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

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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

M1

A1

∫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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A1

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

A1

M1

A1

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