Concept-wise Practice

difference-domain MCQ Questions for Class 11

difference-domain se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.

Practice Questions

3 questions tagged with difference-domain.

यदि (f(x)=\sqrt{x+1}) और (g(x)=\sqrt{x-3}) हैं तो ((f-g)(x)) का प्रांत क्या है?

If (f(x)=\sqrt{x+1}) and (g(x)=\sqrt{x-3}) then what is the domain of ((f-g)(x))?

Explanation opens after your attempt
Correct Answer

A. \([3,\infty))

Step 1

Concept

For both square roots \(x+1\ge 0\) and \(x-3\ge 0\), hence \(x\ge 3\). Even in subtraction both functions must be defined.

Step 2

Why this answer is correct

The correct answer is A. \([3,\infty)). For both square roots \(x+1\ge 0\) and \(x-3\ge 0\), hence \(x\ge 3\). Even in subtraction both functions must be defined.

Step 3

Exam Tip

दोनों वर्गमूलों के लिए \(x+1\ge 0\) और \(x-3\ge 0\), इसलिए \(x\ge 3\)। घटाव में भी दोनों फलन परिभाषित होने चाहिए।

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यदि (f(x)=\frac{1}{x-2-4}) और (g(x)=\frac{1}{x-2}) हों, तो ((f-g)(x)) का प्रांत क्या होगा?

If (f(x)=\frac{1}{x-2-4}) and (g(x)=\frac{1}{x-2}), what is the domain of ((f-g)(x))?

Explanation opens after your attempt
Correct Answer

A. \( \mathbb{R}\setminus{-2,2} \)

Step 1

Concept

The first function gives \(x^2-4\ne 0\), so \(x\ne \pm2\), and the second gives \(x\ne 2\). The common domain is \( \mathbb{R}\setminus{-2,2} \).

Step 2

Why this answer is correct

The correct answer is A. \( \mathbb{R}\setminus{-2,2} \). The first function gives \(x^2-4\ne 0\), so \(x\ne \pm2\), and the second gives \(x\ne 2\). The common domain is \( \mathbb{R}\setminus{-2,2} \).

Step 3

Exam Tip

पहले फलन में \(x^2-4\ne 0\) से \(x\ne \pm2\), और दूसरे में \(x\ne 2\)। संयुक्त प्रांत \( \mathbb{R}\setminus{-2,2} \) है।

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यदि (f(x)=\sqrt{x-2}) और (g(x)=\sqrt{x+2}) हों, तो ((f-g)(x)) का प्रांत क्या होगा?

If (f(x)=\sqrt{x-2}) and (g(x)=\sqrt{x+2}), what is the domain of ((f-g)(x))?

Explanation opens after your attempt
Correct Answer

A. \( [2,\infty\) )

Step 1

Concept

The first root needs \(x\ge 2\) and the second needs \(x\ge -2\). Their intersection is \( [2,\infty\) ).

Step 2

Why this answer is correct

The correct answer is A. \( [2,\infty\) ). The first root needs \(x\ge 2\) and the second needs \(x\ge -2\). Their intersection is \( [2,\infty\) ).

Step 3

Exam Tip

पहले मूल के लिए \(x\ge 2\) और दूसरे के लिए \(x\ge -2\) चाहिए। प्रतिच्छेद \( [2,\infty\) ) है।

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