JAMB Mathematics Past Questions & Answers - Page 153

761.

The minimum point on the curve y = x2 - 6x + 5 is at

A.

(1, 5)

B.

(2, 3)

C.

(-3, -4)

D.

(3, -4)

Correct answer is D

Given the curve \(y = x^{2} - 6x + 5\)

At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)

\(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\)

\(2x - 6 = 0 \implies x = 3\)

Since 3 > 0, it is a minimum point.

When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\)

Hence, the turning point has coordinates (3, -4).

762.

A force of 5 units acts on a particle in the direction to the east and another force of 4 units acts on the particle in the direction north-east. The resultants of the two forces is

A.

\(\sqrt{3}\) units

B.

3 units

C.

\(\sqrt{41 + 20 \sqrt{2}}\) units

D.

\(\sqrt{41 - 20 \sqrt{2}}\) units

Correct answer is C

Force to the east = 5 units

force to the North - east = 4 units.

Resultant of the two forces is the square root

52 + 42 = 41 and plus the sum of its resistance

5 x 4\(\sqrt{2}\) = 20\(\sqrt{2}\)

= \(\sqrt{41 + 20 \sqrt{2}}\) units

763.

Simplify \(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\)

A.

\(\frac{4ab}{a - b}\)

B.

\(\frac{-4ab}{a^2 - b^2}\)

C.

\(\frac{-4ab}{a^{-2} - b}\)

D.

\(\frac{4ab}{a^{-2} - b^{-2}}\)

Correct answer is B

\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\)

applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b)

= (a = b + a = b)(a = b - a = b)

2a(-2b) = -4ab

= \(\frac{-4ab}{(a + b)(a - b)}\)

= \(\frac{-4ab}{a^2 - b^2}\)

764.

If the four interior angles of a quadrilateral are (p + 10)°, (p - 30)°, (2p - 45)°, and (p + 15)°, then p is

A.

125o

B.

82o

C.

135o

D.

105o

E.

60o

Correct answer is B

Sum of interior angles of a polygon = (2n - 4) x 90

where n is the no. of sides

sum of interior angles of the quad. = ({2 x 4} -4) x 90

(8 - 4) x 90 = 4 x 90o

= 360o

(p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o

p + 10o + p - 30o + 2p - 45o + p + 15 = 360o

5p = 360o + 50o

= 82o

765.

If a circular paper disc is trimmed in such a way that its circumference is reduced in the ratio 2:5, In what ratio is the surface area reduced?

A.

8 : 125

B.

2 : 5

C.

8 : 25

D.

4 : 25

E.

4 : 10

Correct answer is D

surface area of formula = πr\(^2\) 

If the radius is reduced then let its radius be x.

Its area is πx\(^2\) .

x : r = 2 : 5,

so \(\frac{x}{r}\) =  \(\frac{2}{5}\)

 → 5x = 2r

Hence, x = 0.4r.

Hence the area of the new circle is π (0.4r)\(^2\)  = 0.16π r\(^2\) .

The ratio of the two areas is

0.16πr\(^2\) : πr\(^2\) 

 = 0.16 : 1 = 16 : 100

= 4 : 25.