Mathematics 2006 Jamb Examination

A. (33, 4)
Mode is score with highest frequency = 4 (frequency 12). Number scoring ≥4 = 12+10+6+4+1 = 33.

A. 15
Number of ways = C(6,4) = 6!/(4!2!) = 15.

C. 32
Five consecutive integers have the middle value equal to the mean = 30, so largest = 30+2 = 32.

D. 6
x/(9+x)=2/5 ⇒ 5x = 18+2x ⇒ 3x = 18 ⇒ x = 6.

A. 2/3
Mean = 2x. Deviations: 0, -1, +1. Variance = (0^2+(-1)^2+1^2)/3 = 2/3.

C. 2/5
Total cards = 5+6+3+6 = 20. GTN or Qtel = 5+3 = 8. Probability = 8/20 = 2/5.

Solution: use the pie chart information and given relations.\[\text{Let housing} = h,\ \text{food} = 2h,\ \text{transport} = t,\ \text{school} = 2t.\] From the chart the school-fees sector is a right angle (90°), so school fees = \(\tfrac{90}{360}=\tfrac14\) of income: \[\text{school} = \tfrac{1}{4}\times30{,}000 = 7{,}500.\] Hence \[t=\tfrac{\text{school}}{2}=\tfrac{7{,}500}{2}=3{,}750.\] The remainder for food+housing is \[30{,}000- (7{,}500+3{,}750)=30{,}000-11{,}250=18{,}750.\] But food+housing = \(2h + h =3h\), so \[3h=18{,}750\Rightarrow h=6{,}250.\] Therefore food = \(2h=2\times6{,}250=12{,}500.\)

D. 3

\[^{n+1}C_3 = 4(^nC_3)\] \[=\dfrac{(n+1)!}{(n+1-3)!3!} = 4\left(\dfrac{n!}{(n-3)!3!}\right)\] \[\dfrac{(n+1)n!}{(n-2)(n-3)!}=4\left(\dfrac{n!}{n-3!}\right)\] \[=\dfrac{n+1}{n-2}=\dfrac{4}{1}\] \[n+1 = 4(n-2)\] \[n+1 = 4n-8\] \[-3n = -9\] \[\dfrac{-9}{-3}=3\]

B. \(y = x^2 + 7x - 18\)
Integrate the gradient to get the curve:\[y'=2x+7\Rightarrow y=\int(2x+7)\,dx = x^2+7x+C.\] Use point (2,0): \[0 = 2^2 +7\cdot2 + C \]\[= 4+14 + C =18 + C\Rightarrow C=-18.\] Hence \[y = x^2 + 7x - 18.\]

C. \(4x^3 - 2 - 2/x^3\)
Let \(u = x^2 - x^{-1}.\) Then \(\dfrac{d}{dx}(u^2)=2u\,u'.\) Compute:\[u'=2x + x^{-2}.\] So\[\dfrac{d}{dx}\left(x^2-\dfrac{1}{x}\right)^2 = 2\left(x^2-\dfrac{1}{x}\right)\left(2x+\dfrac{1}{x^2}\right).\] Expand stepwise:\[\left(x^2-\dfrac{1}{x}\right)\left(2x+\dfrac{1}{x^2}\right)\]\[=2x^3+1-2-\dfrac{1}{x^3}=2x^3-1-\dfrac{1}{x^3}.\] Multiply by 2:\[4x^3-2-\dfrac{2}{x^3}.\] That is \(4x^3-2-2/x^3\).