Mathematics 2021 Jamb Examination

To factorize this expression, we use the method of grouping:
Group the terms in pairs: \((m^3 - m^2) + (2m - 2)\)
Factor out the common term from each group:\[m^2(m - 1) + 2(m - 1)\]Now, factor out the common binomial \((m - 1)\):\[(m^2 + 2)(m - 1)\]

The sum of angles in a quadrilateral is 360^o. We set up an equation:\[(5x - 30) + (4x + 60) + (60 - x) + (3x + 61) = 360\]Combine like terms:\[(5x + 4x - x + 3x) + (-30 + 60 + 60 + 61) = 360\]\[11x + 151 = 360\]\[11x = 360 - 151 = 209\]\[x = \frac{209}{11} = 19\]Now, calculate each angle:
1. \(5(19) - 30 = 95 - 30 = 65^o\)
2. \(4(19) + 60 = 76 + 60 = 136^o\)
3. \(60 - 19 = 41^o\)
4. \(3(19) + 61 = 57 + 61 = 118^o\)
The smallest angle is 41^o, which corresponds to the expression 60 - x.

Let's check the pattern for \(n = 1, 2, 3, 4\):
For \(n = 1: 1^2 + 1 = 2\)
For \(n = 2: 2^2 + 2 = 6\)
For \(n = 3: 3^2 + 3 = 12\)
For \(n = 4: 4^2 + 4 = 20\)
The sequence perfectly matches the formula \(n^2 + n\).

The identity element \(e\) must satisfy the condition \(m \ast e = m\) for any real number \(m\):\[m(e) + m + e = m\]Subtract \(m\) from both sides:\[me + e = 0\]Factor out \(e\):\[e(m + 1) = 0\]For this to be true for all values of \(m\), \(e\) must be 0.

Recognize this as a difference of two squares:\[(9a^2)^2 - (4b^2)^2\]Apply the formula \(x^2 - y^2 = (x - y)(x + y)\):\[(9a^2 - 4b^2)(9a^2 + 4b^2)\]The first term \((9a^2 - 4b^2)\) is also a difference of two squares:\[(3a - 2b)(3a + 2b)(9a^2 + 4b^2)\]

Convert the logarithmic equation into its exponential form \(\log_b a = c \implies b^c = a\):\[9^{1.5} = x\]Since 1.5 is the same as \(\frac{3}{2}\):\[x = 9^{3/2}\]\[x = (\sqrt{9})^3\]\[x = 3^3 = 27\]

Solve the inequality in steps:
Add 6 to all parts of the inequality:\[-2 + 6 < 2x < 4 + 6\]\[4 < 2x < 10\]Divide all parts by 2:\[2 < x < 5\]The integers strictly between 2 and 5 are 3 and 4.

Points Y, X (East), and Z (South of Y) form a right-angled triangle with the right angle at Y. \(YZ = 6\)km and \(ZX = 12\)km.
Calculate angle \(\angle YXZ\) (internally):\[\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{6}{12} = 0.5\]\[\theta = \arcsin(0.5) = 30^o\]X is East of Y. To find the bearing of Z from X, we measure clockwise from the North at X. Moving West from X takes us back toward Y (270^o). Since Z is south, the bearing is \(270^o - 30^o = 240^o\).

Let Y = Yam, P = Plantain, M = Maize.
\(n(Y \cap P \cap M) = 3\).
\(n(P \cap M) = 5\), so only P and M = \(5 - 3 = 2\).
\(n(Y \cap M) = 4\), so only Y and M = \(4 - 3 = 1\).
Given Only Y and P = 2.
Now find single items:
Only Maize = \(12 - (2 + 3 + 1) = 6\).
Only Yam = \(10 - (1 + 3 + 2) = 4\).
Only Plantain = \(14 - (2 + 3 + 2) = 7\).
Total women = \(6 + 4 + 7 + 2 + 1 + 2 + 3 = 25\).

If \((x + 2)\) is a factor, then \(f(-2) = 0\). If \((x - 1)\) is a factor, then \(f(1) = 0\).
For \(f(1) = 0\):\[L(1) + 2k(1)^2 + 24 = 0 \implies L + 2k = -24\] (Equation 1)
For \(f(-2) = 0\):\[L(-2) + 2k(-2)^2 + 24 = 0 \implies -2L + 8k = -24\]Divide by -2: \(L - 4k = 12\) (Equation 2)
Subtract Eq 2 from Eq 1:\[(2k) - (-4k) = -24 - 12 \implies 6k = -36 \implies k = -6\]Substitute \(k = -6\) into Eq 1:\[L + 2(-6) = -24 \implies L - 12 = -24 \implies L = -12\]