A function is defined as \(f(x) = k\sqrt x \), with \(k > 0\) and \(x \geqslant 0\) .

(a)     Sketch the graph of \(y = f(x)\) .

(b)     Show that f is a one-to-one function.

(c)     Find the inverse function, \({f^{ - 1}}(x)\) and state its domain.

(d)     If the graphs of \(y = f(x)\) and \(y = {f^{ - 1}}(x)\) intersect at the point (4, 4) find the value of k .

(e)     Consider the graphs of \(y = f(x)\) and \(y = {f^{ - 1}}(x)\) using the value of k found in part (d).

(i)     Find the area enclosed by the two graphs.

(ii)     The line x = c cuts the graphs of \(y = f(x)\) and \(y = {f^{ - 1}}(x)\) at the points P and Q respectively. Given that the tangent to \(y = f(x)\) at point P is parallel to the tangent to \(y = {f^{ - 1}}(x)\) at point Q find the value of c .

(a)

    A1

Note: Award A1 for correct concavity, passing through (0, 0) and increasing.

Scales need not be there.

 

[1 mark]

 

(b)     a statement involving the application of the Horizontal Line Test or equivalent     A1

[1 mark]

 

(c)     \(y = k\sqrt x \)

for either \(x = k\sqrt y \) or \(x = \frac{{{y^2}}}{{{k^2}}}\)     A1

\({f^{ - 1}}(x) = \frac{{{x^2}}}{{{k^2}}}\)     A1

\({\text{dom}}\left( {{f^{ - 1}}(x)} \right) = \left[ {0,\infty } \right[\)     A1

[3 marks]

 

(d)     \(\frac{{{x^2}}}{{{k^2}}} = k\sqrt x \,\,\,\,\,\)or equivalent method     M1

\(k = \sqrt x \)

\(k = 2\)     A1

[2 marks]

 

(e)     (i)     \(A = \int_a^b {({y_1} - {y_2}){\text{d}}x} \)     (M1)

\(A = \int_0^4 {\left( {2{x^{\frac{1}{2}}} - \frac{1}{4}{x^2}} \right){\text{d}}x} \)     A1

\( = \left[ {\frac{4}{3}{x^{\frac{3}{2}}} - \frac{1}{{12}}{x^3}} \right]_0^4\)     A1

\( = \frac{{16}}{3}\)     A1

 

(ii)     attempt to find either \(f'(x)\) or \(({f^{ - 1}})'(x)\)     M1

\(f'(x) = \frac{1}{{\sqrt x }},{\text{ }}\left( {({f^{ - 1}})'(x) = \frac{x}{2}} \right)\)     A1A1

\(\frac{1}{{\sqrt c }} = \frac{c}{2}\)     M1

\(c = {2^{\frac{2}{3}}}\)     A1

[9 marks]

Total [16 marks]

Many students could not sketch the function. There was confusion between the vertical and horizontal line test for one-to-one functions. A significant number of students gave long and inaccurate explanations for a one-to-one function. Finding the inverse was done very well by most students although the notation used was generally poor. The domain of the inverse was ignored by many or done incorrectly even if the sketch was correct. Many did not make the connections between the parts of the question. An example of this was the number of students who spent time finding the point of intersection in part e) even though it was given in d).