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At x = 1, substituting x = 1 in the equation: ax2 + bx + c = 5; f(1) => a + b + c = 5 .....(1) Taking the first derivative of f(x) in the original equation gives dy/dx = 2ax + b = 2x + 1 (given)....(2) From (2),=> b = 1, and 2ax = 2x, => a = 1. ( comparing coefficients) substituting into (1) 1 + 1 + c = 5, => c = 5 - 2 = 3 Thus a = 1, b = 1 and c = 3

\Delta PQR is isosceles ∴∠PRQ = 30o Also in \Delta PQR ^Q = 180 -(30 + 30) = 120o PQRS is a cyclic quad ∴^S + ^Q = 180(opp ∠s of a cyclic quad) S + 120 = 180 S = 60o In \DeltaPSR x + 50 + 60 = 180(sum of ∠s of a \Delta) x + 110 = 180 x = 180 - 110 x = 70o

8 71 8 8 rem 7 8 1 rem 0 0 rem 1 71\(_8\) = 107\(_8\)

5-3log52 x 22log23 i Let -3log52 = p => log52-3 = p ∴2-3 = 5p ∴5-3log52 = 5log52-3 = 5p ii 22log23 = q => log532 = q ∴32 = 2q ∴222log23 = 2q = 32 = 9 i x ii = 2-3 x 9 = 1/23 x 9 = 1/8 x 9 = 9/8 = 11/8

C.P = N50000 Loss = x% = \(\frac{\text{x}}{100}\) x 50000 = 500x S.P = C.P - Loss = 50,000 - 500x Profit = 40% = \(\frac{40}{100}\) x (50000 - 500x) = 10,000 \(\frac{2}{5}\) x (50000 - 500x) = 10,000 100,000 - 1000x = 50,000 1000x = 50000 x = 50

Given P\(^T\) = -P, b = 1 If P = \(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\), then, P\(^T\) = \(\begin{pmatrix}a & c\\ b & d\end{pmatrix}\) \(\begin{pmatrix}a & c\\ b & d\end{pmatrix}\) = -\(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\) = \(\begin{pmatrix}-a & -1\\ -c & -d\end{pmatrix}\) comparing a = -a ⇒ 0, b = 1, c = -1, d = -d ⇒ 0 \(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\) = \(\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\)

\(\begin{pmatrix}t-4 & 0 & 0\\ -1 & t+1 & 1\\3 & 4 & t-2\end{pmatrix}\) = 0 t - 4\(\begin{pmatrix}t+1 & 1\\4 & t-2\end{pmatrix}\) + 0\(\begin{pmatrix}-1 & 1\\3 & t-2\end{pmatrix}\) + 0\(\begin{pmatrix}-1 & t+1\\3 & 4\end{pmatrix}\) t - 4[(t + 1)(t - 1) - 4] = 0 (t - 4)[ t\(^2\) - t - 6] = 0 = (t - 4)((t - 3)(t + 2) = 0 t = 4 or -2 or 3

\(2x - 2 \leq\) y = \( y \geq\) 2x - 2 when x = 0 , y = -2 and when y = 0, x = 1 \(2y - 2 \leq\) x when x = 0, y = 1 and when y = 0, x = 1 Use the diagram above for the OPTIONS

\(\bigtriangleup\)OEN = \(\bigtriangleup\)OGN OE = OG = r(radii) (r - 8)\(^2\) + 12\(^2\) = r\(^2\) r\(^2\) - 16r + 64 + 144 = r\(^2\) 16r = 64 + 144 16r = 64 + 144 = 208 \(\frac{208}{16}\) = 13cm FH = 2FO = 2OH = 2R = 2 x 13cm = 26cm

The bearing of the ship from its original position is S90ºE

A = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\)16 - x\(^2\)dx = [16 - \(\frac{x^3}{3} + c]^{4}_{0}\) = \(\frac{128}{3}\) square units

If the total number of students represents 100%, then 70% corresponds to the point on the graph where the cumulative frequency reaches 70 From the graph, find where the cumulative frequency line intersects the 70% mark. This typically corresponds to a specific age on the X-axis

The Relation between Power, Force and Velocity is given by the relation: P = F.V , this power is the product of force and velocity. Therefore, The power of the system is reduced to one-quarter of its original value when both the force and the velocity are reduced by half.

As the temperature increases, the molecules in the gas move faster, impacting the gas's container more frequently and exerting a greater force. This increases the pressure. And similarly As the pressure goes up, the temperature also goes up. So the temperature and the pressure is directly proportional to each other. At a fixed point, temperature is kept constant. Hence, at this point, it can be deduced that \(P_{1} = P_{2}\)

In a solenoid, the north pole is determined using the right-hand grip rule: if the current flows in the direction of the arrows, the north pole is at the end where the thumb points when the fingers follow the current. Therefore, the north pole is at the end labeled Q.