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JAMB Mathematics Past Questions & Answers - Page 168

836.

Solve the given equation $(\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0$

$\frac{1}{27}$

View answer & explanation

Correct answer is A

$(\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0$

Let $\log_{3} x = a$ .

$a^{2} - 6a + 9 = 0$

$a^{2} - 3a - 3a + 9 = 0$

$a(a - 3) - 3(a - 3) = 0$

$(a - 3)(a - 3) = 0$

$\implies a = 3 (twice)$

$\log_{3} x = 3 \implies x = 3^{3} = 27$

837.

The sum of the root of a quadratic equation is $\frac{5}{2}$ and the product of its root is 4. The quadratic equation is

x + 8x² - 5 = 0

2x² - 5x + 8 = 0

2x² - 8x + 5 = 0

x² + 8x - 5 = 0

View answer & explanation

Correct answer is B

x2 - (sum of roots) x + (product of roots) = 0

x2 - $\frac{5x}{2}$ + 4 = 0

2x2 - 5x + 8 = 0

838.

What is the size of an exterior angle of a regular pentagon?

36^o

60^o

72^o

120^o

360^o

View answer & explanation

Correct answer is C

Pentagon is a polygon with 5 sides. Sum of interior angles of polygon (regular) = (2n - 4) x 90o

= 5

= (10 - 4) x 90o

= 540o

each interior angle = $\frac{540}{5}$

= 108o

= 180o let x represent ext. angle

x + 108o = 180v

x = 180o - 108o

= 72o

839.

The difference between 4 $\frac{5}{7}$ and 2 $\frac{1}{4}$ greater by

$\frac{23}{28}$

$\frac{24}{28}$

$\frac{50}{56}$

$\frac{27}{28}$

$\frac{48}{56}$

View answer & explanation

Correct answer is C

Difference between 4 $\frac{5}{7}$ and 2 $\frac{1}{4}$

$\frac{33}{7}$ - $\frac{9}{4}$ = $\frac{69}{24}$

The sum of $\frac{1}{14}$ and 1 $\frac{1}{14}$ + $\frac{3}{2}$

= $\frac{11}{7}$

$\frac{69}{28}$ - $\frac{11}{7}$ = $\frac{25}{28}$

= $\frac{50}{56}$