Maths

4k3 - 36k2 + 72k = 32, find k.

4k3 - 36k2 + 72k = 32
4k3 - 36k2 + 72k - 32 = 0
k3 - 9k2 + 18k - 8 = 0

Let f(k) = k3 - 9k2 + 18k - 8
f(2) = (2)3 - 9(2)2 + 18(2) - 8 = 0
(k - 2) is a factor of f(k).
f(k) = (k - 2)(k2 - 7x + 4)

(k - 2)(k2 - 7x + 4) = 0
k = 2 ooro k = {7 + √[72 - 4(1)(4)]}/2 ooro k = {7 - √[72 - 4(1)(4)]}/2
k = 2 ooro k = (7 + √33)/2 ooro k = (7 - √33)/2

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If f(x)=x/(x+1), then f[x/(x+1)]=

f(x) = x/(x + 1)

f[x/(x + 1)]
= [x/(x + 1)]/{[x/(x + 1)] + 1}
= [x/(x + 1)]/{[x/(x + 1)] + [(x + 1)/(x + 1)]}
= [x(x + 1)]/[(2x + 1)/(x + 1)]
= [x(x + 1)] x [(x+1)/(2x + 1)]
= x/(2x + 1)

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Compare x^2006/(x-1) and 6^2006 / (6-1) to find the remainder when 6^2006 is divided by 5.

When x = 6,
x2006/(x - 1) = 62006/(6 - 1)

x2006/(x - 1):
When x2006 is divided by (x - 1),
the remainder is (1)2006 = 1

62006/5 = x2006/(6 - 1):
When 62006 is divided by (6 - 1),
the remainder = 1

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Let k be a non-zero constant. When x3 + kx2 + 2kx + 3k is divided by x + k, the remainder is k. Find k.

f(x) = x3 + kx2 + 2kx + 3k
When f(x) is divided by x + k, remainder = f(-k)

f(-k) = k
(-k)3 + k(-k)2 + 2k(-k) + 3k = k
-k3 + k3 - 2k2 + 3k = k
-2k2 + 2k = 0
-2k(k - 1) = 0
k = 0 or k = 1
=

4k^3 - 36k^2 + 72k = 32
k^3 - 9k^2 + 18k - 8 = 0
(k - 2)(k^2 - 7k + 4) = 0
k=2 or k=[7 +/- √33]/2

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f(x) = x/(x+1)

f(x/(x+1)) = [x/(x+1)] / [x/(x+1) + 1]
= x/(2x+1)

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By Remainder Theorem, when x^2006 is divided by (x-1),
Remainder = 1^2006 = 1

Putting x=6 into the above result,
when 6^2006 is divided by (6-1)=5, the remainder is 1.

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Let f(x) = x^3 + kx^2 + 2kx + 3k

By Remainder Theorem, when f(x) is divided by (x+k),
Remainder = f(-k) = -k^3 + k^3 - 2k^2 + 3k = -2k^2 + 3k

Now given Remainder = k, thus
-2k^2 + 3k = k
k^2 - k = 0
k(k-1) = 0
k=0 (rejected since k is non-zero constant) or k=1
Hence, k=1.