Functions and Graphs: MCQ test, Set 8

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

The pre-image of $27$ under $f(x) = 5 x + 2$ is
A $137$ B $5$ C $\displaystyle \frac{29}{5}$ D $-5$

Answer and solution

Answer: (B) $5$

$f(x) = 27 \Rightarrow 5 x + 2 = 27 \Rightarrow x = 5$
If $A = \{1, 3, 5\}$ , $B = \{1, 3, 5, 7, 9\}$ and $f : A \to B$ is defined by $f(x) = 2 x - 1$ , then the range of $f$ is
A $\{1, 5, 9\}$ B $\{1, 3, 5, 7, 9\}$ C $\{3, 7, 11\}$ D $\{1, 3, 5\}$

Answer and solution

Answer: (A) $\{1, 5, 9\}$

$f(1) = 2 (1) - 1 = 1$ , $f(3) = 2 (3) - 1 = 5$ , $f(5) = 2 (5) - 1 = 9$

Range $= f(A) = \{1, 5, 9\}$ (not the codomain $B$ )
The domain of the polynomial function $f(x) = 5x^2 - 2x + 9$ is
A $(0, \infty)$ B $\mathbb{R} - \{0\}$ C $\mathbb{R} = (-\infty, \infty)$ D $[0, \infty)$

Answer and solution

Answer: (C) $\mathbb{R} = (-\infty, \infty)$

$f(x)$ is defined for every real $x$

Domain of a polynomial function is $\mathbb{R}$
If $I_A$ is the identity function on $A = \{1, 3, 5, 7\}$ , then the range of $I_A$ is
A $\{1\}$ B $\{1, 3, 5, 7\}$ C $\{7\}$ D $\{2, 6, 10, 14\}$

Answer and solution

Answer: (B) $\{1, 3, 5, 7\}$

Each element is its own image

Range $= A = \{1, 3, 5, 7\}$
The line $y = 12 - 8 x$ cuts the y-axis at
A $(12, 0)$ B $\displaystyle (\frac{3}{2}, 0)$ C $(0, -12)$ D $(0, 12)$

Answer and solution

Answer: (D) $(0, 12)$

put $x = 0$ : $y = 12$
The roots of $x^{2} + 3 x - 28 = 0$ are
A $\{4, 7\}$ B $\{-7, -4\}$ C $\{-7, 4\}$ D $\{-4, 7\}$

Answer and solution

Answer: (C) $\{-7, 4\}$

$\Delta = b^2 - 4ac = 3^2 - 4(1)(-28) = 121$

$\displaystyle x = \frac{-3 \pm 11}{2} = -7, 4$
If $\alpha, \beta$ are the roots of $2 x^{2} + 7 x - 4 = 0$ , then $\alpha\beta =$
A $2$ B $-2$ C $\displaystyle - \frac{7}{2}$ D $\displaystyle - \frac{1}{2}$

Answer and solution

Answer: (B) $-2$

$a = 2, c = -4$

$\displaystyle \alpha\beta = \frac{c}{a} = -2$
The minimum value of $x^{2} + 10 x + 28$ is
A $-5$ B $53$ C $3$ D $28$

Answer and solution

Answer: (C) $3$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(-5, \frac{112 - 100}{4}\right) = (-5, 3)$

$a > 0$ : minimum value $= 3$
The graph of $y = 25x^2 + 20x + 4$ touches the x-axis at
A $\displaystyle (- \frac{4}{5}, 0)$ B $\displaystyle (- \frac{2}{5}, 0)$ C $\displaystyle (\frac{2}{5}, 0)$ D $(0, 4)$

Answer and solution

Answer: (B) $\displaystyle (- \frac{2}{5}, 0)$

$\Delta = 0$ , perfect square: $y = (5 x + 2)^{2}$

$\displaystyle y = 0 \Rightarrow x = - \frac{2}{5}$
If $-x^2 - x - 1 \ge 0$ , then $x \in$
A $\displaystyle \mathbb{R} - \{- \frac{1}{2}\}$ B $\mathbb{R}$ C $\phi$ D $\displaystyle \{- \frac{1}{2}\}$

Answer and solution

Answer: (C) $\phi$

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

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

$a < 0$ , $\Delta < 0$ : $y < 0$ for all $x \in \mathbb{R}$

$\therefore x \in \phi$
If $x^2 + 10x + 25 \le 0$ , then $x \in$
A $\{-5\}$ B $\mathbb{R}$ C $\phi$ D $\mathbb{R} - \{-5\}$

Answer and solution

Answer: (A) $\{-5\}$

$a = 1 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 10^2 - 4(1)(25) = 0$

$\Delta = 0 \Rightarrow y = (x + 5)^2$

$y > 0$ for all $x \ne -5$ , and $y = 0$ at $x = -5$

$\therefore x \in \{-5\}$
$\log_{1/6} 216 =$
A $\displaystyle - \frac{1}{3}$ B $-3$ C $3$ D $36$

Answer and solution

Answer: (B) $-3$

$\displaystyle \left(\frac{1}{6}\right)^{-3} = 6^3 = 216$

So the value is $-3$
$\displaystyle \frac{\log 64}{\log 4} =$
A $3$ B $2$ C $16$ D $\displaystyle \frac{1}{3}$

Answer and solution

Answer: (A) $3$

$\displaystyle \frac{\log a}{\log b} = \log_b a$

$\log_4 64 = \log_4 4^3 = 3$

(It is not $\displaystyle \log \frac{64}{4}$ )
Which of the following is positive?
A $\log_{1/6} 40$ B $\log_{1/2} 5$ C $\log_{0.3} 0.6$ D $\log_4 0.9$

Answer and solution

Answer: (C) $\log_{0.3} 0.6$

$\log_a x > 0$ when $a > 1, x > 1$ or $0 < a < 1, 0 < x < 1$

$\log_{0.3} 0.6$ : both between $0$ and $1$ : positive
$\displaystyle \frac{\log (2\sqrt{2}) + \log (5\sqrt{5})}{\log 10} =$
A $\displaystyle \frac{1}{2}$ B $10$ C $3$ D $\displaystyle \frac{3}{2}$

Answer and solution

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

$2\sqrt{2} \cdot 5\sqrt{5} = 2^{3/2} 5^{3/2} = 10^{3/2}$

$\displaystyle \frac{\log 10^{3/2}}{\log 10} = \log_{10} 10^{3/2} = \frac{3}{2}$
$|(-3)(-4)| + |-3 + 4| =$
A $11$ B $-13$ C $19$ D $13$

Answer and solution

Answer: (D) $13$

$|12| + |1| = 12 + 1 = 13$
If $|x + 6| \ge 2$ , then $x \in$
A $[-8, -4]$ B $(-\infty, -8] \cup [-4, \infty)$ C $(-\infty, -8) \cup (-4, \infty)$ D $(-8, -4)$

Answer and solution

Answer: (B) $(-\infty, -8] \cup [-4, \infty)$

$x + 6 \le -2$ (or) $x + 6 \ge 2$

$\Rightarrow x \le -8$ (or) $x \ge -4$

$\Rightarrow x \in (-\infty, -8] \cup [-4, \infty)$
$|5 x - 1| = 5 x - 1$ for $x \in$
A $\displaystyle (-\infty, \frac{1}{5})$ B $\displaystyle (\frac{1}{5}, \infty)$ C $\mathbb{R}$ D $\displaystyle [\frac{1}{5}, \infty)$

Answer and solution

Answer: (D) $\displaystyle [\frac{1}{5}, \infty)$

$|T| = T \Rightarrow T \ge 0$

$\displaystyle 5 x - 1 \ge 0 \Rightarrow x \ge \frac{1}{5}$
For $x > 2$ , $|x + 3| + |x - 2| =$
A $2x + 1$ B $2x - 1$ C $5$ D $-2x - 1$

Answer and solution

Answer: (A) $2x + 1$

Here $x + 3 > 0$ and $x - 2 > 0$

$|x + 3| + |x - 2| = (x + 3) + (x - 2) = 2x + 1$
If $|x^2 - 9| + |x - 3| = 0$ , then $x \in$
A $\phi$ B $\{-3\}$ C $\{-3, 3\}$ D $\{3\}$

Answer and solution

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

Both must be $0$ : $x^2 - 9 = 0$ and $x - 3 = 0$

$x = \pm 3 \text{ and } x = 3 \Rightarrow x = 3$
The domain of $f(x) = \sqrt{x^{2} + 4}$ is
A $[2, \infty)$ B $\mathbb{R} - \{-2, 2\}$ C $[-2, 2]$ D $\mathbb{R}$

Answer and solution

Answer: (D) $\mathbb{R}$

$\sqrt{x^{2} + 4}$ is defined when $x^{2} + 4 \ge 0$

$a > 0, \Delta = -16 < 0$

Domain $= \mathbb{R}$
$[-4] + [4.9] =$
A $0$ B $0.9$ C $1$ D $-1$

Answer and solution

Answer: (A) $0$

$[-4] = -4$ , $[4.9] = 4$

$-4 + 4 = 0$
If $x = 3.2$ , then $[-x] =$
A $3$ B $-3$ C $-4$ D $4$

Answer and solution

Answer: (C) $-4$

$x \notin \mathbb{Z}$ : $[-x] = -[x] - 1 = -3 - 1 = -4$
The integral part and the fractional part of $-2.7$ are
A $-3$ and $0.3$ B $-3$ and $-0.3$ C $-2$ and $-0.7$ D $-2$ and $0.7$

Answer and solution

Answer: (A) $-3$ and $0.3$

$-2.7 = -3 + 0.3$

Integral part $-3$ , fractional part $0.3$
For $x \in \mathbb{R}$ , the set of all values of $x^2$ is
A $[0, \infty)$ B $[1, \infty)$ C $(0, \infty)$ D $\mathbb{R}$

Answer and solution

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

$x^2 \ge 0$ and takes the value $0$ at $x = 0$

Range $= [0, \infty)$

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MCQ test, Set 8