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DP1 Maths AA Revision CH 4.1 - 4.4 MS

This document contains solutions to three math problems involving calculus concepts. Problem 1 involves finding the points of intersection and tangency between a curve and a line. Two methods are provided, both yielding the solutions m = ±2. Problem 2 finds the point of inflection and maximum of a curve. The point of inflection is (-0.803, -2.08) and the maximum occurs at the same x-value. Problem 3 examines the graph of a piecewise function. Part a derives the derivative and finds critical points. Part b sketches the graph and locates points of inflection. Part c determines the range of the original function and its composition with another function.

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79 views3 pages

DP1 Maths AA Revision CH 4.1 - 4.4 MS

This document contains solutions to three math problems involving calculus concepts. Problem 1 involves finding the points of intersection and tangency between a curve and a line. Two methods are provided, both yielding the solutions m = ±2. Problem 2 finds the point of inflection and maximum of a curve. The point of inflection is (-0.803, -2.08) and the maximum occurs at the same x-value. Problem 3 examines the graph of a piecewise function. Part a derives the derivative and finds critical points. Part b sketches the graph and locates points of inflection. Part c determines the range of the original function and its composition with another function.

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武佳萱
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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DP1 Maths AA Revision Ch 4.1 – 4.

4 MS

1. Method 1: y = 4 – x2
dy m
= –2x = m when x = – (M1)
dx 2
 m m 
2
Thus,  − ,4 −  lies on y = mx + 5. (R1)
 2 4 
m2 m2
Then, 4 – =− + 5, so m2 = 4
4 2
m = 2. (A1) (C3)
Method 2: For intersection: mx + 5 = 4 – x2 or x2 + mx + 1 = 0. (M1)
For tangency: discriminant = 0 (M1)
Thus, m2 – 4 = 0
m = 2 (A1) (C3)
[3]

2. METHOD 1
2
f  ( x) = 4 x 3 − (M1)(A1)
x2
4
f  ( x) = 12 x 2 + (A1)
x3
f  ( x) = 0 (M1)

1
 x = − 5 = −0.803 and y = −2.08 (accept −2.07 ) (A1)(A1)
3

  1 5 5 
The point of inflexion is (−0.803, − 2.08)  or  − 5 , − 3 3  (C5)(C1)
  3 
METHOD 2
2
f  ( x) = 4 x 3 − (M1)(A1)
x2
f  ( x) has a maximum when x = −0.803 (M1)(A2) (C5)
y = −2.08 (accept −2.07 ) (A1) (C1)
[6]

(2 x – 1)(x 2 + x + 1) – (2 x + 1)(x 2 – x + 1)
(x )
3. (a) (i) f '(x) = 2
(M1)(A1)
2
+ x +1
(
2 x2 – 1)
(x + x + 1)
= 2 (A1)
2

(ii) f (x) = 0 => x = ±1


 1   1 
A 1,  B(–1, 3)  or A (–1, 3) B1,   (A1)(A1) 5
 3   3 

1
(b) (i)
y

–1 1 x

–2

(G2)
Note: Award (G1) for general shape
and (G1) for indication of scale.
(ii) The points of inflexion can be found by locating the max/min
on the graph of f'.
This gives x = –1.53,–0.347,1.88 . (G3)
OR

f (x) =
(
– 4 x3 – 3x – 1 )
(x )3
(M1)
2
+ x +1
f (x) = 0  x3 – 3x – 1 = 0 (A1)
 x = 1.53, –0.347, 1.88 (G1) 5
(c) The graph of y = f(x) helps:

1/3
–1 +1

2
1 
(i) Range of f is  ,3 . (A1)(A1)
3 
1 
(ii) We require the image set of  ,3 .
3 
1 1
– +1
 
1 7 9 – 3 +1 7
f  = 9 3 = , f(3) = = (M1)
 3 1 + 1 +1 13 9 + 3 +1 13
9 3
1 7 
Range of g is  ,  . (A1)(A1) 5
 3 13 
Note: Since the question did not specify exact ranges accept
open intervals or numerical approximations
(e.g. [0.333, 0.538]).
[15]

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