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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

\(\begin{array}{c|c} \text{Class intervals} & F & \text{Class boundary}\\ \hline 5 - 7 & 17 & 4.5 - 9.5\\ 10 - 14 & 32 & 9.5 - 14.5\\ 15 - 19 & 25 & 14.5 - 19.5\\ 20 - 24 & 24 & 19.5 - 24.5 \end{array}\) mode = 9.5 + \(\frac{D_1}{D_2 + D_1}\) x C = 9.5 + \(\frac{5(32 - 17)}{2(32) - 17 - 25}\) = 9.5 + \(\frac{75}{22}\) = 12.91 \(\approx\) 12.9

mean (x) = \(\frac{\sum x}{N}\) = k + k + 1 + k + 3 = \(\frac{3k + 3}{3}\) = k + 1 \(\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline k & -1 & 1 \\ k + 1 & 0 & 0\\ k + 2 & 1 \\ \hline & & 2\end{array}\) Variance (52) = \(\frac{\sum (x - x)^2}{N}\) = \(\frac{2}{3}\)

mean (x̄) = \(\frac{1 + x + 1 + 2x + 1}{3}\) = \(\frac{3x + 3}{3}\) = 1 + x \(\begin{array}{c|c} X & (X -X) & (X -X)^2\\ \hline 1 & -x & x^2 \\ x + 1 & 0 & 0\\2x + 1 & x & x^2\\ \hline & & 2x^2\end{array}\) S.D = \(\sqrt{\frac{\sum(x - x)^2}{n}}\) = \(\sqrt{(6)}\) ( given ) \(\sqrt{(6)}\) = \(\sqrt{\frac{2x^2}{3}}\) cancel square roots and cross multiply \(2x^2\) = 18 divide both sides by 2 \(x^2\) = 9 ∴ x = \(\pm\) \(\sqrt{9}\) = \(\pm\)3

Number of red balls = 16, Number of blue balls = 20 Let x represent the No of white balls to be added ∴ Total number of balls = 36 + x Pr( red ball) = \(\frac{ 16}{ 36 + x } = \frac{2}{5}\) (remember that Pr( red ball ) = \frac{2}{5}\)) cross multiplying 2(36 + x) = 80 2x = 80 - 72 2x = 8 x = \(\frac{8}{2}\) = 4