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Cos 30 = \(\frac{5}{\text{x}}\) x cos 30 = 5, => x = 5\(\sqrt{3}\) Coordinates of P = - 5\(\sqrt{3}\), 0 Coordinates of Q = 0, 5 Gradient of PQ = (y\(_2\) - y\(_1\)) (X\(_2\) - X\(_1\)) = \(\frac{(5 - 0)}{(0 - 5\sqrt{3}}\)) = \(\frac{5}{5\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\) Equation of PQ = y - y\(_1\) = m (x -x\(_1\)) y - 0 = \(\frac{1}{\sqrt{3}}\) (x -( -5\(\sqrt{3}\))) Thus: \(\sqrt{3}\)y = x + 5\(\sqrt{3}\)

first term (a) = \(\frac{5}{6}\) common difference = \(\frac{8}{6}\) - \(\frac{5}{6}\) → \(\frac{3}{6}\) or \(\frac{1}{2}\) A.P formula → T\(_n\) = a + (n - 1)d T\(_n\) = \(\frac{5}{6}\) + (19 - 1)\(\frac{1}{2}\) T\(_n\) = \(\frac{5}{6}\) + 9 → 9\(\frac{5}{6}\)

The two major lines have their lower portions involved. Since there are no broken lines here,\(\leq\) will hold for the two major lines that is, 2y - x - 2 = 0 becomes 2y - x - 2 \(\leq\) 0 y + 2x + 2 = 0 becomes y + 2x + 2 \(\leq\) 0 Triangle SPT is located at the positive side of the y-axis, so y \(\geq\) 0, and the negative side of the x-axis so, x \(\leq\) Therefore, OPTION C is correct.

An element e of the binary expression * is said to be the identity if for any a * e = a from the given table the identity element is m i.e K ⨁ m = k k ⨁ m = k.

\(T_{n} = 2^{2n - 1}\) \(2^{2n - 1} = 2^9\) \(2n - 1 = 9 \implies 2n = 9 + 1\) \(2n = 10 \implies n = 5\) The 5th term = 2\(^9\)

A quadratic equation is an equation of the second order. The highest power in the equation is 2.

Let the straight line PQ = y 2x\(^2\) - 5x - 1 - y = 2x\(^2\) - 5x + 2 y = 2x\(^2\) - 5x - 1 - 2x\(^2\) + 5x - 2 y = -3

\(t^2\) + kt + \(\frac{25}{4}\) The general form of quadratic expression = ax\(^2\) + bx + c comparing \(t^2\) + kt + \(\frac{25}{4}\) and the general form a = 1, b = k and c = \(\frac{25}{4}\) using: b\(^2\) = 4ac we can find the value of k. ( note: there is/are another way(s) to find the value of k) k\(^2\) = 4 x 1 x \(\frac{25}{4}\) k\(^2\) = 25 taking the square root of both sides k = 5

9\(^x\) = 729 3\(^2\)\(^x\) = 3\(^6\) 2x = 6 x = 3

x\(^2\) - 3x - 10 = 0 Using factorisation method: x\(^2\) - 3x - 10 = 0 becomes x\(^2\) - 5x + 2x - 10 = 0 (x\(^2\) - 5x) + (2x - 10) = 0 x(x - 5) + 2(x -5) = 0 x - 5 = 0 or x + 2 = 0 \(\therefore\) x = 5 or -2

(a) x -3 -2 -1 0 1 2 3 4 \(y = x^{2}\) 9 4 1 0 1 4 9 16 (b) \(y = 2x + 3\) x -3 -2 -1 0 1 2 3 4 2x -6 -4 -2 0 2 4 6 8 3 3 3 3 3 3 3 3 3 y -3 -1 1 3 5 7 9 11 see graph above (c)(i) The roots of the equation are -1 and 3, from the graph. (ii) The gradient of \(y = x^{2}\) at x = -2 : \(\frac{8}{-2.5} = -3.2\)

(a) \(2x^{2} + (p + 1)x + q = 0\) The roots of the equation are 1 and 3, therefore (x - 1)(x - 3) = 0. Equation : \(x^{2} - 3x - x + 3 = 0\) \(x^{2} - 4x + 3 = 0\) \(\therefore 2x^{2} + (p + 1)x + q \equiv x^{2} - 4x + 3\) \(\frac{2x^{2} + (p + 1)x + q}{2} = x^{2} + \frac{(p + 1)}{2} x + \frac{q}{2}\) \(\implies \frac{q}{2} = 3\) \(q = 6\) \(\frac{p + 1}{2} = -4 \implies p + 1 = -8\) \(p = -9\) (p, q) = (-9, 6). (b) \(W \propto \frac{1}{d^{2}} \) \(W = \frac{k}{d^{2}}\) \(k = Wd^{2}\) \(W = \frac{80 \times (6400)^{2}}{d^{2}}\) (equation of variation). \(W = \frac{80 \times (6400)^{2}}{(6400 + 800)^{2}}\) \(W = \frac{80 \times (6400)^{2}}{(7200)^{2}}\) = 63.21 kg

From the constraint y < 3 and y > 0, the possible values for y are 1 and 2 When y = 1, 3y + 5x \(\leq\) 15 3 + 5x \(\leq\) 15 ⇒ 5x \(\leq\) 12 ⇒ 2.4 since x must be an integer greater than 0 but less than or equal to 2.4, the possible values of x are 1 and 2 When y = 2, 3y + 5x \(\leq\) 15 ⇒ 6 + 5x \(\leq\) 15 ⇒ 6 \(\leq\) 15 - 6 = 11 ⇒ x = 2.2 Again, Since x must be an integer greater than 0 and less than or equal to 2.2, the only possible integer value of x is 1 So, the pairs (x, y) that satisfy all the conditions are ( 1, 1), (1, 2) and (2, 1).

\(\begin{vmatrix}-1 & -1 & -1\\3 & 1 & -1\\1 & 2 & 1\end{vmatrix}\) = -\(\begin{vmatrix}1 & -1 \\2 & 1\end{vmatrix}\) + \(\begin{vmatrix}3 & -1\\1 & 1\end{vmatrix}\) - \(\begin{vmatrix}3 & 1\\1 & 2\end{vmatrix}\) = -( 1 + 2) + (3 + 1) - ( 6 - 1) = -3 + 4 - 5 = 4 - 8 = -4.

K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) K\(^2\) = K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) x K = \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) = \(\begin{pmatrix}7 & 6\\18 & 19\end{pmatrix}\) I = \(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\) K\(^2\) + K + I = \(\begin{pmatrix}7 & 6\\18 & 19\end{pmatrix}\) + \(\begin{pmatrix}2 & 1\\3 & 4\end{pmatrix}\) + \(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\) = \(\begin{pmatrix}10 & 7\\21 & 24\end{pmatrix}\)