Domain & Range
Q1โ10 of 25Find the domain of the definition of the function
y = |x|.
$y = \vert{}x\vert{}$
Rule: Absolute value functions accept any real number input.
Calculation: $-\infty < x < +\infty$.
Correct Option: (b)
Find the domain of the definition of the function
y = โx.
$y = \vert{}x\vert{}$
Rule: Absolute value functions accept any real number input.
Calculation: $-\infty < x < +\infty$.
Correct Option: (b)
Find the domain of the definition of the function
y = |โx|.
$y = \vert{}x\vert{}$
Rule: Absolute value functions accept any real number input.
Calculation: $-\infty < x < +\infty$.
Correct Option: (b)
Find the domain of the definition of the function
y = (x โ 2)ยนแยฒ + (8 โ x)ยนแยฒ.
$y = (x-2)^{1/2} + (8-x)^{1/2}$
Rule: Both expressions inside square roots must be $\ge 0$.
Calculation:
$x - 2 \ge 0 \implies x \ge 2$
$8 - x \ge 0 \implies x \le 8$
Intersection: $2 \le x \le 8$
Find the domain of the definition of the function
y = (9 โ xยฒ)ยนแยฒ.
$y = (9-x^2)^{1/2}$
Rule: $9 - x^2 \ge 0$.
Calculation: $(3-x)(3+x) \ge 0 \implies -3 \le x \le 3$.
Correct Option: (a)
Find the domain of the definition of the function
y = 1/(xยฒ โ 4x + 3).
Rule: The denominator cannot be zero.
Calculation: $x^2 - 4x + 3 \ne 0 \implies (x-1)(x-3) \ne 0 \implies x \ne 1, 3$.
Correct Option: (d)
What will be the domain of the definition of the function f(x) = 8โxC5โx for positive values of x?
Rule: For combinations $^{n}\text{C}_{r}$: $n \ge 1$, $r \ge 0$, and $n \ge r$.
Calculation:
$8 - x \ge 1 \implies x \le 7$
$5 - x \ge 0 \implies x \le 5$
$(8-x) \ge (5-x) \implies 8 \ge 5$ (Always true)
Positive integers for $x \le 5$: $\{1, 2, 3, 4, 5\}$.
Correct Option: (c)
Find the domain of the definition of the function
y = 1 / (4 โ xยฒ)ยนแยฒ.
Rule: The expression under the square root in the denominator must be strictly greater than zero.
Calculation: $4 - x^2 > 0 \implies (2-x)(2+x) > 0 \implies -2 < x < 2$ or $(-2, 2)$.
Correct Option: (a)
The domain of definition of the function
\( y=\dfrac{1}{\log_{10}(1-x)}+(x+2)^{1/2} \)
Calculation:
Logarithm argument: $1 - x > 0 \implies x < 1$
Non-zero denominator: $\log_{10}(1-x) \ne 0 \implies 1 - x \ne 1 \implies x \ne 0$
Square root argument: $x + 2 \ge 0 \implies x \ge -2$
Intersection: $x \in [-2, 1)$ excluding $0$.
Correct Option: (d)
The domain of definition of
\( y=\left[\log_{10}\left(\dfrac{5x-x^2}{4}\right)\right]^{1/2} \) is
$y = \left[\log_{10}\left(\frac{5x-x^2}{4}\right)\right]^{1/2}$
Rule: For $\sqrt{\log_{10}(A)}$ to be defined, $A \ge 1$.
Calculation: $\frac{5x-x^2}{4} \ge 1 \implies 5x - x^2 \ge 4 \implies x^2 - 5x + 4 \le 0 \implies (x-1)(x-4) \le 0 \implies x \in [1, 4]$.
Correct Option: (a)