2010 - JAMB Mathematics Past Questions and Answers - page 2

P = (\begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix})

|P| = 2 - 1 x -3 = 5

P^-1 = (\frac{1}{5})(\begin{pmatrix} 1 & 3 \ -1 & 2 \end{pmatrix})

= (\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix})

(x + 15)^o + (2x - 45)^o + (x + 10)^o = (2n - 4)90^o

when n = 4

x + 15^o + 2x - 45^o + x - 30^o + x + 10^o = (2 x 4 - 4) 90^o

5x - 50^o = (8 - 4)90^o

5x - 50^o = 4 x 90^o = 360^o

5x = 360^o + 50^o

5x = 410^o

x = (\frac{410^o}{5})

= 82^o

Hence, the value of the least interior angle is (x - 30^o)

= (82 - 30)^o

= 52^o

Area of (\Delta)PQR = (\frac{1}{2})bh

= (\frac{1}{2}) x 4 x 5

= 10cm²

Total surface area of the cylindrical pipe = area of circular base + curved surface area

= (\pi)r² + 2(\pi)rh

= (\pi) x 7² + 2(\pi) x 7 x 50

= 49(\pi) x 700(\pi)

= 749(\pi)m²

Let D denote the distance between ((\frac{1}{2}), -(\frac{1}{2})) then using

D = (\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2})

= (\sqrt{(-{\frac{1}{2} - \frac{1}{2}})^2 + (-{\frac{1}{2} - \frac{1}{2}})^2})

= (\sqrt{(-1)^2 + (-1)^2})

= (\sqrt{1 + 1})

= √2

Let (x₁, y₁) = (1, 1) and (x₂, y₂)= (2, 5)

then gradient m of (\bar{PQ}) is

m = (\frac{y_2 - y_1}{x_2 - x_1}) = (\frac{5 - 1}{2 - 1})

= (\frac{4}{1})

= 4

cot(\theta) = (\frac{1}{\cos \theta})

= (\frac{8}{15})(given)

tan(\theta) = (\frac{15}{18})

By Pythagoras theorem,

x² = 15² + 8²

x² = 225 + 64 = 289

x = (\sqrt{289})

= 17

Hence sin(\theta) = (\frac{15}{x})

= (\frac{15}{17})

If y = (2x + 1)³, then

Let u = 2x + 1 so that, y = u³

(\frac{dy}{du}) = 3u² and (\frac{dy}{dx}) = 2

Hence by the chain rule,

(\frac{dy}{dx}) = (\frac{dy}{du}) x (\frac{du}{dx})

= 3u² x 2

= 6u²

= 6(2x + 1)²

If y = x sinx, then

Let u = x and v = sinx

(\frac{du}{dx}) = 1 and (\frac{dv}{dx}) = cosx

Hence by the product rule,

(\frac{dy}{dx}) = v (\frac{du}{dx}) + u(\frac{dv}{dx})

= (sin x) x 1 + x cosx

= sinx + x cosx

(\int^{2}{0}(x^3 + x^2))dx = (\int^{2}{0})((\frac{x^4}{4} + {\frac {x^3}{3}}))

= ((\frac{2^4}{4} + {\frac {2^3}{3}})) - ((\frac{0^4}{4} + {\frac {0^3}{3}}))

= ((\frac{16}{4} + {\frac {8}{3}})) - 0

= (\frac{80}{12}

= {\frac {20}{3}}) or 6(\frac{2}{3})