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The following videos will give you the worked solutions and answers for the Edexcel GCE Core Mathematics C3 Advanced June 2012. The questions are given here.

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C3 Edexcel Core Mathematics June 2012 Question 6

Range of a Function

6. The functions f and g are defined by

f: x |→ e^{x} + 2 , x ∈ ℜ

g : x |→ ln x , x > 0

(a) State the range of f.

(b) Find fg(x) , giving your answer in its simplest form.

(c) Find the exact value of x for which f(2x +3) = 6

(d) Find f−1 , the inverse function of f, stating its domain.

(e) On the same axes sketch the curves with equation y = f(x) and y = f^{-1}(x) coordinates of all the points where the curves cross the axes.

7. (a) Differentiate with respect to x,

(i) x^{1/2}ln(3x)

(ii) (1 - 10x)/(2x - 1)^{5} giving your answer in its simplest form.

(b) Given that y = 3 tan 2y, find dy/dx, in terms of x.

7 (a)(i) Product Rule/Chain Rule7 (a)(ii) Quotient Rule/Chain Rule

C3 Edexcel Core Mathematics June 2012 Question 8

8. f(x) = 7 cos 2x - 24 sin 2x

Given that f(x) = R cos(2x + α) , where R > 0 and 0 < α < 90°,

(a) find the value of R and the value of α.

(b) Hence solve the equation

f(x) = 7 cos 2x - 24 sin 2x = 12.5

for 0 ≤ x < 180° , giving your answers to 1 decimal place.

(c) Express 14 cos^{2}x - 14 sinx cosx in the form a cos 2x + b sin 2x + c, where a, b, and c are constants to be found.

(d) Hence, using your answers to parts (a) and (c), deduce the maximum value of

14 cos^{2}x - 14 sinx cos

8 (a) Rcos( ) method

8 (c)

C3 Edexcel Core Mathematics June 2012 Question 9

Transformations of graphs (mod types)

Figure 2 shows part of the curve with equation y = f(x)

The curve passes through the points P(−1.5, 0) and Q(0, 5P) as shown.

On separate diagrams, sketch the curve with equation

(a) y = |f(x)|

(b) y = f(|x|)

(c) y = 2f(3x)

Indicate clearly on each sketch the coordinates of the points at which the curve crosses or meets the axes.

Trig. Identities

5. (a) Express 4cosec

(b) Hence show that

4cosec^{2}2θ cosec^{2}θ = sec^{2}θ

(c) Hence or otherwise solve, for 0 < θ < π,

4cosec^{2}2θ cosec^{2}θ = 4

giving your answers in terms of π.

5 (c)

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