Re: Cambridge specimen papers,Qs
I'm assuming that you meant the Polynomial questions only.
I will do Spec 2 Q7 (a) and you can use the method to try Spec 1 Q7 (a) for yourself.
Show that the remainder when the polynomial P(x) is divided by x² - a² is (1/2a)*{P(a) - P(-a)}x + (1/2)*{P(a) + P(-a)}
Let P(x) = (x² - a²)*Q(x) + R(x) where Q(a) * 0, Q(-a) * 0
[Note: The remainder has to have degree ≤ 2, i.e. R(x) = mx + n)]
When x = a, remainder = P(a) = ma + n (1)
When x = -a, remainder = P(-a) = -ma + n (2)
(1) + (2): 2n = P(a) + P(-a)
.: n = (1/2)*{P(a) + P(-a)}
(1) - (2): 2ma = P(a) - P(-a)
.: m = (1/2a)*{P(a) - P(-a)}
Hence R(x) = (1/2a)*{P(a) - P(-a)}x + (1/2)*{P(a) + P(-a)} as required
Find the remainder when P(x) = x^n - a^n is divided by x² - a² in each of the cases (i) n even (ii) n odd
(i) When n is even, P(-x) = (-x)^n - a^n = x^n - a^n = P(x)
.: P(a) = P(-a)
Hence R(x) = (1/2a)*{P(a) - P(a)}x + (1/2)*{P(a) + P(a)} = P(a) = 0
(ii) When n is odd P(-x) = (-x)^n - a^n = -(x^n + a^n)
.: P(a) = a^n - a^n = 0 & P(-a) = -(a^n + a^n) = -2a^n
Hence R(x) = (1/2a)*{0 + 2a^n}x + (1/2)*{0 - 2a^n}
= [a^(n-1)]x - a^n
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In Spec 1 Q7 (a), the presence of multiple roots must be considered.
Differentiate P(x)= (x - a)²*Q(x) + R(x) where Q(a) * 0, R(x) = mx + n to obtain
P'(x) = 2(x - a)*Q(x) + (x - a)²*Q'(x) + R'(x)
= (x - a){2Q(x) + (x - a)*Q'(x)} + m
= (x - a)*S(x) + m where S(a) * 0.
Then proceed by the same principle as above, etc.
