(a)     Differentiate \(f(x) = \arcsin x + 2\sqrt {1 - {x^2}} \) , \(x \in [ - 1, 1]\) .

(b)     Find the coordinates of the point on the graph of \(y = f (x)\) in \([ - 1, 1]\), where the gradient of the tangent to the curve is zero.

(a)     \(f'(x) = \frac{1}{{\sqrt {1 - {x^2}} }} - \frac{{2x}}{{\sqrt {1 - {x^2}} }}\)   \(\left( { = \frac{{1 - 2x}}{{\sqrt {1 - {x^2}} }}} \right)\)     M1A1A1

Note: Award A1 for first term,
                     M1A1 for second term (M1 for attempting chain rule).

(b)     \(f'(x) = 0\)     (M1)

\(x = 0.5\) , \(y = 2.26\) or \(\frac{\pi }{6} + \sqrt 3 \)   (accept (\(0.500\), \(2.26\))     A1A1     N3

[6 marks]

Most candidates scored well on this question, showing competence at non-trivial differentiation. The follow through rules allowed candidates to recover from minor errors in part (a). Some candidates demonstrated their resourcefulness in using their GDC to answer part (b) even when they had been unable to gain full marks on part (a).