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junior college 1 | H2 Maths
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Elle
Elle

junior college 1 chevron_right H2 Maths chevron_right Singapore

hello can anyone help me with this question?

Date Posted: 4 years ago
Views: 644
J
J
4 years ago
f(x) = (px+q)/(x + r)

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)
J
J
4 years ago
a < b ⇔f(a) < f(b)

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

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Heng Hong Hwee
Heng Hong Hwee's answer
15 answers (A Helpful Person)
1st
for part (a) i think it is like that? since f(x) = f(-x) then just subst x with -x then equate and simplify
J
J
4 years ago
x should not be crossed out on both sides.
That eliminates a possible solution, whereby x = 0
Elle
Elle
4 years ago
thank you !!
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Heng Hong Hwee
Heng Hong Hwee's answer
15 answers (A Helpful Person)
ok idk how to use all the math terms and all but the geist it this. since f(x) is strictly increasing, for a
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Heng Hong Hwee
Heng Hong Hwee's answer
15 answers (A Helpful Person)
oh yaaa !! there is another answer maybe?? since the qn say for some values of k so k is either q or q/r