Algebra Questions and Answers

Answer: C

Both (i) and (ii) are correct.

Compare the two functions by their difference:

\[ \begin{aligned} g(t) - f(t) &= 4t - 0.01t^{2} \\ &= 0.01t\,(400 - t) \end{aligned} \]

For \(0 \lt t \lt 400\) the difference is positive, so \(g(t) \gt f(t)\); for example at t = 1, g = 4 while f = 0.01. That makes statement (i) true.

For \(t \gt 400\) the difference is negative, so taking T = 400 gives \(f(t) \gt g(t)\) for all \(t \gt T\). Statement (ii) is also true: a quadratic with a positive leading coefficient eventually outgrows any linear function, however small its coefficient. The options naming only one statement, or neither, are therefore wrong.

Answer: B

Necessarily, \(f(x) = g(y) =\) constant.

Fix any value \(y_{0}\). Then \(f(x) = g(y_{0})\) for every real \(x\), so \(f\) takes the single value \(c = g(y_{0})\) everywhere: it is constant. Since \(f(x) = g(y)\) for every \(y\) as well, \(g(y) = c\) for all \(y\). Both functions are the same constant.

The constant need not be zero: \(f(x) = g(y) = 5\) satisfies the condition, so 'both equal 0' is not necessary. 'Neither is constant' is the opposite of what was just shown. The last option simplifies to \(2g(y) = 0\), i.e. \(g(y) = 0\), which again is only one special case.

Answer: B

The answer is x = y.

First evaluate the right-hand side: \(2 \odot 2 = \frac{2^{2}}{2} = 2\). Then

\[ \begin{aligned} \frac{x^{2} + y^{2}}{xy} &= 2 \\ x^{2} - 2xy + y^{2} &= 0 \\ (x - y)^{2} &= 0 \end{aligned} \]

so x = y. Checking a distractor: with x = 2y the left side is (4 + 1)/2 = 2.5, and with x = y/2 it is also 2.5, not 2; x = 3y/2 gives 13/6. Only x = y makes the sum of squares equal to twice the product.

Answer: B

The value is 0.

Simplify the arguments first: -(-3 + 2) = 1 and (-2 + 3) = 1. So the numerator is \(\lambda(1, 1)\), and since p = q the first case applies:

\[ \lambda(1,1) = (1 - 1)^{2} = 0 \]

The denominator is -(-2 + 1) = 1, so the expression is 0/1 = 0.

The trap is the case split: with p = q the rule 'if p ≥ q' applies, so the square is used, not the sum. Using the sum by mistake would give 2, which is not offered, and none of -1, 16/3 or 16 can arise once both arguments are seen to equal 1.

Answer: A

The graph is \(f(x) = x^{2}\,2^{-|x|}\).

Read three features from the graph: it is symmetric about the vertical axis (an even function), it is never negative and touches zero at \(x = 0\), and it decays to zero on both sides.

\(x^{2}\,2^{-|x|}\) has all three: \(x^{2}\) and \(|x|\) are both even, the product is non-negative and zero at the origin, and \(2^{-|x|}\) makes it vanish as \(|x|\) grows, with peaks in between.

\(x\,2^{-|x|}\) is odd, so it would be negative for \(x \lt 0\). \(|x|\,2^{-x}\) and \(x\,2^{-x}\) contain \(2^{-x}\), which grows without bound as x becomes large and negative, so they cannot fall back to zero on the left. Only the first option fits.

Answer: B

The correct statement is \(a_1 \lt 50 \lt a_2\).

Both brothers follow the same weight function w, which is strictly increasing up to 50 and strictly decreasing after. If both ages were below 50, \(a_1 \lt a_2\) would give \(w(a_1) \lt w(a_2)\); if both were above 50, it would give \(w(a_1) \gt w(a_2)\). Either way the weights differ.

Equal weights at different ages are only possible when the ages lie on opposite sides of the peak, one rising and one falling. If one of them were exactly 50 he would be at the unique maximum, which the other, at a different age, cannot match. So the younger brother is under 50 and the elder over 50.

Answer: A

The graph is P.

Work out \(g(x) = |x+2| - |x-1|\) on the three intervals set by x = -2 and x = 1:

\[ \begin{aligned} x \le -2:&\quad g = -(x+2)+(x-1) = -3 \\ -2 \le x \le 1:&\quad g = (x+2)+(x-1) = 2x+1 \\ x \ge 1:&\quad g = (x+2)-(x-1) = 3 \end{aligned} \]

Taking the absolute value, f is 3 for x up to -2, falls along |2x + 1| to 0 at x = -1/2, rises back to 3 at x = 1, and stays at 3. That is graph P.

Q is the graph of g itself, without the outer absolute value. R and S rise without bound to the left, which f cannot do, since it never exceeds 3.

Answer: B

The graph is \(y = ||x| - 1| - 1\).

Check the plotted points. At x = 0: |0 - 1| - 1 = 0. At x = ±1: |1 - 1| - 1 = -1. At x = ±2: |2 - 1| - 1 = 0. At x = ±3: |3 - 1| - 1 = 1. These are exactly the corners and ends of the W in the figure, and the function is linear between them.

\(y = ||x| + 1| - 2\) gives -1 at x = 0, and \(y = ||x| + 1| - 1\) gives 0 at x = 0 but then rises steadily with no dips. \(y = ||x - 1| - 1|\) is never negative, while the graph dips to -1.

Answer: C

The solution set is \(x \ge \tfrac{1}{9}\), which is number line (C): a filled dot just to the right of 0 with the bold ray running to the right.

Multiply both sides by 3 (a positive number, so the sign is kept) and collect terms:

\[ \begin{aligned} 3(2 - 5x) &\le -(6x - 5) \\ 6 - 15x &\le -6x + 5 \\ 1 &\le 9x \\ x &\ge \tfrac{1}{9} \end{aligned} \]

The boundary 1/9 is included (the sign is 'less than or equal to'), so the dot is filled, and it lies a little to the right of 0.

Line (A) shows \(x \le 1/9\), the inequality reversed. Line (B) starts at a negative number, as if the boundary were \(-1/9\). Line (D) shows two separate rays, which a linear inequality can never produce.

Answer: C

The area is 2.

Simplify the function first. Since \(\sqrt{e^{x}} = e^{x/2}\),

\[ \begin{aligned} f(x) &= 2\ln\left(e^{x/2}\right) \\ &= 2 \cdot \frac{x}{2} = x \end{aligned} \]

So the region is the triangle under the line \(y = x\) between \(x = 0\) and \(x = 2\):

\[ \int_{0}^{2} x\,dx = \frac{x^{2}}{2}\Big|_{0}^{2} = 2 \]

Dropping the factor 2 in front of the logarithm gives \(f(x) = x/2\) and an area of 1, and halving once more gives 1/2. 4 is \(x^{2}\) at \(x = 2\), the antiderivative without its division by 2.