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# Edexcel maths c3 practice paper 2 question 3 on functions watch

The function f is even and has domain ℝ. For x ≥ 0, f(x) = x^2– 4ax, where a is a positive
constant.

(a) In the space below, sketch the curve with equation y = f(x), showing the coordinates of
all the points at which the curve meets the axes. (3)

(b) Find, in terms of a, the value of f(2a) and the value of f(–2a). (2)

Given that a = 3,

(c) use algebra to find the values of x for which f(x) = 45. (4)
2. Have you actually attempted any of it?

For part (a), start by factorising the equation to find where it crosses the x-axis.

For part (b), simply replace the x values in the equation with the values in brackets.

For part (c), rewrite the equation with 'a' equaling 3, set it equal to 45 and then solve for x from there.
3. a)Equate to zero. factorise out the x. It crosses at (0,0) and 4a.
b)f(2a)= 4a^2 - 8a^2 = -4a^2. Do the same for f(-2a)
c) x^2-12x-45=0 then solve.
Have you actually attempted any of it?

For part (a), start by factorising the equation to find where it crosses the x-axis.

For part (b), simply replace the x values in the equation with the values in brackets.

For part (c), rewrite the equation with 'a' equaling 3, set it equal to 45 and then solve for x from there.
Yeah I have, this is the mark scheme:

Q3:
A)I don't understand why there's a line from (0,0) to (-4a,0)

B)Why is f(2a) and f(-2a) equal to -4a^2? because it's an even function?

C)why is it plus or minus 15?
Attached Files
5. C3 Practice Paper A2 mark scheme.doc (108.5 KB, 150 views)
6. (Original post by sytner9)
a)Equate to zero. factorise out the x. It crosses at (0,0) and 4a.
b)f(2a)= 4a^2 - 8a^2 = -4a^2. Do the same for f(-2a)
c) x^2-12x-45=0 then solve.
7. (Original post by AS777)
Yeah I have, this is the mark scheme:

Q3:
A)I don't understand why there's a line from (0,0) to (-4a,0)

B)Why is f(2a) and f(-2a) equal to -4a^2? because it's an even function?

C)why is it plus or minus 15?
(a) f(x) = x^2– 4ax only applies when x is greater than or equal to zero so you only know what the graph looks like on the right hand side of the y-axis. You're told the function is even, however, so you know it has to be symmetrical about the y-axis, allowing you to fill in the left hand side.

(b) Again, you know it's an even function so f(x)=f(-x)

(c) -3 can't be a solution as the domain only contains x values greater than or equal to zero. Then you just refer to the reflection of the graph in the y-axis to know that if +15 is a solution, -15 must also be a solution.
(a) f(x) = x^2– 4ax only applies when x is greater than or equal to zero so you only know what the graph looks like on the right hand side of the y-axis. You're told the function is even, however, so you know it has to be symmetrical about the y-axis, allowing you to fill in the left hand side.

(b) Again, you know it's an even function so f(x)=f(-x)

(c) -3 can't be a solution as the domain only contains x values greater than or equal to zero. Then you just refer to the reflection of the graph in the y-axis to know that if +15 is a solution, -15 must also be a solution.
Thank you so much!

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