Relations & Functions – Domain, Range, Types of Functions Quiz - ICSE Class 11 Mathematics
Free practice quiz: Relations & Functions – Domain, Range, Types of Functions (ICSE Class 11 Mathematics). 54 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
Play all 54 questions freeSample questions
Let f(x) = \frac{1}{\sqrt{x^2 - 4}}. Find the domain of f(x).
- (-∞, -2) ∪ (2, ∞)
- (-∞, -2] ∪ [2, ∞)
- (-2, 2)
- [-2, 2]
Show answer
Answer: A. (-∞, -2) ∪ (2, ∞)
For f(x) to be defined, the expression under the square root must be positive, i.e., $x^2 - 4 > 0$. This inequality holds when $x^2 > 4$, which means $x < -2$ or $x > 2$. Therefore, the domain is (-∞, -2) ∪ (2, ∞).
Let $f(x) = |x-3|$. What is the range of $f(x)$?
- [0, ∞)
- (0, ∞)
- ℝ
- [3, ∞)
Show answer
Answer: A. [0, ∞)
The absolute value of any real number is always non-negative. The expression $|x-3|$ will be zero when $x=3$. For all other values of $x$, $|x-3|$ will be positive. Therefore, the range of $f(x) = |x-3|$ is all non-negative real numbers, which is the interval [0, ∞).
Determine the range of $f(x) = |x| + 2$.
- [2, ∞)
- (2, ∞)
- [0, ∞)
- ℝ
Show answer
Answer: A. [2, ∞)
The term $|x|$ is always non-negative, meaning $|x| \ge 0$. Adding 2 to $|x|$, we get $|x| + 2 \ge 0 + 2$, so $f(x) \ge 2$. The minimum value of $f(x)$ is 2, which occurs when $x=0$. As $|x|$ increases, $f(x)$ increases without bound. Thus, the range is [2, ∞).
Let $f(x) = \frac{x}{x^2-9}$. Find the domain of $f(x)$.
- ℝ \{-3, 3}
- ℝ \{3}
- ℝ \{-3}
- ℝ \{0, 3, -3}
Show answer
Answer: A. ℝ \{-3, 3}
The function $f(x) = \frac{x}{x^2-9}$ is a rational function. It is defined for all real numbers except where the denominator is zero. The denominator $x^2-9 = (x-3)(x+3)$ is zero when $x=3$ or $x=-3$. Therefore, the domain of $f(x)$ is all real numbers except 3 and -3.
If $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x}$, find the domain of $(f \circ g)(x)$.
- ℝ \{0}
- ℝ \{1}
- ℝ
- ℝ \{-1, 0, 1}
Show answer
Answer: A. ℝ \{0}
The composite function $(f \circ g)(x) = f(g(x))$. The domain of $g(x) = \frac{1}{x}$ is ℝ \{0}. Now we evaluate $f(g(x)) = f(\frac{1}{x}) = \frac{1}{\frac{1}{x}} = x$. The resulting expression is $x$, whose domain is ℝ. However, the domain of the composite function is restricted by the domain of the inner function $g(x)$. Thus, the domain of $(f \circ g)(x)$ is ℝ \{0}.