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Let the first term of the geometric progression (GP) be \( a \) and the common ratio be \( r \). The sum of the first three terms is: a + ar + ar\(^2\) = a(1 + r + r\(^2\)) \(S_3 = a(1 + r + r^2)\) The sum to infinity is: \(S_{\infty} = \frac{a}{1 - r}\) But: \(S_3 = \frac{1}{2} S_{\infty}\) Substituting the expressions gives: \(a(1 + r + r^2) = \frac{1}{2} \cdot \frac{a}{1 - r}\) Dividing by \( a \): \(1 + r + r^2 = \frac{1}{2(1 - r)}\) Cross-multiplying: \(2(1 + r + r^2)(1 - r) = 1\) Expanding and simplifying leads to: \(2(1 - r^3) = 1\) Thus: \(r^3 = \frac{1}{2}\) Taking the cube root: \(r = \sqrt[3]{\frac{1}{2}} = \frac{1}{\sqrt[3]{2}}\) Therefore, the positive common ratio of the progression is: \(\frac{1}{\sqrt[3]{2}}\).

The identity element (e) under an operation, say \(\otimes\), is the element such that for any given element under the operation, say a, \(a \otimes e = e \otimes a = a\) From the table, q is the identity element. \(p \otimes q = q \otimes p = p\) Same as all through.

x \(\ast\) y = xy - y - x, x \(\ast\) 3 = 3x - 3 - x = 2x - 3 2 \(\ast\) x = 2x - x - 2 = x - 2 ∴ 2x - 3 = x - 2 x = -2 + 3 = 1

\(\begin{vmatrix} x & 1 & 0 \\ 1-x & 2 & 3 \\ 1 & 1+x & 4\end{vmatrix}\) = x\(\begin{vmatrix}2 & 3 \\ 1+x & 4\end{vmatrix}\) - \(\begin{vmatrix}1-x & 3 \\ 1 & 4\end{vmatrix}\) = 0 = x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x] = 5x - 3x2 -1 + 4x = -3x2 + 9X - 1

PQ = \(\begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}\)\(\begin{pmatrix} u & 4+u \\ -2v & v \end{pmatrix}\) = \(\begin{pmatrix} (2u-6v & 2(4+u) +3v)\\ 4u-10v & 4(4+u)+5v \end{pmatrix}\) = \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) 2u - 6v = 1.....(i) 4u - 10v = 0.......(ii) 2(4 + u) + 3v = 0......(iii) 4(4 + u) + 5v = 1......(iv) 2u - 6v = 1 .....(i) x 2 4u - 10v = 0......(ii) x 1 \(\frac{\text{4u - 12v = 0}}{\text{-4u - 10v = 0}}\) -2v = 2 = v = -1 2u - 6(-1) = 1 = 2u = 5 u = -\(\frac{5}{2}\) ∴ (U, V) = (-\(\frac{5}{2}\) - 1)

No expanded working has been attached for this question yet. The verified answer is Option B.

d2 = (y - y)2 + (x - x)2 5 = 4x2 + 1 = 25= 4x2 + 1 = 4x2 = 25 - 1= 24 x2 = \(\frac{24}{4}\) x = \(\sqrt{6}\)

y = 4x + 3 when x = 0, y = 3 \(\to\) (0, 3) when y = 0, x = -\(\frac{3}{4}\) \(\to\) (\(\frac{3}{4}\), 0) mid-point \(\frac{0 + (-{\frac{3}{4}})}{2}\), \(\frac{3 + 0}{2}\) -\(\frac{3}{8}\), \(\frac{3}{2}\)

cos x + sin x = \(\frac{1}{cos x - sinx}\) = (cosx + sinx)(cosx - sinx) = 1 = cos\(^2\)x - sin\(^2\)x = 1 cos\(^2\)x - (1 - cos\(^2\)x) = 1 = 2cos\(^2\)x = 2 cos\(^2\)x = 1 = cosx = \(\pm\)1 x = cos\(^{-1}\) (\(\pm\)1) = 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\) (possible solution)

Tan60º = \(\frac{150}{y}\) \(\sqrt{3}\) = \(\frac{150}{y}\) y = \(\frac{150\sqrt3}{3}\) y = 50\(\sqrt3\) Tan45º = \(\frac{150}{z}\) 1 = \(\frac{150}{z}\) z = 150 But x = z + y x = 150 + 50\(\sqrt3\) x = 50(3 +\(\sqrt{3}\))m

y = 243(4x + 5)-2, find \(\frac{dy}{dx}\) = -1944(4x + 5)-3 = 1944(9)-3 \(\frac{dy}{dx}\) when x = 1 = -\(\frac{1944}{9^3}\) = -\(\frac{1944}{729}\) = \(\frac{-8}{3}\)

let y = \(\frac{x}{cosx}\) = x sec x ( since \(\frac{1}{cosx}\) = sec x ) let u = x, v = sec x \(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\) \(\frac{dy}{dx}\) = x [secx tanx] + secx Therefore, \(\frac{dy}{dx}\)\(\frac{x}{cosx}\) = x secx tanx + secx

∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx ∫\(^{\pi}_{2}\) dx = [X]\(^{\pi}_{2}\) = \(\pi\) - 2 + c when c is an arbitrary constant of integration

m = \(\frac{dy}{dv}\) = 6x - 5 ∫dy = ∫(6x - 5)dx y = 3x2 - 5x + C when x = 2, y = 5 ∴ 5 = 3(2)2 - 5(2) +C C = 3 ∴ y = 3x2 - 5x + 3

mean(x) = \(\frac{\sum x}{N}\) = \(\frac{48}{8}\) = 5.875 re-arranging the numbers; 2, 3, 5, 6, 2, 7, 8, 9 median = \(\frac{6 + 7}{2}\) = \(\frac{1}{2}\) = 6.5 m + 2n = 5.875 + (6.5)2 = 13 + 5.875 = 18.875 = \(\approx\) = 19