Functions and Graphs: MCQ test, Set 9

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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.

Let $A = \{-3, -2, -1, 0, 1, 2\}$ and $f(x) = x^{2} + x$ for $x \in A$ . The pre-images of $2$ under $f$ are
A $\{6\}$ B $\{-2, 1\}$ C $\{1\}$ D $\{-2\}$

Answer and solution

Answer: (B) $\{-2, 1\}$

Pre-images of $2$ are the elements of $A$ whose image is $2$

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

Answer and solution

Answer: (C) $\{0, 2\}$

$f(-1) = (-1)^{2} + (-1) = 0$ , $f(0) = (0)^{2} + (0) = 0$ , $f(1) = (1)^{2} + (1) = 2$

Range $= f(A) = \{0, 2\}$ (not the codomain $B$ )
The degree of the polynomial function $f(x) = 2x^6 - 3x^2 + x^7 - 5$ is
A $6$ B $5$ C $4$ D $7$

Answer and solution

Answer: (D) $7$

Degree = highest power of $x$ with non-zero coefficient $= 7$
If $f$ is a constant function on $\mathbb{R}$ with $f(0) = 4$ , then $f(1) + f(2) + f(3) =$
A $6$ B $4$ C $12$ D $24$

Answer and solution

Answer: (C) $12$

$f(x) = 4$ for every $x$

$f(1) + f(2) + f(3) = 4 + 4 + 4 = 12$
The inclination of the line $y = 13 - 11 x$ with the x-axis is
A equal to $90^\circ$ B equal to $0^\circ$ C less than $90^\circ$ D greater than $90^\circ$

Answer and solution

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

Here $a = -11 < 0$

So the inclination is greater than $90^\circ$
For $2 x^{2} - 3 x - 5 = 0$ :
A $\Delta = 49$ , roots are real and distinct B $\Delta = -31$ , roots are not real (imaginary) C $\Delta = 37$ , roots are real and distinct D $\Delta = 49$ , roots are real and equal

Answer and solution

Answer: (A) $\Delta = 49$ , roots are real and distinct

$\Delta = b^2 - 4ac = (-3)^2 - 4(2)(-5) = 49$

So the roots are real and distinct
If $\alpha, \beta$ are the roots of $x^{2} + 6 x - 3 = 0$ , then $|\alpha - \beta| =$
A $2 \sqrt{6}$ B $4 \sqrt{3}$ C $2 \sqrt{3}$ D $6$

Answer and solution

Answer: (B) $4 \sqrt{3}$

$\Delta = b^2 - 4ac = 6^2 - 4(1)(-3) = 48$

$\displaystyle |\alpha - \beta| = \frac{\sqrt{\Delta}}{|a|} = 4 \sqrt{3}$
The vertex of $y = - x^{2} - 6 x + 2$ is
A $(-3, 11)$ B $(3, 2)$ C $(3, 11)$ D $(-3, -11)$

Answer and solution

Answer: (A) $(-3, 11)$

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, \frac{4ac - b^2}{4a}\right) = \left(-3, \frac{-8 - 36}{-4}\right) = (-3, 11)$
The lowest point of the graph of $y = (x + 1)(x - 5)$ is
A $(-2, -9)$ B $(2, -9)$ C $(2, 9)$ D $(2, -5)$

Answer and solution

Answer: (B) $(2, -9)$

It cuts the x-axis at $x = -1, 5$ ; lowest point at $\displaystyle x = \frac{\alpha + \beta}{2} = 2$

$y = (2 + 1)(2 - 5) = -9$
If $(x + 2)^2 - 7 < 0$ , then $x \in$
A $[-2 - \sqrt{7}, -2 + \sqrt{7}]$ B $(-\infty, -2 - \sqrt{7}] \cup [-2 + \sqrt{7}, \infty)$ C $(-\infty, -2 - \sqrt{7}) \cup (-2 + \sqrt{7}, \infty)$ D $(-2 - \sqrt{7}, -2 + \sqrt{7})$

Answer and solution

Answer: (D) $(-2 - \sqrt{7}, -2 + \sqrt{7})$

$y = (x + 2)^2 - 7 = x^{2} + 4 x - 3$

$a = 1 > 0$ (upward parabola)

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

$\displaystyle y = 0 \Rightarrow x = \frac{-4 \pm \sqrt{28}}{2} = -2 \pm \sqrt{7}$

Sign: $+$ outside $-2 - \sqrt{7}$ and $-2 + \sqrt{7}$ , $-$ between them

$\therefore x \in (-2 - \sqrt{7}, -2 + \sqrt{7})$
If $-x^2 + x + 6 \ge 0$ , then $x \in$
A $(-2, 3)$ B $[-2, 3]$ C $(-\infty, -2] \cup [3, \infty)$ D $(-\infty, -2) \cup (3, \infty)$

Answer and solution

Answer: (B) $[-2, 3]$

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

$\Delta = b^2 - 4ac = 1^2 - 4(-1)(6) = 25$

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

Sign: $-$ outside $-2$ and $3$ , $+$ between them

$\therefore x \in [-2, 3]$
If $\log_x 125 = 3$ , then $x =$
A $5$ B $\displaystyle \frac{125}{3}$ C $375$ D $25$

Answer and solution

Answer: (A) $5$

$x^3 = 125 = 5^3 \Rightarrow x = 5$
$\log_5 9 \cdot \log_3 25 =$
A $2$ B $1$ C $4$ D $225$

Answer and solution

Answer: (C) $4$

$\log_5 9 = 2 \log_5 3, \ \log_3 25 = 2 \log_3 5$

$4 \log_5 3 \cdot \log_3 5 = 4 \times 1 = 4$
The graph of $y = \log_{1/2} x$ passes through
A $\displaystyle \left(\frac{1}{4}, 2\right)$ B $(2, 1)$ C $(4, 2)$ D $\displaystyle \left(\frac{1}{2}, -1\right)$

Answer and solution

Answer: (A) $\displaystyle \left(\frac{1}{4}, 2\right)$

$\displaystyle \log_{1/2} \frac{1}{4} = \log_{2^{-1}} 2^{-2} = 2$

So $\displaystyle (\frac{1}{4}, 2)$ is on the graph (each option checked as $y - \log_{1/2} x$ )
$6^{\log_{36} 25} =$
A $\displaystyle \frac{1}{5}$ B $6$ C $25$ D $5$

Answer and solution

Answer: (D) $5$

$6^{\log_{6^2} 5^2} = 6^{\log_6 5} = 5$
$\displaystyle \frac{|-18|}{|-6|} =$
A $12$ B $\displaystyle \frac{1}{3}$ C $-3$ D $3$

Answer and solution

Answer: (D) $3$

$\displaystyle \frac{18}{6} = 3$
If $|5 x - 2| < 8$ , then $x \in$
A $\displaystyle [- \frac{6}{5}, 2]$ B $\displaystyle (- \frac{6}{5}, 2)$ C $\displaystyle (-\infty, - \frac{6}{5}) \cup (2, \infty)$ D $\displaystyle (-\infty, - \frac{6}{5}] \cup [2, \infty)$

Answer and solution

Answer: (B) $\displaystyle (- \frac{6}{5}, 2)$

$-8 < 5 x - 2 < 8$

$\Rightarrow -8 + 2 < 5 x < 8 + 2$

$\displaystyle \Rightarrow - \frac{6}{5} < x < 2$

$\displaystyle \Rightarrow x \in (- \frac{6}{5}, 2)$
If $|x^{2} + 2 x| = 3$ , then $x \in$
A $\{-3, 1\}$ B $\{-1, 3\}$ C $\{1\}$ D $\{-3, -1, 1, 3\}$

Answer and solution

Answer: (A) $\{-3, 1\}$

$x^{2} + 2 x = \pm 3$

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

$x^{2} + 2 x = -3 \Rightarrow x^{2} + 2 x + 3 = 0$ , $\Delta = -8 < 0$ : no real $x$
The critical points for solving $|2 x + 7| + |4 - x| = 10$ are
A $\displaystyle \{- \frac{7}{2}, 4, 10\}$ B $\displaystyle \{-4, \frac{7}{2}\}$ C $\displaystyle \{- \frac{7}{2}, 4\}$ D $\displaystyle \{\frac{7}{2}, 4\}$

Answer and solution

Answer: (C) $\displaystyle \{- \frac{7}{2}, 4\}$

Critical points: where each expression inside $|\,|$ is $0$

$\displaystyle 2 x + 7 = 0 \Rightarrow x = - \frac{7}{2}; \ 4 - x = 0 \Rightarrow x = 4$
If $\displaystyle \left|\frac{8 x}{x^{2} + 16}\right| \le 1$ , then $x \in$
A $[-4, 4]$ B $\mathbb{R}$ C $\mathbb{R} - \{0\}$ D $\phi$

Answer and solution

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

$\displaystyle \frac{8|x|}{x^{2} + 16} \le 1$ (the denominator is positive)

(i) $x < 0$ : $-8 x \le x^{2} + 16 \Rightarrow x^{2} + 8 x + 16 \ge 0 \Rightarrow (x + 4)^2 \ge 0$ : true for all $x < 0$

(ii) $x \ge 0$ : $8 x \le x^{2} + 16 \Rightarrow x^{2} - 8 x + 16 \ge 0 \Rightarrow (x - 4)^2 \ge 0$ : true for all $x \ge 0$

From (i), (ii): $x \in (-\infty, 0) \cup [0, \infty) = \mathbb{R}$
The range of $f(x) = \sqrt{x}$ is
A $[1, \infty)$ B $(0, \infty)$ C $\mathbb{R}$ D $[0, \infty)$

Answer and solution

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

$\sqrt{x}$ is defined for $x \ge 0$ and $\sqrt{x} \ge 0$

Range $= [0, \infty)$
$[-8.6] + [-0.6] =$
A $-9$ B $-10$ C $-8$ D $-9.2$

Answer and solution

Answer: (B) $-10$

$-9 \le -8.6 < -8$ , so $[-8.6] = -9$ ; $[-0.6] = -1$

$-9 + (-1) = -10$
The domain of $\displaystyle f(x) = \frac{1}{\sqrt{[x] - x}}$ is
A $\phi$ B $\mathbb{R}$ C $\mathbb{Z}$ D $\mathbb{R} - \mathbb{Z}$

Answer and solution

Answer: (A) $\phi$

Need $[x] - x > 0$ , i.e. $[x] > x$

$[x] > x$ for no real $x$ : domain $= \phi$
If $x - [x] = 0$ , then $x$ is
A a negative number B a positive number C an integer D between $0$ and $1$

Answer and solution

Answer: (C) an integer

$x - [x] = 0 \Rightarrow [x] = x$ , which holds exactly for $x \in \mathbb{Z}$
The graph of $y = 5^x$ cuts the y-axis at
A $(0, 5)$ B $(1, 0)$ C $(0, 1)$ D $(5, 0)$

Answer and solution

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

put $x = 0$ : $y = 5^0 = 1$

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MCQ test, Set 9