Functions and Graphs: MCQ test, Set 6

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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 $-8$ under $f(x) = 3 x + 4$ is
A $-4$ B $-20$ C $\displaystyle - \frac{4}{3}$ D $4$

Answer and solution

Answer: (A) $-4$

$f(x) = -8 \Rightarrow 3 x + 4 = -8 \Rightarrow x = -4$
If $A = \{0, 1, 2\}$ , $B = \{2, 5, 8, 11\}$ and $f : A \to B$ is defined by $f(x) = 3 x + 2$ , then the range of $f$ is
A $\{2, 5, 8\}$ B $\{2, 5, 8, 11\}$ C $\{0, 1, 2\}$ D $\{-2, 1, 4\}$

Answer and solution

Answer: (A) $\{2, 5, 8\}$

$f(0) = 3 (0) + 2 = 2$ , $f(1) = 3 (1) + 2 = 5$ , $f(2) = 3 (2) + 2 = 8$

Range $= f(A) = \{2, 5, 8\}$ (not the codomain $B$ )
The degree of the polynomial function $f(x) = x - 4x^3 + 9$ is
A $1$ B $2$ C $3$ D $9$

Answer and solution

Answer: (C) $3$

Degree = highest power of $x$ with non-zero coefficient $= 3$
The identity function on $A = \{a, b, c\}$ is
A $\{(a, a), (b, b), (c, c)\}$ B $\{(a, b), (b, c), (c, a)\}$ C $\{(a, c), (b, b), (c, a)\}$ D $\{(a, a)\}$

Answer and solution

Answer: (A) $\{(a, a), (b, b), (c, c)\}$

$I_A(x) = x$ for every $x \in A$

So $I_A = \{(a, a), (b, b), (c, c)\}$
The inclination of the line $y = 11 x - 3$ 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: (B) less than $90^\circ$

Here $a = 11 > 0$

So the inclination is less than $90^\circ$
For $x^{2} + 5 x + 7 = 0$ :
A $\Delta = -23$ , roots are not real (imaginary) B $\Delta = -3$ , roots are real and distinct C $\Delta = -3$ , roots are not real (imaginary) D $\Delta = 53$ , roots are real and distinct

Answer and solution

Answer: (C) $\Delta = -3$ , roots are not real (imaginary)

$\Delta = b^2 - 4ac = 5^2 - 4(1)(7) = -3$

So the roots are not real (imaginary)
$x^{2} - x - 12 =$
A $(x - 3)(x - 4)$ B $(x + 3)(x + 4)$ C $(x - 3)(x + 4)$ D $(x + 3)(x - 4)$

Answer and solution

Answer: (D) $(x + 3)(x - 4)$

$\displaystyle x = \frac{1 \pm 7}{2} = -3, 4$

$x^{2} - x - 12 = (x - \alpha)(x - \beta) = (x + 3)(x - 4)$
The maximum value of $- x^{2} + 2 x + 4$ is
A $3$ B $4$ C $1$ D $5$

Answer and solution

Answer: (D) $5$

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

$a < 0$ : maximum value $= 5$
The graph of $y = (x + 4)(x - 3)$ cuts the x-axis at
A $(-4, 0)$ and $(-3, 0)$ B $(-4, 0)$ and $(3, 0)$ C $(4, 0)$ and $(-3, 0)$ D $(4, 0)$ and $(3, 0)$

Answer and solution

Answer: (B) $(-4, 0)$ and $(3, 0)$

put $y = 0$ : $(x + 4)(x - 3) = 0 \Rightarrow x = -4, 3$
If $(5 - x)(x - 1) \le 0$ , then $x \in$
A $(-\infty, 1] \cup [5, \infty)$ B $(1, 5)$ C $(-\infty, 1) \cup (5, \infty)$ D $[1, 5]$

Answer and solution

Answer: (A) $(-\infty, 1] \cup [5, \infty)$

$y = (5 - x)(x - 1) = - x^{2} + 6 x - 5$

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

$\Delta = b^2 - 4ac = 6^2 - 4(-1)(-5) = 16$

$\displaystyle y = 0 \Rightarrow x = \frac{-6 \pm 4}{-2} = 1, 5$

Sign: $-$ outside $1$ and $5$ , $+$ between them

$\therefore x \in (-\infty, 1] \cup [5, \infty)$
If $x^2 - 9 < 0$ , then $x \in$
A $(-\infty, -3] \cup [3, \infty)$ B $[-3, 3]$ C $(-3, 3)$ D $(-\infty, -3) \cup (3, \infty)$

Answer and solution

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

$a = 1 > 0$ (upward parabola)

$\Delta = b^2 - 4ac = 0^2 - 4(1)(-9) = 36$

$y = 0 \Rightarrow x = \pm \sqrt{9} = -3, 3$

Sign: $+$ outside $-3$ and $3$ , $-$ between them

$\therefore x \in (-3, 3)$
$2^7 = 128$ in logarithmic form is
A $\log_7 128 = 2$ B $\log_2 7 = 128$ C $\log_2 128 = 7$ D $\log_{128} 2 = 7$

Answer and solution

Answer: (C) $\log_2 128 = 7$

$a^x = N \Leftrightarrow x = \log_a N$

Base $2$ , power $7$ : $\log_2 128 = 7$
$\log_3 54 - \log_3 2 =$
A $3$ B $1$ C $27$ D $\log_3 52$

Answer and solution

Answer: (A) $3$

$\displaystyle \log_3 \frac{54}{2} = \log_3 27 = 3$
If $\log_3 (x + 1) \ge 0$ , then $x \in$
A $[0, \infty)$ B $(0, \infty)$ C $(-1, 0]$ D $[-1, \infty)$

Answer and solution

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

Base $3 > 1$ : $\log_a T \ge 0 \Leftrightarrow T \ge 1$

$x + 1 \ge 1 \Rightarrow x \ge 0$
If $\log_2 (x^2 + 3x) - \log_2 (x + 3) = 4$ , then $x =$
A $13$ B $4$ C $8$ D $16$

Answer and solution

Answer: (D) $16$

$\displaystyle \log_2 \frac{x(x + 3)}{x + 3} = 4 \Rightarrow \log_2 x = 4 \Rightarrow x = 2^4 = 16$
If $x = 5$ and $y = -8$ , then $|x + y| =$
A $13$ B $-3$ C $40$ D $3$

Answer and solution

Answer: (D) $3$

$|5 + (-8)| = |-3| = 3$
If $|1 - x| > 4$ , then $x \in$
A $(-3, 5)$ B $(-\infty, -3] \cup [5, \infty)$ C $(-\infty, -3) \cup (5, \infty)$ D $[-3, 5]$

Answer and solution

Answer: (C) $(-\infty, -3) \cup (5, \infty)$

$1 - x < -4$ (or) $1 - x > 4$

$\Rightarrow x < -3$ (or) $x > 5$ (dividing by a negative number turns the signs)

$\Rightarrow x \in (-\infty, -3) \cup (5, \infty)$
If $|x^{2} - 1| = 3$ , then $x \in$
A $\{-2, - \sqrt{2}, \sqrt{2}, 2\}$ B $\{-2, 2\}$ C $\{4\}$ D $\{2\}$

Answer and solution

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

$x^{2} - 1 = \pm 3$

$x^{2} - 1 = 3 \Rightarrow x^{2} - 4 = 0 \Rightarrow x = \pm \sqrt{4} = -2, 2$

$x^{2} - 1 = -3 \Rightarrow x^{2} + 2 = 0$ , $\Delta = -8 < 0$ : no real $x$
The solution set of $|x + 1| + |x - 3| = 4$ is
A $\phi$ B $\{3\}$ C $[-1, 3]$ D $\mathbb{R}$

Answer and solution

Answer: (C) $[-1, 3]$

Case (i) Suppose $x < -1$ ( $x + 1 < 0, x - 3 < 0$ ): $(- x - 1) + (3 - x) = 4 \Rightarrow 2 - 2 x = 4$ $\Rightarrow$ $x = -1$ , not in this case: no solution

Case (ii) Suppose $-1 \le x \le 3$ ( $x + 1 \ge 0, x - 3 \le 0$ ): $(x + 1) + (3 - x) = 4 \Rightarrow 4 = 4$ : true for every $x$ in this case

Case (iii) Suppose $x > 3$ ( $x + 1 > 0, x - 3 > 0$ ): $(x + 1) + (x - 3) = 4 \Rightarrow 2 x - 2 = 4$ $\Rightarrow$ $x = 3$ , not in this case: no solution

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

Answer and solution

Answer: (B) $(-\infty, -3) \cup (-1, 1) \cup (3, \infty)$

$x = 0$ is the critical point

Case (i) Suppose $x < 0$ : $|x| = -x$ , $x^{2} + 4 x + 3 > 0 \Rightarrow (x + 1)(x + 3) > 0 \Rightarrow x \in (-\infty, -3) \cup (-1, \infty)$ ; common values $(-\infty, -3) \cup (-1, 0)$

Case (ii) Suppose $x \ge 0$ : $|x| = x$ , $x^{2} - 4 x + 3 > 0 \Rightarrow (x - 3)(x - 1) > 0 \Rightarrow x \in (-\infty, 1) \cup (3, \infty)$ ; common values $[0, 1) \cup (3, \infty)$

From (i), (ii): $x \in (-\infty, -3) \cup (-1, 1) \cup (3, \infty)$
The domain of $\displaystyle f(x) = \frac{1}{\sqrt{x^{2} - 5 x + 4}}$ is
A $(-\infty, 1] \cup [4, \infty)$ B $(-\infty, 1) \cup (4, \infty)$ C $(-\infty, -4) \cup (-1, \infty)$ D $(1, 4)$

Answer and solution

Answer: (B) $(-\infty, 1) \cup (4, \infty)$

$\displaystyle \frac{1}{\sqrt{x^{2} - 5 x + 4}}$ is defined when $x^{2} - 5 x + 4 > 0$

$\Rightarrow (x - 4)(x - 1) > 0$

Domain $= (-\infty, 1) \cup (4, \infty)$
$[7.1] - [-7.1] =$
A $0$ B $-1$ C $14$ D $15$

Answer and solution

Answer: (D) $15$

$[7.1] = 7$ , $[-7.1] = -8$

$7 - (-8) = 15$
The domain of $f(x) = \sqrt{x - [x]}$ is
A $[0, \infty)$ B $\mathbb{R}$ C $\mathbb{Z}$ D $\phi$

Answer and solution

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

Need $x - [x] \ge 0$ , i.e. $[x] \le x$

This is true for every real $x$ : domain $= \mathbb{R}$
$\{3\} + \{-3\} =$
A $-6$ B $1$ C $0$ D $6$

Answer and solution

Answer: (C) $0$

$\{3\} = 3 - 3 = 0, \ \{-3\} = -3 - (-3) = 0$
The range of $\displaystyle f(x) = x + \frac{1}{x}$ is
A $\mathbb{R} - \{0\}$ B $(-2, 2)$ C $[2, \infty)$ D $(-\infty, -2] \cup [2, \infty)$

Answer and solution

Answer: (D) $(-\infty, -2] \cup [2, \infty)$

Minimum $2$ at $x = 1$ (for $x > 0$ ), maximum $-2$ at $x = -1$ (for $x < 0$ )

Range $= (-\infty, -2] \cup [2, \infty)$

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MCQ test, Set 6