Question Library

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

3 + 6 + 12 + .....18thy term 1st term = 3, common ratio \(\frac{6}{3}\) = 2 n = 18, sum of GP is given by Sn = a\(\frac{(r^n - 1)}{r - 1}\) s18 = 3\(\frac{(2^{18} - 1)}{2 - 1}\) = 3(2^18 - 1)

x\(^2\) + x + 1 \(\frac{dy}{dx}\) = 2x + 1 At the turning point, \(\frac{dy}{dx}\) = 0 2x + 1 = 0 x = -\(\frac{1}{2}\)

Cos \(\theta\) = \(\frac{12}{13}\) x\(^2\) + 12\(^2\) = 13\(^2\) x\(^2\) = 169- 144 = 25 x = \(\sqrt{25}\) = 5 Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\) If cot\(^2\)\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\) = 1 + \(\frac{1}{\frac{(5)^2}{12^2}}\) = 1 + \(\frac{1}{\frac{25}{144}}\) = 1 + \(\frac{144}{25}\) = \(\frac{25 + 144}{25}\) = \(\frac{169}{25}\)

\(\frac{h_1}{h_2}\) = \(\frac{2}{3}\) h2 = \(\frac{2h_1}{3}\) \(\frac{r_1}{r_2}\) = \(\frac{9}{8}\) r2 = \(\frac{9r_1}{8}\) v1 = \(\pi\)(\(\frac{9r_1}{8}\))2(\(\frac{2h_1}{3}\)) = \(\pi\)r1 2h1 x \(\frac{27}{32}\) v = \(\frac{\pi r_1 2h_1 \times \frac{27}{32}}{\pi r_1 2h_1}\) = \(\frac{27}{32}\) v2 : v1 = 27 : 32

\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\) Average = 3.5 3.5 = \(\frac{(1 \times 1) + (2 \times 4) + (3 \times 5) + (4 \times 6) + 5x + (6 \times 2)}{1 + 4 + 5 + 6 + x + 2}\) \(\frac{3.5}{1}\) = \(\frac{1 + 8 + 15 + 24 + 5x + 12}{18 + x}\) \(\frac{3.5}{1}\) = \(\frac{60 + 5x}{18 + x}\) 60 + 5x = 3.5(18 \(\div\) x) 60 + 5x = = 63 + 1.5x 5x - 1.5x = 63 - 60 1.5x = 3 x = \(\frac{3}{1.5}\) \(\frac{30}{15}\) = 2

The required equation is y = x2 - x - 2 i.e. B where the graph touches the graph touches the x-axis y = 0 x2 - x - 2 = 0 = (x + 1)(x - 2) = 0 Hence roots of the equation are -1 and 2 as shown in the graph

Since PS = QS = RS S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius QPR \(\pm\) \(\frac{100^o}{2}\) = 50o

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

yzx + 105o + 30o = 180o yzx = 180 - 155 = 45o \(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\) yz = \(\frac{8 \sin 30}{sin 45}\) = 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\) = 4 \(\div\) \(\frac{1}{\sqrt{2}}\) = 4\(\sqrt{2}\)cm

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

\(\frac{x}{4} = \frac{6 + x}{6}\) 6x = 4(6 + x) = 24 + 4x x = 12cm CSA = \(\pi RL - \pi rl\) = \(\pi (6) \sqrt{{18^2} + 6^2} - \pi \times 4 \times \sqrt{{12^2} + 4^2}\) = 6\(\pi \sqrt{360} - 4 \pi \sqrt{160}\) = 36\(\pi \sqrt{10} - 16 \pi \sqrt{10}\) = 20\(\pi \sqrt{10}\)cm2

(a) \(\frac{0.016 \times 0.084}{0.48}\) = \(\frac{16 \times 10^{-3} \times 84 \times 10^{-3}}{48 \times 10^{-2}}\) = \(28 \times 10^{-3 - 3 - (-2)}\) = \(28 \times 10^{-4} = 2.8 \times 10^{-3}\) (b) \(T_{n} = a + (n - 1)d\) (Terms of an A.P) \(T_{2} = a + d = 2 ... (1)\) \(T_{6} = a + 5d = 5 ... (2)\) \((2) - (1) : 4d = 3 \implies d = \frac{3}{4} = 0.75\) Put the value of d in (1), so we have \(a + 0.75 = 2 \implies a = 1.25\) Therefore, the length of the poles in metres = \(1.25, 2, 2.75, 3.5, 4.25, 5, 5.75, 6.5\)

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

(a) Given any triangle ABC To prove that < A + < B + < C = 180^\circ Construction: Produce BC to X and draw CP // BA. Proof: with the lettering \(a_{1} = a_{2}\) (alternate angles); \(b_{1} = b_{2}\) (corresponding angles) \(< BCA + < A + < B = 180^\circ ; < ACB + < A + < B = 180^\circ\) \(\therefore < A + < B + < C = 180^\circ\) (b) From the given triangle and with reference to letterings; \(\hat{a} + \hat{c} + \hat{c} = \hat{b}\) (exterior opposite angles of a triangle) \(a + c = e\) (exterior opposite angles of a triangle) \(a + b = a + a + c + c \implies a + b = 2(a + c)\) = \(2e \text{where e = a + c}\) i.e \(< CAE + < CBD = 2 < CED\) (Proved)