JAMB Mathematics Past Questions & Answers - Page 101

[1 ÷ (x+1)] + [1 ÷ (x − 1)]

= ((x − 1) + [(x + 1)) ÷ (x+1)(x − 1)]

Using the L.C.M.

= (x − 1 + x + 1) ÷ (x + 1)(x − 1)

= (x + 2 − 1 + 1) ÷ (x + 1)(x − 1)

= 2x ÷ (x + 1)(x − 1) =2x ÷ (x + 1)(x − 1)

Find the slant height

\( l^2 = h^2 + r^2(h = 4cm,r = 3cm)\)

\( l^2 = 4^2 + 3^2 = 16 + 9 = 25 \)

\( l^2 = √ 25 \)

Squaring both sides

l = 5cm

The area of curved surface (s) =π(3)(5)

15π = 15 × 3.14

= 47.1cm²

\(\int (5x^{3} + 7x^{2} - 2x + 5) \mathrm d x\)

= \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - \frac{2x^{2}}{2} + 5x + c\)

= \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - x^{2} + 5x + c\)

\(\log_{5}(y^{2} x^{5} \div 125b^{\frac{1}{2}})\)

= \(\log_{5} y^{2} + \log_{5} x^{5} - [\log_{5} 125 + \log_{5} b^{\frac{1}{2}}\)

= \(2\log_{5} y + 5\log_{5} x - \log_{5} 5^{3} - \frac{1}{2} \log_{5} b\)

= \(2\log_{5} y + 5\log_{5} x - 3 - \frac{1}{2}\log_{5} b\)

Convert the binary to base 10 and they convert back to base two

10011₂ + xxxxx₂ + 11100₂ + 101₂ = 1001111₂


(1 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰) + xxxxx₂ +(1 × 2⁴ + 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2⁰) + (1 × 2² + 0 × 2¹ + 1 × 2⁰)

= (16 + 0 + 0 + 2 + 1) + xxxxx₂ + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)

=(64 + 0 + 0 + 8 + 4 + 2 + 1)

19 + xxxxx₂ + 33 = 79

xxxxx₂ + 52 = 79

xxxxx₂ = 79 − 52

xxxxx₂ = 27₁₀

\( \begin{array}{c|c}
2 & 27 \\
\hline
2 & 13 \text{ rem 1}\\
2 & 6 \text{ rem 1}\\
2 & 3 \text{ rem 0}\\
2 & 1 \text{ rem 1}\\
& 0 \text{ rem 1}\\
\end{array}\uparrow \)

27₁₀ = 11011₂

Therefore xxxxx₂ = 27₁₀ = 11011₂