2015 - JAMB Mathematics Past Questions and Answers - page 6

( [ 1 ÷ 64^{(x + 2)}]= [4^{(x − 3)} ÷ 16^x ] )

( 64^{−(x + 2)} = [4^{(x − 3)}] ÷[16^x] )

breakdown 4,16,64 into a small index no

( 2^{−6(x + 2)} = 2^{2(x − 3)} ÷ 2^{4(x)} )

( 2^{−6x− 12} = 2^{2x − 4x − 6} )

( 2^{−6x −12} = 2^{−2x − 6} )

− 6x − 12 = − 2x − 6

Collect the like term

−6x + 2x = −6 + 12

−4x =6

x = ( \frac{6}{4} )

x = ( \frac{−3}{2} )

The external radius = 11cm

The internal radius = 10cm

The area of cross section = π(π² − 10²)

= π (11 + 10)(11 − 10)

π(21)(1)

= 21πcm²

x² − 2x − 3 = 0

−3x + x = − 2x

− 3x × x = − 3x²

(x² − 3x)(x − 3) = 0

(x − 3)(x + 1) = 0

Either (x − 3) = 0 or (x + 1) = 0

x = 3 or − 1

“Explanation

V = πr²h

V = ?, h = 4cm, r = 5cm

V = 3.14(5)² (4)

V = 3.14(25)(4)

V = 3.14 × 1200

= 314cm²

Let u= x²

du = 2xdx

Let v = sinx

∫dv = ∫ sin x i.e. v = cos x

∫ udv = uv − ∫ vdu

∫ x² Sin xdx = − x² Cos x − ∫ (− Cos x)2 × dx

x² Cos x + 2 ∫ x cos x¹ dx

= − x² Cos x + 2 ∫ x Cos x² dx

2 ∫ u Cos xdx

Let u = x

du = dx

dv = Cos x

∫dv = ∫ Cos x

∫ x² Sin x¹ dx = − x² Cos x + 2Sin x − 2Cos x + C

∫ x² Sin xdx

= − x² Cos x + 2xSin + 2Cos x + C

Prob(A) = ( \frac{1}{5} ) , Prob(B) = ( \frac{1}{4} ), Prob (A ∩ B) = ( \frac{1}{15} ), Prob (A ∪ B) = ?

Note:

I. the probability is either event of A or B or both.

The formula using is prob (A ∪ B) = ∩ prob(A) + prob(A) − prob(A ∩ B)

II. But if the probability of both outcomes A and B

The formula using is Prob(A ∩ B) − prob(A) + prob(B) – prob (A ∪ B)

In this question, A or B, Prob (A ∩ B): A and B, Prob (A ∩ B)

prob (A ∪ B) = prob(A) + prob(B) − Prob (A ∩ B) is used.

prob (A ∪ B) = ( \frac{1}{15} ) + ( \frac{1}{3} ) − ( \frac{1}{15} )

= (3 + 5 + 1) ÷ 15

= ( \frac{7}{15} )

4√(390625T − 8)½

[(390625T − 8)½]¼

[((390625)½ × T − 8)½]¼

(√ 390625 × T − 4 )¼

(625 × T −¼)¼

(54 × T −4)¼

= (54)¼ × (T − 4 )¼

= 5 × T − 1

= 5 × 1/T

= 5/T

(159.75)₁₀ = (y)₆

( \begin{array}{c|c}

6 & 159 \

\hline

6 & 26 \text{ rem 3} \

6 & 4 \text{ rem 2} \

& 0  \text{   rem 4} \\

\end{array}\uparrow )

159<sub>10</sub> = 423<sub>6</sub><br />

Then Convert 0.75 to base 6<br />

0.75 x 6 =4.5 i.e 4.5 - 3.00 = 1.5

0.5 x 6 = 3.00 423 + 1.5 =424.5

 159.75 = (424.5)<sub>6</sub><br />

  (159)<sub>10</sub> = (424.5)<sub>6</sub>

Percentage Loss = (actual loss ) ÷ (Cost price) × 100

Actual loss = Cost Price − Selling Price (sold price)

= N800 − N520 = N280

Percentage loss = (280 ÷ 800) × 100 = 35%

∴ His loss percent = 35%

6½ − 2¾ + 1½

73/12 − 11/4 + 3/2

[(73 − 33 + 18) ÷ (12)]

58/12

29/6

( 4\frac{5}{6})