Mathematics 2006 Jamb Examination

B. 2
Solution: Find critical points and use second derivative test.\[ f(x)=3x^3-9x^2 \]\[ f'(x)=9x^2-18x=9x(x-2) \]Critical points: \(x=0, x=2\).\[ f''(x)=18x-18 \]Evaluate f'':\[ f''(0)=-18<0 \quad (local\ maximum) \]\[ f''(2)=18>0 \quad (local\ minimum) \]Hence the function attains its minimum at \(x=2\).

D. \(\dfrac{x^2}{2} + \sin x + C\)
Solution: Integrate both terms.\[ y=\int (x+\cos x)\,dx = \int x\,dx + \int \cos x\,dx \]\[ y=\dfrac{x^2}{2}+\sin x + C \]

A. -2cos2θ
Solution: Use chain rule.\[ u=\cos\theta-\sin\theta \quad \Rightarrow \quad \dfrac{d}{d\theta}(u^2)=2u\,u' \]\[ u'=-\sin\theta-\cos\theta=-(\sin\theta+\cos\theta) \]\[ \dfrac{d}{d\theta}(\cos\theta-\sin\theta)^2 = 2(\cos\theta-\sin\theta)(-\sin\theta-\cos\theta) \] Note that \((\cos\theta-\sin\theta)(\sin\theta+\cos\theta)=\cos^2\theta-\sin^2\theta=\cos2\theta\).

Thus\[ \dfrac{d}{d\theta}(\cos\theta-\sin\theta)^2 = -2\cos2\theta \]

Find antiderivative and evaluate limits.\[ \int (1-2x)\,dx = x - x^2 + C \]\[ [x-x^2]_{-4}^{0} = (0-0) -((-4)-(-4)^2)\]\[ = 0 -(-4-16)=20 \]

D. \(\dfrac{2}{15}\)
Solution:\[ \text{Numerator}=\dfrac{7}{9}-\dfrac{2}{3}=\dfrac{7}{9}-\dfrac{6}{9}=\dfrac{1}{9} \]\[ \text{Inside denominator: }\dfrac{\tfrac{2}{5}}{\tfrac{4}{5}}=\dfrac{2}{5}\times\dfrac{5}{4}=\dfrac{1}{2} \]\[ \text{Denominator}=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{2+3}{6}=\dfrac{5}{6} \]\[ \text{Expression}=\dfrac{\tfrac{1}{9}}{\tfrac{5}{6}}=\tfrac{1}{9}\times\tfrac{6}{5}=\tfrac{6}{45}=\tfrac{2}{15} \]

B. 1:13
Solution:\[ m:n = 13:11 \Rightarrow m=13k, n=11k \]\[ m^2 - n^2 = (13k)^2 - (11k)^2 = 169k^2 - 121k^2 = 48k^2 \]\[ (m+n)^2 = (13k+11k)^2 = 24^2 k^2 = 576k^2 \]\[ \text{Ratio} = \dfrac{m^2-n^2}{(m+n)^2} = \dfrac{48k^2}{576k^2} = \dfrac{1}{12} \Rightarrow 1:12 \]

C. -2
Solution:\[ 3^{-4} \times 9^2 \times 81^{-1} = 3^{-4} \times (3^2)^2 \times (3^4)^{-1} \]\[ = 3^{-4} \times 3^4 \times 3^{-4} = 3^{-4} \]\[ \log_9(3^{-4}) = \dfrac{\log_3 3^{-4}}{\log_3 9} = \dfrac{-4}{2} = -2 \]

(K2)₆ * 36 = 35(K4)₅

(K*6¹ + 2*6⁰) 3*6⁰ = 3*5⁰(K*5¹ + 4*5⁰)

3(6K + 2) = 3(5K + 4)

6K + 2 = 5K + 4

6K - 5K = 4 - 2

K = 2

A. 30%
Solution:
Let x = number of people who speak both languages.\[ 350 + 200 - x = 500 \Rightarrow x = 50 \]\[ \text{Percentage} = \dfrac{50}{500} \times 100 = 10\% \]

N20,000 - N10,000 = N10,000
1^st tax on N5000 at N50 per N1000
= 5*50 = N250
2^nd tax on the remaining N5000 at N55 per N1000 = 5*55 = N275
Total tax paid = N250 + N275
= N525