JAMB Mathematics Past Questions & Answers - Page 308

1,536.

Solve for x and y if x - y = 2 and x2 - y2 = 8

A.

(-1, 3)

B.

(3, 1)

C.

(-3, 1)

D.

(1, 3)

Correct answer is B

x - y = 2 ...........(1)

x2 - y2 = 8 ........... (2)

x - 2 = y ............ (3)

Put y = x -2 in (2)

x2 - (x - 2)2 = 8

x2 - (x2 - 4x + 4) = 8

x2 - x2 + 4x - 4 = 8

4x = 8 + 4 = 12

x = \(\frac{12}{4}\)

= 3

from (3), y = 3 - 2 = 1

therefore, x = 3, y = 1

1,537.

If 9x2 + 6xy + 4y2 is a factor of 27x3 - 8y3, find the other factor.

A.

2y + 3x

B.

2y - 3x

C.

3x + 2y

D.

3x - 2y

Correct answer is D

27x3 - 8y3 = (3x - 2y)3

But 9x2 + 6xy + 4y2 = (3x +2y)2

So, 27x3 - 8y3 = (3x - 2y)(3x - 2y)2

Hence the other factor is 3x - 2y

1,538.

Make Q the subject of formula if p = \(\frac{M}{5}\)(X + Q) + 1

A.

\(\frac{5P - MX + 5}{M}\)

B.

\(\frac{5P - MX - 5}{M}\)

C.

\(\frac{5P + MX + 5}{M}\)

D.

\(\frac{5P + MX - 5}{M}\)

Correct answer is B

p = \(\frac{M}{5}\)(X + Q) + 1

P - 1 = \(\frac{M}{5}\)(X + Q)

\(\frac{5}{M}\)(p - 1) = X + Q

\(\frac{5}{M}\)(p - 1)- x = Q

Q = \(\frac{5(p -1) - Mx}{M}\)

= \(\frac{5p - 5 - Mx}{M}\)

= \(\frac{5p - Mx - 5}{M}\)

1,539.

Find the equation of a line parallel to y = -4x + 2 passing through (2,3)

A.

y + 4x + 11 = 0

B.

y - 4x - 11 = 0

C.

y + 4x - 11 = 0

D.

y - 4x + 11 = 0

Correct answer is C

By comparing y = mx + c

with y = -4x + 2,

the gradient of y = -4x + 2 is m1 = -4

let the gradient of the line parallel to the given line be m2,

then, m2 = m1 = -4

(condition for parallelism)

using, y - y1 = m2(x - x1)

Hence the equation of the parallel line is

y - 3 = -4(x-2)

y - 3 = -4 x + 8

y + 4x = 8 + 3

y + 4x = 11

y + 4x - 11 = 0

1,540.

At what value of X does the function y = -3 - 2x + X2 attain a minimum value?

A.

-1

B.

14

C.

4

D.

1

Correct answer is D

Given that y = -3 - 2x + X2

then, \(\frac{dy}{dx}\) = -2 + 2x

At maximum value, \(\frac{dy}{dx}\) = O

therefore, -2 + 2x

2x = 2

x = 2/2 = 1