Mathematics 2004 Jamb Examination

Given:
\(y = x^3 + 2x^2 - 5x - 6\)
Try rational root theorem
Possible rational roots: \( \pm1, \pm2, \pm3, \pm6 \)

Test x = -1: 
\((-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0\)
\(\Rightarrow x = -1\) is a root.

Polynomial division
Divide \(x^3 + 2x^2 - 5x - 6\) by \(x + 1\):\[ \begin{array}{r|rrr}
 & x^2 & + x & -6 \\
x + 1 & x^3 & + 2x^2 & -5x & -6 \\
      & x^3 & + x^2 &     & \\
\hline
      &     & x^2 & -5x & \\
      &     & x^2 & + x & \\
\hline
      &     &     & -6x & -6 \\
      &     &     & -6x & -6 \\
\hline
      &     &     & 0 & 0 \\
\end{array}\]

Quotient: \(x^2 + x - 6\)

Factor the quadratic
\(
x^2 + x - 6 = (x - 2)(x + 3)
\)

Final factorized form
\(y = (x + 1)(x - 2)(x + 3)\)

Solve y = 0
\((x + 1)(x - 2)(x + 3) = 0\)

Roots:
\(x = -1, \quad x = 2, \quad x = -3\)
Therefore, the curve crosses the x-axis at:-3, -1, 2

\[\boxed{\text{Correct answer: C. } 2, -1 \text{ and } -3}\]  

\[f(x) = 3x^3 + 5x^2 - 11x + 4\] the remainder when a polynomial \(f(x)\) is divided by \(x + 3\) is: \[f(-3)\] \[f(-3) = 3(-3)^3 + 5(-3)^2 - 11(-3) + 4\] \[= 3(-27) + 5(9) + 33 + 4\] \[= -81 + 45 + 33 + 4 = 1 \] \[ \boxed{1} \]

\[ ac - 2bc - a^2 + 4b^2\] Group terms\[(ac - 2bc) + (-a^2 + 4b^2)\] Factor each group: \[= c(a - 2b) - (a^2 - 4b^2)\] Recall that: \[a^2 - 4b^2 = (a - 2b)(a + 2b)\] So: \[= c(a - 2b) - (a - 2b)(a + 2b)\]

Factor out the common factor (a - 2b)\[= (a - 2b)\left(c - (a + 2b)\right)\] Simplify the expression inside the parentheses:\[ = (a - 2b)(c - a - 2b)\]

Final Answer\[\boxed{(a - 2b)(c - a - 2b)}\]  

Given: \( y \propto \dfrac{1}{x} \Rightarrow y = \dfrac{k}{x} \)

We are told that:\[y = 4 \quad \text{when} \quad x = \dfrac{1}{2}\] Find the constant \(k\) \[ 4 = \dfrac{k}{\dfrac{1}{2}} \Rightarrow k = 4 \cdot \dfrac{1}{2} = 2\] Find  \(x\) when \(y = 10\)\[ y = \dfrac{2}{x}, \quad \text{so} \quad 10 = \dfrac{2}{x}
\Rightarrow x = \dfrac{2}{10} = \dfrac{1}{5} \]

Final Answer: \[ \boxed{\dfrac{1}{5}} \]  

The length \(L\) of a simple pendulum varies directly as the square of its period \(T\), i.e.,\[L \propto T^2 \Rightarrow L = kT^2\]

Find the constant k
We are told that when \(T = 4\) seconds, \(L = 64\) cm:\[64 = k \cdot 4^2 = k \cdot 16 \Rightarrow k = \dfrac{64}{16} = 4\]

Find the length when \(T = 9\) seconds\[L = 4 \cdot 9^2 = 4 \cdot 81 = 324 \text{ cm}\]

Final Answer:\[\boxed{324\ \text{cm}}\]  

The line passes through the points \((0, 6)\) and \((2, 0)\), and the shaded region is below the line.

Find the equation of the line
Calculate the slope:\[\text{slope} = \dfrac{0 - 6}{2 - 0} = \dfrac{-6}{2} = -3\] Use the point-slope form with point \((0, 6)\):\[y - 6 = -3(x - 0) \Rightarrow y = -3x + 6\] Rewriting the equation:\[y + 3x = 6\] Determine the inequality
Since the shaded region lies \textbf{below} the line, we use:\[y + 3x < 6\]

Final Answer:\[\boxed{\{(x, y) : y + 3x < 6\}}\]  

The inequality is  \[-1 < 3 - 2x \leq 5\] Subtract 3 from all parts:  \[-1 - 3 < 3 - 2x - 3 \leq 5 - 3\]\[-4 < -2x \leq 2\] Divide all parts by \(-2\), and reverse the inequality signs:  \[\dfrac{-4}{-2} > x \geq \dfrac{2}{-2}\]\[2 > x \geq -1\] This simplifies to:  \[-1 \leq x < 2\] Integral values of \(x\):  \[\boxed{-1,\ 0,\ 1}\]  

We are given:

\[Q_n = 3 \cdot 2^{n - 2}, \quad U_m = 3 \cdot 2^{2m - 3}\]

We are to find:\[Q_2 \cdot U_2\] Evaluate \(Q_2\):\[Q_2 = 3 \cdot 2^{2 - 2} = 3 \cdot 2^0 = 3 \cdot 1 = 3\] Evaluate \(U_2\):\[U_2 = 3 \cdot 2^{2 \cdot 2 - 3} = 3 \cdot 2^1 = 3 \cdot 2 = 6\] Product:\[Q_2 \cdot U_2 = 3 \cdot 6 = \boxed{18}\]

We are given: \[ a = 6, \quad \text{and} \quad ar^3 = 162 \] We are to find the sum of the first three terms: \[ S = a + ar + ar^2 \] Find the common ratio: \[ ar^3 = 162 \Rightarrow 6r^3 = 162 \Rightarrow r^3 = \frac{162}{6} = 27 \Rightarrow r = \sqrt[3]{27} = 3 \] Compute the sum: \[ S = 6 + 6r + 6r^2 = 6 + 6 \cdot 3 + 6 \cdot 3^2 = 6 + 18 + 54 = \boxed{78} \]

We are given the geometric series:\[\dfrac{1}{2},\ \dfrac{1}{6},\ \dfrac{1}{18},\ \ldots\]Find the first term and common ratio: \[a = \dfrac{1}{2}, \quad r = \dfrac{\dfrac{1}{6}}{\dfrac{1}{2}} = \dfrac{1}{6} \cdot \dfrac{2}{1} = \dfrac{1}{3}\]Since \( |r| < 1 \), the sum to infinity is given by:\[S_\infty = \dfrac{a}{1 - r} = \dfrac{\dfrac{1}{2}}{1 - \dfrac{1}{3}} = \dfrac{\dfrac{1}{2}}{\dfrac{2}{3}} = \dfrac{1}{2} \cdot \dfrac{3}{2} = \dfrac{3}{4}\]\[\boxed{\dfrac{3}{4}}\]