2023 CCEA GCSE Mathematics 2210 模拟试题及答案详解

The highest common factor of \( 4x^2y \) and \( 6xy^2 \) is \( 2xy \). Dividing each term by \( 2xy \) gives \( 4x^2y - 6xy^2 = 2xy(2x-3y) \).

M1: highest common factor \( 2xy \) correctly identified; M1: correct terms inside the bracket, \( (2x-3y) \); A1: fully factorised, \( 2xy(2x-3y) \), with no further common factor. [3]

\( \sqrt{45} = \sqrt{9\times5} = \sqrt{9}\times\sqrt{5} = 3\sqrt{5} \).

M1: \( 45 \) correctly split as \( 9\times5 \); M1: \( \sqrt{9}=3 \) correctly evaluated; A1: final answer \( 3\sqrt{5} \). [3]

5 parts \( = 45 \), so 1 part \( = 45 \div 5 = 9 \). Blue counters \( = 3 \) parts \( = 3\times9 = 27 \).

M1: value of one part found, \( 45\div5=9 \); M1: correct method \( 9\times3 \); A1: answer \( 27 \). [3]

Ordering the data: 9, 12, 14, 15, 15, 18, 21. With 7 values, the median is the 4th value: 15 minutes. The range \( = 21 - 9 = 12 \) minutes.

M1: data correctly ordered; A1: median \( = 15 \); A1: range \( = 12 \). [3]

Class width \( = 25-10 = 15 \). Frequency \( = \) frequency density \( \times \) width \( = 4\times15 = 60 \).

M1: class width correctly found, \( 15 \); M1: correct method, frequency density \( \times \) width; A1: answer \( 60 \). [3]

Add 7 to both sides: \( 3x \le 18 \). Divide both sides by 3: \( x \le 6 \). On a number line this is shown with a filled (closed) circle at 6 and an arrow extending to the left.

M1: \( +7 \) applied correctly, \( 3x\le18 \); A1: correct solution \( x\le6 \); B1: correct number-line representation (closed circle at 6, arrow pointing left). [3]

15% of £64 \( = 0.15\times64 = £9.60 \). Sale price \( = 64 - 9.60 = £54.40 \) (or directly \( 64\times0.85 = £54.40 \)).

M1: 15% of 64 correctly calculated, \( £9.60 \) (or use of multiplier 0.85); M1: correct method to find sale price; A1: answer \( £54.40 \). [3]

There are \( 4+5+3=12 \) counters in total. \( P(\text{blue}) = \dfrac{5}{12} \). \( P(\text{not green}) = \dfrac{4+5}{12} = \dfrac{9}{12} = \dfrac{3}{4} \).

A1: \( P(\text{blue}) = 5/12 \); M1: correct numerator \( 9 \) for not green (12 − 3); A1: \( P(\text{not green}) = 3/4 \), correctly simplified. [3]

Substituting \( n=8 \): \( 4(8)-3 = 32-3 = 29 \).

M1: correct substitution \( n=8 \); A1: answer \( 29 \). [2]

Simple interest \( = \dfrac{P\times R\times T}{100} = \dfrac{800\times2.5\times3}{100} = £60 \).

M1: correct method \( \dfrac{P\times R\times T}{100} \) (or equivalent); A1: answer \( £60 \). [2]

We need two numbers that multiply to give \( -24 \) and add to give \( 2 \): these are \( 6 \) and \( -4 \). So \( x^2+2x-24 = (x+6)(x-4) = 0 \). Setting each factor to zero: \( x=-6 \) or \( x=4 \).

M1: correct pair of factors of \( -24 \) that sum to \( 2 \) identified; M1: correctly factorised as \( (x+6)(x-4) \); A1: \( x=-6 \); A1: \( x=4 \). [4]

Adding the two equations eliminates y: \( (2x+y)+(x-y) = 11+1 \Rightarrow 3x = 12 \Rightarrow x=4 \). Substituting into \( x-y=1 \): \( 4-y=1 \Rightarrow y=3 \).

M1: valid elimination or substitution method shown; A1: \( x=4 \); A1: \( y=3 \); B1: solution checked/consistent in both original equations. [4]

Gradient \( = \dfrac{12-4}{5-1} = \dfrac{8}{4} = 2 \). Using point A(1,4): \( 4 = 2(1)+c \Rightarrow c = 2 \). Equation: \( y = 2x+2 \).

M1: correct gradient method \( \dfrac{\Delta y}{\Delta x} \); A1: gradient \( =2 \); M1: correct substitution of a point to find c; A1: \( c=2 \); A1: final equation \( y=2x+2 \). [5]

Gradient of L \( = 3 \). Perpendicular gradient \( = -\dfrac{1}{3} \) (negative reciprocal). Using \( (6,1) \): \( y-1 = -\dfrac{1}{3}(x-6) \Rightarrow y = -\dfrac{1}{3}x+2+1 = -\dfrac{1}{3}x+3 \).

M1: gradient of L identified as \( 3 \); M1: perpendicular gradient correctly found as \( -1/3 \); M1: correct substitution using point \( (6,1) \); A1: correct simplification; A1: final equation \( y=-\tfrac{1}{3}x+3 \) (accept equivalent forms, e.g. \( x+3y=9 \)). [5]

By Pythagoras: height \( = \sqrt{6.5^2-2.5^2} = \sqrt{42.25-6.25} = \sqrt{36} = 6.0 \text{ m} \). Angle with the ground: \( \cos\theta = \dfrac{2.5}{6.5} = 0.3846 \), so \( \theta = \cos^{-1}(0.3846) = 67.4° \) (1 d.p.).

M1: correct Pythagorean setup \( 6.5^2-2.5^2 \); A1: height \( =6.0 \text{ m} \); M1: correct trig ratio (cosine) set up for the angle; M1: correct evaluation of \( \cos^{-1}(2.5/6.5) \); A1: angle \( =67.4° \) (1 d.p.). [5]

Curved surface area \( = 2\pi rh = 2\pi(4)(10) = 251.3 \text{ cm}^2 \). Area of two circular ends \( = 2\pi r^2 = 2\pi(4)^2 = 100.5 \text{ cm}^2 \). Total surface area \( = 251.3+100.5 = 351.9 \approx 352 \text{ cm}^2 \) (3 s.f.).

M1: correct curved surface area formula used; A1: curved SA \( \approx251 \text{ cm}^2 \); M1: correct area of two circular ends found, \( \approx100.5 \text{ cm}^2 \); M1: correct method adding both parts; A1: final answer \( 352 \text{ cm}^2 \) (3 s.f.). [5]

The angle at the centre of a circle is twice the angle at the circumference when subtended by the same arc. So angle \( ACB = \dfrac{130°}{2} = 65° \).

B1: correct theorem named (angle at centre is twice angle at circumference on the same arc); M1: correct method, halving 130°; A1: answer \( 65° \); B1: correct recognition that C lies on the major arc so subtends the same arc AB as the centre angle. [5]

Using the cosine rule: \( AC^2 = AB^2+BC^2-2(AB)(BC)\cos(ABC) = 8^2+10^2-2(8)(10)\cos55° = 64+100-160(0.5736) = 164-91.78 = 72.2 \). \( AC = \sqrt{72.2} = 8.50 \text{ cm} \) (3 s.f.).

M1: correct cosine rule formula stated; M1: correct substitution of values; A1: \( \cos55° \) term correctly evaluated; A1: \( AC^2 \approx72.2 \); A1: \( AC=8.50 \text{ cm} \) (3 s.f.). [5]

Time \( = 2 \text{ h } 15 \text{ min} = 2.25 \text{ h} \). Speed \( = \dfrac{180}{2.25} = 80 \text{ km/h} \). Converting: \( 80 \text{ km/h} = \dfrac{80\times1000}{3600} = 22.2 \text{ m/s} \) (3 s.f.).

M1: time correctly converted to 2.25 h; M1: correct speed formula, distance/time; A1: speed \( =80 \text{ km/h} \); M1: correct conversion method to m/s; A1: answer \( 22.2 \text{ m/s} \) (3 s.f.). [5]

(a) Length \( = x+3 \), so area \( = x(x+3) = 40 \), giving \( x^2+3x-40=0 \). (b) Using the quadratic formula: \( x = \dfrac{-3\pm\sqrt{3^2-4(1)(-40)}}{2(1)} = \dfrac{-3\pm\sqrt{9+160}}{2} = \dfrac{-3\pm\sqrt{169}}{2} = \dfrac{-3\pm13}{2} \). So \( x=5 \) or \( x=-8 \). Since a width cannot be negative, \( x=-8 \) is rejected, so the width is \( 5 \text{ m} \). (c) Length \( = x+3 = 5+3 = 8 \text{ m} \).

M1: correct equation formed, \( x(x+3)=40 \); A1: rearranged to standard form, \( x^2+3x-40=0 \); M1: quadratic formula correctly stated and substituted; A1: discriminant correctly evaluated, \( \sqrt{169}=13 \); A1: both roots found, \( x=5 \) and \( x=-8 \); B1: negative root rejected with a valid reason (width cannot be negative); A1: length correctly stated as \( 8 \text{ m} \). [7]

First find the cone's perpendicular height using Pythagoras: \( h = \sqrt{10^2-6^2} = \sqrt{100-36} = \sqrt{64} = 8 \text{ cm} \). Volume of cone \( = \tfrac{1}{3}\pi(6)^2(8) = \tfrac{1}{3}\pi(288) = 96\pi \text{ cm}^3 \). Setting this equal to the sphere's volume: \( \tfrac{4}{3}\pi R^3 = 96\pi \Rightarrow R^3 = \dfrac{96\times3}{4} = 72 \). \( R = \sqrt[3]{72} = 4.16 \text{ cm} \) (3 s.f.).

M1: correct use of Pythagoras to find the cone's height, \( 8 \text{ cm} \); M1: correct substitution into the cone volume formula; A1: cone volume \( =96\pi \) (or \( 301.6 \text{ cm}^3\)); M1: sphere volume formula set equal to the cone volume; A1: correctly rearranged, \( R^3=72 \); M1: correct cube-root method; A1: final answer \( R=4.16 \text{ cm} \) (3 s.f.). [7]

(a) Arc length \( = \dfrac{140}{360}\times2\pi(9) = \dfrac{140}{360}\times56.55 = 22.0 \text{ cm} \) (3 s.f.). (b) Sector area \( = \dfrac{140}{360}\times\pi(9)^2 = \dfrac{140}{360}\times254.5 = 99.0 \text{ cm}^2 \) (3 s.f.).

B1: correct fraction \( 140/360 \) identified; M1: correct arc length formula substituted; A1: arc length correctly evaluated; A1: arc length \( =22.0 \text{ cm} \) (3 s.f.); M1: correct sector area formula substituted; A1: correctly evaluated; A1: sector area \( =99.0 \text{ cm}^2 \) (3 s.f.). [7]

(a) Multiplying both sides by 3 and dividing by \( \pi(R^2+Rr+r^2) \): \( h = \dfrac{3V}{\pi(R^2+Rr+r^2)} \). (b) \( R^2+Rr+r^2 = 36+24+16 = 76 \). \( h = \dfrac{3\times500}{\pi\times76} = \dfrac{1500}{238.8} = 6.28 \text{ cm} \) (3 s.f.).

M1: correct rearrangement steps (multiply by 3, divide by \( \pi(R^2+Rr+r^2) \)); A1: correct rearranged formula, \( h=\dfrac{3V}{\pi(R^2+Rr+r^2)} \); M1: correct substitution of \( R=6, r=4 \); A1: \( R^2+Rr+r^2=76 \) correctly evaluated; M1: denominator \( \pi\times76 \) correctly evaluated; M1: correct division method; A1: final answer \( h=6.28 \text{ cm} \) (3 s.f.). [7]

\( 2^3=8 \), \( 3^2=9 \), \( \sqrt{49}=7 \). \( 8\times9-7 = 72-7 = 65 \).

M1: \( 2^3=8 \) correctly evaluated; M1: \( 3^2=9 \) and \( \sqrt{49}=7 \) both correctly evaluated; A1: final answer \( 65 \), with correct order of operations. [3]

The common difference is \( 4 \), so the nth term is \( 4n+c \); using the first term, \( 4(1)+c=7 \Rightarrow c=3 \), giving nth term \( =4n+3 \). Setting \( 4n+3=151 \Rightarrow 4n=148 \Rightarrow n=37 \). Since \( n=37 \) is a positive whole number, 151 is indeed a term in the sequence (the 37th term).

M1: common difference correctly identified as \( 4 \); A1: correct nth term expression, \( 4n+3 \); A1: correct conclusion (\( n=37 \), a whole number, so 151 is a term), shown by valid working. [3]

Each y-coordinate has changed sign while each x-coordinate is unchanged: \( (1,1)\to(1,-1) \), \( (3,1)\to(3,-1) \), \( (1,4)\to(1,-4) \). This is a reflection in the x-axis, i.e. the line \( y=0 \).

B1: transformation type correctly identified as a reflection; B1: correct mirror line stated, \( y=0 \) (the x-axis); B1: description is full and precise (single named transformation with correct mirror line, no unnecessary extra transformations). [3]

A rotation of 180° about the origin maps \( (x,y) \) to \( (-x,-y) \). Applying this to \( P(2,3) \) gives \( P'=(-2,-3) \).

M1: correct rule for a 180° rotation about the origin stated or applied, \( (x,y)\to(-x,-y) \); A1: correct coordinates, \( (-2,-3) \); B1: clear, correct reasoning/method shown. [3]

Rounding to the nearest whole centimetre means the actual length could be up to half a centimetre either side of 15 cm. Lower bound \( = 15-0.5 = 14.5 \text{ cm} \). Upper bound \( = 15+0.5 = 15.5 \text{ cm} \).

A1: lower bound \( =14.5 \text{ cm} \); A1: upper bound \( =15.5 \text{ cm} \). [2]

(a) \( 4.5\times10^{-3} = 0.0045 \). (b) \( (2\times10^5)\times(3\times10^{-2}) = (2\times3)\times10^{5+(-2)} = 6\times10^3 \).

A1: correct ordinary number, \( 0.0045 \); M1: coefficients correctly multiplied, \( 2\times3=6 \); M1: powers of 10 correctly combined, \( 10^{5-2}=10^3 \); A1: final answer in correct standard form, \( 6\times10^3 \). [4]

\( \dfrac{\sqrt{72}}{\sqrt{2}} = \sqrt{\dfrac{72}{2}} = \sqrt{36} = 6 \). To rationalise \( \dfrac{5}{\sqrt{5}} \), multiply top and bottom by \( \sqrt{5} \): \( \dfrac{5\sqrt{5}}{5} = \sqrt{5} \).

M1: correct combination of surds, \( \sqrt{72/2}=\sqrt{36} \); A1: first answer, \( 6 \); M1: correct rationalising method, multiplying by \( \sqrt{5}/\sqrt{5} \); A1: second answer, \( \sqrt{5} \). [4]

\( P(\text{1st red}) = \dfrac{6}{10} \). Since one red counter is removed and not replaced, \( P(\text{2nd red}) = \dfrac{5}{9} \). \( P(\text{both red}) = \dfrac{6}{10}\times\dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3} \).

M1: correct first probability, \( 6/10 \); M1: correct second probability without replacement, \( 5/9 \); M1: correct multiplication of the two probabilities; A1: final answer, simplified to \( 1/3 \). [4]

(a) \( P(\text{not rain}) = 1-0.3 = 0.7 \). (b) Exactly one rainy day occurs either as (rain, then not rain) or (not rain, then rain): \( P = (0.3\times0.7)+(0.7\times0.3) = 0.21+0.21 = 0.42 \).

A1: \( P(\text{not rain})=0.7 \); M1: both relevant cases identified (rain-then-not, not-then-rain); M1: each case correctly calculated (0.21 each) or a valid combined method used; A1: final answer, \( 0.42 \). [4]

(a) Relative frequency \( = \dfrac{35}{150} = 0.2\overline{3} \approx0.233 \). (b) Expected number of sixes in 480 rolls \( = 0.2\overline{3}\times480 = 112 \).

M1: correct relative frequency method, \( 35/150 \); A1: relative frequency \( \approx0.233 \); M1: correct scaling method, \( \times480 \); A1: final answer, \( 112 \) (allow follow-through from candidate's relative frequency). [4]

Substitute \( y=x+1 \) into the circle equation: \( x^2+(x+1)^2=13 \). Expanding: \( x^2+x^2+2x+1=13 \Rightarrow 2x^2+2x-12=0 \Rightarrow x^2+x-6=0 \). Factorising: \( (x+3)(x-2)=0 \), so \( x=-3 \) or \( x=2 \). Since two distinct real solutions for x exist, the line intersects the circle at two points. Using \( y=x+1 \): when \( x=-3 \), \( y=-2 \); when \( x=2 \), \( y=3 \). The points of intersection are \( (-3,-2) \) and \( (2,3) \).

M1: correct substitution of \( y=x+1 \) into the circle equation; A1: \( (x+1)^2 \) correctly expanded; A1: correctly simplified to \( x^2+x-6=0 \); M1: correctly factorised, \( (x+3)(x-2) \); A1: both x-values found, \( x=-3, x=2 \); A1: both y-values correctly found via \( y=x+1 \); B1: both coordinate pairs correctly stated, confirming two distinct intersection points. [7]

(a) Opposite angles of a cyclic quadrilateral sum to 180°: \( (2x+15)+(3x-5)=180 \Rightarrow 5x+10=180 \Rightarrow 5x=170 \Rightarrow x=34 \). (b) Angle DAB \( = 2(34)+15 = 68+15 = 83° \). (c) Since AB is a diameter, angle ADB \( = 90° \), because the angle in a semicircle (subtended by a diameter) is always a right angle.

B1: correct circle theorem stated (opposite angles of a cyclic quadrilateral sum to 180°); M1: correct equation formed, \( (2x+15)+(3x-5)=180 \); A1: correctly simplified, \( 5x+10=180 \); A1: \( x=34 \); M1: correct substitution to find angle DAB; A1: angle DAB \( =83° \); B1: angle ADB \( =90° \) correctly stated; B1: correct reason given (angle in a semicircle). [8]

Multiplying both sides by 5: \( 2x+3 = 35 \). Subtracting 3: \( 2x = 32 \). Dividing by 2: \( x = 16 \).

M1: correct method, multiplying both sides by 5; A1: answer \( x=16 \). [2]

The locus is the perpendicular bisector of the line segment AB. To construct it: open the compasses to a radius greater than half of AB, draw an arc centred at A above and below the line AB, then draw an arc of the same radius centred at B, so that the arcs intersect at two points, one on each side of AB; the straight line joining these two intersection points is the perpendicular bisector.

B1: correct locus named, the perpendicular bisector of AB; B1: correct construction method described (equal-radius arcs from A and B intersecting on both sides, joined by a straight line). [2]

Sum of interior angles \( = (n-2)\times180° = (10-2)\times180° = 1440° \). Each interior angle of a regular decagon \( = 1440°\div10 = 144° \).

M1: correct method, sum of interior angles \( (n-2)\times180 \), divided by \( n \) (or equivalent formula); A1: answer \( 144° \). [2]

Area scale factor \( = (\text{linear scale factor})^2 = 3^2 = 9 \). Area of shape B \( = 8\times9 = 72 \text{ cm}^2 \).

M1: correct area scale factor identified, \( 9 \) (i.e. \( 3^2 \)); A1: answer \( 72 \text{ cm}^2 \). [2]

Actual distance \( = 8.4\times25\,000 = 210\,000 \text{ cm} \). Converting to metres: \( 210\,000\div100 = 2100 \text{ m} \). Converting to kilometres: \( 2100\div1000 = 2.1 \text{ km} \).

M1: correct scaling method, \( 8.4\times25\,000 \); M1: correct conversion from cm to km (\( \div100\,000 \), or via metres); A1: final answer, \( 2.1 \text{ km} \). [3]

(a) On the x-axis, \( y=0 \): \( 2x=12 \Rightarrow x=6 \), giving \( (6,0) \). (b) On the y-axis, \( x=0 \): \( 3y=12 \Rightarrow y=4 \), giving \( (0,4) \).

M1: correct method for the x-intercept (\( y=0 \)); M1: correct method for the y-intercept (\( x=0 \)); A1: both coordinate pairs correctly stated, \( (6,0) \) and \( (0,4) \). [3]

Total parts \( = 2+3+4 = 9 \). Value of 1 part \( = 360\div9 = £40 \). Cara receives \( 4 \) parts \( = 4\times40 = £160 \).

M1: total number of parts correctly identified, \( 9 \); M1: value of one part correctly found, \( £40 \); A1: Cara's share correctly found, \( £160 \). [3]

Let h be the height of the flagpole. \( \tan32° = \dfrac{h}{25} \), so \( h = 25\tan32° = 25\times0.6249 = 15.6 \text{ m} \) (1 d.p.).

M1: correct trigonometric ratio identified (tangent); M1: correct equation set up, \( \tan32°=h/25 \); A1: \( \tan32° \) correctly evaluated; M1: correct rearrangement/multiplication; A1: final answer \( 15.6 \text{ m} \) (1 d.p.). [5]

By the cosine rule: \( PR^2 = PQ^2+QR^2-2(PQ)(QR)\cos(PQR) \). \( 144 = 49+81-2(7)(9)\cos\theta \Rightarrow 144=130-126\cos\theta \Rightarrow 126\cos\theta = 130-144=-14 \Rightarrow \cos\theta = -0.1111 \). \( \theta = \cos^{-1}(-0.1111) = 96.4° \) (1 d.p.).

M1: cosine rule correctly rearranged to find an angle; M1: correct substitution of all three side lengths; A1: \( \cos\theta = -0.111 \) correctly found; M1: correct use of \( \cos^{-1} \) with the negative value; A1: final answer \( 96.4° \) (1 d.p.). [5]

Two orders give exactly one milk chocolate: milk-then-dark, or dark-then-milk. \( P(\text{milk, then dark}) = \dfrac{5}{12}\times\dfrac{7}{11} = \dfrac{35}{132} \). \( P(\text{dark, then milk}) = \dfrac{7}{12}\times\dfrac{5}{11} = \dfrac{35}{132} \). Total \( = \dfrac{35}{132}+\dfrac{35}{132} = \dfrac{70}{132} = \dfrac{35}{66} \).

M1: correct first-branch probabilities, \( 5/12 \) and \( 7/12 \); M1: correct second-branch (conditional, without replacement) probabilities shown; A1: both relevant path probabilities correctly calculated, \( 35/132 \) each; M1: correct addition of both paths; A1: final answer, simplified to \( 35/66 \). [5]

Linear scale factor \( = \dfrac{PQ}{AB} = \dfrac{9}{6} = 1.5 \). Area scale factor \( = 1.5^2 = 2.25 \). Area of PQR \( = 24\times2.25 = 54 \text{ cm}^2 \).

M1: correct linear scale factor method; A1: linear scale factor \( =1.5 \); M1: correct area scale factor method (linear SF squared); A1: area scale factor \( =2.25 \) correctly used; A1: final area of PQR \( =54 \text{ cm}^2 \). [5]

(a) Since y is inversely proportional to \( x^2 \): \( y = \dfrac{k}{x^2} \). Substituting \( x=3, y=8 \): \( 8 = \dfrac{k}{9} \Rightarrow k=72 \). So \( y = \dfrac{72}{x^2} \). (b) When \( x=4 \): \( y = \dfrac{72}{16} = 4.5 \). (c) When \( y=0.5 \): \( 0.5 = \dfrac{72}{x^2} \Rightarrow x^2 = \dfrac{72}{0.5} = 144 \Rightarrow x = 12 \) (taking the positive root).

M1: correct proportionality statement, \( y=k/x^2 \); M1: correct substitution to find k; A1: \( k=72 \) confirmed; M1: correct substitution for part (b); A1: \( y=4.5 \) for part (b); M1: correct rearrangement for part (c), \( x^2=144 \); A1: \( x=12 \) for part (c). [7]

(a) \( x=0: y=7 \); \( x=1: y=1-4+7=4 \); \( x=2: y=4-8+7=3 \); \( x=3: y=9-12+7=4 \); \( x=4: y=16-16+7=7 \). (b) At \( x=2.5: y=6.25-10+7=3.25 \). At \( x=3.5: y=12.25-14+7=5.25 \). Gradient estimate \( = \dfrac{5.25-3.25}{3.5-2.5} = \dfrac{2}{1} = 2 \). (c) Since the coefficient of \( x^2 \) is positive, the parabola opens upwards, so the curve has a minimum turning point. Completing the square: \( x^2-4x+7 = (x-2)^2-4+7 = (x-2)^2+3 \). The minimum occurs when \( (x-2)^2=0 \), i.e. at \( x=2, y=3 \), giving the turning point \( (2,3) \).

A1: all five table values correct (7, 4, 3, 4, 7); M1: y-values at \( x=2.5 \) and \( x=3.5 \) correctly calculated (3.25 and 5.25); M1: correct gradient method, \( \Delta y/\Delta x \); A1: gradient estimate \( =2 \); B1: correctly identified as a minimum, with a valid reason (positive \( x^2 \) coefficient/upward parabola); M1: correct completed-square form, \( (x-2)^2+3 \); A1: correct turning point coordinates, \( (2,3) \); B1: overall coherent, consistent working across all three parts. [8]