# M2 BK1 Ans

Topics: Second derivative test, First derivative test, Fermat's theorem Pages: 234 (8798 words) Published: February 26, 2014
7

New Progress in Senior Mathematics Module 2 Book 1 (Extended Part) Solution Guide

7

dy 1
=
dx x

7.2

Applications of
Differentiation

dy
dx

p.219

=1
x =1

Slope of the normal = −1
The equation of the normal is
y − 4 = (−1)( x − 1)
x+ y −5 = 0

pp.233 – 239
p.233
(a)

Even function

(b)

Odd function

(c)

Odd function

(d)

Neither even nor odd

(e)

Even function

(f)

Odd function

7.3

xy + x 2 = −9
y=

p.220

−9− x
x

2

d
d
(−9 − x 2 ) − (−9 − x 2 ) ( x)
dx
dx
x2
2
9− x
=
x2

dy
=
dx

x

Slope of the line 4 x − 3 y − 12 = 0 is −

p.234
(a) x-intercepts = −6, 1; y-intercept = −6
(b)

x-intercepts = 1, 2; y-intercept = −4

(c)

x-intercept = 3; y-intercept =

(d)

x-intercept = 4; y-intercepts = −4, 4

(e)

3
dy 9 − x 2
=
=−
4
dx
x2
36 − 4 x 2 = −3 x 2

x-intercepts = −6, 6; y-intercepts = −3, 3

9
4

x 2 = 36
x = −6 or 6
When x = −6, y =
When x = 6, y =

(f)

no x-intercept; no y-intercept

p.239
(a)
x = −4

(b)

x = 1, x = −7

(c)

x = 8, x = −4

y = 2x + 7
1
dy
d
(2 x)
=
dx 2 2 x + 7 dx
1
=
2x + 7
dy
dx

1
3

The equation of the tangent is
1
y − 3 = ( x − 1)
3
x − 3y + 8 = 0

© Hong Kong Educational Publishing Co.

15
− 9 − 62
=− .
6
2

15 

The points of tangency are  − 6,
 and
2

15 

 6, − .
2

15 

At the point  6, − , the equation of the tangent is
2

3
 15 
y −  −  = − ( x − 6)
2
4

3 x + 4 y + 12 = 0

2(1) + 7

x =1

=

p.218

1

=

− 9 − (−6) 2 15
= .
−6
2

15 

At the point  − 6,
, the equation of the tangent is
2

3
 15 
y −   = − [ x − (−6)]
4
 2
3 x + 4 y − 12 = 0

pp.218 – 263

7.1

4
4
= .
(−3) 3

178

Applications of Differentiation

7.4

d 2
d
(x + 2 y 2 ) =
(36)
dx
dx
dy
2x + 4 y
=0
dx
dy
x
=−
2y
dx
Let ( x1 , y1 ) be the point of tangency.
dy
Slope of the tangent at ( x1 , y1 ) =
dx

p.221

7.6

dy
= 6 x 2 + 18 x − 24
dx
= 6( x + 4)( x − 1)
dy
=0
dx
6( x + 4)( x − 1) = 0

For stationary points,

x = −4 or 1
When x = −4, y = 2(−4) 3 + 9(−4) 2 − 24(−4) − 96 = 16
When x = 1, y = 2 + 9 − 24 − 96

( x1 , y1 )

x
=− 1
2 y1

= −109

2y
Slope of the normal at ( x1 , y1 ) = 1
x1

x

−4 < x < 1

x >1

+

+

dy
changes from + to – as x
dx
increases through −4,
Since the sign of

2 y1 y1 − 0
=
x1
x1 − 1
2 x1 y1 − 2 y1 = x1 y1

(−4, 16) is a maximum point.
dy
changes from – to + as x
Since the sign of
dx

y1 ( x1 − 2) = 0
y1 = 0 or x1 = 2
When y1 = 0, x1 = ± 36 − 0 = ±6

increases through 1,

(1, −109) is a minimum point.

The equation of the normal is
2(0)
2(0)
y −0 =
( x − 1) and y − 0 =
( x − 1)
−6
6
y=0
When x1 = 2, y1 = ±

x < −4

dy
dx

On the other hand, the slope of the normal is
y −0