The domain of the function y = ^(-1)(x-1) + ^(-1)√(x-1) is: Option 1: [-1, 1] Option 2: [0, 1] Option 3: [1, 2] Option 4: [0, 2]

Answer: [1, 2]. Solution: For the function y = ^(-1)(x-1) + ^(-1)√(x-1) to be defined, all parts must work. The function ^(-1) only accepts inputs between -1 and 1 (inclusive). For ^(-1)(x-1): -1 ≤ (x-1) ≤ 1 -1 + 1 ≤ x - 1 + 1 ≤ 1 + 1 0 ≤ x ≤ 2 From the first term: x [0, 2] For ^(-1)√(x-1), there are two conditions: The square root must be real (non-negative radicand): x - 1 ≥ 0 x ≥ 1 The function ^(-1) only accepts inputs between -1 and 1. Since square roots are always non-negative: 0 ≤ √(x-1) ≤ 1 √(x-1) ≤ 1 x - 1 ≤ 1 x ≤ 2 Combining both conditions: 1 ≤ x ≤ 2 From the second term: x [1, 2] For the entire function to be defined: First term requires: [0, 2] Second term requires: [1, 2] The intersection is: [1, 2] Therefore, the domain is [1, 2].

Algebra

Relations & Functions

Medium

core

The domain of the function $y = \sin^{-1}(x-1) + \cos^{-1}\sqrt{x-1}$ is:

$[-1, 1]$

$[0, 1]$

$[1, 2]$

$[0, 2]$

✅ Correct Option: 3

For the function $y = \sin^{-1}(x-1) + \cos^{-1}\sqrt{x-1}$ to be defined, all parts must work.

The function $\sin^{-1}$ only accepts inputs between $-1$ and $1$ (inclusive).

For $\sin^{-1}(x-1)$ :

$-1 \leq (x-1) \leq 1$

$-1 + 1 \leq x - 1 + 1 \leq 1 + 1$

$0 \leq x \leq 2$

From the first term: $x \in [0, 2]$

For $\cos^{-1}\sqrt{x-1}$ , there are two conditions:

The square root must be real (non-negative radicand):

$x - 1 \geq 0$

$x \geq 1$

The function $\cos^{-1}$ only accepts inputs between $-1$ and $1$ .

Since square roots are always non-negative:

$0 \leq \sqrt{x-1} \leq 1$

$\sqrt{x-1} \leq 1$

$x - 1 \leq 1$

$x \leq 2$

Combining both conditions: $1 \leq x \leq 2$

From the second term: $x \in [1, 2]$

For the entire function to be defined:

First term requires: $[0, 2]$

Second term requires: $[1, 2]$

The intersection is: $[1, 2]$

Therefore, the domain is $[1, 2]$ .

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The graph given below represents which of the following function?

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Let $R : \mathbb{R} \to \mathbb{R}$ , where $\mathbb{R}$ is the set of real numbers and R be a relation. An element $(x, y) \in R$ if $x + y - \sqrt{2}$ belongs to the set of irrational numbers. Then the relation R is :

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If $f:\mathbb{R} \to [-5, \infty)$ is defined as $f(x) = x^2 - 5$ , then the function f is

A. one-one

B. many-one

C. onto

D. into

Which of the above statements are true?

Choose the correct answer from the options given below:

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Given the relation $R=\{(1,2),(2,3)\}$ on a set $A=\{1,2,3\}$ , then which of following statements are TRUE?

A. Minimum number of ordered pairs are added to R so that enlarged relation is reflexive is 3

B. Minimum number of ordered pairs are added to R so that enlarged relation is symmetric is 2

C. Minimum number of ordered pairs are added to R so that enlarged relation is an equivalence relation is 7

D. Minimum number of ordered pair are added to R so that enlarged relation is an equivalence relation is 6.

Choose the correct answer from the options given below:

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Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = 100x + 1$ , where $\mathbb{R}$ is a set of real numbers, then

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Let R be the relation over the set A of all straight lines in a plane such that $l_{1} \mathrm{R} l_{2} \Leftrightarrow l_{1}$ is parallel to $l_{2}$ . Then R is :

The domain of the function y=sin⁡−1(x−1)+cos⁡−1x−1y = \sin^{-1}(x-1) + \cos^{-1}\sqrt{x-1}y=sin−1(x−1)+cos−1x−1​ is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The domain of the function $y = \sin^{-1}(x-1) + \cos^{-1}\sqrt{x-1}$ is: