End of chapter test
10 Analyzing rates of change:
differential calculus
All questions can be done by using a GDC.
1 Show that the tangent y  x2 and y 
2 Let f (x) 
ax  a
,a 
xa
coordinate x0  0 is
*
1
1
at their common point are perpendicular.

2x 2
. Find a such that the gradient of the curve f at the point with x -
1
.
2
3 We consider a quadratic curve f (x)  ax2  bx  c , where a, b, c 
. Find a, b, c such that the
curve passes through point A(1,2) and the line y  x is the tangent to the curve at the origin.
4 Show that the straight line y  3x  2 has two common points with the curve y  x3 . Also,
show that the above straight line is tangent to the curve at one of the above common points.
5 If f has second derivative on
and f (x2 )  xf (x) , find the value of f (1) .
6 Show that the curves with equation:
e x  e x
2
e x  e x

C2 : y 
sin(x), 0  x 
2
2
C1 : y 
have the same tangent at their common point.
7 We consider function f (x)  sin2(x), x 
. Prove that, f (x)  4f (x)  2, x 
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.
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End of chapter test
Answers
1 Finds the point of intersection: x 2 
1
1
  x  1, y  1
2x 2
Finds the tangent line to the y  x2 at point (1,1) : m1 
Finds the tangent line to the y 
 2x x 1  2

x 1
1
1

2x x 1
2
1
)  1 , the tangent lines meet perpendicularly.
2
a(x  a)  (ax  a)
a2  a
.

2
(x  a)
(x  a)2
From hypothesis we have: f (0) 
Obtains:
x 1
1
1
dy

at the point (1,1) : m2 
2x 2
dx
Therefore, because m1  m2  2  (
2 Finds derivative: f (x) 
dy
dx
1
.
2
1
a2  a

 a2  2(a2  a)  a2  2a  0  a(a  2)  0  a  0 or a  2 .
2 (0  a)2
Needs to reject a  0 .
Hence, a  2
3 Since A(1,2) lies on the curve then: f (1)  2  2  a  12  b  1  c  2  a  b  c .
Also, (0,0) lies on the curve. Then: f (0)  0  0  a  02  b  0  c  c  0 .
But, (0,0) is the touching point of the tangent line to the curve, which line has gradient 1.
Hence, f (0)  1 . But, f (x)  2ax  b . So, 1  2a  0  b  b  1 .
Finally, we get: 2  a  1  0  a  1 .
So, f (x)  x2  x is the required curve.
4 Attempts to solve the simultaneous equations:
y  x3
y  3x  2
 3x  2  x3  x3  3x  2  0
The latter equation has x  1 as solution.
x2  x  2
Attempts long division: x  1 x 3  3x  2
Therefore, x3  3x  2  (x  1)(x2  x  2)  (x  1)(x  1)(x  2)  (x  1)2(x  2) .
Hence,
x 1
y 1
or x  2
.
or y  8
Therefore, points: A(1,1), B(2, 8) .
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End of chapter test
Finds tangent lines at both points:
y   3x 2
At point A(1,1) :
m  3  12  3
y  mx  c  y  3x  c  1  3  1  c  c  2
y  3x  2
y   3x 2
At point B(2, 8) :
m  3  (1)2  12
y  mx  c  y  12x  c  8  12  (2)  c  c  16
y  12x  16
Therefore, at point A the straight line is tangent to the curve.
5 Differentiating at both sides once: f (x2 )2x  f (x)  xf (x) - Correct use of Product Rule and
Chain Rule
Differentiating at both sides again: f (x2 )4x2  f (x2 )2  f (x)  f (x)  xf (x) - Correct use of
Product Rule and Chain Rule.
Substituting x  1 in the latter equation, we obtain:
f (12 )  4  1  f (12 )  2  f (1)  f (1)  1  f (1)
4f (1)  2f (1)  f (1)  f (1)  f (1)
3f (1)  0  f (1)  0
6 Finds the common point of the curves:
e x  e x
0
2
or
e x  e x
e x  e x
e x  e x

sin(x) 
(1  sin(x))  0
2
2
2
sin(x)=1
impossible equation
x 

2


e2  e
Obtains: x 
and y 
2
2


2

 e2  e
Hence, common point: A( ,
2


2
2
)

 e2  e
Finds tangent to C1 at point A( ,
2

e2  e
Tangent line: y 
2


2

e2  e

2


2
2


2
(x 

 e2  e
Finds tangent to C2 at point A( ,
2

2


2
2
e x  e x
) : y 
2

x

e2  e

2


2
2
)

e x  e x
ex  e x
e2  e
) : y 
sin(x) 
cos(x)

2
2
2

x


2
2

e2  e
Tangent line: y 
2


2

e2  e

2


2
(x 

2
)
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End of chapter test
Therefore, the curves have the same tangent line at their common point.
7 Finds the first derivative: f (x)  2 sin(x)cos(x) .
Finds the second derivative: f (x)  2 cos(x)cos(x)  2 sin(x)( sin(x))  2 cos2(x)  2 sin2(x) .
LHS: 2 cos2(x)  2 sin2(x)  4 sin2(x)  2 cos2(x)  2 sin2(x)  2 .
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