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**Question 1**
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Given that \(3x + 4y + 6 = 0\) and \(4x - by + 3 = 0\) are perpendicular, find the value of b.

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

**Answer Details**

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) \(x^{2} + y^{2} - 4x + 6y + 9 = 0\).

**Question 3**
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If \(8^{x} ÷ (\frac{1}{4})^{y} = 1\) and \(\log_{2}(x - 2y) = 1\), find the value of (x - y).

**Question 4**
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Evaluate \(\frac{\tan 120° + \tan 30°}{\tan 120° - \tan 60°}\)

**Question 5**
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Express (14N, 240°) as a column vector.

**Answer Details**

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: \(\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}\).

**Question 6**
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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.

**Answer Details**

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).

**Question 7**
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Find the coefficient of \(x^3\) in the binomial expansion of \((3x + 4)^4\) in ascending powers of x.

**Question 8**
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If \(\frac{x + P}{(x - 1)(x - 3)} = \frac{Q}{x - 1} + \frac{2}{x - 3}\), find the value of (P + Q).

**Question 9**
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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.

**Question 10**
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If \(\frac{1}{5^{-y}} = 25(5^{4-2y})\), find the value of y.

**Answer Details**

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.

**Question 11**
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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.

**Answer Details**

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.

**Question 12**
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Find the distance between the points (2, 5) and (5, 9).

**Answer Details**

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).

**Question 13**
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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?

**Answer Details**

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 |

**Question 14**
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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 \((1.98)^{6}\).

**Question 15**
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Forces 90N and 120N act in the directions 120° and 240° respectively. Find the resultant of these forces.

**Question 16**
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A ball is thrown vertically upwards with a velocity of 15\(ms^{-1}\). Calculate the maximum height reached. \([g = 10ms^{-2}]\)

**Answer Details**

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.

**Question 17**
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Find the derivative of \(\sqrt[3]{(3x^{3} + 1}\) with respect to x.

**Answer Details**

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) \(\frac{3x^2}{\sqrt[3]{(3x^3 + 1)^2}}\).

**Question 18**
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A function is defined by \(h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}\). Find \(h^{-1}(\frac{1}{2})\).

**Question 19**
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If \(T = \begin{pmatrix} -2 & -5 \\ 3 & 8 \end{pmatrix}\), find \(T^{-1}\), the inverse of T.

**Question 20**
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Simplify \(\frac{1 + \sqrt{8}}{3 - \sqrt{2}}\).

**Answer Details**

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).

**Question 21**
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Integrate \((x - \frac{1}{x})^{2}\) with respect to x.

**Answer Details**

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).

**Question 22**
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If \(Px^{2} + (P+1)x + P = 0\) has equal roots, find the values of P.

**Answer Details**

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

**Question 23**
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In how many ways can 9 people be seated on a bench if only 3 places are available?

**Answer Details**

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.

**Question 24**
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If the points (-1, t -1), (t, t - 3) and (t - 6, 3) lie on the same straight line, find the values of t.

**Question 25**
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Simplify: \((1 - \sin \theta)(1 + \sin \theta)\).

**Answer Details**

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 \(\cos^{2} \theta\).

**Question 26**
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If \(f(x) = 3x^{3} + 8x^{2} + 6x + k\) and \(f(2) = 1\), find the value of k.

**Answer Details**

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.

**Question 27**
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Given that \(AB = \begin{pmatrix} 4 \\ 3 \end{pmatrix}\) and \(AC = \begin{pmatrix} 2 \\ -3 \end{pmatrix}\), find |BC|.

**Answer Details**

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.

**Question 29**
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If \((x + 2)\) and \((3x - 1)\) are factors of \(6x^{3} + x^{2} - 19x + 6\), find the third factor.

**Answer Details**

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.

**Question 30**
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Find \(\lim\limits_{x \to 3} (\frac{x^{3} + x^{2} - 12x}{x^{2} - 9})\)

**Answer Details**

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 \(\frac{7}{2}\).

**Question 31**
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A particle accelerates at 12\(ms^{-2}\) and travels a distance of 250m in 6 seconds. Find the initial velocity of the particle.

**Answer Details**

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.

**Question 32**
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If \(2, (k+1), 8,...\) form an exponential sequence (GP), find the values of k.

**Question 33**
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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.

**Question 34**
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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.

**Answer Details**

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.

**Question 35**
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Find the variance of 1, 2, 0, -3, 5, -2, 4.

**Answer Details**

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) \(\frac{52}{7}\).

**Question 36**
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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.

**Answer Details**

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.

**Question 37**
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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.

**Answer Details**

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) \(48\pi cm^{2}s^{-1}\).

**Question 39**
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If a fair coin is tossed four times, what is the probability of obtaining at least one head?

**Answer Details**

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) \(\frac{15}{16}\).

**Question 40**
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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?

**Answer Details**

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).

**Question 42**
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(a) Given that \(x = 3i - j, y = 2i + kj\) and the cosine of the angle between x and y is \(\frac{\sqrt{5}}{5}\), find the values of the constant k.

(b) In the quadrilateral ABCD,

\(\overrightarrow{AB} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}, \overrightarrow{AC} = \begin{pmatrix} -6 \\ -9 \end{pmatrix}\)

and \(\overrightarrow{BD} = \begin{pmatrix} 4 \\ -7 \end{pmatrix}\). Show whether or not ABCD is a parallelogram.

None

**Answer Details**

None

**Question 43**
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