Year :
2019
Title :
Mathematics (Core)
Exam :
JAMB Exam
Paper 1 | Objectives
21 - 30 of 80 Questions
| # | Question | Ans |
|---|---|---|
| 21. |
If given two points A(3, 12) and B(5, 22) on a x-y plane. Find the equation of the straight line with intercept at 2. A. y = 5x + 2 B. y = 5x + 3 C. y = 12x + 2 D. y = 22x + 3 Detailed SolutionThe equation of a straight line is given as \(y = mx + b\)where m = the slope of the line b = intercept Given points A(3, 12) and B(5, 22), the slope = \(\frac{22 - 12}{5 - 3}\) = \(\frac{10}{2}\) = 5 Hence, the equation of the line is \(y = 5x + 2\). There is an explanation video available below. |
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| 22. |
If P(2, m) is the midpoint of the line joining Q(m, n) and R(n, -4), find the values of m and n. A. m = 0, n = 4 B. m = 4, n = 0 C. m = 2, n = 2 D. m = -2, n = 4 Detailed SolutionQ(m, n) and R(n, -4)Midpoint : P(2, m) \(\implies (\frac{m + n}{2}, \frac{n - 4}{2}) = (2, m)\) \(m + n = 2 \times 2 \implies m + n = 4 ... (i)\) \(n - 4 = 2 \times m \implies n - 4 = 2m ... (ii)\) Solving (i) and (ii) simultaneously, m = 0 and n = 4. There is an explanation video available below. |
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| 23. |
If \(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\), find the value of x. A. 10 B. 30 C. 14 D. 28 Detailed Solution\(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\)\(\implies (2 \times 9) - (-4 \times x) = 58\) \(18 + 4x = 58 \implies 4x = 58 - 18 = 40\) \(x = 10\) There is an explanation video available below. |
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| 24. |
If \(y = 6x^3 + 2x^{-2} - x^{-3}\), find \(\frac{\mathrm d y}{\mathrm d x}\). A. \(\frac{\mathrm d y}{\mathrm d x} = 15x^2 - 4x^{-2} - 3x^{-2}\) B. \(\frac{\mathrm d y}{\mathrm d x} = 6x + 4x^{-1} - 3x^{-4}\) C. \(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\) D. \(\frac{\mathrm d y}{\mathrm d x} = 12x^2 + 4x^{-1} - 3x^{-2}\) Detailed Solution\(y = 6x^3 + 2x^{-2} - x^{-3}\)\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\) There is an explanation video available below. |
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| 25. |
\(\frac{d}{dx} [\log (4x^3 - 2x)]\) is equal to A. \(\frac{12x - 2}{4x^2}\) B. \(\frac{43x^2 - 2x}{7x}\) C. \(\frac{4x^2 - 2}{7x + 6}\) D. \(\frac{12x^2 - 2}{4x^3 - 2x}\) Detailed Solution\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1)Let u = 4x\(^3\) - 2x. \(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\) \(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\) \(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\) \(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\) = \(\frac{12x^2 - 2}{4x^3 - 2x}\) There is an explanation video available below. |
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| 26. |
If \(f(x) = 3x^3 + 4x^2 + x - 8\), what is the value of f(-2)? A. -24 B. 30 C. -18 D. -50 Detailed Solution\(f(x) = 3x^3 + 4x^2 + x - 8\)\(f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) - 8\) = \(-24 + 16 - 2 - 8\) = -18 There is an explanation video available below. |
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| 27. |
Solve for x in \(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\) A. \(x \leq 1\frac{1}{2}\) B. \(x \leq \frac{21}{2}\) C. \(x \geq \frac{21}{2}\) D. \(x \geq 1\frac{1}{2}\) Detailed Solution\(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)2(4x - 6) \(\leq\) 3(3 + 2x) 8x - 12 \(\leq\) 9 + 6x 8x - 6x \(\leq\) 9 + 12 2x \(\leq\) 21 \(x \leq \frac{21}{2}\) There is an explanation video available below. |
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| 28. |
Solve the inequality: -7 \(\leq\) 9 - 8x < 16 - x A. -1 \(\leq\) x \(\leq\) 2 B. -1 \(\leq\) x < 2 C. -1 < x < 2 D. -1 < x \(\leq\) 2 Detailed Solution-7 \(\leq\) 9 - 8x < 16 - x-7 \(\leq\) 9 - 8x and 9 - 8x < 16 - x -7 - 9 \(\leq\) -8x and -8x + x < 16 - 9 -16 \(\leq\) -8x and -7x < 7 \(\therefore\) x \(\leq\) 2 and -1 < x -1 < x \(\leq\) 2. There is an explanation video available below. |
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| 29. |
The nth term of a sequence is given by 2\(^{2n - 1}\). Find the sum of the first four terms. A. 74 B. 32 C. 42 D. 170 Detailed Solution\(T_n = 2^{2n - 1}\)\(T_1 = 2^{2(1) - 1} \) = 2 \(T_2 = 2^{2(2) - 1}\) = 8 \(T_3 = 2^{2(3) - 1}\) = 32 \(T_4 = 2^{2(4) - 1}\) = 128 \(T_1 + T_2 + T_3 + T_4 = 2 + 8 + 32 + 128\) = 170 There is an explanation video available below. |
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| 30. |
Integrate \(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\) A. \(4\frac{1}{2}\) B. \(3\frac{1}{2}\) C. \(7\frac{1}{2}\) D. \(5\frac{1}{4}\) Detailed Solution\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\) = \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\) = \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\) = \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\) = \(\frac{22}{3} - (-\frac{1}{6})\) = \(\frac{22}{3} + \frac{1}{6}\) = \(\frac{15}{2}\) = \(7\frac{1}{2}\) There is an explanation video available below. |
| 21. |
If given two points A(3, 12) and B(5, 22) on a x-y plane. Find the equation of the straight line with intercept at 2. A. y = 5x + 2 B. y = 5x + 3 C. y = 12x + 2 D. y = 22x + 3 Detailed SolutionThe equation of a straight line is given as \(y = mx + b\)where m = the slope of the line b = intercept Given points A(3, 12) and B(5, 22), the slope = \(\frac{22 - 12}{5 - 3}\) = \(\frac{10}{2}\) = 5 Hence, the equation of the line is \(y = 5x + 2\). There is an explanation video available below. |
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| 22. |
If P(2, m) is the midpoint of the line joining Q(m, n) and R(n, -4), find the values of m and n. A. m = 0, n = 4 B. m = 4, n = 0 C. m = 2, n = 2 D. m = -2, n = 4 Detailed SolutionQ(m, n) and R(n, -4)Midpoint : P(2, m) \(\implies (\frac{m + n}{2}, \frac{n - 4}{2}) = (2, m)\) \(m + n = 2 \times 2 \implies m + n = 4 ... (i)\) \(n - 4 = 2 \times m \implies n - 4 = 2m ... (ii)\) Solving (i) and (ii) simultaneously, m = 0 and n = 4. There is an explanation video available below. |
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| 23. |
If \(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\), find the value of x. A. 10 B. 30 C. 14 D. 28 Detailed Solution\(\begin{vmatrix} 2 & -4 \\ x & 9 \end{vmatrix} = 58\)\(\implies (2 \times 9) - (-4 \times x) = 58\) \(18 + 4x = 58 \implies 4x = 58 - 18 = 40\) \(x = 10\) There is an explanation video available below. |
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| 24. |
If \(y = 6x^3 + 2x^{-2} - x^{-3}\), find \(\frac{\mathrm d y}{\mathrm d x}\). A. \(\frac{\mathrm d y}{\mathrm d x} = 15x^2 - 4x^{-2} - 3x^{-2}\) B. \(\frac{\mathrm d y}{\mathrm d x} = 6x + 4x^{-1} - 3x^{-4}\) C. \(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\) D. \(\frac{\mathrm d y}{\mathrm d x} = 12x^2 + 4x^{-1} - 3x^{-2}\) Detailed Solution\(y = 6x^3 + 2x^{-2} - x^{-3}\)\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\) There is an explanation video available below. |
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| 25. |
\(\frac{d}{dx} [\log (4x^3 - 2x)]\) is equal to A. \(\frac{12x - 2}{4x^2}\) B. \(\frac{43x^2 - 2x}{7x}\) C. \(\frac{4x^2 - 2}{7x + 6}\) D. \(\frac{12x^2 - 2}{4x^3 - 2x}\) Detailed Solution\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1)Let u = 4x\(^3\) - 2x. \(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\) \(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\) \(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\) \(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\) = \(\frac{12x^2 - 2}{4x^3 - 2x}\) There is an explanation video available below. |
| 26. |
If \(f(x) = 3x^3 + 4x^2 + x - 8\), what is the value of f(-2)? A. -24 B. 30 C. -18 D. -50 Detailed Solution\(f(x) = 3x^3 + 4x^2 + x - 8\)\(f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) - 8\) = \(-24 + 16 - 2 - 8\) = -18 There is an explanation video available below. |
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| 27. |
Solve for x in \(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\) A. \(x \leq 1\frac{1}{2}\) B. \(x \leq \frac{21}{2}\) C. \(x \geq \frac{21}{2}\) D. \(x \geq 1\frac{1}{2}\) Detailed Solution\(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)2(4x - 6) \(\leq\) 3(3 + 2x) 8x - 12 \(\leq\) 9 + 6x 8x - 6x \(\leq\) 9 + 12 2x \(\leq\) 21 \(x \leq \frac{21}{2}\) There is an explanation video available below. |
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| 28. |
Solve the inequality: -7 \(\leq\) 9 - 8x < 16 - x A. -1 \(\leq\) x \(\leq\) 2 B. -1 \(\leq\) x < 2 C. -1 < x < 2 D. -1 < x \(\leq\) 2 Detailed Solution-7 \(\leq\) 9 - 8x < 16 - x-7 \(\leq\) 9 - 8x and 9 - 8x < 16 - x -7 - 9 \(\leq\) -8x and -8x + x < 16 - 9 -16 \(\leq\) -8x and -7x < 7 \(\therefore\) x \(\leq\) 2 and -1 < x -1 < x \(\leq\) 2. There is an explanation video available below. |
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| 29. |
The nth term of a sequence is given by 2\(^{2n - 1}\). Find the sum of the first four terms. A. 74 B. 32 C. 42 D. 170 Detailed Solution\(T_n = 2^{2n - 1}\)\(T_1 = 2^{2(1) - 1} \) = 2 \(T_2 = 2^{2(2) - 1}\) = 8 \(T_3 = 2^{2(3) - 1}\) = 32 \(T_4 = 2^{2(4) - 1}\) = 128 \(T_1 + T_2 + T_3 + T_4 = 2 + 8 + 32 + 128\) = 170 There is an explanation video available below. |
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| 30. |
Integrate \(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\) A. \(4\frac{1}{2}\) B. \(3\frac{1}{2}\) C. \(7\frac{1}{2}\) D. \(5\frac{1}{4}\) Detailed Solution\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\) = \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\) = \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\) = \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\) = \(\frac{22}{3} - (-\frac{1}{6})\) = \(\frac{22}{3} + \frac{1}{6}\) = \(\frac{15}{2}\) = \(7\frac{1}{2}\) There is an explanation video available below. |