2006 - JAMB Mathematics Past Questions and Answers - page 4
31
What is the locus of points equidistant from the lines ax + bc + c = 0?
A
A line bx - ay +q = 0
B
A line ax - ay +q = 0
C
A line bx + ay +q = 0
D
A line bx + ay +q = 0
correct option: b
Locus of point equidistant from a given straight is the perpendicular bisector of the straight line
∴Gradient of the line ax + by + c = 0
implies by = -ax - c
y = -a/bx - c
the gradient (m) = a/b
∴Gradient of the perpendicular (m) = b/a
If P(x,y) is the point intersection of the two lines equation of the perpendicular becomes
y - y1 = m(x - x1)
y - y1 = b/a(x - x1)
ay - ay1 = bx - bx1
ay - bx + bx1 - ay1 = 0
ay - bx + b(x1 - y1) = 0
implies ay - bx + q
Users' Answers & Comments∴Gradient of the line ax + by + c = 0
implies by = -ax - c
y = -a/bx - c
the gradient (m) = a/b
∴Gradient of the perpendicular (m) = b/a
If P(x,y) is the point intersection of the two lines equation of the perpendicular becomes
y - y1 = m(x - x1)
y - y1 = b/a(x - x1)
ay - ay1 = bx - bx1
ay - bx + bx1 - ay1 = 0
ay - bx + b(x1 - y1) = 0
implies ay - bx + q
32
PQRSTW is a regular hexagon and QS intersects RT at V. Calculate ∠TVS
A
120o
B
90o
C
30o
D
60o
correct option: d
Each ∠of a regular polygon
\(=\frac{(n-2)180}{n}\\ =\frac{(6-2)180}{6}\\ =\frac{(4)180}{6}\)
= 120o
ΔQSR is isosceles
∴ Q = S = 30o
Also ΔTSR is isosceles
∴ T = R = 30o
∠TSV + ∠VSR = 120
∠TSV + 30 = 120
∠TSV = 120 - 30
∠TSV = 90o
∴∴VTS + ∴VST + ∴TVS = 180 (sum of ∠s of a Δ)
30 + 90 + ∠TVS = 180
120 + ∠TVS = 180
∠TVS = 180 - 120
∠TVS = 60o
Users' Answers & Comments\(=\frac{(n-2)180}{n}\\ =\frac{(6-2)180}{6}\\ =\frac{(4)180}{6}\)
= 120o
ΔQSR is isosceles
∴ Q = S = 30o
Also ΔTSR is isosceles
∴ T = R = 30o
∠TSV + ∠VSR = 120
∠TSV + 30 = 120
∠TSV = 120 - 30
∠TSV = 90o
∴∴VTS + ∴VST + ∴TVS = 180 (sum of ∠s of a Δ)
30 + 90 + ∠TVS = 180
120 + ∠TVS = 180
∠TVS = 180 - 120
∠TVS = 60o
33
If the locus of the points which are equidistant from point P and Q meets line PQ at point N, then PN equals
A
NQ
B
1/4NQ
C
2NQ
D
1/2NQ
correct option: a
Locus of point equidistant from points P and q is a perpendicular bisector of the straight joining P and Q
∴PN = NQ
Users' Answers & Comments∴PN = NQ
34
A
14 cm
B
14√3 cm
C
10 cm
D
10√3 cm
correct option: a
In ΔPST; cos 60 = ST/8
ST = 8cos 60
ST = 8 x 1/2 = 4
TR = 10 cm (opp. sides of a rectangle PQRT)
SR = ST + TR
SR = 4 + 10
SR = 14 cm
Users' Answers & CommentsST = 8cos 60
ST = 8 x 1/2 = 4
TR = 10 cm (opp. sides of a rectangle PQRT)
SR = ST + TR
SR = 4 + 10
SR = 14 cm
35
A
Y ≥ 0, y ≥ x and y + x ≤ 4
B
Y ≤ x, y + x ≤ 4
C
Y + x ≥ 4, y ≤ x
D
Y ≤ x, y + x ≤ 4 and y ≥ 0
correct option: b
Y = x implies y ≤ x
Y + x = 4 implies y = 4 – x
∴y = 4 – x
∴y ≤ x, y + x ≤ 4 and y ≥ 0
Users' Answers & CommentsY + x = 4 implies y = 4 – x
∴y = 4 – x
∴y ≤ x, y + x ≤ 4 and y ≥ 0
36
A binary operation θ defined on the set of real number is such that xθy = xy/6 for all x, y ∈ R. Find the inverse of 20 under this operation when the identity element is 6
A
1/12
B
10/3
C
1/20
D
9/5
correct option: d
X ⊕ y = xy/6 = xe/6 = x
Xe = 6x
e = 6
X ⊕ y = xx1/6 where x1 = 20
Xx20/6 = 6
X = 36/20
X = 9/5
Users' Answers & CommentsXe = 6x
e = 6
X ⊕ y = xx1/6 where x1 = 20
Xx20/6 = 6
X = 36/20
X = 9/5
37
If p varies inversely as the cube of q and q varies directly as the square of r, what is the relationship between p and r?
A
p varies directly as r3
B
p varies inversely as r6
C
p varies inversely as 6√P
D
p varies directly as r6
correct option: c
P ∝ 1/q3
P = K/q3
q3 = K/P
q = K/p1/3
But q ∝ r2
q = Kr2
K/p1/3 = Kr2
r2 = K/p1/3 x 1/K
r2 = 1/p1/3
r = (1/p1/3)2
r = 1/p1/6
r ∝ 1/p1/6
∴ r varies inversely as 6√P
Users' Answers & CommentsP = K/q3
q3 = K/P
q = K/p1/3
But q ∝ r2
q = Kr2
K/p1/3 = Kr2
r2 = K/p1/3 x 1/K
r2 = 1/p1/3
r = (1/p1/3)2
r = 1/p1/6
r ∝ 1/p1/6
∴ r varies inversely as 6√P
38
A binary operation * on the set of rational numbers is defined as \(x \times y = \frac{x^2 - y^2}{2xy}find -5 \times 3\)
A
\(\frac{-8}{15}\)
B
\(\frac{8}{15}\)
C
\(\frac{17}{15}\)
D
\(\frac{-17}{15}\)
correct option: a
\(x \times y = \frac{x^2 - y^2}{2xy}find -5 \times 3\\
=\frac{(x+y)(x-y)}{2xy}\\
=\frac{(-5+3)(-5-3)}{2(-5\times3)}\\
=\frac{-2 \times -8}{2(-5\times3)}\\
=\frac{-8}{15}\)
Users' Answers & Comments39
Find the value of k if the expression kx3 + x2 - 5x - 2 leaves a remainder 2 when it is divided by 2x + 1
A
10
B
8
C
-10
D
-8
correct option: c
F(x) = Q x D + R
Kx3 + x2 - 5x – 2 = Q(2x+1)+R
If 2x+1 = 0 implies x = -1/2
∴k(-1/2)3 + (-1/2)2 -5(-1/2) -2
= Q(2(-1/2) + 1) +2
K(-1/8) + 1/4 + 5/2 - 2 = Q(-1+1)+2
-k/8 + 1/4 + 5/2 - 2 = 0+2
(-k+2+20-16) / 8 = 2
(-k+6) / 8 = 2
-k+6 = 2*8
-k+6 = 16
-k = 16-6
-k = 10
∴k = -10
Users' Answers & CommentsKx3 + x2 - 5x – 2 = Q(2x+1)+R
If 2x+1 = 0 implies x = -1/2
∴k(-1/2)3 + (-1/2)2 -5(-1/2) -2
= Q(2(-1/2) + 1) +2
K(-1/8) + 1/4 + 5/2 - 2 = Q(-1+1)+2
-k/8 + 1/4 + 5/2 - 2 = 0+2
(-k+2+20-16) / 8 = 2
(-k+6) / 8 = 2
-k+6 = 2*8
-k+6 = 16
-k = 16-6
-k = 10
∴k = -10
40
Solve the inequalities for which \(\frac{x+4}{3}-\frac{x-3}{2} < 4\)
A
x < 7
B
x > -7
C
x < -7
D
x > 7
correct option: b
\(\frac{x+4}{3}-\frac{x-3}{2} < 4\\
\frac{(2(x+4))-(3(x-3))}{6}<4\)
2(x+4) – 3(x-3) < 4 x 6
2x + 8 – 3x+9 < 24
-x + 17 < 24
-x < 24 – 17
-x < 7
X > -7
Users' Answers & Comments2(x+4) – 3(x-3) < 4 x 6
2x + 8 – 3x+9 < 24
-x + 17 < 24
-x < 24 – 17
-x < 7
X > -7