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junior college 1 | H2 Maths
3 Answers Below
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hello can anyone help me with this question?
If f(x) is even, f(-x) = f(x)
(p(-x) + q)/((-x) + r) = (px + q)/(x + r)
(q - px)(x + r) = (px + q)(r - x)
qx + qr - px² - pxr = pxr - px² + qr - qx
2qx - 2pxr = 0
qx - prx = 0
x(q - pr) = 0
x = 0 or q - pr = 0 → q = pr
So when x = 0, f(x) = f(0)
= (p(0) + q)/(0 + r)
= q/r
When q = pr
f(x) = (px + pr)/(x + r)
= p(x + r)/(x + r)
= p
(Proved)
when x = a, f(x) = f(a)
When x = b, f(x) = f(b)
When x = f(a), f-¹(x)
= f-¹(f(a))
= a
When x = f(b), f-¹(x)
= f-¹(f(b))
= b
Since a < b,
f-¹(f(a)) < f-¹(f(b))
Since f(a) < f(b),
This means f(a) < f(b) ⇔f-¹(f(a)) < f-¹(f(b))
See 3 Answers
That eliminates a possible solution, whereby x = 0