Question 5:
5 | (i) | s(x) = f(x) + g(x)
= f(x) + g(x)
= (f(x) + g(x))
= s(x) ( so s is odd ) | M1
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[2] | must have s(−x) = …
(ii) | p(x) = f(x)g(x)
=( f(x))  (g(x))
= f(x)g(x) = p(x)
so p is even | M1
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[2] | must have p(−x) = …
Allow SC1 for showing that p(−x) = p(x)
using two specific odd functions, but in this
case they must still show that p is even | e.g. f(x) = x, g(x) = x3, p(x) = x4
p(−x) = (−x)4 = x4 = p(x) , so p even
condone f and g being the same
function
6 (i) f(–x) = f(x)
Symmetrical about Oy. | B1
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[2]
(ii) (A) even
(B)neiher
(C)od | B1
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[3]
7 (i) (A)
−π/2 π/2 π
(B)
−2π −π π | B1
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[4] | Zeros shown every π/2.
Correct shape, from −π to π
Translated in x-direction
π to the left
(ii) f′(x)=− 1 e − 1 5 x sinx+e − 1 5 x cosx
5
1 − 1 x − 1 x
f′(x)=0when − e 5 sinx+e 5 cosx=0
5
⇒ 1 e − 1 5 x (−sinx+5cosx)=0
5
⇒sin x = 5 cos x
⇒ sinx
=5
cosx
⇒tan x = 5*
⇒x = 1.37(34…)
⇒y = 0.75 or 0.74(5…) | B1
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[6] | 1
− x
e 5 cosx
1 − 1 x
...− e 5 sinx
5
1
− x
dividing by e 5
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1.4 or better, must be in radians
0.75 or better
1
− (x+π)
(iii) f(x+π)=e 5 sin(x+π)
1 1
− x − π
= e 5 e 5 sin(x+π)
1 1
− x − π
= −e 5 e 5 sinx
1
= −e − 5 π f(x)*
2π let u = x − π, du = dx
∫ f(x)dx
π
= π
∫ f(u+π)du
0
1
= π − π
∫ −e 5 f(u)du
0
1
= − π π *
−e 5 ∫ f(u)du
0
Area enclosed between π and 2π
1
= (−)e − 5 π× area between 0 and π. | M1
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[8] | 1 1 1
− (x+π) − x − π
e 5 =e 5 .e 5
sin(x + π) = −sin x
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∫ f(u+π)du
limits changed
using above result or repeating work
or multiplied by 0.53 or better