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Let the numbers be x and y x+y = 2p.....(i) x-y = p......(ii) 2x = 3p x = 3p/2

\(a*a^{-1} = aa^{-1} + a + a^{-1} = e\) if e = 0 \(2.2^{-1} + 2 + 2^{-1} = 0\) collecting like terms, we have: \(3.2^{-1} + 2 = 0\) = \(2^{-1}\) = -\(\frac{2}{3}\)

Expand the options and collect like terms to check which gives the presented expression. Expansion of Options A: (x+y+6)(x+y-3) → \(x^2+xy-3x + xy+y^2-3y +6x+6y-18\) Collect like terms: \(x^2 + xy + xy + y^2 +6x - 3x +6y - 3y - 18\) = \(x^2 + 2xy + y^2 + 3x + 3y - 18\)

l\(^2\) = 3 (k + m) Since they are consecutive positive numbers, we have l = k+1, m = k+2. \(\to\) (k+1)\(^2\) = 3(k + k + 2) k\(^2\) + 2k + 1 = 3(2k + 2) k\(^2\) + 2k + 1 = 6k + 6 k\(^2\) + 2k - 6k + 1 - 6 = 0 k\(^2\) - 4k - 5 = 0 k\(^2\) - 5k + k - 5 = 0 k(k - 5) + 1(k - 5) = 0 k = -1 or 5 Since k, l and m are positive, then k = 5. m = k + 2 = 5 + 2 = 7.

\(\frac{1}{x^{3} - 1}\) \(x^{3} - 1 = (x - 1)(x^{2} + x + 1)\) \(\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + x + 1}\) \(\frac{1}{x^{3} - 1} = \frac{A(x^{2} + x + 1) + (Bx + C)(x - 1)}{x^{3} - 1}\) Comparing the two sides of the equation, \(A + B = 0 ... (1)\) \(A - B + C = 0 ... (2)\) \(A - C = 1 ... (3)\) From (3), \(C = A - 1\), putting that in (2), \(A - B = -C \implies A - B = 1 - A\) \(2A - B = 1 ... (4)\) (1) + (4): \(3A = 1 \implies A = \frac{1}{3}\) \(A = -B \implies B = -\frac{1}{3}\) \(C = A - 1 \implies C = \frac{1}{3} - 1 = -\frac{2}{3}\) \(\therefore \frac{1}{x^{3} - 1} = \frac{1}{3(x - 1)} - \frac{x - 2}{3(x^{2} + x + 1)}\) = \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)

Let the common ratio be r so that the first term is 2r. Sum, s = \(\frac{a}{1-r}\) ie. 8 = \(\frac{2r}{1-r}\) 8(1-r) = 2r, 8 - 8r = 2r 8 = 2r + 8r 8 = 10r r = \(\frac{4}{5}\). where common ratio (r) = \(\frac{second term(n_2)}{first term(a)}\), r = \(\frac{n_2}{2r}\) r = \(\frac{4}{5}\) and a = 2r or \(\frac{8}{5}\) \(\frac{4}{5}\) * \(\frac{8}{5}\) = n\(_2\) \(\frac{32}{25}\) = n\(_2\) The sum of the first two terms = a + n\(_2\) = \(\frac{8}{5}\) + \(\frac{32}{25}\) = \(\frac{40 + 32}{25}\) = \(\frac{72}{25}\)

2x\(^2\) + x - 1 2x - 1 4x\(^3\) + 0x\(^2\) - 3x + 1 - 4x\(^3\) - 2x\(^2\) 2x - 1 + 2x\(^2\) - 3x + 1 - -x + 1 2x - 1 - 2x +1 - - 2x + 1 0 0 Thus, Divide 4x\(^3\) - 3x + 1 by 2x - 1 = 2x\(^2\) + x - 1

\(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\) r = 4 units; centre (\(\alpha\), -2). \((x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}\) \((x - \alpha)^{2} + (y - (-2))^{2} = 4^{2}\) \(x^{2} - 2\alpha x + \alpha^{2} + y^{2} + 4y + 4 = 16\) \(x^{2} + y^{2} - 2\alpha x + 4y = 16 - \alpha^{2} - 4\) \(\therefore 12 - \alpha^{2} = \alpha \implies \alpha^{2} + \alpha - 12 = 0\) \(\alpha^{2} - 3\alpha + 4\alpha - 12 = 0\) \(\alpha(\alpha - 3) + 4(\alpha - 3) = 0\) \((\alpha + 4)(\alpha - 3) = 0 \implies +\alpha = 3\)

Hint: Make a sketch forming a right angled triangle. Let x = height of the tree above the man. Such that x/25 = sin 30. x = 12.5m The height of the tree = 12.5 + 1.7 = 14.2 m

bisector theorem: \(\frac{|MN|}{|MO|}\) = \(\frac{|PO|}{|NP|}\) taking the bisected angle:x and y = |ON|=12 : x+y= 12 x = 12 - y |PO| = 12 - y \(\frac{6}{4}\)= \(\frac{12-y}{y}\) 6y = 4 (12-y) 6y = 48 - 4y = 4.8 Recall that x+y= 12 12 - 4.8 =7.2

Let \(\phi\) be the angle between the two lines. tan \(\phi\) = \(\frac{m_1 - m_2}{1 + m_1 m_2}\) where m\(_1\) = slope of line 1; m\(_2\) = slope of line 2. Line 1: 2x + y = 3 \(\implies\) y = 3 - 2x. Line 2: 3x - 2y = 5 \(\implies\) -2y = 5 - 3x. y = \(\frac{3}{2}\)x - \(\frac{5}{2}\). m\(_1\) = -2, m\(_2\) = \(\frac{3}{2}\). tan \(\phi\) = \(\frac{-2 - \frac{3}{2}}{1 + (-2 \times \frac{3}{2})}\) = \(\frac{\frac{-7}{2}}{-2}\) \(\therefore\) Tan \(\phi\) = \(\frac{7}{4}\).

\(\frac{150}{PR}\) = tan45º PR = \(\frac{150}{tan45º}\) = 150m

The locus of a point P(x,y) such that PV = PW where V = (1,1) and W = (3,5). This means that the point P moves so that its distance from V and W are equidistance. PV = PW \(\sqrt{(x-1)^{2} + (y-1)^{2}} = \sqrt{(x-3)^{2} + (y-5)^{2}}\). Squaring both sides of the equation, (x-1)2 + (y-1)2 = (x-3)2 + (y-5)2. x2-2x+1+y2-2y+1 = x2-6x+9+y2-10y+25 Collecting like terms and solving, x + 2y = 8.

\(y = x(2-x) \Rightarrow y= 2x - x^{2}\); \(\int^{2}_{0}(2x-x^{2} = (x^{2}-\frac{x{3}}{3})^{2}\) solving further gives (4 - \(\frac{1}{3} * 8) - (0) = \frac{4}{3} sq\hspace{1 mm}unit\)

\(\int_{0}^{z} (\sin x - \cos x) \, dx\) where \( z = \frac{\pi}{4} \), we have: \(\int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) \, dx = \left[-\cos x - \sin x\right]_{0}^{\frac{\pi}{4}}\) Evaluating at the bounds gives: \(-\sqrt{2} - (-1) = -\sqrt{2} + 1\) Thus, the value of the integral is: \(\int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) \, dx = 1 - \sqrt{2}\)