1998 - JAMB Mathematics Past Questions and Answers - page 2
11
The locus of all points at a distance 8cm from a point N passes through points T and S. If S is equidistant from T and N, find the area of triangle STN.
A
4\(\sqrt{3cm^2}\)
B
16\(\sqrt{3cm^2}\)
C
32cm2
D
64cm2
correct option: b
Users' Answers & Comments12
If the distance between the points (x, 3) and (-x, 2) is 5. Find x
A
6.0
B
2.5
C
\(\sqrt{6}\)
D
\(\sqrt{3}\)
correct option: c
d2 = (y - y)2 + (x - x)2
5 = 4x2 + 1 = 25= 4x2 + 1
= 4x2 = 25 - 1= 24
x2 = \(\frac{24}{4}\)
x = \(\sqrt{6}\)
Users' Answers & Comments5 = 4x2 + 1 = 25= 4x2 + 1
= 4x2 = 25 - 1= 24
x2 = \(\frac{24}{4}\)
x = \(\sqrt{6}\)
13
The midpoint of the segment of the line y = 4x + 3 which lies between the x-ax 1 is and the y-ax 1 is
A
(\(\frac{3}{2}\), \(\frac{3}{2}\))
B
(\(\frac{2}{3}\), \(\frac{3}{2}\))
C
(\(\frac{3}{8}\), \(\frac{3}{2}\))
D
(-\(\frac{3}{8}\), \(\frac{3}{2}\))
correct option: d
y = 4x + 3
when x = 0, y = 3 \(\to\) (0, 3)
when y = 0, x = -\(\frac{3}{4}\) \(\to\) (\(\frac{3}{4}\), 0)
mid-point \(\frac{0 + (-{\frac{3}{4}})}{2}\), \(\frac{3 + 0}{4}\)
-\(\frac{3}{8}\), \(\frac{3}{2}\)
Users' Answers & Commentswhen x = 0, y = 3 \(\to\) (0, 3)
when y = 0, x = -\(\frac{3}{4}\) \(\to\) (\(\frac{3}{4}\), 0)
mid-point \(\frac{0 + (-{\frac{3}{4}})}{2}\), \(\frac{3 + 0}{4}\)
-\(\frac{3}{8}\), \(\frac{3}{2}\)
14
solve the equation cos x + sin x \(\frac{1}{cos x - sinx}\) for values of such that 0 \(\leq\) x < 2\(\pi\)
A
\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)
B
\(\frac{\pi}{3}\), \(\frac{2\pi}{3}\)
C
0, \(\frac{\pi}{3}\)
D
0, \(\pi\)
correct option: d
cos x + sin x \(\frac{1}{cos x - sinx}\)
= (cosx + sinx)(cosx - sinx) = 1
= cos2x + sin2x = 1
cos2x - (1 - cos2x) = 1
= 2cos2x = 2
cos2x = 1
= cosx = \(\pm\)1 = x
= cos-1x (\(\pm\), 1)
= 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\)
(possible solution)
Users' Answers & Comments= (cosx + sinx)(cosx - sinx) = 1
= cos2x + sin2x = 1
cos2x - (1 - cos2x) = 1
= 2cos2x = 2
cos2x = 1
= cosx = \(\pm\)1 = x
= cos-1x (\(\pm\), 1)
= 0, \(\pi\) \(\frac{3}{2}\pi\), 2\(\pi\)
(possible solution)
15
From the top of a vertical mast 150m high., two huts on the same ground level are observed. One due east and the other due west of the mast. Their angles of depression are 60o and 45o respectively. Find the distance between the huts
A
150(1 + \(\sqrt{3}\))m
B
50( \(\sqrt{3}\) - \(\sqrt{3}\))m
C
150 \(\sqrt{3}\)m
D
\(\frac{50}{\sqrt{3}}\)m
correct option: b
\(\frac{150}{Z}\) = tan 60o,
Z = \(\frac{150}{tan 60^o}\)
= \(\frac{150}{3}\)
= 50\(\sqrt{3}\)cm
\(\frac{150}{X x Z}\) = tan45o = 1
X + Z = 150
X = 150 - Z
= 150 - 50\(\sqrt{3}\)
= 50( \(\sqrt{3}\) - \(\sqrt{3}\))m
Users' Answers & CommentsZ = \(\frac{150}{tan 60^o}\)
= \(\frac{150}{3}\)
= 50\(\sqrt{3}\)cm
\(\frac{150}{X x Z}\) = tan45o = 1
X + Z = 150
X = 150 - Z
= 150 - 50\(\sqrt{3}\)
= 50( \(\sqrt{3}\) - \(\sqrt{3}\))m
16
If y = 243(4x + 5)-2, find \(\frac{dy}{dx}\) when x = 1
A
\(\frac{-8}{3}\)
B
\(\frac{3}{8}\)
C
\(\frac{9}{8}\)
D
-\(\frac{8}{9}\)
correct option: a
y = 243(4x + 5)-2, find \(\frac{dy}{dx}\)
= -1944(4x + 5)-3
= 1944(9)-3
\(\frac{dy}{dx}\) when x = 1
= -\(\frac{1944}{9^3}\)
= -\(\frac{1944}{729}\)
= \(\frac{-8}{3}\)
Users' Answers & Comments= -1944(4x + 5)-3
= 1944(9)-3
\(\frac{dy}{dx}\) when x = 1
= -\(\frac{1944}{9^3}\)
= -\(\frac{1944}{729}\)
= \(\frac{-8}{3}\)
17
Differentiate \(\frac{x}{cosx}\) with respect to x
A
1 + x sec x tan x
B
1 + sec2 x
C
cos x + x tan x
D
x sec x tan x + secx
correct option: d
let y = \(\frac{x}{cosx}\) = x sec x
y = u(x) v (x0
\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)
dy x [secx tanx] + secx
x = x secx tanx + secx
Users' Answers & Commentsy = u(x) v (x0
\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)
dy x [secx tanx] + secx
x = x secx tanx + secx
18
Evaluate ∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx
A
\(\frac{\pi}{2}\)
B
\(\pi\) - 2
C
\(\frac{\pi}{3}\)
D
\(\pi\) + 2
correct option: b
∫\(^{\pi}_{2}\)(sec2 x - tan2x)dx
∫\(^{\pi}_{2}\) dx = [X]\(^{\pi}_{2}\)
= \(\pi\) - 2 + c
when c is an arbitrary constant of integration
Users' Answers & Comments∫\(^{\pi}_{2}\) dx = [X]\(^{\pi}_{2}\)
= \(\pi\) - 2 + c
when c is an arbitrary constant of integration
19
find the equation of the curve which passes through by 6x - 5
A
6x2 - 5x + 5
B
6x2 + 5x + 5
C
3x2 - 5x - 5
D
3x2 - 5x + 3
correct option: d
m = \(\frac{dy}{dv}\) = 6x - 5
∫dy = ∫(6x - 5)dx
y = 3x2 - 5x + C
when x = 2, y = 5
∴ 5 = 3(2)2 - 5(2) +C
C = 3
∴ y = 3x2 - 5x + 3
Users' Answers & Comments∫dy = ∫(6x - 5)dx
y = 3x2 - 5x + C
when x = 2, y = 5
∴ 5 = 3(2)2 - 5(2) +C
C = 3
∴ y = 3x2 - 5x + 3
20
If m and n are the mean and median respectively of the set of numbers 2, 3, 9, 7, 6, 7, 8, 5, find m + 2n to the nearest whole number
A
19
B
18
C
13
D
12
correct option: a
mean(x) = \(\frac{\sum x}{N}\)
= \(\frac{48}{8}\)
= 5.875
re-arranging the numbers;
2, 3, 5, 6, 2, 7, 8, 9
median = \(\frac{6 + 7}{2}\)
= \(\frac{1}{2}\)
= 6.5
m + 2n = 5.875 + (6.5)2
= 13 + 5.875
= 18.875
= \(\approx\) = 19
Users' Answers & Comments= \(\frac{48}{8}\)
= 5.875
re-arranging the numbers;
2, 3, 5, 6, 2, 7, 8, 9
median = \(\frac{6 + 7}{2}\)
= \(\frac{1}{2}\)
= 6.5
m + 2n = 5.875 + (6.5)2
= 13 + 5.875
= 18.875
= \(\approx\) = 19