Solution of Cambridge International AS/A Level: May/June 2022

Cambridge International AS and A Level may /June 2022 shahbaz ahmed May 2024 Q 1(a) Express x2 − 8x + 11 in the form (x + p)2 + q where p and q are constants (b) Hence find the exact solutions of the equation x2 − 8x + 11 = 1 Solution (a) x2 − 8x + 11 = x2 − 8x + 16 − 16 + 11 x2 − 8x + 11 = (x2 − 8x + 16) − 16 + 11 x2 − 8x + 11 = (x − 4)2 − 5 Comparing it with the equation (x + p)2 + q =⇒ p = −4, q = −5 (b) x2 − 8x + 11 = 1 x2 − 8x + 11 = (x − 4)2 − 5 = 1 (x − 4)2 − 5 = 1 (x − 4)2 − 5 + 5 = 1 + 5 (x − 4)2 = 6 √ x−4=± 6 √ x=4± 6 1 Q2 The thirteenth term of an arithmetic progression is 12 and the sum of the first 30 terms is −15. Find the sum of the first 50 terms of the progression. Solution an = a + (n − 1)d Put an = 12, n = 13 12 = a + (13 − 1)d 12 = a + 12d a + 12d = 12 ............... eq (1) Formula to find the sum of an arithmetic progression sn = n 2 [2a + (n − 1)d] Putting Sn = −15, n = 30 −15 = 30 2 [2a + (30 − 1)d] −15 = 15[2a + 29d] −1 = [2a + 29d] 2a + 29d = −1...................eq(2) Multiplying eq (1) by 2 and subtraction from equation (2) 2a + 29d − 2a − 24d = −1 − 24 5d = −25 d = −5 Putting in equation (1) a + 12(−5) = 12 a − 60 = 12 a = 72 2 Putting a = 72, and d = −5, n = 50, in the equation sn = s50 = 50 2 [2(72) n 2 [2a + (n − 1)d] + (50 − 1)(−5)] s50 = 25[144 + 49(−5)] s50 = −2525 Q3 The coefficient of x4 in the expansion of 2 (2x2 + kx )5 is a.The coefficient of x2 in the expansion of (2kx − 1)4 is b (a) Find a and b in terms of the constant k. (b) Given that a+b = 216 , find the possible values of k. Solution (r+1)th term in the expansion of (2x2 + k2 5 x ) 2 Tr+1 =n Cr (2x2 )n−r ( kx )r Put n=5 2 Tr+1 =5 Cr (2x2 )5−r ( kx )r =5 Cr 25−r x10−2r k 2r x−r Tr+1 =5 Cr 25−r x10−2r−r k 2r Tr+1 =5 Cr 25−r x10−3r k 2r Now according to the given condition x4 = x10−3r 4 = 10 − 3r 3r = 10 − 4 = 6 r=2 Putting in the equation 3 Tr+1 =5 Cr 25−r x10−3r k 2r =⇒ T2+1 =5 C2 25−2 x10−3(2) k 2(2) T3 = 10(8)x4 k 4 T3 = 80x4 k 4 coefficient of x4 = 80k 4 Hence a=80k 4 (r+1)th term in the expansion of (2kx − 1)4 Tr+1 =n Cr (2kx)4−r (−1)r Tr+1 =n Cr 24−r k 4−r x4−r (−1)r According to the given condition x4−r = x2 =⇒ 4−r =2 r=2 Putting r = 2 in the equation Tr+1 =n Cr 24−r k 4−r x4−r (−1)r T2+1 =4 C2 24−2 k 4−2 x4−2 (−1)2 T3 = 6(4)k 2 x2 = 24k 2 x2 Hence coefficient of x2 = 24k 2 = b (b) a + b = 216 Putting the values of a and b 80k 4 + 24k 2 = 216 10k 4 + 3k 2 = 27 10k 4 + 3k 2 − 27 = 0 4 10k 4 + 18k 2 − 15k 2 − 27 = 0 2k 2 (5k 2 + 9) − 3(5k 2 + 9) (2k 2 − 3)(5k 2 + 9) = 0 2k 2 − 3 = 0 2k 2 = 3 k2 = 3 2 k=± q 3 2 √ =± 6 2 Similarly k2 + 9 = 0 5k 2 = −9 k 2 = − 95 k = ± √3ι5 Q4 (a) Prove the identity sin3 θ sin θ−1 − sin2 θ 1+sin θ ≡ − tan2 θ(1 + sin θ) (b)Hence solve the equation sin3 θ sin θ−1 − sin2 θ 1+sin θ ≡ − tan2 θ(1 + sin θ) for 0 < θ < 2π Solution LHS= sin3 θ sin θ−1 − sin2 θ 1+sin θ θ LHS= sin2 θ[ sinsinθ−1 − 1 1+sin θ ] θ LHS= sin2 θ[ sinsinθ−1 − 1 sin θ+1 ] θ(sin θ+1)−(sin θ−1) ] LHS= sin2 θ[ sin (sin θ−1)(sin θ+1) 5 2 LHS= sin2 θ[ (sin θ+sin θ)−sin θ+1 ] (sin2 θ−12 ) 2 (sin θ+1 LHS= sin2 θ[ (sin 2 θ−12 ) ] 2 (sin θ+1 LHS= − sin2 θ[ 1−(sin 2 θ] 2 θ+1 LHS= − sin2 θ[ (sin cos2 θ ] 2 2 sin θ LHS= − cos 2 θ (sin θ + 1) LHS= − tan2 θ(sin2 θ + 1) = RHS LHS+RHS (b) Let − tan2 θ(1 + sin θ) = 0 for 0 < θ < 2π 2 sin θ − cos 2 θ (1 + sin θ) = 0 for 0 < θ < 2π −sin2 θ(1 + sin θ) = 0 for 0 < θ < 2π sin2 θ(1 + sin θ) = 0 for 0 < θ < 2π =⇒ sin2 θ = 0 sin θ = 0 θ=0 Since 0 < θ < 2π sin θ = 0 has no solution in the given domain 1 + sin θ = 0 sin θ = −1 θ = arcsin −1 = 3 π2 6 Q5 The diagram shows a sector ABC of a circle with centre A and radius r. The line BD is perpendicular to AC. Angle CAB is θ radians. (a) Given that θ = 16 π , find the exact area of BCD in terms of r. √ (b) Given instead that the length of BD is 3 2 r find the exact perimeter of BCD in terms of r 7 , Reference to the right angled triangle ADB π 6 θ= sin π6 = BD r BD = r sin π6 = r 2 Similarly cos π6 = √ 3 2 AD r AD r √ = 23 r = AD Now area of the right angled triangle ADB = 12 (AD)(BD) = 12 ( √ 3 r 2 r)( 2 ) √ = 3r 2 8 Area of the sector ACB= 12 r2 π6 = πr 2 12 Area of BCD=Area of sector ACB-Area of the right angled triangle ADB 8 2 Area of BCD= πr 12 − π − Area of BCD=( 12 √ 3r 2 8 √ 3 2 8 )r (b) If √ BD= 3 2 r =⇒ AD= 2r Now DC=AC-AD=r − Arc BC= 12 r π6 = r 2 = r 2 πr 12 Perimeter of BCD =BD+DC+arc √ = 3 2 r + r 2 + πr 12 Perimeter of BCD =BD+DC+arc BC √ =( 3 2 + 1 2 + π 12 )r Q6 The function f is defined as follows: f (x) = x2 −4 x2 +4 for x > 2. (a) Find an expression for f −1 (x) (b) Show that 1 − 8 x2 +4 can be expressed as and hence state the range of f (c)Explain why the composite function ff cannot be formed. Solution y = f (x) = x2 −4 x2 +4 y(x2 + 4) = x2 − 4 yx2 − x2 = −4y − 4 9 x2 −4 x2 +4 x2 (y − 1) = −4(y + 1) y+1 x2 = −4 y−1 1+y x2 = 4 1−y q q 1+y 1+y x = 4 1−y x = 2 1−y y = f (x) =⇒ x = f −1 (y) =⇒ f −1 (y) = 2 q 1+y 1−y Since y is a dummy variable .Replacing y by x q 1+x f −1 (x) = 2 1−x (b) 1− 8 x2 +4 x2 +4−8 x2 +4 x2 −4 x2 +4 Since f (x) = 1 − 8 x2 +4 As x −→ ±∞ 8 x2 +4 −→ 0 So f (x) = 1 − 0 = 1 as x −→ ±∞ Similarly At x = 0 , f (0) = 1 − 8 02 +4 =1− 8 0+4 So range of f(x)= [-1,1] (c) Since Range of f = [−1, 1] Domain of f = [2, ∞) Since f [−1, 1] ̸= (2, ∞) ff cannot be formed. 10 =1− 8 4 = 1 − 2 = −1 Figure 1: Q7 11 The diagram shows the curve with equation y = 1 (3x − 2) 2 and the line y = 12 x + 1. The curve and the line intersect at points A and B (a) Find the coordinates of A and B. (b) Hence find the area of the region enclosed between the curve and the line. Solution 1 y = (3x − 2) 2 y = 12 x + 1 1 (3x − 2) 2 = 12 x + 1 Squaring 3x − 2 = 14 x2 + x + 1 4(3x − 2) = [ 14 x2 + x + 1] × 4 12x − 8 = x2 + 4x + 4 x2 + 4x + 4 = 12x − 8 x2 + 4x + 4 − 12x + 8 = 0 x2 − 8x + 12 = 0 x2 − 6x − 2x − 12 = 0 x(x − 6) − 2(x − 6) = 0 (x − 2)(x − 6) = 0 x − 2 = 0 =⇒ x = 2 x − 6 = 0 =⇒ x = 6 Now at x = 6 12 y = 12 (6) + 1 = 4 at x = 2 y = 12 (2) + 1 = 2 Hence points A and B are: A(2, 2), B(6, 4)) (b) Area of the region enclosed between the curve and the line. R6 1 = 2 [(3x − 2) 2 − ( 12 x + 1)]dx R6 R6 R6 1 = 2 (3x − 2) 2 dx − 2 12 xdx − 2 dx R6 R6 R6 1 = 31 2 (3x − 2) 2 (3dx) − 12 2 xdx − 2 dx 3 =( 13 )( 23 )|(3x − 2) 2 |62 − ( 12 )( 21 )|x2 |62 − |x|62 3 3 =( 13 )( 23 )|[(3 × 6 − 2) 2 − (3 × 2 − 2) 2 ]| − ( 12 )( 12 )|(62 − 22 )| − |(6 − 2)| 3 3 =( 13 )( 23 )|[(16) 2 − (4) 2 ]| − ( 12 )( 12 )|(36 − 4)| − |4| √ √ 3 =( 13 )( 23 )|[( 16)3 − 4 ]| − ( 21 )( 12 )|(36 − 4)| − |4| √ 3 √ =( 13 )( 23 )|[( 16)3 − 4 ]| − ( 21 )( 12 )|(32)| − |4| = 29 (43 − 23 ) − 8 − 4 = 29 (64 − 8) − 8 − 4 = 29 (56) − 8 − 4 = 112 9 − 12 = 49 Q8 13 Figure 2: (a) The curve y = sinx is transformed to the curve y = 4 sin( x2 − 30◦ ) Describe fully a sequence of transformations that have been combined, making clear the order in which the transformations are applied. (b) Find the exact solutions of the equation y = √ 4 sin( x2 − 30◦ ) = 2 2 for 0◦ ≤ x ≤ 360◦ Solution Note: Figure added in Q8 are neither the part of question nor the part of answer.These figure are drawn just to ensure visual learning. Vertical stretches y = 4 sin x Horizontal Stretches y = 4 sin x2 14 Figure 3: Figure 4: 15 Figure 5: 16 Horizontal Translation y = 4 sin( x2 − 30◦ ) (b) √ 4 sin( x2 − 30◦ ) = 2 2 √ 2 2 4 √ x ◦ sin( 2 − 30 ) = 22 √ sin( x2 − 30◦ ) = 22 √ x ◦ −1 ( 22 ) 2 − 30 = sin sin( x2 − 30◦ ) = x 2 − 30◦ = 45◦ x 2 = 45◦ + 30◦ x 2 = 75◦ x = 75◦ × 2 = 150◦ Q9 The equation of a circle is x2 +y 2 +6x−2y−26 = 0 (a) Find the coordinates of the centre of the circle and the radius. Hence find the coordinates of the lowest point on the circle. (b) Find the set of values of the constant k for which the line with equation y = kx − 5 intersect the circle at two distinct points. Solution X 2 + y 2 + 6x − 2y − 26 = 0.....................eq(1) (x2 + 6x) + (y 2 − 2y) = 2 (x2 + 6x + 9) + (y 2 − 2y + 1) = 26 + 9 + 1 17 2 x − (−3) + (y − 1)2 = 36 =⇒ centre(−3, 1), radius = 6 Taking derivative with respect to x of equation (1) d(x2 +y 2 +6x−2y−26) dx = d(0) dx dy dy + 6 − 2 dx =0 2x + 2y dx dy 2(y − 1) dx = −2x − 6 dy dx = −(x+3) y−1 Put dy dx = −(x+3) y−1 −(x+3) y−1 =0 =0 −(x + 3) = 0 x+3=0 x = −3 Putting in equation (1) (−3)2 + y 2 + 6(−3) − 2y − 26 = 0 9 + y 2 − 18 − 2y − 26 = 0 y 2 − 2y − −35 = 0 y 2 − 7y + 5y − 35 = 0 y(y − 7) + 5(y − 7) = 0 (y + 5)(y − 7) = 0 y + 5 = 0 =⇒ y = −5 y − 7 = 0 =⇒ y = 7 Hence coordinates of the lowest point on the circle are: (−3. − 5) 18 Figure 6: 19 (b) Putting y = kx − 5 in equation (1) x2 + (kx − 5)2 + 6x − 2(kx − 5) − 26 = 0 x2 + k 2 x2 − 10kx + 25 + 6x − 2kx + 10 − 26 = 0 x2 + k 2 x2 − 10kx + 25 + 6x − 2kx + 10 − 26 = 0 (1 + k 2 )x2 − 12kx + 6x + 9 = 0 (1 + k 2 )x2 + 6x(1 − 2k) + 9 = 0 For real roots: Discriminant :> 0 2 6(1 − 2k) − 36(1 + k 2 ) > 0 36(1 − 4k + 4k 2 ) − 36(1 + k 2 ) > 0 36 − 144k + 144k 2 − 36 − 36k 2 > 0 108k 2 − 144k > 0 k(108K − 144) > 0 =⇒ EITHER : k>0 Or 108k − 144 > 0 k>0 Or k> k<0 Or 108k − 144 < 0 k<0 Or 108k < 144 k<0 Or k< 4 3 OR 4 3 Hence k<0 Or k> 4 3 20 Q10 The equation of a curve is such that d2 y dx2 = 6x2 − x43 . The curve has a stationary point at (−1, 92 ). (a) Determine the nature of the stationary point at (−1, 29 ). (b) Find the equation of the curve. (c) Show that the curve has no other stationary points. (d) A point A is moving along the curve and the y-coordinate of A is increasing at a rate of 5 units per second. Find the rate of increase of the x-coordinate of A at the point where x = 1 . Solution (a) Putting x = −1 in the equation : d2 y dx2 = 6x2 − d2 y dx2 = 6(−1)2 − d2 y dx2 = 6 + 4 = 10 > 0. 4 x3 . 4 (−1)3 . So there is minimum at (−1, 29 ). 21 (b) d2 y dx2 = 6x2 − 4 x3 . dy d[ dx ] = [6x2 − 4 x3 ]dx Taking integral R dy R d[ dx ] = [6x2 − 4x−3 ]dx R 2 R dy 6x dx − 4x−3 dx dx = R 2 R −3 dy dx dx = 6 x dx − 4 x −2 3 dy dx = 6( x3 ) − 4( x−2 ) + C dy dx = 2x3 + 2x−2 + C dy dx = 2x3 + 2 x2 +C Since (−1, 92 ) is a stationary point =⇒ dy dx = 0 at x = −1 0 = 2(−1)3 + 2 (−1)2 +C 0 = −2 + 2 + C C=0 Now dy dx = 2x3 + 2 x2 dy = 2x3 dx + 2 x2 dx dy = 2x3 dx + 2x−2 dx Taking integral R R R dy = 2x3 dx + 2x−2 dx R R y = 2 x3 dx + 2 x−2 dx y = 24 x4 + 2 −1 (−1) x +D y = 12 x4 − 2x−1 + D Putting (−1, 92 ). 9 2 = 12 (−1)4 − 2(−1)−1 + D 9 2 = 1 2 9 2 = 12 (−1)4 − 2(−1)−1 + D +2+D 22 9 2 − 1 2 −2=D D=2 Hence y = 12 x4 − 2x−1 + 2 (c) Since y = 12 x4 − 2x−1 + 2 Taking derivative with respect to x dy dx = 2x3 + 2 x2 At stationary points dy dx = 2x3 + 2x3 + 2 x2 2 x2 =0 =0 2x3 = − x22 x3 = − x12 x5 = −1 x5 = ei(π+2kπ) , k=0,1,2,3,4 x = ei( π+2kπ ) 5 ,k=0,1,2,3,4 Roots are: π ei( 5 ) ,ei( 3π 5 ) ,ei( 5π 5 ) ,ei( 7π 5 ) ,ei( 9π 5 ) Only real root: ei( 5π 5 ) = eiπ = cos π + i sin π = −1 Remaining are imaginary roots.So stationary point is only at (−1, 92 ). (d) y = 12 x4 − 2x−1 + 2 Taking derivative with respect to t dy dt −2 dx = 12 (4x3 ) dx dt + 2x dt dy dt −2 dx = 2x3 dx dt + 2x dt dy dt 23 = 4 1 d(x ) 2 dt −1 − 2 d(xdt + d(2) dt dy dt = [2x3 + 2x−2 ] dx dt dy dt = [2x3 + 2 dx x2 ] dt Putting x = 1, 5 = [2(1)3 + dy dt =5 2 dx (1)2 ] dt 5 = [2 + 2] dx dt dx dt = 5 4 24