A-Level AQA Maths Pure General Binomial Expansion: 6 (a) Find the first two terms,

6 (a) Find the first two terms, in ascending powers of x, of the binomial expansion of $$\left(1 - \frac{x}{2}\right)^{\frac{1}{2}}$$ 6 (b) Hence, for small values of x, show that $$\sin 4x + \sqrt{\cos x} \approx A + Bx + Cx^2$$ where A, B and C are constants to be found. - AQA - A-Level Maths Pure - Question 6 - 2022 - Paper 1

Question 6

$6-(a)-Find-the-first-two-terms,-in-ascending-powers-of-x,-of-the-binomial-expansion-of--$$\left(1---\frac{x}{2}\right)^{\frac{1}{2}}$$--6-(b)-Hence,-for-small-values-of-x,-show-that--$$\sin-4x-+-\sqrt{\cos-x}-\approx-A-+-Bx-+-Cx^2$$--where-A,-B-and-C-are-constants-to-be-found.-AQA-A-Level Maths Pure-Question 6-2022-Paper 1.png$

6 (a) Find the first two terms, in ascending powers of x, of the binomial expansion of $$\left(1 - \frac{x}{2}\right)^{\frac{1}{2}}$$ 6 (b) Hence, for small values... show full transcript

Worked Solution & Example Answer:6 (a) Find the first two terms, in ascending powers of x, of the binomial expansion of $$\left(1 - \frac{x}{2}\right)^{\frac{1}{2}}$$ 6 (b) Hence, for small values of x, show that $$\sin 4x + \sqrt{\cos x} \approx A + Bx + Cx^2$$ where A, B and C are constants to be found. - AQA - A-Level Maths Pure - Question 6 - 2022 - Paper 1

Step 1

Find the first two terms, in ascending powers of x, of the binomial expansion of (1 - x/2)^(1/2)

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Answer

To find the first two terms of the binomial expansion for $\left(1 - \frac{x}{2}\right)^{\frac{1}{2}}$ , we can use the binomial theorem:

$(1 + u)^{n} = 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots$

In this case, (u = -\frac{x}{2}) and (n = \frac{1}{2}). Therefore, we have:

First term:
$1$
Second term:
$\frac{1}{2} \left(-\frac{x}{2}\right) = -\frac{x}{4}$

Thus, the first two terms of the expansion are:

$\approx 1 - \frac{x}{4}$

Step 2

Hence, for small values of x, show that sin 4x + √cos x ≈ A + Bx + Cx²

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Answer

Using small angle approximations, we can expand both (\sin 4x) and (\sqrt{\cos x}):

For (\sin 4x), apply the approximation: $\sin kx \approx kx$
Thus,
$\sin 4x \approx 4x$ .
For (\sqrt{\cos x}), noting that for small values of (x):
$\cos x \approx 1 - \frac{x^2}{2}$
Therefore, using the expansion for the square root, $\sqrt{\cos x} \approx \sqrt{1 - \frac{x^2}{2}} \approx 1 - \frac{1}{2}\cdot \frac{x^2}{2} = 1 - \frac{x^2}{4}$ .

Putting it all together:

$\sin 4x + \sqrt{\cos x} \approx 4x + \left(1 - \frac{x^2}{4}\right).$

This simplifies to:

$\approx 1 + 4x - \frac{x^2}{4}$ .

Now, comparing to the form (A + Bx + Cx^2), we identify:

(A = 1)
(B = 4)
(C = -\frac{1}{4}).

Find the first two terms, in ascending powers of x, of the binomial expansion of (1 - x/2)^(1/2)

Hence, for small values of x, show that sin 4x + √cos x ≈ A + Bx + Cx²

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