यदि (f(x)=\frac{1}{x-1}) और डोमेन ([3,9]) है, तो रेंज क्या होगी?

If (f(x)=\frac{1}{x-1}) and the domain is ([3,9]), what is the range?

Explanation opens after your attempt
Correct Answer

A. \([\frac{1}{8},\frac{1}{2}]\)

Step 1

Concept

On this domain the denominator is positive and the reciprocal decreases. (x=3) gives \(\frac{1}{2}\) and (x=9) gives \(\frac{1}{8}\).

Step 2

Why this answer is correct

The correct answer is A. \([\frac{1}{8},\frac{1}{2}]\). On this domain the denominator is positive and the reciprocal decreases. (x=3) gives \(\frac{1}{2}\) and (x=9) gives \(\frac{1}{8}\).

Step 3

Exam Tip

इस domain पर denominator धनात्मक है और reciprocal घटता है। (x=3) से \(\frac{1}{2}\) और (x=9) से \(\frac{1}{8}\) मिलता है।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि (f(x)=\frac{1}{x-1}) और डोमेन ([3,9]) है, तो रेंज क्या होगी? / If (f(x)=\frac{1}{x-1}) and the domain is ([3,9]), what is the range?

Correct Answer: A. \([\frac{1}{8},\frac{1}{2}]\). Explanation: इस domain पर denominator धनात्मक है और reciprocal घटता है। (x=3) से \(\frac{1}{2}\) और (x=9) से \(\frac{1}{8}\) मिलता है। / On this domain the denominator is positive and the reciprocal decreases. (x=3) gives \(\frac{1}{2}\) and (x=9) gives \(\frac{1}{8}\).

Which concept should I revise for this Mathematics MCQ?

On this domain the denominator is positive and the reciprocal decreases. (x=3) gives \(\frac{1}{2}\) and (x=9) gives \(\frac{1}{8}\).

What exam hint can help solve this Mathematics question?

इस domain पर denominator धनात्मक है और reciprocal घटता है। (x=3) से \(\frac{1}{2}\) और (x=9) से \(\frac{1}{8}\) मिलता है।