⁡2θ\dfrac{\sin2\theta}{1+\cos2\theta} in terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta only. State the simplified denominator, in the form kcos⁡2θk\cos^2\theta. State kk.

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Question 301

[1 marks]Differentiation (chain rule)
Given y=ln⁡[cos⁡(x2+1)]y = \ln\left[\cos(x^2+1)\right], find dydx\dfrac{dy}{dx}.
  1. A2xtan⁡(x2+1)2x\tan(x^2+1)
  2. B−2xcos⁡(x2+1)\dfrac{-2x}{\cos(x^2+1)}
  3. C−2xtan⁡(x2+1)-2x\tan(x^2+1)
  4. D−tan⁡(x2+1)-\tan(x^2+1)

Question 302

[2 marks]Differentiation (chain rule)
Using the chain rule on y=ln⁡[cos⁡(x2+1)]y=\ln[\cos(x^2+1)] with u=x2+1u=x^2+1, find dydu\dfrac{dy}{du}, in terms of uu.

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Question 401

[1 marks]Variation
MM is inversely proportional to the cube root of (n−1)(n-1), and M=5M = 5 when n=9n = 9. Find the formula connecting MM and nn.
  1. AM=10n−13M = \dfrac{10}{\sqrt[3]{n-1}}
  2. BM=5n−13M = \dfrac{5}{\sqrt[3]{n-1}}
  3. CM=10n−13M = 10\sqrt[3]{n-1}
  4. DM=40n−13M = \dfrac{40}{\sqrt[3]{n-1}}

Question 402

[1 marks]Variation
Given M=10n−13M = \dfrac{10}{\sqrt[3]{n-1}}, find the exact value of nn when M=25M = 25.
  1. An=25n = \tfrac{2}{5}
  2. Bn=8125n = \tfrac{8}{125}
  3. Cn=1258n = \tfrac{125}{8}
  4. Dn=133125n = \tfrac{133}{125}

Question 403

[2 marks]Variation
MM is inversely proportional to the cube root of (n−1)(n-1), so M=kn−13M=\dfrac{k}{\sqrt[3]{n-1}} for constant kk. Using M=5M=5 when n=9n=9, find n−13=83\sqrt[3]{n-1}=\sqrt[3]{8} first, then solve for kk.

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Question 404

[1 marks]Variation
Using M=10n−13=25M=\dfrac{10}{\sqrt[3]{n-1}}=25, find n−13\sqrt[3]{n-1} (before cubing to find nn).

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Question 501

[1 marks]Completing the square
Express 2x2+3x+12x^2 + 3x + 1 in the form a(x+b)2+ca(x+b)^2 + c.
  1. A2(x+34)2−182\left(x + \tfrac{3}{4}\right)^2 - \tfrac{1}{8}
  2. B2(x+34)2+182\left(x + \tfrac{3}{4}\right)^2 + \tfrac{1}{8}
  3. C2(x+32)2−182\left(x + \tfrac{3}{2}\right)^2 - \tfrac{1}{8}
  4. D2(x−34)2−182\left(x - \tfrac{3}{4}\right)^2 - \tfrac{1}{8}

Question 502

[1 marks]Completing the square
Solve the equation 2x2+3x+1=02x^2 + 3x + 1 = 0.
  1. Ax=−14x = -\tfrac{1}{4} or x=−1x = -1
  2. Bx=12x = \tfrac{1}{2} or x=1x = 1
  3. Cx=−34x = -\tfrac{3}{4} only
  4. Dx=−12x = -\tfrac{1}{2} or x=−1x = -1

Question 503

[2 marks]Completing the square
Completing the square, 2x2+3x+1=2(x+b)2+c2x^2+3x+1=2(x+b)^2+c. State the values of bb and cc, as exact fractions.

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Question 504

[2 marks]Completing the square
Factorise 2x2+3x+12x^2+3x+1, stating both linear factors.

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Question 601

[1 marks]Circular measure and cosine rule
AA and BB lie on a circle of centre OO and radius rr with angle AOB=θAOB = \theta radians, and CC is the midpoint of OAOA. Express BC2BC^2 in terms of rr and θ\theta.
  1. Ar24−r2cos⁡θ\tfrac{r^2}{4} - r^2\cos\theta
  2. B5r24−r2cos⁡θ\tfrac{5r^2}{4} - r^2\cos\theta
  3. C5r24+r2cos⁡θ\tfrac{5r^2}{4} + r^2\cos\theta
  4. Dr2−r24cos⁡θr^2 - \tfrac{r^2}{4}\cos\theta

Question 602

[1 marks]Circular measure and cosine rule
If y2=5r24−r2cos⁡θy^2 = \tfrac{5r^2}{4} - r^2\cos\theta and θ\theta is small, with t=rθt = r\theta the arc length, then y2≈y^2 \approx
  1. A14r2−12t2\tfrac{1}{4}r^2 - \tfrac{1}{2}t^2
  2. B14r2+12t2\tfrac{1}{4}r^2 + \tfrac{1}{2}t^2
  3. C54r2+12t2\tfrac{5}{4}r^2 + \tfrac{1}{2}t^2
  4. D12r2+14t2\tfrac{1}{2}r^2 + \tfrac{1}{4}t^2

Question 603

[2 marks]Circular measure and cosine rule
For small θ\theta, cos⁡θ≈1−θ22\cos\theta\approx1-\dfrac{\theta^2}{2}. Substitute this into y2=5r24−r2cos⁡θy^2=\dfrac{5r^2}{4}-r^2\cos\theta and simplify to the form y2≈r24+Kr2θ2y^2\approx\dfrac{r^2}{4}+Kr^2\theta^2. State KK.

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Question 604

[2 marks]Circular measure and cosine rule
Since θ=tr\theta=\dfrac{t}{r} (arc length t=rθt=r\theta), express r2θ2r^2\theta^2 in terms of tt only.

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Question 701

[1 marks]Proof by induction (matrices)
In a proof by induction that An=(2n+1−n4n1−2n)A^n = \begin{pmatrix} 2n+1 & -n \\ 4n & 1-2n \end{pmatrix} for A=(3−14−1)A = \begin{pmatrix} 3 & -1 \\ 4 & -1 \end{pmatrix}, what is the inductive step?
  1. AShow Ak+1=AkAA^{k+1} = A^k A matches the formula with n=k+1n = k+1, assuming it holds for n=kn = k
  2. BShow that det⁡An=(det⁡A)n\det A^n = (\det A)^n
  3. CShow that An→0A^n \to 0 as n→∞n \to \infty
  4. DVerify the formula holds for n=1n = 1 and n=2n = 2

Question 702

[2 marks]Proof by induction (matrices)
As part of verifying the base case n=2n=2, compute A2=AAA^2=AA for A=(3−14−1)A=\begin{pmatrix}3&-1\\4&-1\end{pmatrix}. State the bottom-left entry of A2A^2.

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Question 703

[2 marks]Proof by induction (matrices)
In the inductive step, assuming Ak=(2k+1−k4k1−2k)A^k=\begin{pmatrix}2k+1&-k\\4k&1-2k\end{pmatrix}, compute Ak+1=AkAA^{k+1}=A^kA. State the top-left entry of Ak+1A^{k+1}, in terms of kk.

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Question 704

[2 marks]Proof by induction (matrices)
Continuing Ak+1=AkAA^{k+1}=A^kA with Ak=(2k+1−k4k1−2k)A^k=\begin{pmatrix}2k+1&-k\\4k&1-2k\end{pmatrix} and A=(3−14−1)A=\begin{pmatrix}3&-1\\4&-1\end{pmatrix}, state the bottom-left entry of Ak+1A^{k+1}, in terms of kk.

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Question 801

[1 marks]Exponential equations and modulus inequalities
Solve the equation 2x+12x−1=5\dfrac{2^x + 1}{2^x - 1} = 5, giving your answer correct to 3 significant figures.
  1. Ax=1.50x = 1.50
  2. Bx=0.405x = 0.405
  3. Cx=0.585x = 0.585
  4. Dx=1.58x = 1.58

Question 802

[1 marks]Exponential equations and modulus inequalities
Solve the inequality ∣x−3∣>2∣3x+1∣|x-3| > 2|3x+1|.
  1. A−1<x<17-1 < x < \tfrac{1}{7}
  2. Bx<−1x < -1 or x>17x > \tfrac{1}{7}
  3. C−17<x<1-\tfrac{1}{7} < x < 1
  4. D−13<x<3-\tfrac{1}{3} < x < 3

Question 803

[2 marks]Exponential equations and modulus inequalities
Let u=2xu=2^x. Solve u+1u−1=5\dfrac{u+1}{u-1}=5 for uu.

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Question 804

[2 marks]Exponential equations and modulus inequalities
Squaring both sides of ∣x−3∣>2∣3x+1∣|x-3|>2|3x+1| and simplifying gives an inequality of the form 7x2+bx+c<07x^2+bx+c<0. State bb and cc.

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Question 805

[2 marks]Exponential equations and modulus inequalities
Factorise 7x2+6x−17x^2+6x-1, stating both linear factors.

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Question 901

[1 marks]Series expansions
Given that xx is small enough for x4x^4 and higher powers to be neglected, expand ex1+x\dfrac{e^x}{1+x}.
  1. A1+x22+x331 + \tfrac{x^2}{2} + \tfrac{x^3}{3}
  2. B1+x22−x331 + \tfrac{x^2}{2} - \tfrac{x^3}{3}
  3. C1+x+x22−x331 + x + \tfrac{x^2}{2} - \tfrac{x^3}{3}
  4. D1−x22+x331 - \tfrac{x^2}{2} + \tfrac{x^3}{3}

Question 902

[1 marks]Series expansions
Use the expansion ex1+x≈1+x22−x33\dfrac{e^x}{1+x} \approx 1 + \dfrac{x^2}{2} - \dfrac{x^3}{3} to evaluate e0.011.01\dfrac{e^{0.01}}{1.01} correct to 7 decimal places.
  1. A0.99995030.9999503
  2. B1.00004971.0000497
  3. C1.00005001.0000500
  4. D1.01000001.0100000

Question 903

[2 marks]Series expansions
State the Maclaurin expansion of exe^x up to the term in x3x^3.

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Question 904

[2 marks]Series expansions
State the binomial expansion of (1+x)−1(1+x)^{-1} up to the term in x3x^3.

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Question 905

[2 marks]Series expansions
Multiplying the series for exe^x and (1+x)−1(1+x)^{-1}, state the coefficient of x1x^1 in the product ex(1+x)−1e^x(1+x)^{-1} (before collecting further terms).

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Question 1001

[1 marks]Modulus functions and graphs
Find the coordinates of the points of intersection of f(x)=2−∣x∣f(x) = 2 - |x| and g(x)=13x+1g(x) = \tfrac{1}{3}x + 1.
  1. A(−32,12)\left(-\tfrac{3}{2}, \tfrac{1}{2}\right) and (34,54)\left(\tfrac{3}{4}, \tfrac{5}{4}\right)
  2. B(−34,34)\left(-\tfrac{3}{4}, \tfrac{3}{4}\right) and (32,32)\left(\tfrac{3}{2}, \tfrac{3}{2}\right)
  3. C(32,12)\left(\tfrac{3}{2}, \tfrac{1}{2}\right) and (−34,54)\left(-\tfrac{3}{4}, \tfrac{5}{4}\right)
  4. D(−3,0)(-3, 0) and (2,0)(2, 0)

Question 1002

[1 marks]Modulus functions and graphs
State the yy-intercept of f(x)=2−∣x∣f(x)=2-|x|.

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Question 1003

[1 marks]Modulus functions and graphs
State the yy-intercept of g(x)=13x+1g(x)=\tfrac13x+1.

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Question 1004

[2 marks]Modulus functions and graphs
Find both xx-intercepts of f(x)=2−∣x∣f(x)=2-|x|.

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Question 1005

[1 marks]Modulus functions and graphs
Find the xx-intercept of g(x)=13x+1g(x)=\tfrac13x+1.

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Question 1006

[2 marks]Modulus functions and graphs
For x≥0x\ge0, solve 2−x=13x+12-x=\tfrac13x+1 to find the xx-coordinate of the intersection point in this region (before finding yy).

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Question 1101

[1 marks]Partial fractions and rational inequalities
Express 2x(x+2)(x−2)(x−1)\dfrac{2x}{(x+2)(x-2)(x-1)} in partial fractions.
  1. A−23(x+2)+1x−2−13(x−1)-\dfrac{2}{3(x+2)} + \dfrac{1}{x-2} - \dfrac{1}{3(x-1)}
  2. B13(x+2)+1x−2+23(x−1)\dfrac{1}{3(x+2)} + \dfrac{1}{x-2} + \dfrac{2}{3(x-1)}
  3. C−13(x+2)+1x−2−23(x−1)-\dfrac{1}{3(x+2)} + \dfrac{1}{x-2} - \dfrac{2}{3(x-1)}
  4. D−13(x+2)−1x−2+23(x−1)-\dfrac{1}{3(x+2)} - \dfrac{1}{x-2} + \dfrac{2}{3(x-1)}

Question 1102

[1 marks]Partial fractions and rational inequalities
Solve the inequality 2x(x+2)(x−2)(x−1)<0\dfrac{2x}{(x+2)(x-2)(x-1)} < 0.
  1. Ax<−2x < -2 or 0<x<10 < x < 1
  2. B−2<x<0-2 < x < 0 or 1<x<21 < x < 2
  3. C0<x<10 < x < 1 or x>2x > 2
  4. D−2<x<2-2 < x < 2

Question 1103

[2 marks]Partial fractions and rational inequalities
Using the cover-up rule, evaluate A=2x(x−2)(x−1)A=\dfrac{2x}{(x-2)(x-1)} at x=−2x=-2, the coefficient of 1x+2\dfrac{1}{x+2} in the partial fraction decomposition of 2x(x+2)(x−2)(x−1)\dfrac{2x}{(x+2)(x-2)(x-1)}.

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Question 1104

[2 marks]Partial fractions and rational inequalities
Using the cover-up rule, evaluate B=2x(x+2)(x−1)B=\dfrac{2x}{(x+2)(x-1)} at x=2x=2, the coefficient of 1x−2\dfrac{1}{x-2}.

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Question 1105

[2 marks]Partial fractions and rational inequalities
Using the cover-up rule, evaluate C=2x(x+2)(x−2)C=\dfrac{2x}{(x+2)(x-2)} at x=1x=1, the coefficient of 1x−1\dfrac{1}{x-1}.

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Question 1201

[1 marks]Complex numbers
Given p=−4+3ip = -4 + 3i and q=−1+3iq = -1 + \sqrt{3}i, calculate ∣p∣|p| and ∣q∣|q|.
  1. A∣p∣=5|p| = 5, ∣q∣=2|q| = \sqrt{2}
  2. B∣p∣=25|p| = 25, ∣q∣=4|q| = 4
  3. C∣p∣=7|p| = 7, ∣q∣=2|q| = 2
  4. D∣p∣=5|p| = 5, ∣q∣=2|q| = 2

Question 1202

[1 marks]Complex numbers
Find the argument of q=−1+3iq = -1 + \sqrt{3}i.
  1. Aπ3\tfrac{\pi}{3}
  2. B2π3\tfrac{2\pi}{3}
  3. C−2π3-\tfrac{2\pi}{3}
  4. D5π6\tfrac{5\pi}{6}

Question 1203

[1 marks]Complex numbers
Given p=−4+3ip = -4 + 3i and q=−1+3iq = -1 + \sqrt{3}i, find pq2pq^2 in the form a+bia + bi.
  1. A(8−63)+(83+6)i(8 - 6\sqrt{3}) + (8\sqrt{3} + 6)i
  2. B(8+63)−(83−6)i(8 + 6\sqrt{3}) - (8\sqrt{3} - 6)i
  3. C(−2−23)i(-2 - 2\sqrt{3})i
  4. D(8+63)+(83−6)i(8 + 6\sqrt{3}) + (8\sqrt{3} - 6)i

Question 1204

[2 marks]Complex numbers
Find q2q^2 for q=−1+3iq=-1+\sqrt3i, in the form a+bia+bi.

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Question 1205

[2 marks]Complex numbers
To find pq\dfrac{p}{q} for p=−4+3ip=-4+3i, q=−1+3iq=-1+\sqrt3i, multiply numerator and denominator by the conjugate −1−3i-1-\sqrt3i. Find the denominator (−1+3i)(−1−3i)(-1+\sqrt3i)(-1-\sqrt3i).

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Question 1206

[2 marks]Complex numbers
Find the numerator (−4+3i)(−1−3i)(-4+3i)(-1-\sqrt3i), in the form a+bia+bi (in terms of 3\sqrt3).

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Question 1207

[2 marks]Complex numbers
Hence state pq\dfrac{p}{q} for p=−4+3ip=-4+3i, q=−1+3iq=-1+\sqrt3i, correct to 3 decimal places, in the form a+bia+bi.

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Question 1208

[1 marks]Complex numbers
State the real part of pq2pq^2, in the form 8+k38+k\sqrt3. State kk.

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Question 1301

[1 marks]Coordinate geometry
The curve y=10−5xy = 10 - \dfrac{5}{x} meets the line y+x=6y + x = 6 at PP and QQ. Find their coordinates.
  1. AP(−1,7)P(-1, 7) and Q(5,1)Q(5, 1)
  2. BP(−5,11)P(-5, 11) and Q(1,5)Q(1, 5)
  3. CP(5,1)P(5, 1) and Q(−1,7)Q(-1, 7)
  4. DP(−5,1)P(-5, 1) and Q(1,11)Q(1, 11)

Question 1302

[1 marks]Coordinate geometry
Find the equation of the perpendicular bisector of the segment joining P(−5,11)P(-5, 11) and Q(1,5)Q(1, 5).
  1. Ay−x−10=0y - x - 10 = 0
  2. By+x−10=0y + x - 10 = 0
  3. Cy+x−6=0y + x - 6 = 0
  4. Dy−x+10=0y - x + 10 = 0

Question 1303

[2 marks]Coordinate geometry
Substituting y=6−xy=6-x into y=10−5xy=10-\dfrac{5}{x} and simplifying (multiply through by xx) gives a quadratic in the form x2+bx+c=0x^2+bx+c=0. State bb and cc.

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Question 1304

[1 marks]Coordinate geometry
Factorise x2+4x−5x^2+4x-5.

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Question 1305

[2 marks]Coordinate geometry
Find the coordinates of the midpoint of PQPQ, where P(−5,11)P(-5,11) and Q(1,5)Q(1,5).

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Question 1306

[1 marks]Coordinate geometry
Find the gradient of the line segment PQPQ, where P(−5,11)P(-5,11) and Q(1,5)Q(1,5).

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Question 1307

[2 marks]Coordinate geometry
State the gradient of the perpendicular bisector of PQPQ (perpendicular to a line of gradient −1-1).

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Question 1401

[1 marks]Differentiation, tangents and normals
The curve y=x2+px+qy = x^2 + px + q has a turning point at (−1,−5)(-1, -5). Find pp and qq.
  1. Ap=−2p = -2, q=−4q = -4
  2. Bp=2p = 2, q=4q = 4
  3. Cp=2p = 2, q=−4q = -4
  4. Dp=−2p = -2, q=4q = 4

Question 1402

[1 marks]Differentiation, tangents and normals
The curve y=x2+2x−4y = x^2 + 2x - 4 cuts the yy-axis at (0,−4)(0, -4). Find the equations of the tangent and normal there, in the form ay+bx+c=0ay + bx + c = 0.
  1. Atangent y−2x−4=0y - 2x - 4 = 0; normal 2y+x−8=02y + x - 8 = 0
  2. Btangent 2y−x+8=02y - x + 8 = 0; normal y−2x+4=0y - 2x + 4 = 0
  3. Ctangent y−2x+4=0y - 2x + 4 = 0; normal 2y+x+8=02y + x + 8 = 0
  4. Dtangent y+2x+4=0y + 2x + 4 = 0; normal 2y−x+8=02y - x + 8 = 0

Question 1403

[2 marks]Differentiation, tangents and normals
Using dydx=2x+p=0\dfrac{dy}{dx}=2x+p=0 at the turning point x=−1x=-1 of y=x2+px+qy=x^2+px+q, find pp.

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Question 1404

[2 marks]Differentiation, tangents and normals
Substituting the turning point (−1,−5)(-1,-5) into y=x2+2x+qy=x^2+2x+q, find qq.

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Question 1405

[2 marks]Differentiation, tangents and normals
For the curve y=x2+2x−4y=x^2+2x-4, find dydx\dfrac{dy}{dx} at x=0x=0 (the gradient of the tangent at the yy-axis crossing).

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Question 1406

[2 marks]Differentiation, tangents and normals
State the gradient of the normal to y=x2+2x−4y=x^2+2x-4 at x=0x=0 (perpendicular to the tangent of gradient 2).

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Question 1501

[1 marks]Matrix transformations
A shear parallel to the xx-axis maps (1,2)(1, 2) onto (7,2)(7, 2). Find the shear factor.
  1. A33
  2. B77
  3. C13\tfrac{1}{3}
  4. D66

Question 1502

[1 marks]Matrix transformations
Triangle ABCABC has vertices A(−3,1)A(-3, 1), B(3,1)B(3, 1) and C(3,5)C(3, 5). Find its area.
  1. A1212 units2^2
  2. B2424 units2^2
  3. C66 units2^2
  4. D1010 units2^2

Question 1503

[1 marks]Matrix transformations
Triangle ABCABC of area 12 units2^2 is transformed by M=(4123)M = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}. Find the area of the image triangle.
  1. A1212 units2^2
  2. B120120 units2^2
  3. C144144 units2^2
  4. D1010 units2^2

Question 1504

[2 marks]Matrix transformations
Find A1A_1, the image of A(−3,1)A(-3,1) under M=(4123)M=\begin{pmatrix}4&1\\2&3\end{pmatrix}, in the form (x,y)(x,y).

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Question 1505

[2 marks]Matrix transformations
Find B1B_1, the image of B(3,1)B(3,1) under M=(4123)M=\begin{pmatrix}4&1\\2&3\end{pmatrix}, in the form (x,y)(x,y).

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Question 1506

[2 marks]Matrix transformations
Find C1C_1, the image of C(3,5)C(3,5) under M=(4123)M=\begin{pmatrix}4&1\\2&3\end{pmatrix}, in the form (x,y)(x,y).

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Question 1507

[2 marks]Matrix transformations
Write the shear matrix parallel to the xx-axis with shear factor k=3k=3, in the form (1k01)\begin{pmatrix}1&k\\0&1\end{pmatrix}. State its determinant (confirming shears preserve area).

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Question 1601

[1 marks]Differential equations
The gradient function of a curve is directly proportional to 20−y20 - y. Given that the gradient is 1 when y=0y = 0, find the differential equation.
  1. Adydx=20(20−y)\dfrac{dy}{dx} = 20(20 - y)
  2. Bdydx=0.05y\dfrac{dy}{dx} = 0.05y
  3. Cdydx=20−y\dfrac{dy}{dx} = 20 - y
  4. Ddydx=0.05(20−y)\dfrac{dy}{dx} = 0.05(20 - y)

Question 1602

[1 marks]Differential equations
Solve dydx=0.05(20−y)\dfrac{dy}{dx} = 0.05(20 - y) given that y=0y = 0 when x=0x = 0.
  1. Ay=20e−0.05xy = 20e^{-0.05x}
  2. By=20+20e−0.05xy = 20 + 20e^{-0.05x}
  3. Cy=20−20e0.05xy = 20 - 20e^{0.05x}
  4. Dy=20−20e−0.05xy = 20 - 20e^{-0.05x}

Question 1603

[1 marks]Differential equations
For y=20−20e−0.05xy = 20 - 20e^{-0.05x}, describe what happens to yy as xx becomes very large.
  1. Ayy approaches 0
  2. Byy oscillates about 20
  3. Cyy increases without limit
  4. Dyy approaches 20 from below

Question 1604

[2 marks]Differential equations
Separating variables in dydx=0,05(20−y)\dfrac{dy}{dx}=0{,}05(20-y) gives ∫120−y dy=∫0,05 dx\displaystyle\int\dfrac{1}{20-y}\,dy=\int0{,}05\,dx. Evaluate the left-hand integral (ignore the constant of integration), in terms of yy.

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Question 1605

[2 marks]Differential equations
Using y=0y=0 when x=0x=0 in −ln⁡(20−y)=0,05x+C-\ln(20-y)=0{,}05x+C, find the constant CC.

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Question 1606

[2 marks]Differential equations
Using y=20−20e−0.05xy=20-20e^{-0.05x}, find yy when x=5x=5, correct to 2 decimal places.

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Question 1607

[3 marks]Differential equations
Using y=20−20e−0.05xy=20-20e^{-0.05x}, find yy when x=10x=10, correct to 2 decimal places.

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