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A1 Doubles Tracking Test 4 Part A(36 marks: 44 minutes)
1.
1
2.
Figure 2 shows part of the curve with equation y=f ( x ) .
The curve passes through the points P(-1.5, 0) and Q(0, 5) as shown.
On separate diagrams, sketch the curve with equation
(a) y=|f ( x )| (2)
(b) y=f (|x|) (2)
Indicate clearly on each sketch the coordinates of the points at which the curve crosses or meets the axes.
3. Find the integral
∫ cosxsin2x dx (3)
4. Prove, from first principles, that the derivative of cos3 x is −3sin 3 x.You may assume that as h→0 , sin 3h
h→3 and cos3h−1
h→0
(5)2
5.
(8)
6.
(5)
END OF TEST
3
Mark Scheme TT4 Part A1.
2.
4
3.
4.
5.
6.
5
∫cos x¿¿¿¿y=¿¿
dydx
=3 cos x¿¿M1A1
¿ 13
sin3 x+cA1
Let f ( x )=cos (3 x )
f ' ( x )= limh→0 ( f ( x+h )−f ( x )
h )¿ limh→ 0 ( cos (3 x+3h)−cos (3 x )
h )¿ limh→0 ( cos3 xcos3h−sin 3xsin3h−cos 3 x
h )¿ limh→ 0
(( cos 3h−1h )¿cos 3x−( sin 3h
h )sin 3x )¿
As h→0 ,( sin 3hh )→3and ( cos3h−1
h )→0 ,
So the expression in side the limit tends to0xcos3 x−3xsin 3 x=−3sin 3 xHence the derivative of cos (3 x )is−3sin (3 x )
M1A1
M1
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∫1+sin 2 xdx=¿[ x−12
cos2 x]0
π3 ¿
¿( π3 + 14 )−(−1
2)
¿ π3
+ 34
When x=π3, y=2+√3
2
Area of triangle ¿ 12
x π3
x2+√32
= 2π+√3π12
Shaded region = π3
+ 34
−2π+√3 π12
=
4 π+9−2π+√3 π12
¿ 112
(2π+9−π √3 ) *
M1 A1
M1A1
B1
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y= k x2−a
kx2+a
dydx
=(kx2+a ) (2kx )−(k x2−a)(2kx)
(kx2+a )2
dydx
= 4 akx(kx2+a )2
Turning point where dydx
=0 ,therefore 4 akx=0
So when x=0 , y=−aa
=−1
Single turning point occurs at (0 ,−1)
M1A1
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