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 y = x 4 − x y = x^{4} - x y = x 4 − x , find d 2 y d x 2 \dfrac{d^{2}y}{dx^{2}} d x 2 d 2 y .
Answer [2]
2 Given that y = 4 3 x + 5 y = \dfrac{4}{3x + 5} y = 3 x + 5 4 , find d y d x \dfrac{dy}{dx} d x d y .
Answer [3]
3 Given that y = 3 x − 1 2 x − 1 y = \dfrac{3x - 1}{2x - 1} y = 2 x − 1 3 x − 1 , find d y d x \dfrac{dy}{dx} d x d y , simplifying your answer.
Answer [3]
4 Given that y = x 3 ( 3 x + 2 ) 2 y = x^{3}(3x + 2)^{2} y = x 3 ( 3 x + 2 ) 2 , find d y d x \dfrac{dy}{dx} d x d y , giving your answer in factorised form.
Answer [3]
5 (a) Given that y = ( x + 2 ) ( 2 x + 1 ) 3 y = (x + 2)(2x + 1)^{3} y = ( x + 2 ) ( 2 x + 1 ) 3 , find d y d x \dfrac{dy}{dx} d x d y , giving your answer in factorised form.
Answer (a) [3]
(b) Given that y = ( x − 4 ) ( x − 3 ) 4 y = (x - 4)(x - 3)^{4} y = ( x − 4 ) ( x − 3 ) 4 , find d y d x \dfrac{dy}{dx} d x d y , giving your answer in factorised form.
Answer (b) [3]
6 (a) Given that y = x 2 ( x − 2 ) 4 y = x^{2}(x - 2)^{4} y = x 2 ( x − 2 ) 4 , find d y d x \dfrac{dy}{dx} d x d y , giving your answer in factorised form.
Answer (a) [3]
(b) Given that y = x 5 x + 2 y = x\sqrt{5x + 2} y = x 5 x + 2 , find d y d x \dfrac{dy}{dx} d x d y , giving your answer in factorised form.
Answer (b) [3]
Section B 7 The equation of a curve is y = − x − 17 x − 1 y = \dfrac{-x - 17}{x - 1} y = x − 1 − x − 17 , x ≠ 1 x \neq 1 x = 1 . (a) Find d y d x \dfrac{dy}{dx} d x d y , simplifying your answer.
Answer (a) [2]
(b) Explain why the gradient of the curve is positive for all values of x x x for which it is defined.
[1]
(c) Find the x x x -coordinates of the points on the curve at which the gradient is 2 2 2 .
Answer (c) [3]
(d) Find d 2 y d x 2 \dfrac{d^{2}y}{dx^{2}} d x 2 d 2 y .
Answer (d) [2]
8 The equation of a curve is y = ( x + 2 ) 3 x + 9 y = (x + 2)\sqrt{3x + 9} y = ( x + 2 ) 3 x + 9 , x > − 3 x > -3 x > − 3 . (a) Show that d y d x = 9 x + 24 2 3 x + 9 \dfrac{dy}{dx} = \dfrac{9x + 24}{2\sqrt{3x + 9}} d x d y = 2 3 x + 9 9 x + 24 .
[3]
(b) Find the gradient of the curve at the point where x = 0 x = 0 x = 0 .
Answer (b) [2]
(c) Find the x x x -coordinate of the point on the curve at which the gradient is zero.
Answer (c) [2]
(d) Find the range of values of x x x for which the gradient of the curve is positive.
Answer (d) [2]
END OF PAPER
Question 1
Answer: = 12 x 2 = 12x^{2} = 12 x 2 [2] d y d x = 4 x 3 − 1 \dfrac{dy}{dx} = 4x^{3} - 1 d x d y = 4 x 3 − 1 d 2 y d x 2 = 4 ( 3 ) x 2 \dfrac{d^{2}y}{dx^{2}} = 4(3)x^{2} d x 2 d 2 y = 4 ( 3 ) x 2 = 12 x 2 = 12x^{2} = 12 x 2 Question 2
Answer: = − 12 ( 3 x + 5 ) 2 = -\dfrac{12}{(3x + 5)^{2}} = − ( 3 x + 5 ) 2 12 [3] y = 4 ( 3 x + 5 ) − 1 y = 4(3x + 5)^{-1} y = 4 ( 3 x + 5 ) − 1 u = 3 x + 5 u = 3x + 5 u = 3 x + 5 d u d x = 3 \dfrac{du}{dx} = 3 d x d u = 3 d y d u = − 4 u − 2 \dfrac{dy}{du} = -4u^{-2} d u d y = − 4 u − 2 d y d x = d y d u × d u d x \dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx} d x d y = d u d y × d x d u d y d x = − 4 ( 3 x + 5 ) − 2 × 3 \dfrac{dy}{dx} = -4(3x + 5)^{-2} \times 3 d x d y = − 4 ( 3 x + 5 ) − 2 × 3 = − 12 ( 3 x + 5 ) − 2 = -12(3x + 5)^{-2} = − 12 ( 3 x + 5 ) − 2 = − 12 ( 3 x + 5 ) 2 = -\dfrac{12}{(3x + 5)^{2}} = − ( 3 x + 5 ) 2 12 Question 3
Answer: = − 1 ( 2 x − 1 ) 2 = \dfrac{-1}{(2x - 1)^{2}} = ( 2 x − 1 ) 2 − 1 [3] u = 3 x − 1 u = 3x - 1 u = 3 x − 1 v = 2 x − 1 v = 2x - 1 v = 2 x − 1 d u d x = 3 \dfrac{du}{dx} = 3 d x d u = 3 d v d x = 2 \dfrac{dv}{dx} = 2 d x d v = 2 d y d x = v d u d x − u d v d x v 2 \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^{2}} d x d y = v 2 v d x d u − u d x d v d y d x = ( 2 x − 1 ) ⋅ 3 − ( 3 x − 1 ) ⋅ 2 ( 2 x − 1 ) 2 \dfrac{dy}{dx} = \dfrac{(2x - 1) \cdot 3 - (3x - 1) \cdot 2}{(2x - 1)^{2}} d x d y = ( 2 x − 1 ) 2 ( 2 x − 1 ) ⋅ 3 − ( 3 x − 1 ) ⋅ 2 = − 1 ( 2 x − 1 ) 2 = \dfrac{-1}{(2x - 1)^{2}} = ( 2 x − 1 ) 2 − 1 Question 4
Answer: = 3 x 2 ( 3 x + 2 ) ( 5 x + 2 ) = 3x^{2}(3x + 2)(5x + 2) = 3 x 2 ( 3 x + 2 ) ( 5 x + 2 ) [3] u = x 3 u = x^{3} u = x 3 v = ( 3 x + 2 ) 2 v = (3x + 2)^{2} v = ( 3 x + 2 ) 2 d u d x = 3 x 2 \dfrac{du}{dx} = 3x^{2} d x d u = 3 x 2 d v d x = 6 ( 3 x + 2 ) \dfrac{dv}{dx} = 6(3x + 2) d x d v = 6 ( 3 x + 2 ) d y d x = u d v d x + v d u d x \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} d x d y = u d x d v + v d x d u d y d x = x 3 ⋅ 6 ( 3 x + 2 ) + ( 3 x + 2 ) 2 ⋅ 3 x 2 \dfrac{dy}{dx} = x^{3} \cdot 6(3x + 2) + (3x + 2)^{2} \cdot 3x^{2} d x d y = x 3 ⋅ 6 ( 3 x + 2 ) + ( 3 x + 2 ) 2 ⋅ 3 x 2 = 3 x 2 ( 3 x + 2 ) ( 5 x + 2 ) = 3x^{2}(3x + 2)(5x + 2) = 3 x 2 ( 3 x + 2 ) ( 5 x + 2 ) Question 5
(a) Answer: = ( 2 x + 1 ) 2 ( 8 x + 13 ) = (2x + 1)^{2}(8x + 13) = ( 2 x + 1 ) 2 ( 8 x + 13 ) [3] u = x + 2 u = x + 2 u = x + 2 v = ( 2 x + 1 ) 3 v = (2x + 1)^{3} v = ( 2 x + 1 ) 3 d u d x = 1 \dfrac{du}{dx} = 1 d x d u = 1 d v d x = 6 ( 2 x + 1 ) 2 \dfrac{dv}{dx} = 6(2x + 1)^{2} d x d v = 6 ( 2 x + 1 ) 2 d y d x = u d v d x + v d u d x \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} d x d y = u d x d v + v d x d u d y d x = ( x + 2 ) ⋅ 6 ( 2 x + 1 ) 2 + ( 2 x + 1 ) 3 ⋅ 1 \dfrac{dy}{dx} = (x + 2) \cdot 6(2x + 1)^{2} + (2x + 1)^{3} \cdot 1 d x d y = ( x + 2 ) ⋅ 6 ( 2 x + 1 ) 2 + ( 2 x + 1 ) 3 ⋅ 1 = ( 2 x + 1 ) 2 ( 8 x + 13 ) = (2x + 1)^{2}(8x + 13) = ( 2 x + 1 ) 2 ( 8 x + 13 ) (b) Answer: = ( x − 3 ) 3 ( 5 x − 19 ) = (x - 3)^{3}(5x - 19) = ( x − 3 ) 3 ( 5 x − 19 ) [3] u = x − 4 u = x - 4 u = x − 4 v = ( x − 3 ) 4 v = (x - 3)^{4} v = ( x − 3 ) 4 d u d x = 1 \dfrac{du}{dx} = 1 d x d u = 1 d v d x = 4 ( x − 3 ) 3 \dfrac{dv}{dx} = 4(x - 3)^{3} d x d v = 4 ( x − 3 ) 3 d y d x = u d v d x + v d u d x \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} d x d y = u d x d v + v d x d u d y d x = ( x − 4 ) ⋅ 4 ( x − 3 ) 3 + ( x − 3 ) 4 ⋅ 1 \dfrac{dy}{dx} = (x - 4) \cdot 4(x - 3)^{3} + (x - 3)^{4} \cdot 1 d x d y = ( x − 4 ) ⋅ 4 ( x − 3 ) 3 + ( x − 3 ) 4 ⋅ 1 = ( x − 3 ) 3 ( 5 x − 19 ) = (x - 3)^{3}(5x - 19) = ( x − 3 ) 3 ( 5 x − 19 ) Question 6
(a) Answer: = 2 x ( x − 2 ) 3 ( 3 x − 2 ) = 2x(x - 2)^{3}(3x - 2) = 2 x ( x − 2 ) 3 ( 3 x − 2 ) [3] u = x 2 u = x^{2} u = x 2 v = ( x − 2 ) 4 v = (x - 2)^{4} v = ( x − 2 ) 4 d u d x = 2 x \dfrac{du}{dx} = 2x d x d u = 2 x d v d x = 4 ( x − 2 ) 3 \dfrac{dv}{dx} = 4(x - 2)^{3} d x d v = 4 ( x − 2 ) 3 d y d x = u d v d x + v d u d x \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} d x d y = u d x d v + v d x d u d y d x = x 2 ⋅ 4 ( x − 2 ) 3 + ( x − 2 ) 4 ⋅ 2 x \dfrac{dy}{dx} = x^{2} \cdot 4(x - 2)^{3} + (x - 2)^{4} \cdot 2x d x d y = x 2 ⋅ 4 ( x − 2 ) 3 + ( x − 2 ) 4 ⋅ 2 x = 2 x ( x − 2 ) 3 ( 3 x − 2 ) = 2x(x - 2)^{3}(3x - 2) = 2 x ( x − 2 ) 3 ( 3 x − 2 ) (b) Answer: = 15 x + 4 2 5 x + 2 = \dfrac{15x + 4}{2\sqrt{5x + 2}} = 2 5 x + 2 15 x + 4 [3] u = x u = x u = x v = 5 x + 2 v = \sqrt{5x + 2} v = 5 x + 2 d u d x = 1 \dfrac{du}{dx} = 1 d x d u = 1 d v d x = 5 2 5 x + 2 \dfrac{dv}{dx} = \dfrac{5}{2\sqrt{5x + 2}} d x d v = 2 5 x + 2 5 d y d x = u d v d x + v d u d x \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} d x d y = u d x d v + v d x d u d y d x = x ⋅ 5 2 5 x + 2 + 5 x + 2 ⋅ 1 \dfrac{dy}{dx} = x \cdot \dfrac{5}{2\sqrt{5x + 2}} + \sqrt{5x + 2} \cdot 1 d x d y = x ⋅ 2 5 x + 2 5 + 5 x + 2 ⋅ 1 = 15 x + 4 2 5 x + 2 = \dfrac{15x + 4}{2\sqrt{5x + 2}} = 2 5 x + 2 15 x + 4 Question 7
(a) Answer: d y d x = 18 ( x − 1 ) 2 \dfrac{dy}{dx} = \dfrac{18}{(x - 1)^{2}} d x d y = ( x − 1 ) 2 18 [2] d y d x = ( x − 1 ) ( − 1 ) − ( − x − 17 ) ( 1 ) ( x − 1 ) 2 \dfrac{dy}{dx} = \dfrac{(x - 1)(-1) - (-x - 17)(1)}{(x - 1)^{2}} d x d y = ( x − 1 ) 2 ( x − 1 ) ( − 1 ) − ( − x − 17 ) ( 1 ) = 18 ( x − 1 ) 2 = \dfrac{18}{(x - 1)^{2}} = ( x − 1 ) 2 18 (b) [1]
( x − 1 ) 2 > 0 (x - 1)^{2} > 0 ( x − 1 ) 2 > 0 for every x ≠ 1 x \neq 1 x = 1 . The numerator 18 18 18 is positive, so d y d x > 0 \dfrac{dy}{dx} > 0 d x d y > 0 for all such x x x . (c) Answer: x = − 2 x = -2 x = − 2 or x = 4 x = 4 x = 4 [3] 18 ( x − 1 ) 2 = 2 \dfrac{18}{(x - 1)^{2}} = 2 ( x − 1 ) 2 18 = 2 ( x − 1 ) 2 = 9 (x - 1)^{2} = 9 ( x − 1 ) 2 = 9 x − 1 = 3 x - 1 = 3 x − 1 = 3 or x − 1 = − 3 x - 1 = -3 x − 1 = − 3 x = − 2 x = -2 x = − 2 or x = 4 x = 4 x = 4 (d) Answer: d 2 y d x 2 = − 36 ( x − 1 ) 3 \dfrac{d^{2}y}{dx^{2}} = -\dfrac{36}{(x - 1)^{3}} d x 2 d 2 y = − ( x − 1 ) 3 36 [2] d y d x = 18 ( x − 1 ) − 2 \dfrac{dy}{dx} = 18(x - 1)^{-2} d x d y = 18 ( x − 1 ) − 2 d 2 y d x 2 = 18 ( − 2 ) ( x − 1 ) − 3 ( 1 ) \dfrac{d^{2}y}{dx^{2}} = 18(-2)(x - 1)^{-3}(1) d x 2 d 2 y = 18 ( − 2 ) ( x − 1 ) − 3 ( 1 ) = − 36 ( x − 1 ) 3 = -\dfrac{36}{(x - 1)^{3}} = − ( x − 1 ) 3 36 Question 8
(a) [3]
d y d x = 3 x + 9 + ( x + 2 ) ⋅ 3 2 3 x + 9 \dfrac{dy}{dx} = \sqrt{3x + 9} + (x + 2) \cdot \dfrac{3}{2\sqrt{3x + 9}} d x d y = 3 x + 9 + ( x + 2 ) ⋅ 2 3 x + 9 3 = 2 ( 3 x + 9 ) + 3 ( x + 2 ) 2 3 x + 9 = \dfrac{2(3x + 9) + 3(x + 2)}{2\sqrt{3x + 9}} = 2 3 x + 9 2 ( 3 x + 9 ) + 3 ( x + 2 ) = 9 x + 24 2 3 x + 9 = \dfrac{9x + 24}{2\sqrt{3x + 9}} = 2 3 x + 9 9 x + 24 (shown) At x = 0 x = 0 x = 0 , d y d x = 9 ( 0 ) + 24 2 9 \dfrac{dy}{dx} = \dfrac{9(0) + 24}{2\sqrt{9}} d x d y = 2 9 9 ( 0 ) + 24 = 4 = 4 = 4 (c) Answer: x = − 8 3 x = -\dfrac{8}{3} x = − 3 8 [2] d y d x = 0 \dfrac{dy}{dx} = 0 d x d y = 0 where the numerator is zero: 9 x + 24 = 0 9x + 24 = 0 9 x + 24 = 0 x = − 8 3 x = -\dfrac{8}{3} x = − 3 8 (d) Answer: x > − 8 3 x > -\dfrac{8}{3} x > − 3 8 [2] 2 3 x + 9 > 0 2\sqrt{3x + 9} > 0 2 3 x + 9 > 0 for every x > − 3 x > -3 x > − 3 , so the sign of d y d x \dfrac{dy}{dx} d x d y is the sign of the numerator. 9 x + 24 > 0 9x + 24 > 0 9 x + 24 > 0 x > − 8 3 x > -\dfrac{8}{3} x > − 3 8