फलन (f(x)=\frac{2x+1}{x-3}) का परिसर क्या है?

What is the range of (f(x)=\frac{2x+1}{x-3})?

Explanation opens after your attempt
Correct Answer

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

Step 1

Concept

If \(y=\frac{2x+1}{x-3}\), then \(x=\frac{3y+1}{y-2}\), so \(y\ne 2\). Hence the range is \(\mathbb{R}-{2}\).

Step 2

Why this answer is correct

The correct answer is A. \(\mathbb{R}-{2}\). If \(y=\frac{2x+1}{x-3}\), then \(x=\frac{3y+1}{y-2}\), so \(y\ne 2\). Hence the range is \(\mathbb{R}-{2}\).

Step 3

Exam Tip

यदि \(y=\frac{2x+1}{x-3}\), तो \(x=\frac{3y+1}{y-2}\), अतः \(y\ne 2\)। इसलिए परिसर \(\mathbb{R}-{2}\) है।

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Mathematics Answer, Explanation and Revision Hints

फलन (f(x)=\frac{2x+1}{x-3}) का परिसर क्या है? / What is the range of (f(x)=\frac{2x+1}{x-3})?

Correct Answer: A. \(\mathbb{R}-{2}\). Explanation: यदि \(y=\frac{2x+1}{x-3}\), तो \(x=\frac{3y+1}{y-2}\), अतः \(y\ne 2\)। इसलिए परिसर \(\mathbb{R}-{2}\) है। / If \(y=\frac{2x+1}{x-3}\), then \(x=\frac{3y+1}{y-2}\), so \(y\ne 2\). Hence the range is \(\mathbb{R}-{2}\).

Which concept should I revise for this Mathematics MCQ?

If \(y=\frac{2x+1}{x-3}\), then \(x=\frac{3y+1}{y-2}\), so \(y\ne 2\). Hence the range is \(\mathbb{R}-{2}\).

What exam hint can help solve this Mathematics question?

यदि \(y=\frac{2x+1}{x-3}\), तो \(x=\frac{3y+1}{y-2}\), अतः \(y\ne 2\)। इसलिए परिसर \(\mathbb{R}-{2}\) है।