1. A sector has a perimeter of 22 cm and an angle of θ radians. The radius is 8 cm.
(a) Find the value of θ[2 marks] 16+8θ=22⇒θ=43
(b) Find the area of the sector. [2 marks] Using A=21r2θ,
A=21×82×43=24
2. A triangle ABC has an angle BAC of 120∘. There are sides of 6 cm and 16 cm.
(a) Show that the angle ACB is 19∘ to the nearest degree [3 marks] Using the sine rule: 6sin(ACB)=16sin120⇒ACB=arcsin(166sin120)=18.95…, which is 19∘ to the nearest degree.
(b) Find the area of the triangle. [3 marks] Other angle is 180−(120+19)=41∘. Hence the area is 21×16×6×sin41=31.5cm2
3. (a) Express 32x−127x in the form 3p where p is an expression in x. [3 marks]
32x−127x=32x−1323x=31−2x
(b) Hence solve the equation 32x−127x=381. [2 marks]
31−2x=334⇒1−2x=34⇒x=−32
4. A geometric series has the nth term un=162(32)n. (a) Find u1 and u2. [2 marks] u1=108,u2=72.
(b) Find the sum to infinity of the series [3 marks] First term a=108, and common ratio r=32. Hence sum to infinity is 1−ra=1−32108=324
(c) Find the smallest value of k such that ∑n=k∞un<2.5. [3 marks] ∑n=k∞un=486(32)k
(32)k<4862.5⇒k>log324862.5. The signs change here as log32x is a decreasing function so if a<b, then log32a>log32b.
Hence k>12.997 so k is at least 13.
5. A curve satisfies dxdy=x23−2x, where x>0. (a) Show that there is only one value of x for which there is a stationary point. [2 marks] At stationary points, dxdy=0⇒x(x21−2)=0⇒x1/2=2⇒x=4 (discarding x=0 as x>0). Hence there is only one value of x for which there is a stationary point.
(b) Find dx2d2y and show that this is a minimum point. [3 marks] dx2d2y=23x−2. When x=4, this is equal to 1. Hence as the second derivative is positive, the point is a minimum.
(c) The line y=2 is a tangent to the curve. Find the equation of the curve. [4 marks] Must be a tangent at the stationary point. Hence the point (4, 2) lies on the curve (the stationary point). Integrating gives y=52x25−x2+c. As y(4)=2, c=526 so y=52x25−x2+526.
6. There is a curve y=23x. (a)(i) Use the trapezium rule with five ordinates (four strips) to approximate ∫0123xdx to two decimal places. [4 marks] Call the integral I. Then I≈21h(y0+y4+2(y1+y2+y3)).
Now, h=41 and yr=243r.
Hence I≈81(20+23+2(243+223+249))=3.44, to two decimal places.
(a) (ii) State how the approximation may be improved. [1 mark] Use more strips/ordinates.
(a) (iii) The point (1, k) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line x=0 and the line y=k, to two decimal places. [3 marks] k=8 so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is 8−3.44=4.56 to two decimal places.
(b) The curve y=23x may be mapped onto the curve of y=23x−4 by a translation and a stretch. Describe: (i) the translation [2 marks]
Translation in the vector (034)
(ii) the stretch [2 marks]
Stretch parallel to the y-axis scale factor 161.
(c) Use logarithms to solve the equation 23x−4=7, giving your answer to three significant figures. [2 marks]
x=31(4+log27)=2.27 to three significant figures.
7. A curve has the equation y=7x+6−x21. (a) The region bounded by the curve and the lines x=1 and x=2 is above the x-axis. Show that the area of this region is 16. [5 marks]
Area (as curve is above the x-axis) is: ∫127x+6−x21dx=[27x2+6x+x1]12 =(27(22)+6(2)+21)−(27(12)+6(1)+1) =(14+12+21)−(27+6+1) =253−221 =232=16
(b) The normal to the curve is parallel to the line 2y+8x=3. Find the equation of the normal to the curve. [6 marks] Differentiating gives dxdy=7+x32. Now, the normal has a gradient of −4. Hence the gradient of the tangent to the curve is 41. So
7+x32=41⇒x38=−27⇒x=−32. Hence y=−1211.
So the equation of the normal is y+1211=−4(x+32).
8. (a) Solve the equation cosθ=32 in the range 0∘⩽θ⩽360∘. Give your answers to the nearest degree. [2 marks]
θ={arccos32,360−arccos32}={48∘,312∘}, to the nearest degree.
(b)(i) Given that 4sinθtanθ=4−cosθ, show that 3cos2θ+4cosθ−4=0. [3 marks]
4sinθcosθsinθ=4−cosθ
4cosθsin2θ=4−cosθ
4sin2θ=4cosθ−cos2θ
4−4cos2θ=4cosθ−cos2θ
3cos2θ+4cosθ−4=0
(ii) Show that there is only one possible value of cosθ. [2 marks] Factorising gives (3cosθ−2)(cosθ+2)=0. As ∣cosθ∣⩽1, the only possible value of cosθ is 32.
(c) Hence solve the equation 4sin4xtan4x=4−cos4x in the range 0∘⩽x⩽180∘, giving your answer to the nearest degree. [4 marks]
Solving cos4x=32 in the range 0∘⩽4x⩽720∘.
Use part (a) and then add on 360∘ to both solutions to give 4x={48∘,312∘,408∘,672∘}, to the nearest degree.
So x={12∘,78∘,102∘,168∘}, to the nearest degree.
9. Given that 3log2(c+2)−log2(2c3+k)=1, express (c+1)2 in terms of k. [7 marks]
(a) (iii) The point (1, ) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line and the line , to two decimal places. [3 marks] so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is to two decimal places.
(a) (iii) The point (1, ) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line and the line , to two decimal places. [3 marks] so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is to two decimal places.
Do you think slightly different methods for 7 will still be fully credited? I did the binomial expansion while it was still in log form and found an equation equal to k-1, then factored out 3
Do you think slightly different methods for 7 will still be fully credited? I did the binomial expansion while it was still in log form and found an equation equal to k-1, then factored out 3
I think that's fine, whether it is in log form or not
1. A sector has a perimeter of 22cm and an angle of θ radians. The radius is 8cm. (a) Find the value of θ[2 marks] 16+8θ=22 θ=43
(b) Find the area of the sector [2 marks] 21×82×43=24
2. A triangle ABC has an angle BAC of
Unparseable latex formula:
120\textdegree
. There are sides of 6cm and 16cm.
(a) Show that the angle ACB is
Unparseable latex formula:
19\textdegree
to the nearest degree [3 marks] 6sin(ACB)=16sin120⇒ACB=arcsin(166sin120)=18.95…, which is 19 degrees to the nearest degree.
(b) Find the area of the triangle. [3 marks] Other angle is 41 degrees. Hence the area is 21×16×6×sin41=31.5cm2
3. (a) Express 32x−127x in the form 3p where p is an expression in x. [3 marks]
32x−127x=32x−1323x=31−2x
(b) Hence solve the equation 32x−127x=381. [2 marks]
31−2x=334⇒1−2x=34⇒x=−32
4. A geometric series has the nth term un=162(32)n. (a) Find u1 and u2. [2 marks] u1=108,u2=72.
(b) Find the sum to infinity of the series [3 marks] First term is 108, common ratio is 2/3. Hence sum to infinity is 1−32108=324
(c) Find the smallest value of k such that ∑n=k∞un<2.5. [3 marks] ∑n=k∞un=486(32)k
486(32)k<2.5⇒k>log324862.5.
Hence k>12.997 so k is at least 13.
5. A curve satisfies dxdy=x23−2x, where x>0. (a) Show that there is only one value of x for which there is a stationary point. [2 marks] dxdy=0⇒x(x21−2)=0⇒x1/2=2⇒x=4. Hence there is only one value of x for which there is a stationary point.
(b) Find dx2d2y and show that this is a minimum point. [3 marks] dx2d2y=23x−2. When x=4, this is equal to 1. Hence it is positive, so the point is a minimum.
(c) The line y=2 is a tangent to the curve. Find the equation of the curve. [4 marks] Must be a tangent at the stationary point. Hence the point (4, 2) lies on the curve (the stationary point). Integrating gives y=52x25−x2+c. As y(4)=2, c=526 so y=52x25−x2+526.
6. There is a curve y=23x. (a)(i) Use the trapezium rule with five ordinates (four strips) to approximate ∫0123xdx to two decimal places. [4 marks] Use the trapezium rule formula to find that it is 3.44 to two decimal places.
(a) (ii) State how the approximation may be improved. [1 mark] Use more strips/ordinates.
(a) (iii) The point (1, k) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line x=0 and the line y=k, to two decimal places. [3 marks] k=8 so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is 8−3.44=4.56 to two decimal places.
(b) The curve y=23x may be mapped onto the curve of y=23x−4 by a translation and a stretch. (i) the translation [2 marks]
Translation in the vector (−34, 0)
(ii) stretch [2 marks]
Stretch parallel to the y-axis scale factor 161.
(c) Use logarithms to solve the equation 23x−4=7, using your answer to three significant figures. [2 marks]
x=31(4+log27)=2.27 to three significant figures.
7. A curve has the equation y=7x+6−x21. (a) The region bounded by the curve and the lines x=1 and x=2 is above the x-axis. Show that the area of this region is 16. [5 marks]
Area (as curve is above the x-axis) is: ∫127x+6−x21dx=[27x2+6x+x1]12. Substitute in the limits to show that the area is 16.
(b) The normal to the curve is parallel to the line 2y+8x=3. Find the equation of the normal to the curve. [6 marks] Differentiating gives dxdy=7−x32. Now, the normal has a gradient of −4. Hence the gradient of the tangent to the curve is 41. So
7−x32=41⇒x38=−27⇒x=−32. Hence y=−1211.
So the equation of the normal is y+1211=−4(x+32).
8. (a) Solve the equation cosθ=32 in the range 0⩽θ⩽360. Give your answers to the nearest degree. [2 marks]
The values of θ are 48 and 312.
(b)(i) Given that 4sinθtanθ=4−cosθ, show that 3cos2θ+4cosθ−4=0. [3 marks]
4cosθsinθsinθ=4−cosθ
4cosθsin2θ=4−cosθ
4sin2θ=4cosθ−cos2θ
4−4cos2θ=4cosθ−cos2θ
3cos2θ+4cosθ−4=0
(ii) Show that there is only one possible value of cosθ. [2 marks] Factorising gives (3cosθ−2)(cosθ+2)=0. As ∣cosθ∣⩽1, the only possible value of cosθ is 32.
(c) Hence solve the equation 4sin4xtan4x=4−cos4x in the range 0⩽x⩽180. [4 marks]
Solving cos4x=32 in the range 0⩽4x⩽720.
Use part (a) and then add on 360 to both solutions to give 4x={48,312,408,672}
So x={12,78,102,168}
9. Given that 3log2(c+2)−log2(2k+c3)=1, express (c+1)2 in terms of k. [7 marks]
log2((c+2)3)−log2(22k+c3)=log22
log2(2k+c32(c+2)3)=log22
2k+c32(c+2)3=2
2(c+2)3=4k+2c3
2c3+12c2+24c+16=4k+2c3
12c2+24c+16=4k
12(c+1)2=4k−4
(c+1)2=3k−1
For 6)b)I) I'm pretty sure the translation was with the vector (4/3,0) and for the last question it asked to fine k in the form (c+1)^2 therefore k=3(c+1)^(2) +1
I used log laws for the first bit and ended up crossing out correct working so 5 marks down drain ? Also how much do you lose for putting gradient of Q as 1/4. I got x and y fine and used equation right just will it be one accuracy mark gone? And for trapezium rule for some reason I have no idea I halved the rectangle coz I'm used to doing triangles hahahahha, will I lose 2 or 3 as I still got 8 but then subtracted 3.44 from 4.
For 6)b)I) I'm pretty sure the translation was with the vector (4/3,0) and for the last question it asked to fine k in the form (c+1)^2 therefore k=3(c+1)^(2) +1
For 6)b)I) I'm pretty sure the translation was with the vector (4/3,0) and for the last question it asked to fine k in the form (c+1)^2 therefore k=3(c+1)^(2) +1