Let f be a function defined by \(f(x) = x - \arctan x\) , \(x \in \mathbb{R}\) .

(a)     Find \(f(1)\) and \(f\left( { - \sqrt 3 } \right)\).

(b)     Show that \(f( - x) = - f(x)\) , for \(x \in \mathbb{R}\) .

(c)     Show that \(x - \frac{\pi }{2} < f(x) + \frac{\pi }{2}\) , for \(x \in \mathbb{R}\) .

(d)     Find expressions for \(f'(x)\) and \(f''(x)\) . Hence describe the behaviour of the graph of f at the origin and justify your answer.

(e)     Sketch a graph of f , showing clearly the asymptotes.

(f)     Justify that the inverse of f is defined for all \(x \in \mathbb{R}\) and sketch its graph.

(a)     \(f(1) = 1 - \arctan 1 = 1 - \frac{\pi }{4}\)     A1

\(f( - \sqrt 3 ) = - \sqrt 3 - \arctan ( - \sqrt 3 ) = - \sqrt 3 + \frac{\pi }{3}\)     A1

[2 marks]

 

(b)     \(f( - x) = - x - \arctan ( - x)\)     M1

\( = - x + \arctan x\)     A1

\( = - (x - \arctan x)\)

\( = - f(x)\)     AG     N0

[2 marks]

 

(c)     as \( - \frac{\pi }{2} < \arctan x < \frac{\pi }{2}\), for any \(x \in \mathbb{R}\)     A1

\( \Rightarrow - \frac{\pi }{2} <  - \arctan x < \frac{\pi }{2}\), for any \(x \in \mathbb{R}\)

then by adding x (or equivalent)     R1

we have \(x - \frac{\pi }{2} < x - \arctan x < x + \frac{\pi }{2}\)     AG     N0

[2 marks]

 

(d)     \(f'(x) = 1 - \frac{1}{{1 + {x^2}}}{\text{ or }}\frac{{{x^2}}}{{1 + {x^2}}}\)     A1A1

\(f''(x) = \frac{{2x(1 + {x^2}) - 2{x^3}}}{{{{(1 + {x^2})}^2}}}{\text{or }}\frac{{2x}}{{{{(1 + {x^2})}^2}}}\)     M1A1

\(f'(0) = f''(0) = 0\)     A1A1

EITHER

as \(f'(x) \geqslant 0\) for all values of \(x \in \mathbb{R}\)

(\((0,{\text{ 0)}}\)is not an extreme of the graph of f (or equivalent ))     R1

OR

as \(f''(x) > 0\)for positive values of x and \(f''(x) < 0\) for negative values of x     R1

THEN

(0, 0) is a point of inflexion of the graph of f (with zero gradient)     A1     N2

[8 marks]

 

(e)

    A1A1A1

Note: Award A1 for both asymptotes.

A1 for correct shape (concavities) \(x < 0\) .

A1 for correct shape (concavities) \(x > 0\) .

 

[3 marks]

 

(f)     (see sketch above)

as f is increasing (and therefore one-to-one) and its range is \(\mathbb{R}\) , 

\({f^{ - 1}}\) is defined for all \(x \in \mathbb{R}\)     R1

use the result that the graph of \(y = {f^{ - 1}}(x)\) is the reflection

in the line y = x of the graph of \(y = f(x)\) to draw the graph of \({f^{ - 1}}\)     (M1)A1

[3 marks]

Total [20 marks]

Parts of this question were answered quite well by many candidates. A few candidates had difficulties with domain of arctan in part (a) and in justifying their reasoning in parts (b) and (c). In part (d) although most candidates were successful in finding the expressions of the derivatives and their values at x = 0, many were unable to use the results to find the nature of the curve at the origin. Very few candidates were successful in answering parts (e) and (f).