JAMB Mathematics Past Questions & Answers - Page 336

1,676.

The remainder when 6p3 - p2 - 47p + 30 is divided by p - 3 is

A.

21

B.

42

C.

63

D.

18

Correct answer is B

Let f(p) = 6p3 - p2 - 47p + 30

Then by the remainder theorem,

(p - 3): f(3) = remainder R,

i.e. f(3) = 6(3)3 - (3)2 - 47(3) + 30 = R

162 - 9 - 141 + 30 = R

192 - 150 = R

R = 42

1,677.

If x - 4 is a factor of x2 - x - k, then k is

A.

4

B.

12

C.

20

D.

2

Correct answer is B

Let f(x) = x2 - x - k
Then by the factor theorem,

(x - 4): f(4) = (4)2 - (4) - k = 0

16 - 4 - k = 0

12 - k = 0

k = 12

1,678.

If S = \(\sqrt{t^2 - 4t + 4}\), find t in terms of S

A.

S2 - 2

B.

S + 2

C.

S - 2

D.

S2 + 2

Correct answer is B

S = \(\sqrt{t^2 - 4t + 4}\)

S2 = t2 - 4t + 4

t2 - 4t + 4 - S2 = 0

Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substituting, we have;

Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\)

\(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\)

\(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\)

\(t = \frac{4 \pm \sqrt{4S^2}}{2}\)

\(t = \frac{2(2 \pm S)}{2}\)

Hence t = 2 + S or t = 2 - S

1,679.

if P = {x:x is odd, \(-1 < x \leq 20\)} and Q is {y:y is prime, \(-2 < y \leq 25\), find P \(\cap\) Q

A.

{3,5,7,11,17,19}

B.

{3,5,11,13,17,19}

C.

{3,5,7,11,13,17,19}

D.

{2,3,5,7,11,13,17,19}

Correct answer is C

P = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

Q = {-1, 3, 5, 7, 11, 13, 17, 19, 23}

P \(\cap\) Q = {3, 5, 7, 11, 13, 17, 19}

1,680.

Simplify \(\frac{\sqrt{5}(\sqrt{147} - \sqrt{12}}{\sqrt{15}}\)

A.

5

B.

\(\frac{1}{5}\)

C.

\(\frac{1}{9}\)

D.

9

Correct answer is A

\(\frac{\sqrt{5}(\sqrt{147} - \sqrt{12}}{\sqrt{15}}\)

\(\frac{\sqrt{5}(\sqrt{49 \times 3} - \sqrt{4 \times 3}}{\sqrt{5 \times 3}}\)

\(\frac{\sqrt{5}(7\sqrt{3} - 2\sqrt{3}}{\sqrt{5} \times \sqrt{3}}\)

\(\frac{\sqrt{3} (7 - 2}{\sqrt{3}}\)

= 5