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Sum of interior angle of hexagon = [2(6) - 4]90o = 720o sum of central angle = 360o Each central angle = \(\frac{360}{6}\) = 60o Area of Hexagon = \(\frac{1}{2}\) x 6 x 6 sin 60o \(\frac{36 \times 6\sqrt{3}}{2 \times 2}\) = \(54 \sqrt{3}\)cm2

By re-arranging the goals in ascending order 0. 0. 0. 0. 0, 1. 1. 1. 1. 1. 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5. Mean = \(\frac{36}{20}\) = 1.8 Median = \(\frac{1 + 2}{2}\) = \(\frac{3}{2}\) = 1.5 Mode = 1 = 1.8, 1.5 and 1

sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3 Where 1 + tan\(^2\) \(\theta\) = sec\(^2\) \(\theta\) 1 + tan\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3 2 tan\(^2\) \(\theta\) = 2 tan\(^2\) \(\theta\) = 1 tan\(\theta\) = \sqrt1 where \sqrt1 = 1 tan\(\theta\) = 1 And tan 45^\circ = 1 ∴ \(\theta\) = 45^\circ

The rise of water is equivalent to the volume of the sphere of radius \(\frac{1}{2}\)cm immersed x \(\frac{1}{\text{No. of sides sq. root 3}}\) Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))3 = \(\frac{1}{8}\) = \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\) = \(\frac{\pi}{24\sqrt{3}}\)

x2 - y - 1 = 0.......(i) y - 2x + 2 = 0......(ii) By re-arranging eqn. (ii) y = 2x - 2........(iii) Subst. eqn. (iii) in eqn (i) x2 - (2x - 2) - 1 = 0 x2 - 2x + 1 = 0 = (x - 1) = 0 When x - 1 = 0 x = 1 Sub. for x = 1 in eqn. (iii) y = 2 - 2 = 0 x = 1, y = 0

No expanded working has been attached for this question yet. The verified answer is Option C.

(x + y) is a factor 2x3 + 2x2y - xy + 3x - y^2

Two triangles have the same area if three sides in one triangle equal to three in the other.

(25)\(^{x - 1}\) = 64(\(\frac{5}{2}\))\(^6\) (\(\frac{5}{2}\))\(^6\) = (\(\frac{5^6}{64}\)) 25\(^{x - 1}\) = 64 x (\(\frac{5^6}{64}\)) 5\(^{2x - 2}\) = 5\(^6\) 2x - 2 = 6 = \(\frac{8}{2}\) = 4

Assuming loge 4.4 = 1.4816 and loge 7.7 = 2.0142 then value of loge \(\frac{7}{4}\) = loge7.7 - loge \(\frac{7.7}{4.4}\) = loge\(\frac{7}{4}\) = 2.0142 -1.4816 = 0.5326

The locus of all points having a distance of 1 unit from each of the two fixed points. a and b is: a perpendicular to the (ab) through the mid-points ab

No expanded working has been attached for this question yet. The verified answer is Option C.

\(\frac{(a - \frac{1}{a}) (a^{\frac{4}{3}} + \frac{1}{a^{\frac{2}{3}}})}{a^2 - \left(\frac{1}{a}\right)^2}\) \( = \frac{(a - \frac{1}{a}) \left(a^{\frac{4}{3}} + \frac{1}{a^{\frac{2}{3}}}\right)}{\left(a - \frac{1}{a}\right)\left(a + \frac{1}{a}\right)} \) \(= \frac{a^{\frac{4}{3}} + \frac{1}{a^{\frac{2}{3}}}}{a + \frac{1}{a}}\) \(= \frac{\frac{a^{2} + 1}{a^{\frac{2}{3}}}}{\frac{a^2 + 1}{a}}\) \(= \frac{1}{a^{\frac{2}{3}}} \cdot a\) \(= a^{1 - \frac{2}{3}} = a^{\frac{1}{3}}\) \(\boxed{a^{\frac{1}{3}}}\)

If k = -7 is put as l = -15, the equation x\(^4\) - kx\(^3\) + 10x\(^2\) + lx - 3 becomes x\(^4\) - (7x\(^3\)) + 10x\(^2\) + (15) -3 = x\(^4\) + 7x\(^3\) + 10x\(^2\) - 15x - 3 This equation is divisible by (x - 1) and (x + 2) with the remainder as 27 k = -7, l = -15

In < AFT: \(\frac{TF}{AF}\) = tan60º therefore, TF = xtan60º ..................1 In < BFT: \(\frac{TF}{BF}\) = tan45º TF = ( 2 + x )tan45º .......................2 equating equations 1 and 2 xtan60º = ( 2 + x )tan45º But tan60º = \(\sqrt3\) and tan45º = 1 x \(\sqrt3\) = 2 + x x \(\sqrt3\) - x = 2 x ( \(\sqrt3\) - 1) = 2 x = \(\frac{2}{(\sqrt3\)) - 1}\) rationalizing the denominator x = \(\frac{2(\sqrt3 + 2)}{3 - 1}\) = \(\frac{2(\sqrt3 + 1)}{2}\) = \(\sqrt3\) + 1