फलन (f(x)=\frac{x-2+3}{x-2+7}) की रेंज क्या है?

What is the range of (f(x)=\frac{x-2+3}{x-2+7})?

Explanation opens after your attempt
Correct Answer

A. \([\frac{3}{7},1\))

Step 1

Concept

\(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\). At (x=0), \(\frac{3}{7}\) is obtained and (1) is never obtained.

Step 2

Why this answer is correct

The correct answer is A. \([\frac{3}{7},1\)). \(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\). At (x=0), \(\frac{3}{7}\) is obtained and (1) is never obtained.

Step 3

Exam Tip

\(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\) है। (x=0) पर \(\frac{3}{7}\) मिलता है और (1) कभी नहीं मिलता।

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Mathematics Answer, Explanation and Revision Hints

फलन (f(x)=\frac{x-2+3}{x-2+7}) की रेंज क्या है? / What is the range of (f(x)=\frac{x-2+3}{x-2+7})?

Correct Answer: A. \([\frac{3}{7},1\)). Explanation: \(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\) है। (x=0) पर \(\frac{3}{7}\) मिलता है और (1) कभी नहीं मिलता। / \(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\). At (x=0), \(\frac{3}{7}\) is obtained and (1) is never obtained.

Which concept should I revise for this Mathematics MCQ?

\(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\). At (x=0), \(\frac{3}{7}\) is obtained and (1) is never obtained.

What exam hint can help solve this Mathematics question?

\(\frac{x^2+3}{x^2+7}=1-\frac{4}{x^2+7}\) है। (x=0) पर \(\frac{3}{7}\) मिलता है और (1) कभी नहीं मिलता।