फलन (f(x)=2\sqrt{x+1}-5) का परिसर क्या है?

What is the range of (f(x)=2\sqrt{x+1}-5)?

Explanation opens after your attempt
Correct Answer

A. \([-5,\infty\))

Step 1

Concept

Since \(\sqrt{x+1}\ge 0\), \(2\sqrt{x+1}-5\ge -5\). Hence the range is \([-5,\infty\)).

Step 2

Why this answer is correct

The correct answer is A. \([-5,\infty\)). Since \(\sqrt{x+1}\ge 0\), \(2\sqrt{x+1}-5\ge -5\). Hence the range is \([-5,\infty\)).

Step 3

Exam Tip

\(\sqrt{x+1}\ge 0\), इसलिए \(2\sqrt{x+1}-5\ge -5\)। अतः परिसर \([-5,\infty\)) 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

फलन (f(x)=2\sqrt{x+1}-5) का परिसर क्या है? / What is the range of (f(x)=2\sqrt{x+1}-5)?

Correct Answer: A. \([-5,\infty\)). Explanation: \(\sqrt{x+1}\ge 0\), इसलिए \(2\sqrt{x+1}-5\ge -5\)। अतः परिसर \([-5,\infty\)) है। / Since \(\sqrt{x+1}\ge 0\), \(2\sqrt{x+1}-5\ge -5\). Hence the range is \([-5,\infty\)).

Which concept should I revise for this Mathematics MCQ?

Since \(\sqrt{x+1}\ge 0\), \(2\sqrt{x+1}-5\ge -5\). Hence the range is \([-5,\infty\)).

What exam hint can help solve this Mathematics question?

\(\sqrt{x+1}\ge 0\), इसलिए \(2\sqrt{x+1}-5\ge -5\)। अतः परिसर \([-5,\infty\)) है।