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Transcript 
C3: Starters
Revise formulae and develop
problem solving skills.
1
2
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4
5
6
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24
DMO’L.St Thomas More
Starter 1
Solve the equation
3 cos 2  sin  1  0
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for
0    2
Starter 1
Solve the equation
3 cos 2  sin  1  0
for
3(1  2 sin  )  sin   1  0
2
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0    2
Starter 1
Solve the equation
for 0    2
3 cos 2  sin  1  0
2
3(1  2 sin  )  sin   1  0
6 sin   sin   2  0
(3 sin   2)( 2 sin   1)  0
2
1
sin    or sin  
3
2
2
  3.87,5.55, 6 ,

DMO’L.St Thomas More
5
6
Back
Starter 2
Prove the identity
sec   tan   sec   tan 
4
4
2
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2
Starter 2
Prove the identity
LHS  sec   tan 
4
4
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Starter 2
Prove the identity
LHS  sec 4   tan 4 
2
2
2
2
 (sec   tan  )(sec   tan  )
2
2
2
2
 (sec   tan  )(1  tan   tan  )
 (sec   tan  )
 RHS
2
2
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Back
Starter 3
Prove the identity
sin 3  sin 
 tan 
cos 3  cos 
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Starter 3
sin 3  sin 
LHS 
cos 3  cos 
2 cos( 32 ) sin( 32 )

cos 3  cos 
2 cos( 32 ) sin( 32 )

2 cos( 32 ) cos( 32 )
2 cos( 2 ) sin(  )

2 cos( 2 ) cos( )
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Starter 3
sin 3  sin 
LHS 
cos 3  cos 
2 cos( 32 ) sin( 32 )

cos 3  cos 
2 cos( 32 ) sin( 32 )

2 cos( 32 ) cos( 32 )
Back
2 cos( 2 ) sin(  )

 tan   RHS
2 cos( 2 ) cos( )
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Starter 4
Given that tan A  34 and sin B  135
where A is acute and B is obtuse,
find
cos ec( A  B )
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Starter 4
tan A 
3
4
sin B  135
By Pythagoras
sin A 
3
5
tan B  125 125
cos A 
4
5
12
cos B  12
13 13
A is
acute
B is
obtuse
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Starter 4
1
cos ec( A  B) 
sin( A  B)
1

sin A cos B  cos A sin B
1
 3
5
12
4




5
13
5
13
1
 36 20
 65  65
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Starter 4
1
cos ec( A  B) 
sin( A  B)
Back
1

sin A cos B  cos A sin B
1
 3
5
12
4




5
13
5
13
1
65
 36 20  
 65  65
16
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Starter 5
Differentiate
y  e sin x
2x
ye
sin x cos x
y
ln x
sin 2 x
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Starter 5
Differentiate
y  e sin x
2x
ue
du
dx
2x
 2e
v  sin x
2 x dv
dx
dy
2x
2x
 2e sin x  e cos x
dx
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 cos x
Starter 5
Differentiate
ye
sin 2 x
du
dx
u  sin 2x
ye
 2 cos 2 x
dy
du
e
dy
u
sin 2 x
 e .2 cos 2 x  2e
cos 2 x
dx
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u
u
Starter 5
Differentiate
sin x cos x
y
ln x
u  sin x cos x
2
2
du
dx  cos x  sin x
du
dx  cos 2 x
dy ln x cos 2 x  1x sin x cos x

2
dx
(ln x)
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v  ln x
dv
1

dx
x
Starter 5
Differentiate
sin x cos x
y
ln x
u  sin x cos x
2
2
du
dx  cos x  sin x
du
dx  cos 2 x
dy ln x cos 2 x  21x sin 2 x

2
dx
(ln x)
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v  ln x
dv
1

dx
x
Starter 5
Differentiate
sin x cos x
y
ln x
u  sin x cos x
2
2
du
dx  cos x  sin x
du
dx  cos 2 x
dy 2 x ln x cos 2 x  sin 2 x

2
dx
2 x(ln x)
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v  ln x
dv
1

dx
x
Back
Starter 6
Differentiate
y  ln(sec x)
y  (1  sin 3x)
cos 2 x
y  sin x
e
7
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Starter 6
Differentiate
y  ln(sec) x u  sec x  (cos x)
du
dx
du
dx
2
 (cos x) sin x

sin x
cos2 x
dy 1 sin x
 .
2
dx u cos x
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1
y  ln u
dy
du

1
u
Starter 6
Differentiate
y  ln(sec x) u  sec x  (cos x)
du
dx
du
dx
2
 (cos x) sin x

1
y  ln u
dy
du

sin x
cos2 x
dy 1 sin x sin x cos x
 .

 tan x
2
2
dx u cos x
cos x
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1
u
Starter 6
Differentiate
y  (1  sin 3x)
7
u  1 sin 3x
du
dx
 3 cos 3x
y u
dy
du
7
 7u
6
dy
6
6
 7u .3 cos 3x  21(1  sin 3x) cos 3x
dx
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Starter 6
Differentiate
cos 2 x
y  sin x
e
dy e

dx
sin x
u  cos 2x
du
dx  2 sin 2 x
ve
sin x
dv
dx  cos x.e
sin x
(2 sin 2 x)  cos 2 x(cos x.e
2 sin x
e
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sin x
)
Starter 6
Differentiate
cos 2 x
y  sin x
e
u  cos 2x
du
dx  2 sin 2 x
ve
sin x
dv
dx  cos x.e
sin x
Back
dy
sin x
 e
(2 sin 2 x  cos 2 x cos x)
dx
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Starter 7
Solve the following equations, giving exact
solutions
e
2x
7
ln( 2 x  1) 2  5
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Starter 7
Solve the following equations, giving exact
solutions
e 7
2x  ln 7
x  12 ln 7
2x
Back
ln( 2 x  1)  5
2 ln( 2 x  1)  5
5
ln( 2 x  1) 
2
5
2x 1  e 2
5
e 2 1
x
2
2
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Starter 8
Show that x 2  5 x  3  0 can in be written in the
form x  5 x  3
Use the iteration xn1  5xn  3
starting with x0  5 to generate x1 , x2 , x3 , x4 , x5 , x6
Show that 5.5 is a root of the equation to
one decimal place.
x  5x  3
2
x  5x  3
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Starter 8
x0  5
Use the iteration xn1  5xn  3
5.2915
x1 
x0  5 to generate x1 , x2 , x3 , x4 , x5 , x6
starting
with
Show that
is a root of the equation to
5.4275
x2  5.5
one decimal place.
Calculator:
x3  5.489
x0  5
x4  5.518
x1  5.2915
x5  5.530
x6  5.537
5=
5Ans+3====
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Starter 8
Show that 5.5 is a root of the equation to
one decimal place.
f ( x)  x  5 x  3
2
f (5.55)  0.525
f (5.45)  0.547
Change of sign  Root between 5.55 and 5.45
Hence, x = 5.5 is a root to 1 decimal place.
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Back
Starter 9
Sketch the graph y  2 x  5
Hence, or otherwise, solve 2 x  5  5
y=5
when x = 3
x = 0 or 5
y  2x  5
y=2x-5
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Back
Starter 10
By differentiating find the coordinates of
the turning point on the curve y  x 3  81ln x
State the nature of the turning point (i.e.
maximum or minimum).
dy
81
2
 3x   0
For turning points
dx
xBack
when xd =2 y3
2
d y81
81
 6 x 2 2 6 x  2  0
2
dx
dx x
x
3x 3  81
Hence, minmum point at (3,-61.99)
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x3
Starter 11
Solve the following equations, giving exact
solutions
e
2x
x 2 x4
e e
ln( 5 x  3)  12
2
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Starter 11
Solve the following equations, giving exact
solutions
e
2x
e
2x
x 2 x4
e e
e
3x4
Take logs base e
2x  3x  4
x  4
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Starter 11
Solve the following equations, giving exact
solutions
ln( 5 x  3)  12
2 ln( 5 x  3)  12
ln( 5 x  3)  6
2
e to the power of
5x  3  e
6
e 3
x
5
6
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Back
Starter 12
kx
Complete the table:
Back
dy
dx
y
n
nkx
n 1
sin x
cos x
cos x
 sin x
tan x
sec 2 x
sec x
sec x tan x
ln x
1
x
ex
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a
x
e
x
x
a ln a
Starter 13
Complete the table:
(3 x  1)
2
sin 2 x
3
cos x
Back
dy
dx
y
n
6nx(3 x  1)
2
n 1
2 cos 2 x
 3 cos x sin x
2
2
tan 4 x
4 sec 4 x
sec 7 x
7 sec 7 x tan 7 x
ln 6 x
1
x
e
6x
35 x
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6e
6x
5(35 x ) ln 3
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