Algebra Review Test 3
Q1–10 of 20Directions for the Questions 1 to 3:Refer to the data given below and answer the questions.
Given \( \dfrac{a}{b}=\dfrac{1}{2},\quad \dfrac{c}{d}=\dfrac{1}{3} \) and \( z=\dfrac{a+c}{b+d} \), answer the questions below on limits of z.
Q. If y ≥ 0 and p ≥ 0 then the limits of ‘z’ are:
Analysis: Since both $a, c \ge 0$, we analyze the extreme ratio cases:
If $a = 0, c > 0$: $z = \frac{c}{3c} = \frac{1}{3}$
If $c = 0, a > 0$: $z = \frac{a}{2a} = \frac{1}{2}$
Since $a$ and $c$ are non-negative, any linear combination gives a weighted average between these two extremes:
Correct Answer: (b) $1/3 \le z \le 1/2$
Directions for the Questions 1 to 3: Refer to the data given below and answer the questions.
Given
\( \dfrac{a}{b}=\dfrac{1}{2},\quad \dfrac{c}{d}=\dfrac{1}{3} \) and \( z=\dfrac{a+c}{b+d} \) , answer the questions below on limits of z.
Q. c ≤ 0 and 1/3≤ z ≤1/2 only if:
For $z \le \frac{1}{2}$:
$$\frac{a+c}{2a+3c} \le \frac{1}{2} \implies 2a + 2c \le 2a + 3c \implies c \ge 0$$(Assuming denominator $2a + 3c > 0$).
If $c \le 0$, for $z$ to lie within $[1/3, 1/2]$, the denominator $2a + 3c$ must remain positive:
$$2a + 3c > 0 \implies 2a > -3c \implies a > -1.5c$$
Correct Answer: (d) $a > -1.5c$
Directions for the Questions 1 to 3: Refer to the data given below and answer the questions. Given
\( \dfrac{a}{b}=\dfrac{1}{2},\quad \dfrac{c}{d}=\dfrac{1}{3} \) and \( z=\dfrac{a+c}{b+d} \) , answer the questions below on limits of z.
Q. If a = –31, which of the following value of ‘d’ gives the highest value of ‘z’?
Analysis:
Given $a = -31 \implies b = -62$.
Since $d = 3c \implies c = \frac{d}{3}$, substitute into $z$:
$$z = \frac{-31 + \frac{d}{3}}{-62 + d} = \frac{-93 + d}{3(d - 62)} = \frac{d - 93}{3d - 186}$$Evaluating options:
(a) $d = 72$: $z = \frac{72 - 93}{3(72) - 186} = \frac{-21}{30} = -0.7$
(b) $d = 721$: $z = \frac{721 - 93}{3(721) - 186} = \frac{628}{1977} \approx 0.317$
(c) $d = -31$: $z = \frac{-31 - 93}{3(-31) - 186} = \frac{-124}{-279} \approx 0.444$
(d) $d = 0$: $z = \frac{-93}{-186} = 0.5$
Comparing values: $0.5$ is the maximum.
Correct Answer: (d) $d = 0$
Find the integral solution of:
5y – 1 < (y + 1)² < (7y – 3)
Step 1: First inequality $5y - 1 < y^2 + 2y + 1$:
Step 2: Second inequality $y^2 + 2y + 1 < 7y - 3$:
Step 3: Intersection of both conditions:
Step 4: Integral values in $2 < y < 4$ $\implies y = 3$.
Correct Answer: (d) 3
If \( f(a)=\dfrac{a-1}{a+1},\quad x\ge0 \) and if \( y=f\!\left(\dfrac{1}{a}\right) \), then
Analysis:
$$y = f\left(\frac{1}{a}\right) = \frac{\frac{1}{a} - 1}{\frac{1}{a} + 1} = \frac{1 - a}{1 + a}$$Derivative with respect to $a$:
$$\frac{dy}{da} = \frac{-(1+a) - (1-a)}{(1+a)^2} = \frac{-2}{(1+a)^2} < 0$$Since the derivative is negative, as $a$ increases, $y$ decreases.
Correct Answer: (b) As ‘a’ increases, ‘y’ decreases
If f and g are real functions defined by f(a) = a + 2 and g(a) = 2a² + 5, then fog is equal to
Analysis:
Correct Answer: (a) $2a^2 + 7$
If ‘p’ and ‘q’ are the roots of the equation x² – 10x + 16 = 0, the value of (1 – p) (1 – q) is
Analysis:
Sum of roots $p + q = 10$, product of roots $pq = 16$.
$$(1 - p)(1 - q) = 1 - (p + q) + pq = 1 - 10 + 16 = 7$$
Correct Answer: (b) 7
Given that ‘a’ and ‘b’ are positive real numbers such that a + b = 1, then what is the minimum value of
\( \sqrt{12+\dfrac{1}{a^2}}+\sqrt{12+\dfrac{1}{b^2}} \)
Analysis:
Expression simplifies to: $4 + \frac{1}{a^2} + \frac{1}{b^2}$
By symmetry, minimum occurs when $a = b = \frac{1}{2}$:
$$\text{Min value} = 4 + \frac{1}{(1/2)^2} + \frac{1}{(1/2)^2} = 4 + 4 + 4 = 12$$
(Exact minimum is 12)
Let p, q and r be distinct positive integers satisfying p < q < r and p + q + r = k. What is the smallest value of k that does not determine p, q, r uniquely?
Analysis:
For $k = 6$: $(1, 2, 3) \implies$ unique.
For $k = 7$: $(1, 2, 4) \implies$ unique.
For $k = 8$:
Partition 1: $(1, 2, 5)$
Partition 2: $(1, 3, 4)$
Thus, $k = 8$ is the smallest integer with multiple valid sets.
Correct Answer: (d) 8
Given odd positive integers p, q and r which of the following is not necessarily true?
Analysis:
(a) $\text{Odd} \times \text{Odd} \times \text{Odd} = \text{Odd}$ (True).
(b) $p^2 + q^3 = \text{Odd} + \text{Odd} = \text{Even}$. $\text{Even} \times 3 \times \text{Odd} = \text{Even}$ (True).
(c) $5p + q + r^4 = \text{Odd} + \text{Odd} + \text{Odd} = \text{Odd}$ (True).
(d) $p^4 + q^4 = (2k+1)^4 + (2m+1)^4 \equiv 1 + 1 = 2 \pmod 4$. Dividing by $2$ leaves an odd quotient. Multiplied by $r^2$ (odd) yields an odd number (Not necessarily even).
Correct Answer: (d) $r^2 (p^4 + q^4)/2$ is even