JAMB Mathematics Past Questions & Answers - Page 10

\(x^2 - x - 4 ≤ 2\)
Subtract two from both sides to rewrite it in the quadratic standard form:
= \(x^2 - x - 4 - 2 ≤ 2 - 2\)
= \(x^2 - x - 6 ≤ 0\)
Now set it = 0 and factor and solve like normal.
= \(x^2 - x\) - 6=0
= \((x - 3)(x + 2)\)=0
\(x\) + 2 = 0 or \(x\) - 3 = 0
\(x\) = -2 or \(x\) = 3
So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, \(x\) = 0) and evaluate the original inequality.
= \(x2 - x - 4 ≤ 2\)
= \((0)^2 - (0) - 4 ≤ 2\)
= \(0 - 0 - 4 ≤ 2\)
\(−4 ≤ 2\)
Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.
∴ \(-2 ≤ x ≤ 3.\)

A binary operation '*' defined on a set A is said to be commutative only if a*b=b*a, ∀a, b∈A. If (a*b)*c=a*(b*c), then the operation is said to associative ∀ a, b∈ A. If (b ο c)*a=(b*a) ο (c*a), then the operation is said to be distributive ∀ a, b, c ∈ A.

Let \(p(x) = -2x^3 + 6x^2 + 17x - 21\)

Using the remainder theorem

Let \(x + 1 = 0\)

∴ \(x = -1\)

Since, \((x + 1)\) divides \(p(x)\), then, remainder will be p(-1)

⇒ p(-1) = -2(-1)\(^3 + 6(-1)^2\) + 17(-1) - 21

∴ p(-1) = -30

Number of students who scored atleast 25% = 5 + 3 + 8 = 16

Let A = \(\begin {bmatrix} a & b\\ c & d \end {bmatrix}\)

i.e \(\begin{bmatrix} a & b\\c & d \end{bmatrix}\) \(\begin{bmatrix} 0 & 1\\2 & -1 \end{bmatrix}\) = \(\begin{bmatrix} 2 & -1\\1 & 0 \end{bmatrix}\)

\(\implies \begin {bmatrix} a(0) + b(2) & a(1) + b(-1)\\ c(0) + d(2) & c(1) + d(-1)\end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\ 1 & 0 \end {bmatrix}\)

\(\implies \begin {bmatrix} 2b & a - b\\2d & c - d\end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\ 1 & 0 \end {bmatrix}\)

By comparing

2b = 2

a - b = -1

2d = 1 and

c - d = 0

∴ b = \(^2/_2\) = 1

a - b = -1

⇒ a - 1 = -1

∴ a = 0

∴ d = \(^1/_2\)

⇒ c = d

∴ c = \(^1/_2\)

∴The matrice A = \(\begin {bmatrix} 0 & 1\\^1/_2 & ^1/_2 \end {bmatrix}\)