Functions and Graphs: MCQ test, Set 3

Download the PDF (with key sheet and solutions, 88.4 KB)

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 $9$ under $f(x) = 2 x - 5$ is
A $-7$ B $13$ C $7$ D $2$

Answer and solution

Answer: (C) $7$

$f(x) = 9 \Rightarrow 2 x - 5 = 9 \Rightarrow x = 7$
If $A = \{-1, 1, 2\}$ , $B = \{1, 2, 4\}$ and $f : A \to B$ is defined by $f(x) = x^{2}$ , then the range of $f$ is
A $\{-1, 1, 2\}$ B $\{1, 2, 4\}$ C $\{1, 4\}$ D $\{-4, -1\}$

Answer and solution

Answer: (C) $\{1, 4\}$

$f(-1) = (-1)^{2} = 1$ , $f(1) = (1)^{2} = 1$ , $f(2) = (2)^{2} = 4$

Range $= f(A) = \{1, 4\}$ (not the codomain $B$ )
The domain of the polynomial function $f(x) = x^3 - 4x + 1$ is
A $[0, \infty)$ B $\mathbb{R} = (-\infty, \infty)$ C $(0, \infty)$ D $\mathbb{R} - \{0\}$

Answer and solution

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

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

Domain of a polynomial function is $\mathbb{R}$
The graph of the identity function $y = x$ makes an angle of ____ with the x-axis.
A $135^\circ$ B $0^\circ$ C $90^\circ$ D $45^\circ$

Answer and solution

Answer: (D) $45^\circ$

$y = x$ passes through $(0, 0), (1, 1), (2, 2)$

It rises $1$ for every $1$ across: angle $45^\circ$
The inclination of the line $y = 14 - 9 x$ with the x-axis is
A greater than $90^\circ$ B less than $90^\circ$ C equal to $90^\circ$ D equal to $0^\circ$

Answer and solution

Answer: (A) greater than $90^\circ$

Here $a = -9 < 0$

So the inclination is greater than $90^\circ$
For $9 x^{2} - 30 x + 25 = 0$ :
A $\Delta = -930$ , roots are not real (imaginary) B $\Delta = 0$ , roots are real and equal C $\Delta = 0$ , roots are real and distinct D $\Delta = 1800$ , roots are real and distinct

Answer and solution

Answer: (B) $\Delta = 0$ , roots are real and equal

$\Delta = b^2 - 4ac = (-30)^2 - 4(9)(25) = 0$

So the roots are real and equal
If $\alpha, \beta$ are the roots of $3 x^{2} - 4 x - 7 = 0$ , then $|\alpha - \beta| =$
A $\displaystyle \frac{4}{3}$ B $10$ C $\displaystyle \frac{10}{3}$ D $\displaystyle \frac{2 \sqrt{17}}{3}$

Answer and solution

Answer: (C) $\displaystyle \frac{10}{3}$

$\Delta = b^2 - 4ac = (-4)^2 - 4(3)(-7) = 100$

$\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = \frac{10}{3}$
The vertex of $y = 3(x - 2)^2 - 5$ is
A $(2, -5)$ B $(-2, -5)$ C $(-2, 5)$ D $(2, 5)$

Answer and solution

Answer: (A) $(2, -5)$

$y = a(x - h)^2 + k$ has vertex $(h, k)$

$\text{Vertex} = (2, -5)$
The graph of $y = 4x^2 - 12x + 9$ touches the x-axis at
A $(3, 0)$ B $(0, 9)$ C $\displaystyle (- \frac{3}{2}, 0)$ D $\displaystyle (\frac{3}{2}, 0)$

Answer and solution

Answer: (D) $\displaystyle (\frac{3}{2}, 0)$

$\Delta = 0$ , perfect square: $y = (2 x - 3)^{2}$

$\displaystyle y = 0 \Rightarrow x = \frac{3}{2}$
If $x^2 + x + 3 < 0$ , then $x \in$
A $\mathbb{R}$ B $\displaystyle \{- \frac{1}{2}\}$ C $\displaystyle \mathbb{R} - \{- \frac{1}{2}\}$ D $\phi$

Answer and solution

Answer: (D) $\phi$

$a = 1 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 1^2 - 4(1)(3) = -11$

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

$\therefore x \in \phi$
If $-x^2 + 6x - 9 < 0$ , then $x \in$
A $\mathbb{R}$ B $\mathbb{R} - \{3\}$ C $\phi$ D $\{3\}$

Answer and solution

Answer: (B) $\mathbb{R} - \{3\}$

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

$\Delta = b^2 - 4ac = 6^2 - 4(-1)(-9) = 0$

$\Delta = 0 \Rightarrow y = -(x - 3)^2$

$y < 0$ for all $x \ne 3$ , and $y = 0$ at $x = 3$

$\therefore x \in \mathbb{R} - \{3\}$
If $\log_4 x = -2$ , then $x =$
A $16$ B $-16$ C $-8$ D $\displaystyle \frac{1}{16}$

Answer and solution

Answer: (D) $\displaystyle \frac{1}{16}$

$\displaystyle x = 4^{-2} = \frac{1}{16}$
$\log_3 7 \cdot \log_7 9 =$
A $9$ B $1$ C $2$ D $\log_{21} 63$

Answer and solution

Answer: (C) $2$

$\log_7 9 \cdot \log_3 7 = \log_3 9 = 2$
If $\log_{1/2} x > 0$ , then $x \in$
A $(-\infty, 1)$ B $(0, 1)$ C $(1, \infty)$ D $(0, \infty)$

Answer and solution

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

Base $\displaystyle 0 < \frac{1}{2} < 1$ : $\log_a x > 0 \Leftrightarrow 0 < x < 1$
$(1.5)^{\log_{1.5} 4 + 2} =$
A $6$ B $9$ C $16$ D $8$

Answer and solution

Answer: (B) $9$

$(1.5)^{\log_{1.5} 4} \cdot (1.5)^2 = 4 \times 2.25 = 9$
If $|x| > x$ , then
A no real $x$ B $x > 0$ C $x \ge 0$ D $x < 0$

Answer and solution

Answer: (D) $x < 0$

For $x \ge 0$ , $|x| = x$ ; for $x < 0$ , $|x| = -x > x$

So $|x| > x \Rightarrow x < 0$
If $|3 x - 2| > 4$ , then $x \in$
A $\displaystyle (-\infty, - \frac{2}{3}) \cup (2, \infty)$ B $\displaystyle (- \frac{2}{3}, 2)$ C $\displaystyle (-\infty, - \frac{2}{3}] \cup [2, \infty)$ D $\displaystyle [- \frac{2}{3}, 2]$

Answer and solution

Answer: (A) $\displaystyle (-\infty, - \frac{2}{3}) \cup (2, \infty)$

$3 x - 2 < -4$ (or) $3 x - 2 > 4$

$\displaystyle \Rightarrow x < - \frac{2}{3}$ (or) $x > 2$

$\displaystyle \Rightarrow x \in (-\infty, - \frac{2}{3}) \cup (2, \infty)$
$|3 x - 4| = 3 x - 4$ for $x \in$
A $\mathbb{R}$ B $\displaystyle (\frac{4}{3}, \infty)$ C $\displaystyle (-\infty, \frac{4}{3})$ D $\displaystyle [\frac{4}{3}, \infty)$

Answer and solution

Answer: (D) $\displaystyle [\frac{4}{3}, \infty)$

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

$\displaystyle 3 x - 4 \ge 0 \Rightarrow x \ge \frac{4}{3}$
For $1 \le x \le 4$ , $|x - 1| + |x - 4| =$
A $3$ B $2x - 5$ C $5 - 2x$ D $-3$

Answer and solution

Answer: (A) $3$

Here $x - 1 \ge 0$ and $x - 4 \le 0$

$|x - 1| + |x - 4| = (x - 1) + (4 - x) = 3$
If $x^2 - 3|x| \le 0$ , then $x \in$
A $[-3, 0]$ B $[0, 3]$ C $[-3, 3]$ D $(-3, 3)$

Answer and solution

Answer: (C) $[-3, 3]$

$x = 0$ is the critical point

Case (i) Suppose $x < 0$ : $|x| = -x$ , $x^{2} + 3 x \le 0 \Rightarrow x (x + 3) \le 0 \Rightarrow x \in [-3, 0]$ ; common values $[-3, 0)$

Case (ii) Suppose $x \ge 0$ : $|x| = x$ , $x^{2} - 3 x \le 0 \Rightarrow x (x - 3) \le 0 \Rightarrow x \in [0, 3]$ ; common values $[0, 3]$

From (i), (ii): $x \in [-3, 3]$
The domain of $\displaystyle f(x) = \frac{1}{\sqrt{2 x + 5}}$ is
A $\displaystyle (\frac{5}{2}, \infty)$ B $\displaystyle [- \frac{5}{2}, \infty)$ C $\displaystyle (-\infty, - \frac{5}{2}]$ D $\displaystyle (- \frac{5}{2}, \infty)$

Answer and solution

Answer: (D) $\displaystyle (- \frac{5}{2}, \infty)$

$\displaystyle \frac{1}{\sqrt{2 x + 5}}$ is defined when $2 x + 5 > 0$

$\Rightarrow 2 x > -5$

$\displaystyle \Rightarrow x > - \frac{5}{2}$

Domain $\displaystyle = (- \frac{5}{2}, \infty)$
$2[3.999] - [-3.999] =$
A $10$ B $2$ C $9$ D $12$

Answer and solution

Answer: (A) $10$

$3 \le 3.999 < 4$ , so $[3.999] = 3$ ; $-4 \le -3.999 < -3$ , so $[-3.999] = -4$

$2(3) - (-4) = 6 + 4 = 10$
If $[x + 3] > 6$ , then $x \in$
A $[4, \infty)$ B $(4, \infty)$ C $(3, \infty)$ D $[3, \infty)$

Answer and solution

Answer: (A) $[4, \infty)$

$[x + 3] = [x] + 3$

$[x] + 3 > 6 \Rightarrow [x] > 3 \Rightarrow [x] \ge 4$

$x \in [4, \infty)$
$\{6.4\} - \{-6.4\} =$
A $0$ B $0.2$ C $-0.2$ D $1$

Answer and solution

Answer: (C) $-0.2$

$\{6.4\} = 6.4 - 6 = 0.4, \ \{-6.4\} = -6.4 - (-7) = 0.6$

$0.4 - 0.6 = -0.2$
The range of $f(x) = x^3$ is
A $(0, \infty)$ B $\mathbb{R}$ C $\mathbb{R} - \{0\}$ D $[0, \infty)$

Answer and solution

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

$x^3$ is negative for $x < 0$ and positive for $x > 0$

Range $= \mathbb{R}$

Menu

MCQ test, Set 3