T-Leng Tuition · mugcornersg.com
Total: 40 marksSuggested time: 1 hour
Name: ______________________________ Class: ____________ Date: ____________
Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.
Section A 1 Given that d d x ( x sin 4 x ) = 4 x cos 4 x + sin 4 x \dfrac{d}{dx}\left(x\sin 4x\right) = 4x\cos 4x + \sin 4x d x d ( x sin 4 x ) = 4 x cos 4 x + sin 4 x , find ∫ x cos 4 x d x \int x\cos 4x \, dx ∫ x cos 4 x d x .
Answer [2]
2 Given that d d x ( x 2 + 4 ) = x x 2 + 4 \dfrac{d}{dx}\left(\sqrt{x^{2} + 4}\right) = \dfrac{x}{\sqrt{x^{2} + 4}} d x d ( x 2 + 4 ) = x 2 + 4 x , find ∫ 3 x x 2 + 4 d x \int \dfrac{3x}{\sqrt{x^{2} + 4}} \, dx ∫ x 2 + 4 3 x d x .
Answer [2]
3 Find ∫ ( 3 x + 2 x ) d x \int \left(3\sqrt{x} + 2x\right) \, dx ∫ ( 3 x + 2 x ) d x .
Answer [2]
4 (a) Find ∫ ( 6 x 2 − 4 x − 1 ) d x \int \left(6x^{2} - 4x - 1\right) \, dx ∫ ( 6 x 2 − 4 x − 1 ) d x .
Answer (a) [1]
(b) Find ∫ ( 2 x 3 − 9 x ) d x \int \left(2x^{3} - \dfrac{9}{\sqrt{x}}\right) \, dx ∫ ( 2 x 3 − x 9 ) d x .
Answer (b) [2]
5 (a) Find ∫ ( 2 x + 1 ) 4 d x \int (2x + 1)^{4} \, dx ∫ ( 2 x + 1 ) 4 d x .
Answer (a) [1]
(b) Find ∫ 4 ( 6 x + 1 ) 3 d x \int \dfrac{4}{(6x + 1)^{3}} \, dx ∫ ( 6 x + 1 ) 3 4 d x .
Answer (b) [2]
6 (a) Find ∫ 4 e 3 x − 4 d x \int 4e^{3x - 4} \, dx ∫ 4 e 3 x − 4 d x .
Answer (a) [2]
(b) Find ∫ e − x d x \int e^{-x} \, dx ∫ e − x d x .
Answer (b) [2]
7 (a) The gradient of a curve is given by d y d x = 2 x − 5 \dfrac{dy}{dx} = 2x - 5 d x d y = 2 x − 5 . The curve passes through the point ( 2 , 3 ) (2, 3) ( 2 , 3 ) . Find the equation of the curve.
Answer (a) [2]
(b) The gradient of a curve is given by d y d x = 10 e 5 x + 3 \dfrac{dy}{dx} = 10e^{5x} + 3 d x d y = 10 e 5 x + 3 . The curve passes through the point ( 0 , 4 ) (0, 4) ( 0 , 4 ) . Find the equation of the curve.
Answer (b) [2]
Section B 8 It is given that y = sin 3 2 x y = \sin^{3} 2x y = sin 3 2 x . (a) Find d y d x \dfrac{dy}{dx} d x d y .
Answer (a) [2]
(b) Hence find ∫ 4 sin 2 2 x cos 2 x d x \int 4\sin^{2} 2x\cos 2x \, dx ∫ 4 sin 2 2 x cos 2 x d x .
Answer (b) [2]
(c) Hence find the exact value of ∫ 0 π 4 4 sin 2 2 x cos 2 x d x \int_{0}^{\dfrac{\pi}{4}} 4\sin^{2} 2x\cos 2x \, dx ∫ 0 4 π 4 sin 2 2 x cos 2 x d x .
Answer (c) [2]
9 It is given that f ( x ) = x cos 2 x f(x) = x\cos 2x f ( x ) = x cos 2 x . (b) Hence find ∫ 4 x sin 2 x d x \int 4x\sin 2x \, dx ∫ 4 x sin 2 x d x .
Answer (b) [3]
(c) A curve has gradient function d y d x = 4 x sin 2 x \dfrac{dy}{dx} = 4x\sin 2x d x d y = 4 x sin 2 x and passes through the point ( 0 , 2 ) (0, 2) ( 0 , 2 ) . Find the equation of the curve.
Answer (c) [2]
10 The gradient function of a curve is d y d x = k x + 3 x 2 \dfrac{dy}{dx} = kx + \dfrac{3}{x^{2}} d x d y = k x + x 2 3 , where k k k is a constant. The curve has a stationary point at ( − 1 , 10 ) (-1, 10) ( − 1 , 10 ) . (a) Find the value of k k k .
Answer (a) [2]
(b) Find the equation of the curve.
Answer (b) [3]
(c) Determine the nature of the stationary point.
Answer (c) [2]
END OF PAPER
Question 1
Answer: ∫ x cos 4 x d x = 1 4 x sin 4 x + 1 16 cos 4 x + c \int x\cos 4x \, dx = \dfrac{1}{4}x\sin 4x + \dfrac{1}{16}\cos 4x + c ∫ x cos 4 x d x = 4 1 x sin 4 x + 16 1 cos 4 x + c [2] x sin 4 x = ∫ 4 x cos 4 x d x + ∫ sin 4 x d x x\sin 4x = \int 4x\cos 4x \, dx + \int \sin 4x \, dx x sin 4 x = ∫ 4 x cos 4 x d x + ∫ sin 4 x d x ∫ sin 4 x d x = − cos 4 x 4 \int \sin 4x \, dx = -\dfrac{\cos 4x}{4} ∫ sin 4 x d x = − 4 cos 4 x ∫ 4 x cos 4 x d x = x sin 4 x − ( − cos 4 x 4 ) \int 4x\cos 4x \, dx = x\sin 4x - \left(-\dfrac{\cos 4x}{4}\right) ∫ 4 x cos 4 x d x = x sin 4 x − ( − 4 cos 4 x ) ∫ x cos 4 x d x = 1 4 x sin 4 x + 1 16 cos 4 x + c \int x\cos 4x \, dx = \dfrac{1}{4}x\sin 4x + \dfrac{1}{16}\cos 4x + c ∫ x cos 4 x d x = 4 1 x sin 4 x + 16 1 cos 4 x + c Question 2
Answer: ∫ 3 x x 2 + 4 d x = 3 x 2 + 4 + c \int \dfrac{3x}{\sqrt{x^{2} + 4}} \, dx = 3\sqrt{x^{2} + 4} + c ∫ x 2 + 4 3 x d x = 3 x 2 + 4 + c [2] ∫ x x 2 + 4 d x = x 2 + 4 + c \int \dfrac{x}{\sqrt{x^{2} + 4}} \, dx = \sqrt{x^{2} + 4} + c ∫ x 2 + 4 x d x = x 2 + 4 + c 3 x x 2 + 4 = 3 × ( x x 2 + 4 ) \dfrac{3x}{\sqrt{x^{2} + 4}} = 3 \times \left(\dfrac{x}{\sqrt{x^{2} + 4}}\right) x 2 + 4 3 x = 3 × ( x 2 + 4 x ) ∫ 3 x x 2 + 4 d x = 3 x 2 + 4 + c \int \dfrac{3x}{\sqrt{x^{2} + 4}} \, dx = 3\sqrt{x^{2} + 4} + c ∫ x 2 + 4 3 x d x = 3 x 2 + 4 + c Question 3
Answer: = 2 x x + x 2 + c = 2x\sqrt{x} + x^{2} + c = 2 x x + x 2 + c [2] ∫ ( 3 x + 2 x ) d x = ∫ ( 3 x 1 2 + 2 x ) d x \int \left(3\sqrt{x} + 2x\right) \, dx = \int \left(3x^{\frac{1}{2}} + 2x\right) \, dx ∫ ( 3 x + 2 x ) d x = ∫ ( 3 x 2 1 + 2 x ) d x ∫ ( 3 x 1 2 + 2 x ) d x = 3 x 3 2 3 2 + 2 x 2 2 + c \int \left(3x^{\frac{1}{2}} + 2x\right) \, dx = \dfrac{3x^{\frac{3}{2}}}{\dfrac{3}{2}} + \dfrac{2x^{2}}{2} + c ∫ ( 3 x 2 1 + 2 x ) d x = 2 3 3 x 2 3 + 2 2 x 2 + c = 2 x 3 2 + x 2 + c = 2x^{\frac{3}{2}} + x^{2} + c = 2 x 2 3 + x 2 + c = 2 x x + x 2 + c = 2x\sqrt{x} + x^{2} + c = 2 x x + x 2 + c Question 4
(a) Answer: = 2 x 3 − 2 x 2 − x + c = 2x^{3} - 2x^{2} - x + c = 2 x 3 − 2 x 2 − x + c [1] ∫ ( 6 x 2 − 4 x − 1 ) d x = 6 x 3 3 − 4 x 2 2 − x + c \int \left(6x^{2} - 4x - 1\right) \, dx = \dfrac{6x^{3}}{3} - \dfrac{4x^{2}}{2} - x + c ∫ ( 6 x 2 − 4 x − 1 ) d x = 3 6 x 3 − 2 4 x 2 − x + c = 2 x 3 − 2 x 2 − x + c = 2x^{3} - 2x^{2} - x + c = 2 x 3 − 2 x 2 − x + c (b) Answer: = 1 2 x 4 − 18 x + c = \dfrac{1}{2}x^{4} - 18\sqrt{x} + c = 2 1 x 4 − 18 x + c [2] ∫ ( 2 x 3 − 9 x ) d x = ∫ ( 2 x 3 − 9 x − 1 2 ) d x \int \left(2x^{3} - \dfrac{9}{\sqrt{x}}\right) \, dx = \int \left(2x^{3} - 9x^{-\frac{1}{2}}\right) \, dx ∫ ( 2 x 3 − x 9 ) d x = ∫ ( 2 x 3 − 9 x − 2 1 ) d x ∫ ( 2 x 3 − 9 x − 1 2 ) d x = 2 x 4 4 − 9 x 1 2 1 2 + c \int \left(2x^{3} - 9x^{-\frac{1}{2}}\right) \, dx = \dfrac{2x^{4}}{4} - \dfrac{9x^{\frac{1}{2}}}{\dfrac{1}{2}} + c ∫ ( 2 x 3 − 9 x − 2 1 ) d x = 4 2 x 4 − 2 1 9 x 2 1 + c = 1 2 x 4 − 18 x 1 2 + c = \dfrac{1}{2}x^{4} - 18x^{\frac{1}{2}} + c = 2 1 x 4 − 18 x 2 1 + c = 1 2 x 4 − 18 x + c = \dfrac{1}{2}x^{4} - 18\sqrt{x} + c = 2 1 x 4 − 18 x + c Question 5
(a) Answer: = 1 10 ( 2 x + 1 ) 5 + c = \dfrac{1}{10}(2x + 1)^{5} + c = 10 1 ( 2 x + 1 ) 5 + c [1] = ( 2 x + 1 ) 5 5 × 2 + c = \dfrac{(2x + 1)^{5}}{5 \times 2} + c = 5 × 2 ( 2 x + 1 ) 5 + c = 1 10 ( 2 x + 1 ) 5 + c = \dfrac{1}{10}(2x + 1)^{5} + c = 10 1 ( 2 x + 1 ) 5 + c (b) Answer: = − 1 3 ( 6 x + 1 ) − 2 + c = -\dfrac{1}{3}(6x + 1)^{-2} + c = − 3 1 ( 6 x + 1 ) − 2 + c [2] ∫ 4 ( 6 x + 1 ) − 3 d x \int 4(6x + 1)^{-3} \, dx ∫ 4 ( 6 x + 1 ) − 3 d x = 4 ( 6 x + 1 ) − 2 − 2 × 6 + c = \dfrac{4(6x + 1)^{-2}}{-2 \times 6} + c = − 2 × 6 4 ( 6 x + 1 ) − 2 + c = − 1 3 ( 6 x + 1 ) − 2 + c = -\dfrac{1}{3}(6x + 1)^{-2} + c = − 3 1 ( 6 x + 1 ) − 2 + c Question 6
(a) Answer: = 4 3 e 3 x − 4 + c = \dfrac{4}{3}e^{3x - 4} + c = 3 4 e 3 x − 4 + c [2] The integral of e to the power x is e to the power x itself. \text{The integral of e to the power x is e to the power x itself.} The integral of e to the power x is e to the power x itself. ∫ 4 e 3 x − 4 d x = 4 3 e 3 x − 4 + c \int 4e^{3x - 4} \, dx = \dfrac{4}{3}e^{3x - 4} + c ∫ 4 e 3 x − 4 d x = 3 4 e 3 x − 4 + c = 4 3 e 3 x − 4 + c = \dfrac{4}{3}e^{3x - 4} + c = 3 4 e 3 x − 4 + c (b) Answer: = − e − x + c = -e^{-x} + c = − e − x + c [2] The integral of e to the power x is e to the power x itself. \text{The integral of e to the power x is e to the power x itself.} The integral of e to the power x is e to the power x itself. ∫ e − x d x = 1 ( − 1 ) e − x + c \int e^{-x} \, dx = \dfrac{1}{(-1)}e^{-x} + c ∫ e − x d x = ( − 1 ) 1 e − x + c = − e − x + c = -e^{-x} + c = − e − x + c Question 7
(a) Answer: y = x 2 − 5 x + 9 y = x^{2} - 5x + 9 y = x 2 − 5 x + 9 [2] y = 2 x 2 2 − 5 x + c y = \dfrac{2x^{2}}{2} - 5x + c y = 2 2 x 2 − 5 x + c 3 = ( 2 ) 2 − 5 ( 2 ) + c 3 = (2)^{2} - 5(2) + c 3 = ( 2 ) 2 − 5 ( 2 ) + c 3 = − 6 + c 3 = -6 + c 3 = − 6 + c c = 9 c = 9 c = 9 y = x 2 − 5 x + 9 y = x^{2} - 5x + 9 y = x 2 − 5 x + 9 (b) Answer: y = 2 e 5 x + 3 x + 2 y = 2e^{5x} + 3x + 2 y = 2 e 5 x + 3 x + 2 [2] y = 2 e 5 x + 3 x + c y = 2e^{5x} + 3x + c y = 2 e 5 x + 3 x + c 4 = 2 e 5 ( 0 ) + 3 ( 0 ) + c 4 = 2e^{5(0)} + 3(0) + c 4 = 2 e 5 ( 0 ) + 3 ( 0 ) + c 4 = 2 + c 4 = 2 + c 4 = 2 + c c = 2 c = 2 c = 2 y = 2 e 5 x + 3 x + 2 y = 2e^{5x} + 3x + 2 y = 2 e 5 x + 3 x + 2 Question 8
(a) Answer: d y d x = 6 sin 2 2 x cos 2 x \dfrac{dy}{dx} = 6\sin^{2} 2x\cos 2x d x d y = 6 sin 2 2 x cos 2 x [2] y = ( sin 2 x ) 3 y = (\sin 2x)^{3} y = ( sin 2 x ) 3 d y d x = 3 ( sin 2 x ) 2 × 2 cos 2 x \dfrac{dy}{dx} = 3(\sin 2x)^{2} \times 2\cos 2x d x d y = 3 ( sin 2 x ) 2 × 2 cos 2 x = 6 sin 2 2 x cos 2 x = 6\sin^{2} 2x\cos 2x = 6 sin 2 2 x cos 2 x (b) Answer: 2 3 sin 3 2 x + c \dfrac{2}{3}\sin^{3} 2x + c 3 2 sin 3 2 x + c [2] From (a), ∫ 6 sin 2 2 x cos 2 x d x = sin 3 2 x + c 1 \int 6\sin^{2} 2x\cos 2x \, dx = \sin^{3} 2x + c_{1} ∫ 6 sin 2 2 x cos 2 x d x = sin 3 2 x + c 1 ∫ sin 2 2 x cos 2 x d x = 1 6 sin 3 2 x + c 2 \int \sin^{2} 2x\cos 2x \, dx = \dfrac{1}{6}\sin^{3} 2x + c_{2} ∫ sin 2 2 x cos 2 x d x = 6 1 sin 3 2 x + c 2 ∫ 4 sin 2 2 x cos 2 x d x = 2 3 sin 3 2 x + c \int 4\sin^{2} 2x\cos 2x \, dx = \dfrac{2}{3}\sin^{3} 2x + c ∫ 4 sin 2 2 x cos 2 x d x = 3 2 sin 3 2 x + c (c) Answer: 2 3 \dfrac{2}{3} 3 2 [2] ∫ 0 π 4 4 sin 2 2 x cos 2 x d x \int_{0}^{\dfrac{\pi}{4}} 4\sin^{2} 2x\cos 2x \, dx ∫ 0 4 π 4 sin 2 2 x cos 2 x d x = [ 2 3 sin 3 2 x ] 0 π 4 = \left[\dfrac{2}{3}\sin^{3} 2x\right]_{0}^{\frac{\pi}{4}} = [ 3 2 sin 3 2 x ] 0 4 π = 2 3 ( sin 3 π 2 − sin 3 0 ) = \dfrac{2}{3}\left(\sin^{3} \dfrac{\pi}{2} - \sin^{3} 0\right) = 3 2 ( sin 3 2 π − sin 3 0 ) = 2 3 ( 1 − 0 ) = \dfrac{2}{3}\left(1 - 0\right) = 3 2 ( 1 − 0 ) = 2 3 = \dfrac{2}{3} = 3 2 Question 9
(a) Answer: f ′ ( x ) = cos 2 x − 2 x sin 2 x f'(x) = \cos 2x - 2x\sin 2x f ′ ( x ) = cos 2 x − 2 x sin 2 x [2] f ′ ( x ) = ( 1 ) cos 2 x + x ( − 2 sin 2 x ) f'(x) = (1)\cos 2x + x(-2\sin 2x) f ′ ( x ) = ( 1 ) cos 2 x + x ( − 2 sin 2 x ) = cos 2 x − 2 x sin 2 x = \cos 2x - 2x\sin 2x = cos 2 x − 2 x sin 2 x (b) Answer: sin 2 x − 2 x cos 2 x + c \sin 2x - 2x\cos 2x + c sin 2 x − 2 x cos 2 x + c [3] ∫ ( cos 2 x − 2 x sin 2 x ) d x = x cos 2 x + c 1 \int (\cos 2x - 2x\sin 2x) \, dx = x\cos 2x + c_{1} ∫ ( cos 2 x − 2 x sin 2 x ) d x = x cos 2 x + c 1 ∫ 2 x sin 2 x d x = ∫ cos 2 x d x − x cos 2 x \int 2x\sin 2x \, dx = \int \cos 2x \, dx - x\cos 2x ∫ 2 x sin 2 x d x = ∫ cos 2 x d x − x cos 2 x ∫ 4 x sin 2 x d x = sin 2 x − 2 x cos 2 x + c \int 4x\sin 2x \, dx = \sin 2x - 2x\cos 2x + c ∫ 4 x sin 2 x d x = sin 2 x − 2 x cos 2 x + c (c) Answer: y = sin 2 x − 2 x cos 2 x + 2 y = \sin 2x - 2x\cos 2x + 2 y = sin 2 x − 2 x cos 2 x + 2 [2] y = sin 2 x − 2 x cos 2 x + c y = \sin 2x - 2x\cos 2x + c y = sin 2 x − 2 x cos 2 x + c At ( 0 , 2 ) (0, 2) ( 0 , 2 ) : 2 = 0 + c 2 = 0 + c 2 = 0 + c c = 2 c = 2 c = 2 y = sin 2 x − 2 x cos 2 x + 2 y = \sin 2x - 2x\cos 2x + 2 y = sin 2 x − 2 x cos 2 x + 2 Question 10
(a) Answer: k = 3 k = 3 k = 3 [2] At a stationary point, d y d x = 0 \dfrac{dy}{dx} = 0 d x d y = 0 : k ( − 1 ) + 3 ( − 1 ) 2 = 0 k(-1) + \dfrac{3}{(-1)^{2}} = 0 k ( − 1 ) + ( − 1 ) 2 3 = 0 − k = − 3 -k = -3 − k = − 3 k = 3 k = 3 k = 3 (b) Answer: y = 3 2 x 2 − 3 x + 11 2 y = \dfrac{3}{2}x^{2} - \dfrac{3}{x} + \dfrac{11}{2} y = 2 3 x 2 − x 3 + 2 11 [3] y = ∫ ( 3 x + 3 x − 2 ) d x y = \int \left(3x + 3x^{-2}\right) dx y = ∫ ( 3 x + 3 x − 2 ) d x = 3 2 x 2 − 3 x + c = \dfrac{3}{2}x^{2} - \dfrac{3}{x} + c = 2 3 x 2 − x 3 + c 10 = 3 2 + 3 + c 10 = \dfrac{3}{2} + 3 + c 10 = 2 3 + 3 + c c = 11 2 c = \dfrac{11}{2} c = 2 11 y = 3 2 x 2 − 3 x + 11 2 y = \dfrac{3}{2}x^{2} - \dfrac{3}{x} + \dfrac{11}{2} y = 2 3 x 2 − x 3 + 2 11 (c) Answer: Minimum point [2]
d 2 y d x 2 = 3 − 6 x 3 \dfrac{d^{2}y}{dx^{2}} = 3 - \dfrac{6}{x^{3}} d x 2 d 2 y = 3 − x 3 6 When x = − 1 x = -1 x = − 1 , d 2 y d x 2 = 3 + 6 \dfrac{d^{2}y}{dx^{2}} = 3 + 6 d x 2 d 2 y = 3 + 6 = 9 = 9 = 9 , so the point is a minimum.