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line = y = 2x + 1, curve = y = \(2x^2\) + 5x - 1 2x + 1 = \(2x^2\) + 5x - 1 \(2x^2\) + 3x - 2 = 0 factorizing (2x - 1)( x + 2) = 0 x = 1/2 or -2 when x = 1/2 y = 2 * 1/2 + 1 = 2 ( using the equation of a straight line ) when x = -2, y = -3 therefore, the points are ( 1/2, 2) and ( -2, -3)

\(\frac{2}{2r - 1}\) - \(\frac{5}{3}\) = \(\frac{1}{r + 2}\) \(\frac{2}{2r - 1}\) - \(\frac{1}{r + 2}\) = \(\frac{5}{3}\) \(\frac{2r + 4 - 2r + 1}{2r - 1 (r + 2)}\) = \(\frac{5}{3}\) \(\frac{5}{(2r + 1)(r + 2)}\) = \(\frac{5}{3}\) 5(2r - 1)(r + 2) = 15 (10r - 5)(r + 2) = 15 \(10r^2\) + 20r - 5r - 10 = 15 \(10r^2\) + 15r = 25 \(10r^2\) + 15r - 25 = 0 \(2r^2\) + 3r - 5 = 0 (\(2r^2 + 5r)\)(2r + 5) = r(2r + 5) - 1(2r + 5) (r - 1)(2r + 5) = 0 r = 1 or \(\frac{-5}{2}\)

2x - 3y = - 10; 10x - 6y = -5 2x - 3y = -10 x 2 10x - 6y = -5 4x - 6y = -20 .......(i) 10x - 6y = -5.......(ii) eqn(ii) - eqn(1) 6x = 15 x = \(\frac{15}{6}\) = \(\frac{5}{2}\) x = 2\(\frac{1}{2}\) Sub. for x in equ.(ii) 10(\(\frac{5}{2}\)) - 6y = -5 6y = 25 + 5 → 30 y = \(\frac{30}{6}\) y = 5

-2 \(\leq\) 2x - 6 < 4 = 2x - 6 < 4 = 2x < 10 = x < 5 2x \(\geq\) -2 + 6 \(\geq\) = x \(\geq\) 2 ∴ 2 \(\leq\) x < 5 [2, 3, 4]

Equal roots implies b\(^2\) - 4ac = 0 a = 1, b = - (p - 2), c = 2p + 1 [-(p - 2)]\(^2\) - 4 x 1 x (2p + 1) = 0 p\(^2\) - 4p + 4 - 4(2p + 1) = 0 p\(^2\) - 4p = 4 - 8p - 4 = 0 p\(^2\) - 12p = 0 p(p - 12) = 0 p = 0 or 12

y = x3 + x and y = x2 + 1 \(\begin{array}{c|c} x & -2 & -1 & 0 & 1 & 2 \\ \hline Y = x^3 + x & -10 & -2 & 0 & 2 & 10 \\ \hline y = x^2 + 1 & 5 & 2 & 1 & 2 & 5\end{array}\) The curves intersect at x = 1

3x2 - 7x + R. Computing the square, we have x2 - \(\frac{7}{3}\) = -\(\frac{R}{3}\) (\(\frac{x}{1} - \frac{7}{6}\))2 = -\(\frac{R}{3}\) + \(\frac{49}{36}\) \(\frac{-R}{3}\) + \(\frac{49}{36}\) = 0 R = \(\frac{49}{36}\) x \(\frac{3}{1}\) = \(\frac{49}{12}\)

2x + y = 4......(i) x^2 + xy = -12........(ii) from eqn (i), y = 4 - 2x = x2 + x(4 - 2x) = -12 x2 + 4x - 2x2 = -12 4x - x2 = -12 x2 - 4x - 12 = 0 (x - 6)(x + 2) = 0 sub. for x = 6, in eqn (i) y = -8, 8 =(6,-8); (-2, 8)

1 - \(\sqrt{13}\) and 1 + \(\sqrt{13}\) sum of roots - \(1 + \sqrt{13} + 1 - \sqrt{13} = 2\) Product of roots = (1 - \(\sqrt{13}\)) (1 + \(\sqrt{13}\)) = -12 x2 - (sum of roots) x + (product of roots) = 0 x2 - 2x - 12 = 0

Numbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c.To get the number of figures divisible by 4, We solve by method of A.P Let x represent numbers divisible by 4, nth term = a + (n - 1)d a = 4, d = 4 Last term = 4 + (n - 1)4 296 = 4 + 4n - 4 = \(\frac{296}{4}\) = 74 rn(Note: 296 is the last Number divisible by 4 between 1 and 300) Prob. of x = \(\frac{74}{298}\) = \(\frac{37}{149}\)

Total distance covered by Musa in 2 hrs = x + 10 + 5x = 6x + 10 Ngozi = 118 km If they are equal, 6x + 10 = 118 but 6x + 10 < 118 6x < 108 = x < 18 0 < x < 18 = 0 \(\leq\) x < 18

Tunde and Shola can do the work in 18 days. Both will do the work in \(\frac{1}{18}\) days. But Tunde can do the whole work in x days; Hence he does \(\frac{1}{x}\) of the work in 1 day. Shola does the work in (x + 15) days; hence, he does \(\frac{1}{x + 15}\) of the work in 1 day. \(\frac{1}{x} + \frac{1}{x + 15} = \frac{1}{18}\) \(\frac{2x + 15}{x^{2} + 15x} = \frac{1}{18}\) \(x^{2} + 15x = 36x + 270\) \(x^{2} + 15x - 36x - 270 = 0\) \(x^{2} - 21x - 270 = 0\)

x3 + px2 + qx + 1 = (x - 1) Q(x) + R x - 2 = 0, x = 2, R = 0, 4p + 2p = -9........(i) x3 + px2 + qx + 1 = (x - 1)Q(x) + R -1 + p - q + 1 = 0 p - q = 0.......(ii) Solve the equation simultaneously p = \(\frac{-3}{2}\) q = \(\frac{-3}{2}\) p + q = \(\frac{3}{2}\) - \(\frac{3}{2}\) = \(\frac{-6}{2}\) = -3

Pq + 1 = q2......(i) t = \(\frac{1}{p}\) - \(\frac{1}{pq}\).........(ii) p = \(\frac{q^2 - 1}{q}\) Sub for p in equation (ii) t = \(\frac{1}{q^2 - \frac{1}{q}}\) - \(\frac{1}{\frac{q^2 - 1}{q} \times q}\) t = \(\frac{q}{q^2 - 1}\) - \(\frac{1}{q^2 - 1}\) t = \(\frac{q - 1}{q^2 - 1}\) = \(\frac{q - 1}{(q + 1)(q - 1)}\) = \(\frac{1}{q + 1}\)

If y = cf(x) cf(5) = 30 + 32 + 30 = 92