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JAMB Mathematics Past Questions & Answers - Page 12

56.

Find the value of y, if log (y + 8) + log (y - 8) = 2log 3 + 2log 5

A.

y = ±5

B.

y = ±10

C.

y = ±17

D.

y = ±13

Correct answer is C

log (y + 8) + log (y - 8) = 2log 3 + 2log 5

⇒ log (y + 8)(y - 8) = 2log 3 + 2log 5 (log ab = log a + log b)

⇒ log (y2 -8y + 8y - 64) = log 32 + log 52

⇒ log (y2 - 64) = log 32 + log 52

⇒ log (y2 - 64) = log 9 + log 25

⇒ log (y2 - 64) = log (9 × 25)

⇒ log(y2 - 64) = log 225

⇒ y2 - 64 = 225

⇒ y2 = 225 + 64

⇒ y2 = 289

⇒ y = ±√289

∴ y = ±17

57.

Find the value of y if 402y=102ten

A.

4

B.

2

C.

5

D.

3

Correct answer is C

402y=102ten

⇒ 4 x y2 + 0 x y1 + 2 x y0 = 102

⇒ 4y2 + 0 + 2 x 1 = 102

⇒ 4y2 + 2 = 102

⇒ 4y2 = 102 - 2

⇒ 4y2 = 100

⇒ y2=1004

⇒ y2 = 25

∴ y = √25 = 5

58.

Determine the area of the region bounded by y = 2x2 + 10 and Y = 4x+16.

A.

18

B.

103

C.

443

D.

643

Correct answer is D

To find the point of intersection, equate the two equations

⇒ 2x2 + 10 = 4x + 16

⇒ 2x2 + 10 - 4x - 16 = 0

⇒ 2x2 - 4x - 6 = 0

Factorize

⇒ 2(x + 1)(x - 3) = 0

∴ x = -1 or 3

The curves will intersect at -1 and 3

Open photo
Area = ba [upper function] - [lower function] dx

 

= 314x+16(2x2+10)dx

= 312x2+4x+6dx

= (2x33+2x2+6x)31

= (2(3)33+2(3)2+6(3))(2(1)33+2(1)2+6(1))

= 18 - (103)

= 18 + 103

= 643

59.

Calculate the mean deviation of the first five prime numbers.

A.

2.72

B.

5.6

C.

5.25

D.

13.6

Correct answer is A

The first five prime numbers are 2, 3, 5, 7 and 11

M.D = (xx̄)n

n = 5

x̄=xn=2+3+5+7+115=285=5.6

M.D = (25.6)+(35.6)+(55.6)+(75.6)+(115.6)5

= (3.6)+(2.6)+(0.6)+(1.4)+(5.4)5=13.65

Take the absolute value of all numbers in bracket

= 3.6+2.6+0.6+1.4+5.45=13.65

60.

Differentiate the function y = \sqrt[3]{x^2}(2x - x^2)

A.

\frac {dy}{dx} = \frac {10x^{5/3}}{3} - \frac {8x^{2/3}}{3}

B.

\frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}

C.

\frac {dy}{dx} = \frac {10x^{5/3}}{3} - \frac {8x^{5/3}}{3}

D.

\frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{2/3}}{3}

Correct answer is B

y = \sqrt[3]{x^2(2x - x^2)} = x^{2/3} (2x - x^2)

= 2x^{5/3} - x^{8/3}

Now, we can differentiate the function

\therefore \frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}