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U = {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30} P = {6,12,18,24,30} Q = {4,8,12,16,20,24,28} P∪Q = {4,6,8,12,16,18,20,24,28,30} (P∪Q)c = {2,10,14,22,26}

\(x\propto \frac{uv}{u+v}\) x = \(\frac{kuv}{u+v}\) K = \(\frac{x(u+v)}{uv}\) \(\frac{3(3+1)}{3\times1}\) \(\frac{3\times4}{3}\) k = 4 x =\(\frac{4uv}{u+v}\) x = \(\frac{4\times 3\times3}{3+3}\) x = \(\frac{36}{6}\) x = 6

5 + x \(\leq\) 8 and 13 + x \(\geq\) 7 5 + x \(\leq\) 8 X \(\leq\) 8 – 5 X \(\leq\) 3 And 13 + x \(\geq\) 7 X \(\geq\) 7 – 13 X \(\geq\) -6 Combining the two -6 \(\leq\) x \(\leq\) 3

x\(^2\) + 4 = x\(^2\) - 3x + 2 3x + 2 = y

L ∝ 1/W L = K/W K = WL = 1.5 * 20 30

3x - 5y + 5 = 0 → eqn1 4x - 7y + 8 = 0 → eqn2 eqn1 * 4; 12x - 20y + 20 = 0 → eqn3 eqn2 * 3; 12x - 21y + 24 = 0 → eqn4 eqn3 - eqn4 = y - 4 = 0 ∴ y = 4 From eqn1, 3x - 5y + 5 = 0 3x - 5(4) + 5 = 0 3x - 20 + 5 = 0 3x - 15 = 0 3x = 15 x = 5 x and y = 5, 4 respectively

To find the common ratio of the geometric progression with terms \( n-2 \), \( n \), and \( n+3 \), we use the property that the square of the middle term is equal to the product of the other two terms: \(n^2 = (n - 2)(n + 3)\) Expanding the right side gives: \(n^2 = n^2 + 3n - 2n - 6\) Simplifying leads to: \(0 = n - 6 \implies n = 6\) The terms of the progression are: - First term: \( n - 2 = 4 \) - Second term: \( n = 6 \) - Third term: \( n + 3 = 9 \) The common ratio \( r \) is calculated as follows: \(r = \frac{n}{n-2} = \frac{6}{4} = \frac{3}{2}\) We can verify it with the third term: \(r = \frac{n+3}{n} = \frac{9}{6} = \frac{3}{2}\) Thus, the common ratio of the geometric progression is: \(\boxed{\frac{3}{2}}\)

Lines bounding \Delta OPQ OQ; y - x = 0 y - x \ge 0 PQ; x + 1 = 0 x + 1 \ge = 0 PO; y + x = 0 y + x \le 0 ∴ x + 1 \ge 0, y + x \le 0, y - x \ge 0

4abx - 2axy - 12b2x + 6bxy = (4abx - 2axy) - (12b2x - 6bxy) = 2ax(2b - y) -6bx(2b - y) = (2ax - 6bx)(2b - y) = 2x(a - 3b)(2b - y)

Using: \(S_n = \frac{n}{2} (a + l)\) where:- \(S_n\) is the sum of the first \(n\) terms, \(a\) is the first term, \(l\) is the last term. Given: \(S_n = 252\), a = -16, l = 72 Step 1: Substitute the Known Values Substituting the values into the sum formula: \(252 = \frac{n}{2} (-16 + 72)\) \(252 = \frac{n}{2} \cdot 56\) \(504 = n \cdot 56\) Now, divide both sides by 56: n = \(\frac{504}{56}\) = 9 Thus, the number of terms in the series is: 9

Area of Trapezium = 1/2(sum of parralel sides) * ht 21 = 1/2(5 + 9)h 42 = 14h h = 42/14 h = 3cm

The locus is the "semicircle with diameter XY" on the given side. By Thales’ theorem, ∠XPY = 90^\circ means P lies on the circle with diameter XY. “One side only” of line XY restricts it to a semicircle.

y = 2x\(^2\) + 5x - 3 dy/dx = 4x + 5 = 4 + 5 = 9

\(\int^{3} _{2}(x^2 - 2x)\)dx =\(\left[\frac{x^3}{3}-\frac{2x^2}{2}\right ]^{3}_{2}\\\left[\frac{x^3}{3}-x^2 + C\right ]^{3}_{2}\\\left[\frac{3^3}{3}-3^2 + C \right ]-\left[\frac{2^3}{3}-2^2 + C \right ]\\9-9-\left[\frac{8}{3}-4 \right ]\) =\(\frac{-8}{3}\) + 4 = \(\frac{4}{3}\)

y = 3 sin(-4x) dy/dx = 3 x - 4 cos (-4x) = -12 cos (-4x)