C3 Sample Paper Old Edexcel Textbook Question

Hi, I am really stuck on this. Please can someone show me how to do it!

f(x) = a/x
g(x) = ax + k

The line y = g(x) is a tangent to the curve y = |f(x)| at the point P, and cuts the curve y = |f(x)| at the point Q.

(b) Show that k = 2a.

Thank you!!

Reply 1

Zacken

22

Original post by 1a2b3c4d

Hi, I am really stuck on this. Please can someone show me how to do it!

f(x) = a/x
g(x) = ax + k

The line y = g(x) is a tangent to the curve y = |f(x)| at the point P, and cuts the curve y = |f(x)| at the point Q.

(b) Show that k = 2a.

Thank you!!

Can you show us the full question, please?

Reply 2

1a2b3c4d

OP

Original post by Zacken

Can you show us the full question, please?

That is the full question. Part (a) asks you to sketch the curve 'with equation y = |f(x)|.

Reply 3

1a2b3c4d

OP

Original post by Zacken

Can you show us the full question, please?

Reply 4

Zacken

22

Original post by 1a2b3c4d

That is the full question. Part (a) asks you to sketch the curve 'with equation y = |f(x)|.

*shrugs*

|f(x)| = g(x) \iff ax + k = \pm \frac{a}{x}

Now consider the two equations:

ax + k = \frac{a}{x} \Rightarrow ax^2 + kx - a = 0

or

ax + k = -\frac{a}{x} \Rightarrow ax^2 + kx + a = 0

For |f(x)| and g(x) to be tangent, one of these has to have a zero discriminant.

(edited 8 years ago)

Reply 5

1a2b3c4d

OP

Sorry could you take it one step further, I still can't quite get it..

Reply 6

Zacken

22

Original post by 1a2b3c4d

Sorry could you take it one step further, I still can't quite get it..

What is the discriminant of those two quadratic equations?

Reply 7

HapaxOromenon

2

Original post by 1a2b3c4d

Hi, I am really stuck on this. Please can someone show me how to do it!

f(x) = a/x
g(x) = ax + k

The line y = g(x) is a tangent to the curve y = |f(x)| at the point P, and cuts the curve y = |f(x)| at the point Q.

(b) Show that k = 2a.

Thank you!!

Following on from Zacken's post, the discriminants are:
(1) k^2-4a(-a) = k^2+4a^2, and (2) k^2-4a(a)=k^2-4a^2.
So in case 1, k^2=-4a^2, which is impossible as the LHS is positive and the RHS is negative.
Thus we go to case 2: k^2=4a^2 -> k=+/-2a, and since k and a are positive, k=2a, as required.

C3 Sample Paper Old Edexcel Textbook Question

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