Mathematics 2008 Jamb Examination
Mathematics 2008 Jamb Examination
Question 61
The locus of a point equidistant from two points P(6, 2) and R(4, 2) is a perpendicular bisector of PR passing through
The midpoint of P(6,2) and R(4,2) is:\[ \left( \dfrac{6+4}{2}, \dfrac{2+2}{2} \right ) = (5,2) \]The perpendicular bisector of PR must pass through this midpoint.
Hence the required point is (5,2).
Hence the required point is (5,2).
Not clear enough?
Question 62
Find the gradient of a line perpendicular to the line with the equation 3x + 2y + 1 = 0.
Rewrite the line in slope form:\[ 2y = -3x - 1 \]\[ y = -\dfrac{3}{2}x - \dfrac{1}{2} \]The gradient is therefore \(-\dfrac{3}{2}\).
The gradient of a line perpendicular to this is the negative reciprocal:\[ m_{\perp} = \dfrac{2}{3} \]Thus the required gradient is 2/3.
The gradient of a line perpendicular to this is the negative reciprocal:\[ m_{\perp} = \dfrac{2}{3} \]Thus the required gradient is 2/3.
Not clear enough?
Question 63
If \( \sin\theta = \dfrac{3}{5} \). Find \( \tan\theta \)
\(\sin\theta = \dfrac{3}{5}\) implies a right-angled triangle with opposite = 3 and hypotenuse = 5.
Using Pythagoras:\[ \text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4 \]Therefore,\[ \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{4} \]
Using Pythagoras:\[ \text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4 \]Therefore,\[ \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{4} \]
Not clear enough?
Question 64
Find the derivative of
\[ y = \dfrac{x^7 - x^5}{x^4} \]
\[ y = \dfrac{x^7 - x^5}{x^4} \]
First simplify the function:\[ y = x^{7-4} - x^{5-4} = x^3 - x \]Differentiating term by term:\[ \dfrac{dy}{dx} = 3x^2 - 1 \]Thus the derivative is \(3x^2 - 1\).
Not clear enough?
Question 65
Differentiate \( \sin x - x \cos x \)
Differentiate each term:
\(\sin x \rightarrow \cos x\).
For \(x\cos x\), apply product rule:\[ \dfrac{d}{dx}(x\cos x) = \cos x - x\sin x \]So the whole derivative becomes:\[ \cos x - (\cos x - x\sin x) = x\sin x \]
\(\sin x \rightarrow \cos x\).
For \(x\cos x\), apply product rule:\[ \dfrac{d}{dx}(x\cos x) = \cos x - x\sin x \]So the whole derivative becomes:\[ \cos x - (\cos x - x\sin x) = x\sin x \]
Not clear enough?
Question 66
Find the minimum value of the function \( y = x(1+x) \)
Expand the function:
\( y = x + x^2 \).
Differentiate:
\( y' = 2x + 1 \).
Set to zero:\[ 2x + 1 = 0 \Rightarrow x = -\dfrac{1}{2} \]Substitute back:\[ y = -\dfrac{1}{2} + \dfrac{1}{4} = -\dfrac{1}{4} \]This is the minimum value since the parabola opens upward.
\( y = x + x^2 \).
Differentiate:
\( y' = 2x + 1 \).
Set to zero:\[ 2x + 1 = 0 \Rightarrow x = -\dfrac{1}{2} \]Substitute back:\[ y = -\dfrac{1}{2} + \dfrac{1}{4} = -\dfrac{1}{4} \]This is the minimum value since the parabola opens upward.
Not clear enough?
Question 67
Evaluate
\[ \int_1^2 (6x^2 - 2x)\, dx \]
\[ \int_1^2 (6x^2 - 2x)\, dx \]
Integrate term by term:\[ \int 6x^2 dx = 2x^3 \]\[ \int -2x dx = -x^2 \]Evaluate from 1 to 2:
At 2:
\(2(8) - 4 = 16 - 4 = 12\)
At 1:
\(2(1) - 1 = 1\)
Difference:\[ 12 - 1 = 11 \]
At 2:
\(2(8) - 4 = 16 - 4 = 12\)
At 1:
\(2(1) - 1 = 1\)
Difference:\[ 12 - 1 = 11 \]
Not clear enough?
Question 68
Evaluate \( \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx \)
The integral of cosine is sine:\[ \sin x \Big|_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \]\(\sin(\dfrac{\pi}{2}) = 1\) and \(\sin(-\dfrac{\pi}{2}) = -1\).
Thus:\[ 1 - (-1) = 2 \]
Thus:\[ 1 - (-1) = 2 \]
Not clear enough?
Question 69
The bar chart above shows the number of times the words a, and, in, it, the, to appear in a paragraph in a book. What is the ratio of the least frequent word?
Ratio of least to most = 2:12
= 2/12
= 1/6
= 2/12
= 1/6
Not clear enough?
Question 70
What is the mean of the data
t, 2t − 1, t − 2, 2t − 1, 4t, 2t + 2?
t, 2t − 1, t − 2, 2t − 1, 4t, 2t + 2?
Sum all terms:\[ t + (2t-1) + (t-2) + (2t-1) + 4t + (2t+2) \]Combine like terms:\[ (1+2+1+2+4+2)t = 12t \]Constants:\[ -1 -2 -1 + 2 = -2 \]Total sum:\[ 12t - 2 \]Mean = divide by 6:\[ \dfrac{12t - 2}{6} = 2t - \dfrac{1}{3} \]
Not clear enough?
You might like
A man exerts a pressure of 2.6 x 103Nm-3 on the ground and has 4 x 10-2m3 of his feet in contact with the ground. The weight of the man is
This question is based on Buchi Emecheta's Second Class Citizen.
For her act of child neglect, Adah's mother was forced by the police to
Calculate the mass of Magnesium that will be liberated from its salt by the same quantity of electricity that liberated 16.0 g of Silver.
[Mg = 24.0, Ag = 108 ]