2010 - JAMB Mathematics Past Questions and Answers - page 2
11
If P = \(\begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix}\)
A
\(\begin{pmatrix} -{\frac{1}{5}} & -{\frac{3}{5}} \ -{\frac{1}{5}} & -{\frac{2}{5}} \end{pmatrix}\)
B
\(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \ {\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)
C
\(\begin{pmatrix} -{\frac{1}{5}} & {\frac{3}{5}} \ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)
D
\(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)
correct option: d
P = \(\begin{pmatrix} 2 & -3 \ 1 & 1 \end{pmatrix}\)
|P| = 2 - 1 x -3 = 5
P-1 = \(\frac{1}{5}\)\(\begin{pmatrix} 1 & 3 \ -1 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)
Users' Answers & Comments|P| = 2 - 1 x -3 = 5
P-1 = \(\frac{1}{5}\)\(\begin{pmatrix} 1 & 3 \ -1 & 2 \end{pmatrix}\)
= \(\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}\)
12
The interior angles of a quadrilateral are (x + 15)o, (2x - 45)o and (x + 10)o. Find the value of the least interior angle.
A
112o
B
102o
C
82o
D
52o
correct option: d
(x + 15)o + (2x - 45)o + (x + 10)o = (2n - 4)90o
when n = 4
x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o
5x - 50o = (8 - 4)90o
5x - 50o = 4 x 90o = 360o
5x = 360o + 50o
5x = 410o
x = \(\frac{410^o}{5}\)
= 82o
Hence, the value of the least interior angle is (x - 30o)
= (82 - 30)o
= 52o
Users' Answers & Commentswhen n = 4
x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o
5x - 50o = (8 - 4)90o
5x - 50o = 4 x 90o = 360o
5x = 360o + 50o
5x = 410o
x = \(\frac{410^o}{5}\)
= 82o
Hence, the value of the least interior angle is (x - 30o)
= (82 - 30)o
= 52o
13
If the two smaller sides of a right angled triangle are 4cm and 5cm, find its area.
A
24cm2
B
10cm2
C
8cm2
D
6cm2
correct option: b
Area of \(\Delta\)PQR = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) x 4 x 5
= 10cm2
Users' Answers & Comments= \(\frac{1}{2}\) x 4 x 5
= 10cm2
14
A cylindrical pipe 50cm long with radius 7m has one end open. What is the total surface area of the pipe?
A
749\(\pi\)m2
B
700\(\pi\)m2
C
350\(\pi\)m2
D
98\(\pi\)m2
correct option: a
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
Users' Answers & Comments= \(\pi\)r2 + 2\(\pi\)rh
= \(\pi\) x 72 + 2\(\pi\) x 7 x 50
= 49\(\pi\) x 700\(\pi\)
= 749\(\pi\)m2
15
Find the distance between the points (\(\frac{1}{2}\), -\(\frac{1}{2}\)).
A
1
B
o
C
√3
D
√2
correct option: d
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
Users' Answers & CommentsD = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
= \(\sqrt{(-{\frac{1}{2} - \frac{1}{2}})^2 + (-{\frac{1}{2} - \frac{1}{2}})^2}\)
= \(\sqrt{(-1)^2 + (-1)^2}\)
= \(\sqrt{1 + 1}\)
= √2
16
Find the gradient of the line passing through the points P(1, 1) and Q(2, 5).
A
3
B
2
C
5
D
4
correct option: d
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
Users' Answers & Commentsthen 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
17
If cot\(\theta\) = \(\frac{8}{15}\), where \(\theta\) is acute, find sin\(\theta\)
A
\(\frac{8}{17}\)
B
\(\frac{15}{17}\)
C
\(\frac{16}{17}\)
D
\(\frac{13}{17}\)
correct option: b
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\(\theta\) = \(\frac{15}{x}\)
= \(\frac{15}{17}\)
Users' Answers & Comments= \(\frac{8}{15}\)(given)
tan\(\theta\) = \(\frac{15}{18}\)
By Pythagoras theorem,
x2 = 152 + 82
x2 = 225 + 64 = 289
x = \(\sqrt{289}\)
= 17
Hence sin\(\theta\) = \(\frac{15}{x}\)
= \(\frac{15}{17}\)
18
If y = (2x + 1)3, find \(\frac{dy}{dx}\)
A
6(2x + 1)
B
3(2x + 1)
C
6(2x + 1)2
D
2(2x + 1)2
correct option: c
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
Users' Answers & CommentsLet 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
19
If y = x sinx, find \(\frac{dy}{dx}\)
A
sin x - x cosx
B
sinx + x cosx
C
sinx - cosx
D
sinx + cosx
correct option: b
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
Users' Answers & CommentsLet 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
20
Evaluate \(\int^{2}_{0}(x^3 + x^2)\)dx.
A
4\(\frac{5}{6}\)
B
6\(\frac{2}{3}\)
C
1\(\frac{5}{6}\)
D
2\(\frac{5}{6}\)
correct option: b
\(\int^{2}_{0}(x^3 + x^2)\)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\(\frac{2}{3}\)
Users' Answers & Comments= (\(\frac{2^4}{4} + {\frac {2^3}{3}}\)) - (\(\frac{0^4}{4} + {\frac {0^3}{3}}\))
= (\(\frac{16}{4} + {\frac {8}{3}}\)) - 0
= \(\frac{80}{12}
= {\frac {20}{3}}\) or 6\(\frac{2}{3}\)