Algebra Questions and Answers

Answer: A

The graph is P.

Work out \(g(x) = |x + 2| - |x - 1|\) piece by piece:

• For \(x \le -2\): \(g = -(x + 2) + (x - 1) = -3\).
• For \(-2 \le x \le 1\): \(g = (x + 2) + (x - 1) = 2x + 1\).
• For \(x \ge 1\): \(g = (x + 2) - (x - 1) = 3\).

Taking the absolute value, \(f(x) = 3\) for \(x \le -2\) and for \(x \ge 1\), and \(f(x) = |2x + 1|\) in between, which falls to 0 at \(x = -0.5\) and rises back to 3. That is the flat-V-flat shape of P.

Q is \(g(x)\) itself, without the outer absolute value. R and S rise to 9 at \(x = -5\), which would need \(f\) to grow without limit on the left, but \(f\) is constant at 3 there.

Answer: A

The value is \(\sqrt{2}\).

Work down from the fourth powers using the identity \((x \pm \tfrac{1}{x})^{2} = x^{2} + \tfrac{1}{x^{2}} \pm 2\):

\[ \begin{aligned} \left(x^{2} + \tfrac{1}{x^{2}}\right)^{2} &= 14 + 2 = 16 \\ x^{2} + \tfrac{1}{x^{2}} &= 4 \\ \left(x - \tfrac{1}{x}\right)^{2} &= 4 - 2 = 2 \end{aligned} \]

So \(x - \tfrac{1}{x} = \sqrt{2}\), taking the positive root as the options do.

\(\sqrt{3}\), \(\sqrt{5}\) and \(\sqrt{7}\) do not square to 2. Note that \(x + \tfrac{1}{x}\) would be \(\sqrt{6}\), since its square is 4 + 2; the question asks for the difference, not the sum.

Answer: C

The value is 0.

Write \(x = p + q\), where \(p\) and \(q\) are the two cube roots. Then:

\[ \begin{aligned} p^{3} + q^{3} &= 2a \\ pq &= \left(a^{2} - (a^{2} + b^{3})\right)^{1/3} \\ &= (-b^{3})^{1/3} = -b \end{aligned} \]

Cubing \(x = p + q\):

\[ \begin{aligned} x^{3} &= p^{3} + q^{3} + 3pq(p + q) \\ &= 2a - 3bx \end{aligned} \]

Hence \(x^{3} + 3bx - 2a = 0\). The other options, \(a^{2}\), \(a^{2} + b^{3}\) and 1, would leave terms that the identity cancels completely.

Answer: D

The value is 50.

The expression \(xy + z(x + y)\) is \(xy + yz + zx\). Use the identity for the square of a sum:

\[ \begin{aligned} (x+y+z)^{2} &= x^{2}+y^{2}+z^{2} + 2(xy+yz+zx) \\ 169 &= 69 + 2(xy+yz+zx) \end{aligned} \]

So \(2(xy + yz + zx) = 100\) and the expression equals 50. 100 would be the answer if one forgot to halve, and 60, 40 and 70 do not follow.

Answer: C

The value is 14.

Since \((2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1\), the reciprocal is \(1/x = 2 - \sqrt{3}\). So \(x + 1/x = 4\):

\[ \begin{aligned} x^{2} + \tfrac{1}{x^{2}} &= \left(x + \tfrac{1}{x}\right)^{2} - 2 \\ &= 16 - 2 = 14 \end{aligned} \]

16 would be the result of forgetting to subtract 2. 6, 15 and 5 do not follow from the identity.

Answer: B

The difference is \(2x^{2} - 4xy + 12y^{2} + 5x - 9y + 3\).

Subtract the second polynomial term by term, changing the sign of each of its terms:

\[ \begin{aligned} x^{2}&: 7 - 5 = 2 \\ xy&: -4 - 0 = -4 \\ y^{2}&: 8 - (-4) = 12 \\ x&: 5 - 0 = 5 \\ y&: -3 - 6 = -9 \\ \text{const}&: 0 - (-3) = 3 \end{aligned} \]

The near miss gives \(-12y^{2}\), forgetting that subtracting \(-4y^{2}\) adds \(4y^{2}\). The options starting \(4x^{2} - 2xy\) get the \(x^{2}\) and \(xy\) terms wrong.

Answer: D

The true option is H.C.F. = 3xy, L.C.M. = \(30x^{2}y^{2}\).

Write \(f_{1} = 6x^{2}y = 2 \times 3 \times x^{2}y\) and \(f_{2} = 15xy^{2} = 3 \times 5 \times xy^{2}\).

The H.C.F. takes the common factors with the lowest powers: \(3 \times x \times y = 3xy\). The L.C.M. takes every factor with the highest power: \(2 \times 3 \times 5 \times x^{2}y^{2} = 30x^{2}y^{2}\). Check: \(3xy \times 30x^{2}y^{2} = 90x^{3}y^{3} = f_{1}f_{2}\).

45xy and 30xy cannot be common factors of 6 and 15, \(30x^{3}y^{3}\) uses too high powers, and \(90x^{2}y^{2}\) multiplies the numbers instead of taking their L.C.M.

Answer: A

The factors are \((x + 2)\) and \((x - 3)\).

Find two numbers whose product is \(-6\) and whose sum is \(-1\): they are \(+2\) and \(-3\).

\[ x^{2} - x - 6 = (x + 2)(x - 3) \]

Check: \(x^{2} - 3x + 2x - 6 = x^{2} - x - 6\). The pair \((x - 2)(x + 3)\) gives \(x^{2} + x - 6\), with the wrong sign of the middle term, and \((x + 1)(x - 6)\) and \((x - 1)(x + 6)\) give middle terms of \(-5x\) and \(+5x\).

Answer: B

The H.C.F. is \(x - 3\).

Factorise both expressions:

\[ \begin{aligned} 2x^2 - 3x - 9 &= (2x + 3)(x - 3) \\ 2x^3 - 54 &= 2(x^3 - 27) \\ &= 2(x - 3)(x^2 + 3x + 9) \end{aligned} \]

The only common factor is \(x - 3\). \(2x + 3\) divides only the first expression, and \(x + 3\) and \(2x - 3\) divide neither.

Check by substitution: at \(x = 3\) the first expression is \(18 - 9 - 9 = 0\) and the second is \(54 - 54 = 0\), so both vanish and share the factor. At \(x = -3/2\) the second expression is \(-27/4 - 54\), not zero, so \(2x + 3\) is not common.

Answer: A

The value of k is -3.

By the factor theorem, x - 1 is a factor of p(x) exactly when p(1) = 0:

\[ \begin{aligned} p(1) &= 4 + 3 - 4 + k = 0 \\ k &= -3 \end{aligned} \]

Check: with k = -3, \(4x^{3} + 3x^{2} - 4x - 3 = (x - 1)(4x^{2} + 7x + 3)\). The value -4 forgets the 3 from the \(3x^{2}\) term, and 0 or -1 leave a non-zero remainder when divided by x - 1.