JAMB Mathematics Past Questions & Answers - Page 26

log\(_{10}\) 24 = log\(_{10}\) 8 \(\times\) log\(_{10}\) 3

where log\(_{10}\) 8 = 3 log\(_{10}\) 2 = 3 \(\times\) m

and log\(_{10}\) 3 = n

: log\(_{10}\) 24 = 3m + n

2\(^{√2x + 1}\) = 32

2\(^{√2x + 1}\) = 2\(^5\)

√2x + 1 = 5

square both sides

2x + 1 = 5\(^2\)

2x + 1 = 25

2x = 25 - 1

2x = 24

x = \(\frac{24}{2}\)

x = 12

(11\(_{two}\))\(^2\) = (11\(_2\) \(\times\) (11\(_2\))

= \({1 \times 2^1 + 1 \times 2^0} \times ({1 \times 2^1 + 1 \times 2^0})\)   

= \({1 \times 2 + 1 \times 1} \times ({1 \times 2 + 1 \times 1})\)

= \({2 + 1} \times ({2 + 1})\)

= 3 \(\times\) 3 

= 9\(_{10}\) or

1001\(_2\) from  

1. To round to three significant figures, look at the fourth significant figure. It's a 5 , so round up or round down if below 5.
2. To round to two significant figures, look at the third significant figure. It's an 8 , so round up.

Given (x + 2) and (x - 1), i.e. x = -2 or +1

when x = -2

L(-2) + 2k(-2)\(^2\) + 24 = 0

f(-2) = -2L + 8k = -24...(i)

And x = 1
L(1) + 2k(1) + 24 = 0

f(1):L + 2k = -24...(ii)

Subst, L = -24 - 2k in eqn (i)

-2(-24 - 2k) + 8k = -24

+48 + 4k + 8k = -24

12k = -24 - 48 = -72

k = \(frac{-72}{12}\)

k = -6

where L = -24 - 2k

L = -24 - 2(-6)

L = -24 + 12

L = -12

That is; K = -6 and L = -12