JAMB Mathematics Past Questions & Answers - Page 284

1,416.

Factorize completely the expression \(abx^2 + 6y - 3ax - 2byx\)

A.

(ax - 2y)(bx - 3)

B.

(bx + 3)(2y - ax)

C.

(bx + 3)(ax - 2y)

D.

(ax - 2y)(ax - b)

Correct answer is A

\(abx^{2} + 6y - 3ax - 2byx\)

Collecting like terms, we have

\(abx^{2} - 3ax + 6y - 2byx\)

= \(ax(bx - 3) + 2y(3 - bx)\)

= \(ax(bx - 3) - 2y(bx - 3)\)

= \((ax - 2y)(bx - 3)\)

1,417.

The graph of f(x) = x2 - 5x + 6 crosses the x-axis at the points

A.

(-6, 0), (-1, 0)

B.

(-3, 0), (-2,0)

C.

(-6, 0),(1, 0)

D.

(2, 0), (3, 0)

Correct answer is D

When X = 3, Y = 0(3, 0),

When x = 2, y = 0(2, 0)

1,418.

Finds a positive value of p if the expression 2x2 - px + p leaves a remainder 6 when divided by x - p and q

A.

1

B.

2

C.

3

D.

4

Correct answer is B

x - p,x = p

2p2 - p2 + p = 6

= p2 + p - 6

= 0

p = 3, 2

1,419.

If x - 1 and x + 1 are both factors of the equation x\(^3\) + px\(^2\) + qx + 6 = 0, evaluate p and q

A.

-6, -1

B.

6, 1

C.

1, -1

D.

6, -6

Correct answer is A

\(x^{3} + px^{2} + qx + 6 = 0\)

f(x - 1) = 0; f(1) = 0.

\(1^{3} + p(1^{2}) + q(1) + 6 = 0 \implies p + q = -7 ... (1)\)

f(x + 1) = 0; f(-1) = 0.

\((-1)^{3} + p(-1^{2}) + q(-1) + 6 = 0 \implies p - q = -5 ... (2)\)

Subtract (2) from (1).

\(2q = -2 \implies q = -1\)

\(p - (-1) = -5 \implies p = -5 - 1 = -6\)

\((p, q) = (-6, -1)\)

1,420.

If S = (x : x\(^2\) = 9, x > 4), then S is equal to

A.

4

B.

{0}

C.

\(\emptyset\)

D.

{\(\emptyset\)}

Correct answer is C

No explanation has been provided for this answer.