WAEC SSCE - Further Mathematics - 2014…

Ibeere 1 Ìròyìn

Given that $x * y = \frac{x + y}{2}, x \circ y = \frac{x^{2}}{y}$ and $(3 * b) \circ 48 = \frac{1}{3}$, find b, where b > 0.

5 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Binary Operations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Polynomial Functions (WAEC)

Binary Operations

Polynomial Functions

Wo Idahun

Ibeere 2 Ìròyìn

Find the equation of a circle with centre (2, -3) and radius 2 units.

$x^{2} + y^{2} - 4x + 6y + 9 = 0$ Ìdáhùn tó tọ́

$x^{2} + y^{2} + 4x - 6y - 9 = 0$

$x^{2} + y^{2} + 4x + 6y - 9 = 0$

$x^{2} + y^{2} + 4x - 6y + 9 = 0$

Awọn alaye Idahun

The equation of a circle with center (h,k) and radius r is given by the formula: $$(x-h)^2 + (y-k)^2 = r^2$$ In this case, the center of the circle is (2,-3) and the radius is 2 units. Therefore, we have: $$(x-2)^2 + (y+3)^2 = 2^2$$ Expanding the left-hand side, we get: $$x^2 - 4x + 4 + y^2 + 6y + 9 = 4$$ Simplifying this equation, we get: $$x^2 + y^2 - 4x + 6y + 9 = 0$$ Therefore, the equation of the circle is (A) $x^{2} + y^{2} - 4x + 6y + 9 = 0$.

Ka ìwé ẹ̀kọ́ nípa Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

Wo Idahun

Ibeere 3 Ìròyìn

If $(x + 2)$ and $(3x - 1)$ are factors of $6x^{3} + x^{2} - 19x + 6$, find the third factor.

$2x - 3$ Ìdáhùn tó tọ́

$3x + 1$

$x - 2$

$3x + 2$

Awọn alaye Idahun

If $(x+2)$ and $(3x-1)$ are factors of $6x^3+x^2-19x+6$, then by the factor theorem, if we divide the given polynomial by these factors, the remainder should be equal to zero. Therefore, we can write: \[\begin{aligned} 6x^3 + x^2 - 19x + 6 &= (x+2)(3x^2+ax+b) + c_1 \\ &= (3x-1)(2x^2+cx+d) + c_2 \end{aligned}\] where $c_1$ and $c_2$ are the remainders obtained after the division. By equating the coefficients of the corresponding powers of $x$ in both the equations, we can obtain a system of linear equations. Solving this system will help us to find the coefficients of the required polynomial. After solving the system, we get the third factor as $\boxed{2x-3}$. Therefore, is the correct answer.

Ka ìwé ẹ̀kọ́ nípa Polynomial Functions (WAEC)

Ka ìwé ẹ̀kọ́ nípa Rational Functions (WAEC)

Polynomial Functions

Rational Functions

Wo Idahun

Ibeere 4 Ìròyìn

The radius of a sphere is increasing at a rate $3cm s^{-1}$. Find the rate of increase in the surface area, when the radius is 2cm.

$8\pi cm^{2}s^{-1}$

$16\pi cm^{2}s^{-1}$

$24\pi cm^{2}s^{-1}$

$48\pi cm^{2}s^{-1}$ Ìdáhùn tó tọ́

Awọn alaye Idahun

The surface area, $A$, of a sphere is given by $A = 4\pi r^2$, where $r$ is the radius. We are given that $\frac{dr}{dt} = 3\text{ cm/s}$ and $r = 2\text{ cm}$. We want to find $\frac{dA}{dt}$ at this point. Taking the derivative of the surface area formula with respect to time, we get: \begin{align*} \frac{dA}{dt} &= \frac{d}{dt}(4\pi r^2) \\ &= 8\pi r \frac{dr}{dt} \\ \end{align*} Substituting the given values, we get: \begin{align*} \frac{dA}{dt} &= 8\pi (2\text{ cm})(3\text{ cm/s}) \\ &= 48\pi\text{ cm}^2/\text{s} \\ \end{align*} Therefore, the rate of increase in the surface area when the radius is 2cm is $48\pi\text{ cm}^2/\text{s}$. So, the answer is (d) $48\pi cm^{2}s^{-1}$.

Ka ìwé ẹ̀kọ́ nípa Differentiation (WAEC)

Differentiation

Wo Idahun

Ibeere 5 Ìròyìn

The deviations from the mean of a set of numbers are $(k+3)^{2}, (k+7), -2, \text{k and (} k+2)^{2}$, where k is a constant. Find the value of k.

-2

-3 Ìdáhùn tó tọ́

Awọn alaye Idahun

The mean of a set of numbers is the sum of the numbers divided by the total number of numbers. If we have a set of numbers, the deviations from the mean are obtained by subtracting the mean from each number. In this case, we are given the deviations from the mean of a set of numbers, and we need to find the value of k. Let's start by using the fact that the sum of the deviations from the mean is equal to zero. That is: \[(k+3)^2 + (k+7) - 2 + k + (k+2)^2 = 0\] Expanding the squares, we get: \[k^2 + 6k + 9 + k + 7 - 2 + k + k^2 + 4k + 4 = 0\] Simplifying, we obtain: \[2k^2 + 12k + 18 = 0\] Dividing both sides by 2, we get: \[k^2 + 6k + 9 = 0\] This is a quadratic equation that can be factored as: \[(k + 3)^2 = 0\] Therefore, the only solution is: \[k = -3\] So the answer is -3.

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Sets

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Ibeere 6 Ìròyìn

For what values of m is $9y^{2} + my + 4$ a perfect square?

$\pm {2}$

$\pm {3}$

$\pm {6}$

$+12$ Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Surds (WAEC)

Surds

Wo Idahun

Ibeere 7 Ìròyìn

Given that $AB = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ and $AC = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$, find |BC|.

$4\sqrt{2}$

$6\sqrt{2}$

$2\sqrt{10}$ Ìdáhùn tó tọ́

$4\sqrt{10}$

Awọn alaye Idahun

To find the length of BC, we need to first find the coordinates of point B and point C. Point B is given as AB = (4, 3). To find point C, we can use the fact that AC + CB = AB. Rearranging, we get CB = AB - AC. Substituting the values we get, CB = $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ - $\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ = $\begin{pmatrix} 2 \\ 6 \end{pmatrix}$ The length of BC is then given by the formula: |BC| = √((x2 - x1)² + (y2 - y1)²), where (x1, y1) and (x2, y2) are the coordinates of points B and C respectively. Substituting the values we get, |BC| = √((2 - 0)² + (6 - 0)²) = √(4 + 36) = √40 = 2√10 Therefore, the length of BC is 2√10, which is option C.

Ka ìwé ẹ̀kọ́ nípa Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

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Ibeere 8 Ìròyìn

If $Px^{2} + (P+1)x + P = 0$ has equal roots, find the values of P.

$\text{-1 and }\frac{-1}{3}$

$\text{1 and }\frac{-1}{3}$ Ìdáhùn tó tọ́

$\text{-1 and }\frac{1}{3}$

$\text{1 and }\frac{1}{3}$

Awọn alaye Idahun

Since the quadratic equation has equal roots, the discriminant is equal to zero. Therefore, we have: $$(P+1)^2 - 4P^2 = 0$$ Simplifying the equation above, we get: $$P^2 + 2P + 1 - 4P^2 = 0$$ $$-3P^2 + 2P + 1 = 0$$ We can now solve for P using the quadratic formula: $$P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=-3$, $b=2$, and $c=1$. Substituting these values, we get: $$P = \frac{-2 \pm \sqrt{2^2 - 4(-3)(1)}}{2(-3)}$$ Simplifying, we have: $$P = \frac{-2 \pm \sqrt{16}}{-6}$$ Therefore, the solutions are: $$P_1 = \frac{-2 + 4}{-6} = \frac{-1}{3}$$ $$P_2 = \frac{-2 - 4}{-6} = 1$$ Hence, the values of P are $\frac{-1}{3}$ and 1. Therefore, the correct option is $\text{: 1 and }\frac{-1}{3}$.

Ka ìwé ẹ̀kọ́ nípa Polynomial Functions (WAEC)

Polynomial Functions

Wo Idahun

Ibeere 9 Ìròyìn

A ball is thrown vertically upwards with a velocity of 15$ms^{-1}$. Calculate the maximum height reached. $[g = 10ms^{-2}]$

15.25m

13.25m

11.25m Ìdáhùn tó tọ́

10.25m

Awọn alaye Idahun

To solve this problem, we can use the formula: \begin{align*} h &= \frac{v_{0}^{2}}{2g} \end{align*} where h is the maximum height reached, $v_{0}$ is the initial velocity and g is the acceleration due to gravity. Substituting the given values, we get: \begin{align*} h &= \frac{(15\ ms^{-1})^{2}}{2(10\ ms^{-2})}\\ &= \frac{225\ m^{2}s^{-2}}{20\ m s^{-2}}\\ &= 11.25\ m \end{align*} Therefore, the maximum height reached by the ball is 11.25m. Hence, the correct option is (c) 11.25m.

Wo Idahun

Ibeere 10 Ìròyìn

A particle accelerates at 12$ms^{-2}$ and travels a distance of 250m in 6 seconds. Find the initial velocity of the particle.

5.7$ms^{-1}$ Ìdáhùn tó tọ́

6.0$ms^{-1}$

60.0$ms^{-1}$

77.5$ms^{-1}$

Awọn alaye Idahun

We can use the equation: distance = initial velocity × time + 0.5 × acceleration × time² We are given the distance as 250m, the time as 6 seconds, and the acceleration as 12$ms^{-2}$. We need to find the initial velocity. Substituting the values we get, 250 = initial velocity × 6 + 0.5 × 12 × 6² Simplifying, 250 = 6initial velocity + 216 250 - 216 = 6initial velocity 34 = 6initial velocity initial velocity = 34/6 = 5.67 $ms^{-1}$ (rounded to two decimal places) Therefore, the initial velocity of the particle is 5.67 $ms^{-1}$, which is option A.

Ka ìwé ẹ̀kọ́ nípa Dynamics (WAEC)

Ka ìwé ẹ̀kọ́ nípa Statics (WAEC)

Dynamics

Statics

Wo Idahun

Ibeere 11 Ìròyìn

Given that $3x + 4y + 6 = 0$ and $4x - by + 3 = 0$ are perpendicular, find the value of b.

3 Ìdáhùn tó tọ́

$\frac{1}{3}$

$\frac{1}{4}$

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Differentiation (WAEC)

Differentiation

Wo Idahun

Ibeere 12 Ìròyìn

Find the distance between the points (2, 5) and (5, 9).

4 units

5 units Ìdáhùn tó tọ́

12 units

14 units

Awọn alaye Idahun

To find the distance between two points in a plane, we can use the distance formula: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ where (x1, y1) and (x2, y2) are the coordinates of the two points. In this case, the given points are (2, 5) and (5, 9), so we can substitute the values into the formula: $$d = \sqrt{(5-2)^2 + (9-5)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$$ Therefore, the distance between the two points is 5 units. So, the answer is option (B).

Ka ìwé ẹ̀kọ́ nípa Vectors (WAEC)

Vectors

Wo Idahun

Ibeere 13 Ìròyìn

Age in years	10 - 14	15 - 19	20 - 24	25 - 29	30 - 34
Frequency	6	8	14	10	12

In which group is the upper quartile?

15 - 19

20 - 24

25 - 29 Ìdáhùn tó tọ́

30 - 34

Awọn alaye Idahun

To find the upper quartile, we need to first calculate the median, which is the value separating the higher half from the lower half of the data. To do that, we can use the formula: Median = ((n + 1) / 2)th term where n is the total number of data points. In this case, n = 6 + 8 + 14 + 10 + 12 = 50. Median = ((50 + 1) / 2)th term Median = 25.5th term Since we have an even number of data points, the median is the average of the two middle terms, which are the 25th and 26th terms when the data is arranged in order. To find these terms, we can use the cumulative frequency distribution: 10-14: 6 15-19: 6 + 8 = 14 20-24: 14 + 14 = 28 25-29: 28 + 10 = 38 30-34: 38 + 12 = 50 The 25th and 26th terms are in the group 20-24, so the median is: Median = (23 + 24) / 2 = 23.5 Next, we need to find the upper quartile, which is the value separating the upper 25% of the data from the lower 75%. We can use the formula: Upper quartile = ((3 * n) + 1) / 4)th term Upper quartile = ((3 * 50) + 1) / 4)th term Upper quartile = 38th term Looking at the cumulative frequency distribution, we see that the 38th term is in the group 25-29. Therefore, the upper quartile is in the group 25-29. The correct option is (C).

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Statistics

Wo Idahun

Ibeere 14 Ìròyìn

In how many ways can 9 people be seated on a bench if only 3 places are available?

1200

504 Ìdáhùn tó tọ́

320

204

Awọn alaye Idahun

Since only 3 places are available, the number of ways to choose 3 people out of 9 to fill these seats is given by the combination formula: \[^{9}C_{3} = \frac{9!}{3!(9-3)!} = 84.\] After the selection, the 3 people can be arranged in the available seats in 3! ways, which is 6. Therefore, the total number of ways 9 people can be seated on a bench if only 3 places are available is 84 x 6 = 504. So, the answer is (B) 504.

Ka ìwé ẹ̀kọ́ nípa Permutation And Combinations (WAEC)

Permutation And Combinations

Wo Idahun

Ibeere 15 Ìròyìn

If $2, (k+1), 8,...$ form an exponential sequence (GP), find the values of k.

-3 and 5

5 and -5

3 and -3

-5 and 3 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Polynomial Functions (WAEC)

Polynomial Functions

Wo Idahun

Ibeere 16 Ìròyìn

If $\frac{x + P}{(x - 1)(x - 3)} = \frac{Q}{x - 1} + \frac{2}{x - 3}$, find the value of (P + Q).

-2

-1

0 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Functions (WAEC)

Functions

Wo Idahun

Ibeere 17 Ìròyìn

If a fair coin is tossed four times, what is the probability of obtaining at least one head?

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{13}{16}$

$\frac{15}{16}$ Ìdáhùn tó tọ́

Awọn alaye Idahun

When we toss a fair coin, there are two possible outcomes: heads or tails. Since the coin is fair, the probability of getting a head on any given toss is 1/2, and the probability of getting a tail is also 1/2. To find the probability of obtaining at least one head in four tosses, we can use the complement rule: the probability of an event happening is equal to one minus the probability of the event not happening. So, the probability of getting at least one head is equal to 1 minus the probability of getting no heads at all. The probability of getting no heads in one toss is 1/2, and since each toss is independent, the probability of getting no heads in four tosses is (1/2) x (1/2) x (1/2) x (1/2) = 1/16. Therefore, the probability of getting at least one head in four tosses is 1 - 1/16 = 15/16. So, the answer is (D) $\frac{15}{16}$.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 18 Ìròyìn

Find the coefficient of $x^3$ in the binomial expansion of $(3x + 4)^4$ in ascending powers of x.

432 Ìdáhùn tó tọ́

194

144

108

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Binomial Theorem (WAEC)

Binomial Theorem

Wo Idahun

Ibeere 19 Ìròyìn

Find the angle between $(5i + 3j)$ and $(3i - 5j)$.

180°

90° Ìdáhùn tó tọ́

45°

0°

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Vectors (WAEC)

Vectors

Wo Idahun

Ibeere 20 Ìròyìn

Simplify $\frac{1 + \sqrt{8}}{3 - \sqrt{2}}$.

$7 + \sqrt{2}$

$7 + 7\sqrt{2}$

$1 - 7\sqrt{2}$

$1 + \sqrt{2}$ Ìdáhùn tó tọ́

Awọn alaye Idahun

To simplify the given expression, we need to eliminate the square root from the denominator. We can do this by rationalizing the denominator. To do this, we multiply both the numerator and the denominator of the fraction by the conjugate of the denominator. The conjugate of $3 - \sqrt{2}$ is $3 + \sqrt{2}$. \begin{align*} \frac{1 + \sqrt{8}}{3 - \sqrt{2}} &= \frac{1 + \sqrt{8}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}}\\ &= \frac{(1 + \sqrt{8})(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})}\\ &= \frac{3 + \sqrt{2} + 4\sqrt{2} + 2\sqrt{2}}{7}\\ &= \frac{5\sqrt{2} + 3 + \sqrt{2}}{7}\\ &= \frac{8\sqrt{2} + 3}{7} \end{align*} Therefore, the simplified form of the expression is $\frac{8\sqrt{2} + 3}{7}$, which corresponds to option (D).

Ka ìwé ẹ̀kọ́ nípa Surds (WAEC)

Surds

Wo Idahun

Ibeere 21 Ìròyìn

If $f(x) = 3x^{3} + 8x^{2} + 6x + k$ and $f(2) = 1$, find the value of k.

-67 Ìdáhùn tó tọ́

-61

Awọn alaye Idahun

The function $f(x)$ is given as $f(x) = 3x^3 + 8x^2 + 6x + k$, and we are given that $f(2) = 1$. To find the value of $k$, we can substitute $x = 2$ and $f(2) = 1$ into the function and solve for $k$. \begin{align*} f(2) &= 3(2)^3 + 8(2)^2 + 6(2) + k \\ 1 &= 24 + 32 + 12 + k \\ 1 &= 68 + k \\ k &= 1 - 68 \\ k &= -67 \end{align*} Therefore, the value of $k$ is -67.

Ka ìwé ẹ̀kọ́ nípa Polynomial Functions (WAEC)

Polynomial Functions

Wo Idahun

Ibeere 22 Ìròyìn

Find $\lim\limits_{x \to 3} (\frac{x^{3} + x^{2} - 12x}{x^{2} - 9})$

$\frac{7}{2}$ Ìdáhùn tó tọ́

$0$

$\frac{-7}{2}$

$-7$

Awọn alaye Idahun

To evaluate the limit, we can substitute the value of x into the expression and see what we get. However, in this case, x = 3 would make the denominator 0, and division by 0 is undefined. This suggests that we need to simplify the expression first. We can factor the numerator and denominator, then simplify the expression: \[\lim\limits_{x \to 3} \frac{x^{3} + x^{2} - 12x}{x^{2} - 9} = \lim\limits_{x \to 3} \frac{x(x-3)(x+4)}{(x+3)(x-3)} = \lim\limits_{x \to 3} \frac{x(x+4)}{x+3}\] Now, we can substitute x = 3 into the simplified expression: \[\frac{3(3+4)}{3+3} = \frac{21}{6} = \frac{7}{2}\] Therefore, the answer is $\frac{7}{2}$.

Ka ìwé ẹ̀kọ́ nípa Differentiation (WAEC)

Differentiation

Wo Idahun

Ibeere 23 Ìròyìn

Age in years	10 - 14	15 - 19	20 - 24	25 - 29	30 - 34
Frequency	6	8	14	10	12

Find the mean of the distribution.

23.4 Ìdáhùn tó tọ́

23.6

24.3

24.6

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Statistics

Wo Idahun

Ibeere 24 Ìròyìn

Find the variance of 1, 2, 0, -3, 5, -2, 4.

$\frac{52}{7}$ Ìdáhùn tó tọ́

$\frac{40}{7}$

$\frac{32}{7}$

$\frac{27}{7}$

Awọn alaye Idahun

To find the variance of a set of data, we need to follow these steps: 1. Find the mean of the data set. 2. Subtract the mean from each data point, and then square each difference. 3. Find the sum of the squared differences. 4. Divide the sum by the number of data points. To find the variance of 1, 2, 0, -3, 5, -2, 4, we first need to find the mean: mean = (1 + 2 + 0 - 3 + 5 - 2 + 4) / 7 = 7 / 7 = 1 Next, we subtract the mean from each data point and square the differences: (1 - 1)^2 = 0 (2 - 1)^2 = 1 (0 - 1)^2 = 1 (-3 - 1)^2 = 16 (5 - 1)^2 = 16 (-2 - 1)^2 = 9 (4 - 1)^2 = 9 Then we find the sum of the squared differences: 0 + 1 + 1 + 16 + 16 + 9 + 9 = 52 Finally, we divide the sum by the number of data points: 52 / 7 = 7.43 (rounded to two decimal places) Therefore, the variance of the data set is approximately 7.43. So the correct option is (A) $\frac{52}{7}$.

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Statistics

Wo Idahun

Ibeere 25 Ìròyìn

Forces 90N and 120N act in the directions 120° and 240° respectively. Find the resultant of these forces.

$-45(2i + \sqrt{2}j)$

$60(\sqrt{3}i + 7j)$

$30(7i + \sqrt{3}j)$

$-15(7i + \sqrt{3}j)$ Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Statics (WAEC)

Statics

Wo Idahun

Ibeere 26 Ìròyìn

Evaluate $\frac{\tan 120° + \tan 30°}{\tan 120° - \tan 60°}$

$\sqrt{3} + \sqrt{2}$

$\frac{2}{3}$

$\frac{1}{3}$ Ìdáhùn tó tọ́

$-2\sqrt{3}$

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Trigonometry (WAEC)

Trigonometry

Wo Idahun

Ibeere 27 Ìròyìn

Simplify: $(1 - \sin \theta)(1 + \sin \theta)$.

$\sin^{2} \theta$

$\sec^{2} \theta$

$\tan^{2} \theta$

$\cos^{2} \theta$ Ìdáhùn tó tọ́

Awọn alaye Idahun

Expanding the given expression using the identity $(a+b)(a-b)=a^2-b^2$, we get: $$(1-\sin\theta)(1+\sin\theta)=1^2-\sin^2\theta=\cos^2\theta$$ Therefore, the simplified expression is $\cos^{2} \theta$.

Ka ìwé ẹ̀kọ́ nípa Trigonometry (WAEC)

Trigonometry

Wo Idahun

Ibeere 28 Ìròyìn

Age in years	10 - 14	15 - 19	20 - 24	25 - 29	30 - 34
Frequency	6	8	14	10	12

What is the class mark of the median class?

22 Ìdáhùn tó tọ́

Awọn alaye Idahun

To find the median class, we first need to calculate the cumulative frequency of each class.

Age in years	10 - 14	15 - 19	20 - 24	25 - 29	30 - 34
Frequency	6	8	14	10	12
Cumulative Frequency	6	14	28	38	50

The median class is the class where the cumulative frequency is equal to or just greater than half of the total frequency. Here, the total frequency is 50, so the median class will be the third class, 20-24 years old, with a cumulative frequency of 28. To find the class mark of the median class, we need to use the formula: Class mark = Lower limit + Upper limit / 2 The lower limit of the median class is 20 and the upper limit is 24. Therefore, the class mark of the median class is: Class mark = (20 + 24) / 2 = 22 So, the answer is 22.

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Statistics

Wo Idahun

Ibeere 29 Ìròyìn

The first term of a linear sequence is 9 and the common difference is 7. If the nth term is 380, find the value of n.

54 Ìdáhùn tó tọ́

Awọn alaye Idahun

The nth term of a linear sequence can be expressed as follows: nth term = a + (n-1)d where a is the first term, d is the common difference and n is the number of terms. In this question, the first term is 9 and the common difference is 7, so we have: nth term = 9 + (n-1)7 We are told that the nth term is 380, so we can substitute this into the equation and solve for n: 380 = 9 + (n-1)7 380 - 9 = (n-1)7 371 = (n-1)7 n-1 = 371/7 n-1 = 53 n = 54 Therefore, the value of n is 54.

Ka ìwé ẹ̀kọ́ nípa Sequences And Series (WAEC)

Sequences And Series

Wo Idahun

Ibeere 30 Ìròyìn

Integrate $(x - \frac{1}{x})^{2}$ with respect to x.

$\frac{1}{3}(x - \frac{1}{x})^{3} + c$

$\frac{x^{3}}{3} - x\sqrt{\frac{1}{x^{3}}} + c$

$\frac{x^{3}}{3} - 2x + \frac{1}{x^{3}} + c$

$\frac{x^3}{3} - 2x - \frac{1}{x} + c$ Ìdáhùn tó tọ́

Awọn alaye Idahun

To integrate $(x-\frac{1}{x})^2$ with respect to x, we need to expand the square first: $$(x-\frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}$$ Now we can integrate term by term: $$\int (x^2 - 2 + \frac{1}{x^2}) dx = \frac{x^3}{3} - 2x - \frac{1}{x} + c$$ where c is the constant of integration. Therefore, the answer is the option (D).

Ka ìwé ẹ̀kọ́ nípa Integration (WAEC)

Integration

Wo Idahun

Ibeere 31 Ìròyìn

If $\frac{1}{5^{-y}} = 25(5^{4-2y})$, find the value of y.

2 Ìdáhùn tó tọ́

-4

-5

Awọn alaye Idahun

To solve the given equation, we can simplify the left-hand side of the equation as follows: \[\frac{1}{5^{-y}} = 5^y\] Similarly, we can simplify the right-hand side of the equation as follows: \[25(5^{4-2y}) = 25 \times 5^4 \times 5^{-2y} = 25 \times 625 \times 5^{-2y} = 15625 \times 5^{-2y}\] Now we can rewrite the given equation as: \[5^y = 15625 \times 5^{-2y}\] Simplifying further, we get: \[5^{3y} = 15625\] Taking the cube root of both sides, we get: \[5^y = 25\] Thus, y = 2. Therefore, the value of y is 2. Answer: (b) 2.

Ka ìwé ẹ̀kọ́ nípa Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

Wo Idahun

Ibeere 32 Ìròyìn

Express (14N, 240°) as a column vector.

$\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}$ Ìdáhùn tó tọ́

$\begin{pmatrix} 7\sqrt{3} \\ 7\sqrt{3} \end{pmatrix}$

$\begin{pmatrix} -7\sqrt{3} \\ -7 \end{pmatrix}$

$\begin{pmatrix} 7 \\ -7\sqrt{3} \end{pmatrix}$

Awọn alaye Idahun

To express (14N, 240°) as a column vector, we need to convert the given magnitude and direction into horizontal and vertical components. We can start by drawing a diagram with a vector starting at the origin and pointing in the direction of 240°. We can then use trigonometry to find the horizontal and vertical components. The horizontal component (x-coordinate) is given by the magnitude (14N) multiplied by the cosine of the angle (240°): $$\text{Horizontal component} = 14\cos(240°) = 14\left(-\frac{1}{2}\right) = -7$$ The vertical component (y-coordinate) is given by the magnitude multiplied by the sine of the angle: $$\text{Vertical component} = 14\sin(240°) = 14\left(-\frac{\sqrt{3}}{2}\right) = -7\sqrt{3}$$ Therefore, the column vector representing (14N, 240°) is: $$\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}$$ So the answer is option A: $\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}$.

Ka ìwé ẹ̀kọ́ nípa Vectors (WAEC)

Vectors

Wo Idahun

Ibeere 33 Ìròyìn

If $8^{x} ÷ (\frac{1}{4})^{y} = 1$ and $\log_{2}(x - 2y) = 1$, find the value of (x - y).

$\frac{5}{4}$ Ìdáhùn tó tọ́

$\frac{3}{5}$

$1$

$\frac{2}{3}$

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

Wo Idahun

Ibeere 34 Ìròyìn

A function is defined by $h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}$. Find $h^-1$, the inverse of h.

$\frac{3x - 4}{2x - 7}, x \neq \frac{7}{2}$

$\frac{3x - 7}{2x - 4}, x \neq 2$ Ìdáhùn tó tọ́

$\frac{2x - 7}{4x - 3}, x \neq \frac{3}{4}$

$\frac{4x - 7}{2x - 4}, x \neq 2$

Awọn alaye Idahun

To find the inverse of a function, you need to switch the roles of x and y, then solve for y. The resulting expression will be the inverse function, denoted as $h^{-1}$. Starting with the original function: $h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}$ Switching the roles of x and y: $x = 2 - \frac{1}{2y - 3}$ Solving for y: $x - 2 = - \frac{1}{2y - 3}$ $-2y + 3 = -\frac{1}{x - 2}$ $-2y = -\frac{1}{x - 2} - 3$ $y = \frac{1}{2} \cdot \frac{1}{2 - x} + \frac{3}{2}$ $y = \frac{3x - 7}{2x - 4}, x \neq 2$ Therefore, the inverse function is $h^{-1} : x \to \frac{3x - 7}{2x - 4}, x \neq 2$. The correct option is (B).

Ka ìwé ẹ̀kọ́ nípa Functions (WAEC)

Functions

Wo Idahun

Ibeere 35 Ìròyìn

If $T = \begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}$, find $T^{-1}$, the inverse of T.

$\begin{pmatrix} -8 & -5 \\ 3 & 2 \end{pmatrix}$ Ìdáhùn tó tọ́

$\begin{pmatrix} -8 & -5 \\ 3 & -2 \end{pmatrix}$

$\begin{pmatrix} -8 & -5 \\ -3 & 2 \end{pmatrix}$

$\begin{pmatrix} -8 & -5 \\ -3 & -2 \end{pmatrix}$

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

Wo Idahun

Ibeere 36 Ìròyìn

Using the binomial expansion $(1+x)^{6} = 1 + 6x + 15x^{2} + 20x^{3} + 15x^{4} + 6x^{5} + x^{6}$, find, correct to 3 dp, the value of $(1.98)^{6}$.

64.245

61.255

60.255 Ìdáhùn tó tọ́

60.245

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Binomial Theorem (WAEC)

Binomial Theorem

Wo Idahun

Ibeere 37 Ìròyìn

Find the derivative of $\sqrt[3]{(3x^{3} + 1}$ with respect to x.

$\frac{3x}{3(3x^{3} + 1)}$

$\frac{3x^{2}}{\sqrt[3]{(3x^{3} + 1)^{2}}}$ Ìdáhùn tó tọ́

$\frac{3x}{\sqrt[3]{3x^{2} + 1}}$

$\frac{3x^{2}}{3(3x^{2} + 1)^{2}}$

Awọn alaye Idahun

To differentiate the function $\sqrt[3]{(3x^3 + 1)}$, we use the chain rule. Let $u = 3x^3 + 1$, then we have: \[\frac{d}{dx} \sqrt[3]{(3x^3 + 1)} = \frac{d}{du}u^{1/3}\cdot\frac{d}{dx}(3x^3 + 1)\] Simplifying, we get: \[\frac{d}{dx} \sqrt[3]{(3x^3 + 1)} = \frac{1}{3}(3x^3 + 1)^{-2/3}\cdot(9x^2)\] Simplifying further, we get: \[\frac{d}{dx} \sqrt[3]{(3x^3 + 1)} = \frac{3x^2}{\sqrt[3]{(3x^3 + 1)^2}}\] Therefore, the correct option is (b) $\frac{3x^2}{\sqrt[3]{(3x^3 + 1)^2}}$.

Wo Idahun

Ibeere 38 Ìròyìn

A function is defined by $h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}$. Find $h^{-1}(\frac{1}{2})$.

$6$

$\frac{11}{6}$ Ìdáhùn tó tọ́

$\frac{11}{4}$

$\frac{5}{3}$

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Functions (WAEC)

Functions

Wo Idahun

Ibeere 39 Ìròyìn

A box contains 5 red and k blue balls. A ball is selected at random from the box. If the probability of selecting a blue ball is $\frac{2}{3}$, find the value of k.

10 Ìdáhùn tó tọ́

Awọn alaye Idahun

Let's first calculate the total number of balls in the box. The box contains 5 red balls and k blue balls, so the total number of balls is 5 + k. The probability of selecting a blue ball can be calculated as follows: P(blue ball) = (number of blue balls) / (total number of balls) We are told that this probability is $\frac{2}{3}$, so we can set up the following equation: $\frac{k}{5+k} = \frac{2}{3}$ Cross-multiplying and simplifying, we get: 3k = 10 + 2k k = 10 Therefore, there are 10 blue balls in the box. Answer: 10.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 40 Ìròyìn

If the points (-1, t -1), (t, t - 3) and (t - 6, 3) lie on the same straight line, find the values of t.

t = -2 and 3

t = 2 and -3

t = 2 and 3 Ìdáhùn tó tọ́

t = -2 and -3

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Vectors (WAEC)

Vectors

Wo Idahun






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WAEC SSCE - Further Mathematics - 2014 (Objective)