Download AQA AS MATHEMATICS 7356/2 Paper 2 Version: 1.0 Final Question Paper & Mark scheme [MERGED] and more Exams Mathematics in PDF only on Docsity!
AQA AS MATHEMATICS
7356/2 Paper 2 Version 1
Question 1: Solve the equation 3x2โ5xโ2=03x^2 - 5x - 2 = 03x2โ5xโ2=0. Answer: Using the quadratic formula x=โbยฑb2โ4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2aโbยฑb2โ4ac: a=3a = 3a=3, b=โ5b = -5b=โ5, c=โ2c = -2c=โ Discriminant: b2โ4ac=(โ5)2โ4(3)(โ2)=25+24=49b^2 - 4ac = (-5)^2 - 4(3)(-2) = 25 + 24 = 49b2โ4ac=(โ5)2โ4(3)(โ2)=25+24= Roots: x=โ(โ5)ยฑ492(3)=5ยฑ76x = \frac{-(-5) \pm \sqrt{49}}{2(3)} = \frac{5 \pm 7} {6}x=2(3)โ(โ5)ยฑ49=65ยฑ x=5+76=2x = \frac{5 + 7}{6} = 2x=65+7=
x=5โ76=โ13x = \frac{5 - 7}{6} = -\frac{1}{3}x=65โ7=โ Solution: x=2x = 2x=2 and x=โ13x = -\frac{1}{3}x=โ31. Question 2: Differentiate f(x)=x3โ4x2+6xโ5f(x) = x^3 - 4x^2 + 6x - 5f(x)=x3โ4x2+6xโ5. Answer: Using f (x)=ddx(x3โ4x2+6xโ5)f'(x) = \frac{d}{dx}(x^3 - 4x^2 + 6x - 5)f (x)=dxdโฒ โฒ (x3โ4x2+6xโ5): f (x)=3x2โ8x+6f'(x) = 3x^2 - 8x + 6f (x)=3x2โ8x+6โฒ โฒ Solution: f (x)=3x2โ8x+6f'(x) = 3x^2 - 8x + 6f (x)=3x2โ8x+6.โฒ โฒ Question 3: Evaluate โซ(3x2โ4x+1) dx\int (3x^2 - 4x + 1) , dxโซ(3x2โ4x+1)dx. Answer: Using integration: โซ3x2 dx=x3\int 3x^2 , dx = x^3โซ3x2dx=x โซโ4x dx=โ2x2\int -4x , dx = -2x^2โซโ4xdx=โ2x โซ1 dx=x\int 1 , dx = xโซ1dx=x Solution: x3โ2x2+x+Cx^3 - 2x^2 + x + Cx3โ2x2+x+C, where CCC is the constant of integration. Question 4: Solve sin(2x)=3/2\sin(2x) = \sqrt{3}/2sin(2x)=3/2 for 0โคxโคฯ0 \leq x \leq \pi0โคxโคฯ. Answer: sin(2x)=3/2\sin(2x) = \sqrt{3}/2sin(2x)=3/2 implies 2x=ฯ3,2ฯ32x = \frac{\pi}{3},
frac{2\pi}{3}2x=3ฯ,32ฯ. Dividing by 2: x=ฯ6,ฯ3x = \frac{\pi}{6}, \frac{\pi}{3}x=6ฯ,3ฯ. Solution: x=ฯ6,ฯ3x = \frac{\pi}{6}, \frac{\pi}{3}x=6ฯ,3ฯ. Section B: Statistics Question 5: The heights of students are normally distributed with a mean of 170 cm170 ,
text{cm}170cm and a standard deviation of 10 cm10 , \text{cm}10cm. Find the
Using s=ut+12at2s = ut + \frac{1}{2}at^2s=ut+21at2: u=0u = 0u=0, v=20v = 20v=20, t=10t = 10t=10, a=vโut=2010=2a = \frac{v-u}{t} = \frac{20}{10} = 2a=tvโu=1020=2. s=0 10+12 2 102=100s = 0 \cdot 10 + \frac{1}{2} \cdot 2 \cdot 10^2 = 100s=0 10+21โ
โ
โ
โ
โ
โ
2 102=100. Solution: Distance =100 m= 100 , \text{m}=100m. Question 9: Find the equation of the tangent to the curve y=x2โ4x+7y = x^2 - 4x + 7y=x2โ4x+ at the point where x=3x = 3x=3. Answer:
Differentiate y=x2โ4x+7y = x^2 - 4x + 7y=x2โ4x+7: dydx=2xโ4\frac{dy}{dx} = 2x - 4dxdy=2xโ4.
Slope at x=3x = 3x=3: dydx=2(3)โ4=6โ4=2\frac{dy}{dx} = 2(3) - 4 = 6 - 4 = 2dxdy=2(3)โ4=6โ4=2.
Coordinates of the point: When x=3x = 3x=3, y=(3)2โ4(3)+7=9โ12+7=4y = (3)^2 - 4(3) + 7 = 9 - 12 + 7 = 4y=(3)2โ4(3)+7=9โ12+7=4.
Equation of the tangent (using yโy1=m(xโx1)y - y_1 = m(x - x_1)yโy =m(xโx1)): yโ4=2(xโ3)y - 4 = 2(x - 3)yโ4=2(xโ3). Simplify: y=2xโ6+4y = 2x - 6 + 4y=2xโ6+4. y=2xโ2y = 2x - 2y=2xโ2.
Solution: The equation of the tangent is y=2xโ2y = 2x - 2y=2xโ2. Question 10: Expand (2xโ3)3(2x - 3)^3(2xโ3)3 fully. Answer: Using the binomial theorem: (aโb)3=a3โ3a2b+3ab2โb3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3(aโb)3=a3โ3a2b+3ab2โb Substitute a=2xa = 2xa=2x and b=3b = 3b=3: (2xโ3)3=(2x)3โ3(2x)2(3)+3(2x)(32)โ(3)3(2x - 3)^3 = (2x)^3 - 3(2x)^2(3) + 3(2x)(3^2) - (3)^3(2xโ3)3=(2x)3โ3(2x)2(3)+3(2x) (32)โ(3)3 =8x3โ36x2+54xโ27= 8x^3 - 36x^2 + 54x - 27=8x3โ36x2+54xโ Solution: (2xโ3)3=8x3โ36x2+54xโ27(2x - 3)^3 = 8x^3 - 36x^2 + 54x - 27(2xโ3)3=8x3โ36x2+54xโ27. Question 11: Solve the inequality 2x2โ5xโ3<02x^2 - 5x - 3 < 02x2โ5xโ3<0. Answer:
Question 12: Find the sum of the first 20 terms of the arithmetic sequence where a=5a = 5a=5 and d=3d = 3d=3. Answer: Using the formula Sn=n2(2a+(nโ1)d)S_n = \frac{n}{2}(2a + (n-1)d)Sn=2n (2a+(nโ1)d): S20=202(2(5)+(20โ1)(3))=10(10+57)=10โ
67=670S_{20} =
frac{20}{2}(2(5) + (20-1)(3)) = 10(10 + 57) = 10 \cdot 67 = 670S20=220(2(5)+(20โ1)(3))=10(10+57)=10โ
67= Solution: The sum of the first 20 terms is 670670670. Section B: Statistics (Continued) Question 13: A die is rolled 6 times. Find the probability of rolling exactly two sixes. Answer: Binomial distribution: P(X=2)=(62)(p)2(1โp)4P(X = 2) = \binom{6}{2} (p)^2(1-p)^4P(X=2)=(26)(p)2(1โp)4, where p=16p = \frac{1}{6}p=61: P(X=2)=(62)(16)2(56)4P(X = 2) = \binom{6}{2} \left(\frac{1}{6}
right)^2 \left(\frac{5}{6}\right)^4P(X=2)=(26)(61)2(65)4 (62)=6! 2!(6โ2)!=15\binom{6}{2} = \frac{6!}{2!(6-2)!} = 15 (26)=2! (6โ2)!6!=15 P(X=2)=15โ
136โ
6251296=937546656โ0.201P(X = 2) = 15 \cdot \frac{1}{36} \cdot \frac{625}{1296} = \frac{9375} {46656} \approx 0.201P(X=2)=15โ
361โ
1296625= โ0. Solution: Probability โ0.201\approx 0.201โ0.201.
Question 14: The mean of a data set is 12, and the standard deviation is 3. A new data set is formed by doubling each value. Find the new mean and standard deviation. Answer:
- New mean: 2ร12=242 \times 12 = 242ร12=24.
- New standard deviation: 2ร3=62 \times 3 = 62ร3=6. Solution: New mean = 242424, new standard deviation = 666. Section C: Mechanics (Continued) Question 15: A particle is projected with an initial velocity of 15 m/s15 , \text{m/s}15m/s at an angle of 30 30^\circ30โ โto the horizontal. Find the time of flight. (Take g=9.8 m/s2g = 9.8 , \text{m/s}^2g=9.8m/s2). Answer: Using t=2usinฮธgt = \frac{2u\sin\theta}{g}t=g2usinฮธ: t=2(15)sin(30โ)9.8=30โ
0.59.8=159.8โ1.53 st = \frac{2(15)
sin(30^\circ)}{9.8} = \frac{30 \cdot 0.5}{9.8} = \frac{15}{9.8}
approx 1.53 , \text{s}t=9.82(15)sin(30โ)=9.830โ
0.5=9.815โ1.53s Solution: Time of flight โ1.53 s\approx 1.53 , \text{s}โ1.53s. Question 16: A block of mass 5 kg5 , \text{kg}5kg rests on a rough plane inclined at 30 30^\โ circ30 โto the horizontal. The coefficient of friction is 0.40.40.4. Find the force required to move the block up the plane. (Take g=9.8 m/s2g = 9.8 , \text{m/s}^2g=9.8m/s2).
Find the exact value of โซ02(3x2โ4x+1) dx\int_0^2 (3x^2 - 4x + 1) , dxโซ (3x2โ4x+1)dx. Answer: Integrate term by term: โซ(3x2โ4x+1) dx=x3โ2x2+x+C\int (3x^2 - 4x + 1) , dx = x^3 - 2x^2 + x + Cโซ(3x2โ4x+1)dx=x3โ2x2+x+C Evaluate from x=0x = 0x=0 to x=2x = 2x=2: [x3โ2x2+x]02=(23โ2(22)+2)โ(03โ2(02)+0)\left[ x^3 - 2x^2 + x
right]_0^2 = \left( 2^3 - 2(2^2) + 2 \right) - \left( 0^3 - 2(0^2) + 0 \right)[x3โ2x2+x]02=(23โ2(22)+2)โ(03โ2(02)+0) =(8โ8+2)โ(0)=2= (8 - 8 + 2) - (0) = 2=(8โ8+2)โ(0)= Solution: 222. Question 19: Solve ln(x+2)=1\ln(x + 2) = 1ln(x+2)=1. Answer:
Exponentiate both sides: ln (x+2)=1 โนeln(x+2)=e1\ln(x + 2) = 1 \implies e^{\ln(x + 2)} = e^1ln(x+2)=1 โนeln(x+2)=e1.
Simplify: x+2=ex + 2 = ex+2=e.
Solve for xxx: x=eโ2x = e - 2x=eโ2.
Solution: x=eโ2x = e - 2x=eโ2. Question 20: Find the modulus and argument of the complex number z=3+4iz = 3 + 4iz=3+4i. Answer:
Modulus: โฃ โฃ z =Re(z)2+Im(z)2=32+42=9+16=5|z| = \sqrt{\text{Re}(z)^2 + \text{Im} (z)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 โฃ โฃz =Re(z)2+Im(z)2 =32+ =9+16=5.
Argument: arg (z)=tanโ1(Im(z)Re(z))=tanโ1(43)\arg(z) = \tan^{-1}\left(\frac{\text{Im} (z)}{\text{Re}(z)}\right) = \tan^{-1}\left(\frac{4}{3}
right)arg(z)=tanโ1(Re(z)Im(z))=tanโ1(34).
Question 22: A sample of 50 students has a mean score of 707070, and a standard deviation of 888. Use the normal approximation to estimate the number of students scoring above
Answer:
Standardize x=78x = 78x=78: Z=xโฮผฯ=78โ708=1Z = \frac{x - \mu}{\sigma} = \frac{78 - 70}{8} = 1Z=ฯxโฮผ=878โ70=1.
From standard normal tables: P(Z>1)=1โP(Zโค1)โ1โ0.8413=0.1587P(Z > 1) = 1 - P(Z \leq 1) \approx 1 - 0.8413 = 0.1587P(Z>1)=1โP(Zโค1)โ1โ0.8413=0.1587.
Expected number of students: 0.1587ร50=7.9350.1587 \times 50 = 7.9350.1587ร50=7.935.
Solution: Approximately 888 students. Section C: Mechanics (Continued) Question 23:
A car of mass 1000 kg1000 , \text{kg}1000kg accelerates from rest to 20 m/s20 ,
text{m/s}20m/s in 5 s5 , \text{s}5s. Find the force exerted by the engine. Answer:
Acceleration: a=vโut=20โ05=4 m/s2a = \frac{v - u}{t} = \frac{20 - 0}{5} = 4 , \text{m/s}^2a=tvโu=520โ0=4m/s2.
Force: F=ma=1000 4=4000 NF = ma = 1000 \cdot 4 = 4000 , \โ
text{N}F=ma=1000 4=4000N.โ
Solution: 4000 N4000 , \text{N}4000N. Question 24: A ball is dropped from a height of 20 m20 , \text{m}20m. Find the time it takes to hit the ground. (Take g=9.8 m/s2g = 9.8 , \text{m/s}^2g=9.8m/s2). Answer: Using s=12gt2s = \frac{1}{2}gt^2s=21gt2: 20=12(9.8)t220 = \frac{1}{2}(9.8)t^220=21(9.8)t2 20=4.9t2 โน t2=204.9โ4.0820 = 4.9t^2 \implies t^2 = \frac{20}{4.9} \approx 4.0820=4.9t2โนt2=4.920โ4.08 tโ4.08โ2.02 st \approx \sqrt{4.08} \approx 2.02 , \text{s}tโ4.08โ2.02s
Express 4cos2(x)โ14\cos^2(x) - 14cos2(x)โ1 in terms of cos(2x)\cos(2x)cos(2x). Answer: Using the double-angle identity cos(2x)=2cos2(x)โ1\cos(2x) = 2\cos^2(x) - 1cos(2x)=2cos2(x)โ1: 4cos2(x)โ1=2(2cos2(x))โ1=2cos(2x)4\cos^2(x) - 1 = 2(2\cos^2(x))
- 1 = 2\cos(2x)4cos2(x)โ1=2(2cos2(x))โ1=2cos(2x) Solution: 4cos2(x)โ1=2cos(2x)4\cos^2(x) - 1 = 2\cos(2x)4cos2(x)โ1=2cos(2x). Question 27: Find the area under the curve y=x3+2x2y = x^3 + 2x^2y=x3+2x2 between x=1x = 1x=1 and x=2x = 2x=2. Answer:
- Integrate y=x3+2x2y = x^3 + 2x^2y=x3+2x2: โซ(x3+2x2) dx=x44+2x33+C\int (x^3 + 2x^2) , dx = \frac{x^4} {4} + \frac{2x^3}{3} + Cโซ(x3+2x2)dx=4x4+32x3+C
- Evaluate: [x44+2x33]12=(244+2(23)3)โ(144+2(13)3)\left[ \frac{x^4}{4}
- \frac{2x^3}{3} \right]_1^2 = \left( \frac{2^4}{4} +
frac{2(2^3)}{3} \right) - \left( \frac{1^4}{4} + \frac{2(1^3)} {3} \right)[4x4+32x3]12=(424+32(23))โ(414+32(13)) =(4+163)โ(14+23)= \left( 4 + \frac{16}{3} \right) - \left( \frac{1} {4} + \frac{2}{3} \right)=(4+316)โ(41+32) =123+163โ312โ812=283โ1112=9512= \frac{12}{3} +
frac{16}{3} - \frac{3}{12} - \frac{8}{12} = \frac{28}{3} -
frac{11}{12} = \frac{95}{12}=312+316โ123โ128=328โ =
Solution: 9512\frac{95}{12}1295. Question 28: Solve x3โ3x2+4=0x^3 - 3x^2 + 4 = 0x3โ3x2+4=0. Answer:
Use synthetic or trial division to find a root. Test x=1x = 1x=1: 13โ3(12)+4=1โ3+4=21^3 - 3(1^2) + 4 = 1 - 3 + 4 = 213โ3(12)+4=1โ3+4= (not a root). Test x=โ1x = -1x=โ1: (โ1)3โ3(โ1)2+4=โ1โ3+4=0(-1)^3 - 3(-1)^2 + 4 = -1 - 3 + 4 = 0(โ1)3โ3(โ1)2+4=โ1โ3+4=0 (x=โ1x = -1x=โ1 is a root).
Factorize: Divide x3โ3x2+4x^3 - 3x^2 + 4x3โ3x2+4 by x+1x + 1x+1: Quotient: x2โ4x+4x^2 - 4x + 4x2โ4x+4.
x3โ3x2+4=(x+1)(x2โ4x+4)x^3 - 3x^2 + 4 = (x + 1)(x^2 - 4x + 4)x3โ3x2+4=(x+1)(x2โ4x+4).
Factor further: x2โ4x+4=(xโ2)2x^2 - 4x + 4 = (x - 2)^2x2โ4x+4=(xโ2)2.
Arrange the data (already sorted): 3,7,7,10,143, 7, 7, 10, 143,7,7,10,14.
Median: 777 (middle value).
Lower quartile (Q1Q_1Q1): Median of 3,73, 73,7: Q1=5Q_1 = 5Q1=5.
Upper quartile (Q3Q_3Q3): Median of 10,1410, 1410,14: Q3=12Q_3 = 12Q3=12.
IQR: Q3โQ1=12โ5=7Q_3 - Q_1 = 12 - 5 = 7Q3โQ1=12โ5=7.
Solution: IQR = 777. Section C: Mechanics (Continued) Question 31: A particle moves along a straight line such that its displacement s(t)=t3โ6t2+9ts(t) = t^3 - 6t^2 + 9ts(t)=t3โ6t2+9t. Find its velocity and acceleration at t=2t = 2t=2. Answer:
Velocity: v(t)=dsdt=3t2โ12t+9v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9v(t)=dtds =3t2โ12t+9. At t=2t = 2t=2: v(2)=3(22)โ12(2)+9=12โ24+9=โ3 m/sv(2) = 3(2^2) - 12(2) + 9 = 12 - 24 + 9 = -3 , \text{m/s}v(2)=3(22)โ12(2)+9=12โ24+9=โ3m/s.
Acceleration: a(t)=dvdt=6tโ12a(t) = \frac{dv}{dt} = 6t - 12a(t)=dtdv=6tโ12. At t=2t = 2t=2: a(2)=6(2)โ12=12โ12=0 m/s2a(2) = 6(2) - 12 = 12 - 12 = 0 , \text{m/s}^2a(2)=6(2)โ12=12โ12=0m/s2.
Solution: Velocity =โ3 m/s= -3 , \text{m/s}=โ3m/s, Acceleration =0 m/s2= 0 ,
text{m/s}^2=0m/s2. Question 32: A projectile is fired at a speed of 50 m/s50 , \text{m/s}50m/s at an angle of 60 60^\โ circ60 โto the horizontal. Find the maximum height reached. (Take g=9.8 m/s2g = 9.8 , \text{m/s}^2g=9.8m/s2). Answer:
Vertical component of velocity: uy=50sin(60 )=50 32=253 m/su_y = 50\sin(60^\circ) = 50 \cdot \frac{\โ โ
sqrt{3}}{2} = 25\sqrt{3} , \text{m/s}uy=50sin(60 )=50 23โ โ
=253m/s.
Maximum height: Using h=uy22gh = \frac{u_y^2}{2g}h=2guy2:
- h=(253)22 9.8=625 319.6=187519.6โ95.66 mh = \frac{(25\sqrt{3})^2}โ
โ
{2 \cdot 9.8} = \frac{625 \cdot 3}{19.6} = \frac{1875}{19.6} \approx 95.66 , \text{m}h=2 9.8(253โ
)2=19.6625 3โ
=19.61875โ95.66m Solution: Maximum height โ95.66 m\approx 95.66 , \text{m}โ95.66m.
