f(x) = (2 + kx)^3, |x| < 2, where k is a positive constant The binomial expansion of f(x), in ascending powers of x, up to and including the term in x^2 is A + Bx + \frac{243}{16}x^2 where A and B are constants - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 5

To find k, we need to look at the coefficient of the x^2 term in the binomial expansion. Expanding (2 + kx)^3 gives:

$\binom{3}{2}(2)^{3-2} (kx)^2 = 3(2)(kx)^2 = 6k^2 x^2$

Setting this equal to \frac{243}{16}, we have:

$6k^2 = \frac{243}{16}$

Thus,

$k^2 = \frac{243}{96} = \frac{81}{32}$

Taking the positive root (since k is positive):

$k = \sqrt{\frac{81}{32}} = \frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}$

The coefficient B is found from the linear term in the binomial expansion:

$\binom{3}{1}(2)^{3-1}(kx) = 3(2^2)(kx) = 12k x$

From the previous calculation, substituting k yields:

$B = 12 \cdot \frac{9}{4\sqrt{2}} = \frac{108}{4\sqrt{2}} = \frac{27\sqrt{2}}{2}$