Functions and Graphs: MCQ test, Set 4

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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.

If $f(x) = 5 - 3 x$ , then the image of $2$ under $f$ is
A $1$ B $-1$ C $-11$ D $11$

Answer and solution

Answer: (B) $-1$

$f(2) = 5 - 3 (2) = -1$
If $A = \{1, 2, 3\}$ , $B = \{1, 2, 3, 4\}$ and $f : A \to B$ is defined by $f(x) = 5 - x$ , then the range of $f$ is
A $\{1, 2, 3, 4\}$ B $\{2, 3, 4\}$ C $\{1, 2, 3\}$ D $\{-4, -3, -2\}$

Answer and solution

Answer: (B) $\{2, 3, 4\}$

$f(1) = 5 - (1) = 4$ , $f(2) = 5 - (2) = 3$ , $f(3) = 5 - (3) = 2$

Range $= f(A) = \{2, 3, 4\}$ (not the codomain $B$ )
The degree of the polynomial function $f(x) = 8x^2 + 5x^4 - x + 6$ is
A $4$ B $2$ C $8$ D $3$

Answer and solution

Answer: (A) $4$

Degree = highest power of $x$ with non-zero coefficient $= 4$
Which point lies on the graph of the identity function?
A $(0, 4)$ B $(4, 0)$ C $(-4, -4)$ D $(-4, 4)$

Answer and solution

Answer: (C) $(-4, -4)$

On $y = x$ both coordinates are equal: $(-4, -4)$
The line $y = 7 x - 14$ cuts the x-axis at
A $(0, -14)$ B $(-2, 0)$ C $(14, 0)$ D $(2, 0)$

Answer and solution

Answer: (D) $(2, 0)$

put $y = 0$ : $7 x - 14 = 0 \Rightarrow x = 2$
The discriminant of $2 x^{2} + 5 x - 3 = 0$ is
A $1$ B $29$ C $49$ D $-1$

Answer and solution

Answer: (C) $49$

$\Delta = b^2 - 4ac = 5^2 - 4(2)(-3) = 49$
If $\alpha, \beta$ are the roots of $2 x^{2} - 9 x + 5 = 0$ , then $\alpha + \beta =$
A $\displaystyle \frac{9}{2}$ B $\displaystyle - \frac{5}{2}$ C $\displaystyle \frac{5}{2}$ D $\displaystyle - \frac{9}{2}$

Answer and solution

Answer: (A) $\displaystyle \frac{9}{2}$

$a = 2, b = -9$

$\displaystyle \alpha + \beta = -\frac{b}{a} = \frac{9}{2}$
The range of $f(x) = x^{2} - 4 x + 1$ , $x \in \mathbb{R}$ , is
A $[-3, \infty)$ B $(-\infty, -3]$ C $[2, \infty)$ D $[1, \infty)$

Answer and solution

Answer: (A) $[-3, \infty)$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(2, \frac{4 - 16}{4}\right) = (2, -3)$

Range $= [-3, \infty)$
The lowest point of the graph of $y = (x - 3)(x - 11)$ is
A $(-7, -16)$ B $(7, 33)$ C $(7, 16)$ D $(7, -16)$

Answer and solution

Answer: (D) $(7, -16)$

It cuts the x-axis at $x = 3, 11$ ; lowest point at $\displaystyle x = \frac{\alpha + \beta}{2} = 7$

$y = (7 - 3)(7 - 11) = -16$
If $-2x^2 + 3x - 5 < 0$ , then $x \in$
A $\displaystyle \mathbb{R} - \{\frac{3}{4}\}$ B $\phi$ C $\displaystyle \{\frac{3}{4}\}$ D $\mathbb{R}$

Answer and solution

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

$a = -2 < 0$ (downward parabola)

$\Delta = b^2 - 4ac = 3^2 - 4(-2)(-5) = -31$

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

$\therefore x \in \mathbb{R}$
If $25x^2 - 10x + 1 > 0$ , then $x \in$
A $\phi$ B $\mathbb{R}$ C $\displaystyle \mathbb{R} - \{\frac{1}{5}\}$ D $\displaystyle \{\frac{1}{5}\}$

Answer and solution

Answer: (C) $\displaystyle \mathbb{R} - \{\frac{1}{5}\}$

$a = 25 > 0$ (upward parabola)

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

$\displaystyle \Delta = 0 \Rightarrow y = 25(x - \frac{1}{5})^2$

$y > 0$ for all $\displaystyle x \ne \frac{1}{5}$ , and $y = 0$ at $\displaystyle x = \frac{1}{5}$

$\displaystyle \therefore x \in \mathbb{R} - \{\frac{1}{5}\}$
If $\log_5 (x + 2) = 2$ , then $x =$
A $10$ B $27$ C $23$ D $8$

Answer and solution

Answer: (C) $23$

$x + 2 = 5^2 = 25 \Rightarrow x = 23$
$\displaystyle \frac{1}{\log_4 2} =$
A $\displaystyle \frac{1}{2}$ B $2$ C $4$ D $-2$

Answer and solution

Answer: (B) $2$

$\displaystyle \frac{1}{\log_4 2} = \log_2 4 = 2$
Which of the following is negative?
A $\log_5 1.2$ B $\log_{1/4} 9$ C $\log_{0.2} 0.5$ D $\log_3 7$

Answer and solution

Answer: (B) $\log_{1/4} 9$

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

$\log_{1/4} 9$ : base $< 1$ , $9 > 1$ : negative
$\log_7 2 \cdot \log_{13} 7 \cdot \log_8 13 =$
A $\displaystyle \frac{2}{3}$ B $3$ C $\displaystyle \frac{1}{3}$ D $\displaystyle \frac{1}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{3}$

$\displaystyle = \log_8 2 = \log_{2^3} 2 = \frac{1}{3}$
If $|x| < x$ , then $x \in$
A $\phi$ B $(-\infty, 0)$ C $\mathbb{R}$ D $(0, \infty)$

Answer and solution

Answer: (A) $\phi$

$|x| \ge x$ for every real $x$

So $|x| < x$ has no solution: $x \in \phi$
If $|x| \ge 6$ , then $x \in$
A $(-6, 6)$ B $(-\infty, -6) \cup (6, \infty)$ C $(-\infty, -6] \cup [6, \infty)$ D $[-6, 6]$

Answer and solution

Answer: (C) $(-\infty, -6] \cup [6, \infty)$

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

$\Rightarrow x \in (-\infty, -6] \cup [6, \infty)$
If $|x - 2| = |2 x + 1|$ , then $x \in$
A $\displaystyle \{- \frac{1}{3}, 3\}$ B $\displaystyle \{-3, \frac{1}{3}\}$ C $\displaystyle \{\frac{1}{3}\}$ D $\{-3\}$

Answer and solution

Answer: (B) $\displaystyle \{-3, \frac{1}{3}\}$

$x - 2 = \pm (2 x + 1)$

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

$\displaystyle x - 2 = - 2 x - 1 \Rightarrow 3 x - 1 = 0 \Rightarrow x = \frac{1}{3}$
The solution set of $|2 x - 1| + x = 5$ is
A $\{-4, 2\}$ B $\{2\}$ C $\{-2, 4\}$ D $\{-4\}$

Answer and solution

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

Case (i) Suppose $\displaystyle x < \frac{1}{2}$ ( $2 x - 1 < 0$ ): $(1 - 2 x) + x = 5 \Rightarrow 1 - x = 5$ $\Rightarrow$ $x = -4$ , which is in this case

Case (ii) Suppose $\displaystyle x \ge \frac{1}{2}$ ( $2 x - 1 \ge 0$ ): $(2 x - 1) + x = 5 \Rightarrow 3 x - 1 = 5$ $\Rightarrow$ $x = 2$ , which is in this case

So $x \in \{-4, 2\}$
If $|x^{2} + 2 x| < 8$ , then $x \in$
A $(-2, 4)$ B $[-4, 2]$ C $\mathbb{R}$ D $(-4, 2)$

Answer and solution

Answer: (D) $(-4, 2)$

$-8 < x^{2} + 2 x < 8$

$\Rightarrow x^{2} + 2 x + 8 > 0 \text{ and } x^{2} + 2 x - 8 < 0$

$x^{2} + 2 x + 8 > 0$ : $a > 0, \Delta = -28 < 0 \Rightarrow x \in \mathbb{R}$

$x^{2} + 2 x - 8 < 0$ : $\displaystyle \Delta = 36, \ x = \frac{-2 \pm 6}{2} = -4, 2 \Rightarrow x \in (-4, 2)$

Common values: $x \in (-4, 2)$
The domain of $f(x) = \sqrt{x^{2} - 1}$ is
A $[-1, 1]$ B $(-\infty, -1] \cup [1, \infty)$ C $[1, \infty)$ D $(-\infty, -1) \cup (1, \infty)$

Answer and solution

Answer: (B) $(-\infty, -1] \cup [1, \infty)$

$\sqrt{x^{2} - 1}$ is defined when $x^{2} - 1 \ge 0$

$\Rightarrow (x - 1)(x + 1) \ge 0$

Domain $= (-\infty, -1] \cup [1, \infty)$
The range of the step function $f(x) = [x]$ is
A $\mathbb{N}$ B $\mathbb{R}$ C $[0, \infty)$ D $\mathbb{Z}$

Answer and solution

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

$[x]$ is always an integer and every integer $n$ is $[n]$

Range $= \mathbb{Z}$
If $[x - 1] < -3$ , then $x \in$
A $(-\infty, -1)$ B $(-\infty, -2]$ C $(-\infty, -3)$ D $(-\infty, -2)$

Answer and solution

Answer: (D) $(-\infty, -2)$

$[x - 1] = [x] - 1$

$[x] - 1 < -3 \Rightarrow [x] < -2 \Rightarrow [x] \le -3$

$[x] = -3$ for $-3 \le x < -2$ : $x \in (-\infty, -2)$
$[-1.2] + \{-1.2\} =$
A $-0.8$ B $1.2$ C $-1.2$ D $-2.8$

Answer and solution

Answer: (C) $-1.2$

$[-1.2] = -2, \ \{-1.2\} = -1.2 - (-2) = 0.8$

$-2 + 0.8 = -1.2$

(always $[x] + \{x\} = x$ )
The domain of $f(x) = \log_2 x$ is
A $(0, \infty)$ B $(1, \infty)$ C $\mathbb{R}$ D $[0, \infty)$

Answer and solution

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

$\log_a N$ needs $N > 0$

Domain $= (0, \infty)$

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MCQ test, Set 4