If f(x) = √x, g(x) = 2x - 3, then domain of fog(x) is : Option 1: [3/2, ∞) Option 2: [1/2, ∞) Option 3: (3/2, ∞) Option 4: [5/2, ∞)

Answer: [3/2, ∞). Solution: f g(x)=f(2x-3)=√(2x-3). Domain requires 2x-3≥ 0, i.e., x≥ 3/2. Domain =[3/2,∞).

Algebra

Relations & Functions

Easy

core

If $f(x) = \sqrt{x}$ , $g(x) = 2x - 3$ , then domain of $fog(x)$ is :

$\left[\frac{3}{2}, \infty\right)$

$\left[\frac{1}{2}, \infty\right)$

$\left(\frac{3}{2}, \infty\right)$

$\left[\frac{5}{2}, \infty\right)$

✅ Correct Option: 1

$f\circ g(x)=f(2x-3)=\sqrt{2x-3}$ .

Domain requires $2x-3\geq 0$ , i.e., $x\geq 3/2$ .

Domain $=[3/2,\infty)$ .

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Let S be the set of all real numbers and let R be a relation on S, defined by $aRb \Leftrightarrow |a-b| \leq 1$ . Then, R is

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Let A = {1, 2, 3}. The number of equivalence relations containing (1, 3) is

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If a relation R is defined on the set $X = \{1, 2, 3, 4\}$ as $R = \{(1,1), (2,2), (3,4), (4,3)\}$ , then R is

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A relation R on the set $A = \{1, 2, 3, \ldots, 13, 14\}$ defined as $R = \{(x,y): 3x - y = 0\}$ is

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Match the LIST-I with LIST-II

[Where R and Z denotes set of real numbers and set of integers respectively]

	LIST-I Function		LIST-II Nature
A.	$f: R \to R, f(x) = \sin^2 x + \cos^2 x$	I.	Bijective
B.	$f: Z \to Z, f(x) = x^3$	II.	Surjective but not injective
C.	$f: R \to R, f(x) = x^3 - x$	III.	Injective but not surjective
D.	$f: R \to R, f(x) = 3 - 4x$	IV.	Neither injective nor surjective

Choose the correct answer from the options given below:

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The function $f: [-1, 1] \rightarrow R$ (set of real numbers) given by $f(x) = \frac{x}{x+3}$ is