MCQ test, Set 2

Answer: (D) $\{-3, 3\}$

Pre-images of $9$ are the elements of $A$ whose image is $9$

$f(-3) = 9$, $f(3) = 9$

Answer: (B) $\{-2, 0, 4\}$

$f(-1) = 2 (-1) = -2$, $f(0) = 2 (0) = 0$, $f(2) = 2 (2) = 4$

Range $= f(A) = \{-2, 0, 4\}$ (not the codomain $B$)

Answer: (C) $6$

$f(-1) = 2 (-1)^{2} - 3 (-1) + 1 = 6$

Answer: (C) $\{9\}$

Constant function: every element has the same image $9$

Range $= \{9\}$

Answer: (D) $(0, 9)$

put $x = 0$: $y = 9$

Answer: (D) $\{2, 5\}$

$\Delta = b^2 - 4ac = (-7)^2 - 4(1)(10) = 9$

$\displaystyle x = \frac{7 \pm 3}{2} = 2, 5$

Answer: (B) $\displaystyle - \frac{8}{3}$

$a = 3, c = -8$

$\displaystyle \alpha\beta = \frac{c}{a} = - \frac{8}{3}$

Answer: (D) $5$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(1, \frac{56 - 16}{8}\right) = (1, 5)$

$a > 0$: minimum value $= 5$

Answer: (A) $(0, -10)$

put $x = 0$: $y = -10$

Answer: (B) $(-1, 4)$

$y = (4 - x)(x + 1) = - x^{2} + 3 x + 4$

$a = -1 < 0$ (downward parabola)

$\Delta = b^2 - 4ac = 3^2 - 4(-1)(4) = 25$

$\displaystyle y = 0 \Rightarrow x = \frac{-3 \pm 5}{-2} = -1, 4$

Sign: $-$ outside $-1$ and $4$, $+$ between them

$\therefore x \in (-1, 4)$

Answer: (A) $\displaystyle \{\frac{3}{4}\}$

$y = (4x - 3)^2 = 16 x^{2} - 24 x + 9$

$a = 16 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = (-24)^2 - 4(16)(9) = 0$

$\displaystyle \Delta = 0 \Rightarrow y = 16(x - \frac{3}{4})^2$

$y > 0$ for all $\displaystyle x \ne \frac{3}{4}$, and $y = 0$ at $\displaystyle x = \frac{3}{4}$

$\displaystyle \therefore x \in \{\frac{3}{4}\}$

Answer: (C) $\displaystyle \frac{1}{2}$

$9^{1/2} = \sqrt{9} = 3$

So $\displaystyle \log_9 3 = \frac{1}{2}$

Answer: (A) $\displaystyle \frac{3}{2}$

$\displaystyle \log_{5^2} 5^3 = \frac{3}{2} \log_5 5 = \frac{3}{2}$

Answer: (C) $(5, 6)$

Base $3 > 1$: $0 < x - 5 < 1$

$5 < x < 6$

Answer: (D) $3$

$2^{\log_{2^3} 3^3} = 2^{\log_2 3} = 3$

Answer: (A) $[0, \infty)$

$|x| \ge 0$ for all $x$ and every value $\ge 0$ is taken

Range $= [0, \infty)$

Answer: (D) $(3, 7)$

$-2 < x - 5 < 2$

$\Rightarrow -2 + 5 < x < 2 + 5$

$\Rightarrow 3 < x < 7$

$\Rightarrow x \in (3, 7)$

Answer: (B) $\{-4, 1\}$

$2 x + 3 = \pm 5$

$2 x + 3 = 5 \Rightarrow 2 x - 2 = 0 \Rightarrow x = 1$

$2 x + 3 = -5 \Rightarrow 2 x + 8 = 0 \Rightarrow x = -4$

Answer: (D) $\{-3, 2\}$

Case (i) Suppose $x < -2$ ($x - 1 < 0, x + 2 < 0$): $(1 - x) + (- x - 2) = 5 \Rightarrow - 2 x - 1 = 5$ $\Rightarrow$ $x = -3$, which is in this case

Case (ii) Suppose $-2 \le x \le 1$ ($x - 1 \le 0, x + 2 \ge 0$): $(1 - x) + (x + 2) = 5 \Rightarrow 3 = 5$: not possible, no solution

Case (iii) Suppose $x > 1$ ($x - 1 > 0, x + 2 > 0$): $(x - 1) + (x + 2) = 5 \Rightarrow 2 x + 1 = 5$ $\Rightarrow$ $x = 2$, which is in this case

So $x \in \{-3, 2\}$

Answer: (B) $\phi$

The sum is $0$ only if $|x - 3| = 0$ and $|x - 5| = 0$ together

No $x$ makes both zero: $x \in \phi$

Answer: (A) $\displaystyle [- \frac{7}{3}, \infty)$

$\sqrt{3 x + 7}$ is defined when $3 x + 7 \ge 0$

$\Rightarrow 3 x \ge -7$

$\displaystyle \Rightarrow x \ge - \frac{7}{3}$

Domain $\displaystyle = [- \frac{7}{3}, \infty)$

Answer: (A) $-1$

$[-0.7] = -1$, $[0.7] = 0$

$-1 + 0 = -1$

Answer: (C) $(-\infty, 1)$

$[x - 2] = [x] - 2$

$[x] - 2 \le -2 \Rightarrow [x] \le 0$

$[x] = 0$ for $0 \le x < 1$: $x \in (-\infty, 1)$

Answer: (B) $[0, 1)$

$0 \le x - [x] < 1$ for every $x$ (it is $0$ at integers and never reaches $1$)

Answer: (C) falls from left to right and passes through $(0, 1)$

$\displaystyle a = \frac{1}{3}$ lies between $0$ and $1$, so $a^x$ decreases

$x = 0 \Rightarrow y = 1$: it passes through $(0, 1)$