Cambridge International Examination 9709/21

Cambridge International Examination 9709/21 May/June 2022 Selected Questions shahbaz ahmed June 2024 Selected Questions with the proofs of relevant formulas and skills Q1 1 The variables x and y satisfy the equation y = 42x−a , where a is an integer. As shown in the diagram, the graph of lny against x is a straight line passing through the point (0, −20.8), where the second coordinate is given correct to 3 significant figures. (a) Show that the gradient of the straight line is ln 16. (b) Determine the value of a. Solution(a) y = 42x−a Where a is an integer ln y = ln 42x−a ln y = (2x − a) ln 4 Taking derivative with respect to x 2 d((ln 4)(2x−a) dx = (ln 4) d((2x−a) dx d(ln y) dx = d(ln y) dx = (ln 4)[ d(2x) dx − d(ln y) dx = 2(ln 4)[ dx dx − 0] d(ln y) dx = 2(ln 4) d(ln y) dx = ln 42 = ln 16 da dx ] Hence y) Gradient= d(ln = ln 16 dx (b) Since ln y = (2x − a) ln 4 Putting (0, −20.8) = (x, ln y) in the above equation −20.8 = [2 × 0 − a] ln 4 −20.8 = (−a) ln 4 20.8 = a ln 4 a ln 4 = 20.8 a= 20.8 ln 4 ........................................................................... Q2(a) Express the equation 7 tan θ + 4 cot θ − 13 sec θ = 0 in terms of sin θ only (b) Hence solve the equation 7 tan θ+4 cot θ−13 sec θ = 0 for 0◦ < θ < 360◦ Solution (a)7 tan θ + 4 cot θ − 13 sec θ = 0 sin θ cos θ 1 7 cos θ + 4 sin θ − 13( cos θ ) = 0 sin θ cos θ 1 [7 cos θ + 4 sin θ − 13( cos θ )] sin θ cos θ = 0 7 sin2 θ + 4 cos2 θ − 13 sin θ = 0 7 sin2 θ + 4(1 − sin2 θ) − 13 sin θ = 0 7 sin2 θ + 4 − 4 sin2 θ − 13 sin θ = 0 3 sin2 θ − 13 sin θ + 4 = 0 3 (b)Since 7 tan θ + 4 cot θ − 13 sec θ = 3 sin2 θ − 13 sin θ + 4 = 0 3 sin2 θ − 13 sin θ + 4 = 0 3 sin2 θ − (12 + 1) sin θ + 4 = 0 3 sin2 θ − 12 sin θ − sin θ + 4 = 0 3 sin θ(sin θ − 4) − 1(sin θ − 4) = 0 (sin θ − 4)(3 sin θ − 1) = 0 (sin θ − 4) = 0 θ = arcsin 4 is undefined 3 sin θ − 1 = 0 =⇒ θ = arcsin 31 = 19◦ 28′ 16.3” ..................................................................... Q3 The diagram shows the curve with equation y = 3 sin x − 3 sin 2x for 0 ≤ x ≤ π. The curve meets the x-axis at the origin and at the points with x-coordinates a and π. 4 (a) Find the exact value of a. (b) Find the area of the shaded region. Putting y = 3 sin x − 3 sin 2x = 0 3 sin x − 3 sin 2x = 0 3[sin x − sin 2x] = 0 [sin x − sin 2x] = 0 3 sin x − sin 2x = 0 sin x − (2 sin x cos x) = 0 sin x[1 − 2 cos x] = 0 Either sin x = 0 Or [1 − 2 cos x] = 0 Now sin x = 0 =⇒ x = 0, π Similarly [1 − 2 cos x] = 0 =⇒ cos x = 1 2 =⇒ x = arccos 12 = π 3 At x = 0 is origin and this point is excluded from the domain of shaded region. Hence a = π 3 Domain of the function for shaded region=[ π3 , π] It can be verified that at x = a = At x = a = π 3 π 3, y=0 , y = 3 sin π3 − 3 sin 2( π3 ) = √ 3 3 2 − √ 3 3 2 =0 That is x = a = π3 satisfies the equation in the shaded region . Rπ (b)Area= π ydx 3 Rπ Area= π [3 sin x − 3 sin 2x ]dx 3 Rπ Rπ Area= π 3 sin x dx − π 3 sin 2x dx 3 3 d(cos x) = − sin x dx −d(cos x) = sin x dx d(cos 2x) = − sin 2x d(2x) = −2 sin 2x dx − 12 d(cos 2x) = sin 2x dx 5 Putting in the above integral Rπ Rπ Area=3 π [−d(cos x)] − 3 π [− 21 d(cos 2x)] 3 3 Rπ Rπ Area=-3 π [d(cos x)] − 3[− 12 ] π [d(cos 2x)] 3 3 Rπ Rπ Area=-3 π [d(cos x)] + 23 π [d(cos 2x)] 3 3 Area= (−3)[cos x]ππ + 32 [cos 2x]ππ 3 Area= (−3)[cos π − cos 3 π 3] + 3 2 [cos 2π − cos 2π 3 ] Area= (−3)[−1 − 12 ] + 32 [1 − (− 21 ] Area= (−3)[− 32 ] + 32 [1 + 12 ] Area= 9 2 + 9 4 = 27 4 .................................................................. Q4 6 A curve has equation x2 y + 2y 3 = 48 . Find the equation of the normal to the curve at the point (4, 2) , giving your answer in the form ax + by + c = 0 where a, b and c are integers. Solution x2 y + 2y 3 = 48 Taking derivative with respect to x d(x2 y+2y 3 ) dx d(x2 y) dx 2 + = d(48) dx d(2y 3 ) dx =0 3 ) d(y ) 2 dy y d(x dx + x dx + 2 dx = 0 dy dy y(2x) + x2 dx + 2(3)y 2 dx =0 7 dy dy 2xy + x2 dx + 6y 2 dx =0 dy =0 2xy + [x2 + 6y 2 ] dx Putting (4, 2) = (x, y) dy 2 × 4 × 2 + [42 + 6 × 22 ] dx =0 dy =0 16 + [40] dx dy [40] dx = −16 dy dx 2 = − 16 40 = − 5 = m Where m is the slope of tangent passing through the point(4, 2) Let m1 be the slope of the normal through the same point. Then mm1 = −1 Putting − 25 = m − 25 m1 = −1 m1 = 5 2 Using the equation of line through the point (x1 , y1 ) with slope m1 y − y1 = m1 (x − x1 ) Equation of the normal passing through (4, 2) = (x1 , y1 ) with m1 = y − 2 = 25 (x − 4) 2y − 4 = 5x − 20 5x − 20 = 2y − 4 5x − 2y − 20 + 4 = 0 5x − 2y − 16 = 0 8 5 2