फलन (f(x)=\sqrt{x-2+4x+7}) का परिसर क्या है?

What is the range of (f(x)=\sqrt{x-2+4x+7})?

Explanation opens after your attempt
Correct Answer

A. \( [\sqrt{3},\infty\) )

Step 1

Concept

Since (x-2+4x+7=(x+2)2+3), the inside minimum is (3). Taking the square root gives the minimum \(\sqrt{3}\).

Step 2

Why this answer is correct

The correct answer is A. \( [\sqrt{3},\infty\) ). Since (x-2+4x+7=(x+2)2+3), the inside minimum is (3). Taking the square root gives the minimum \(\sqrt{3}\).

Step 3

Exam Tip

(x-2+4x+7=(x+2)2+3), इसलिए अंदर का न्यूनतम (3) है। वर्गमूल लेने पर न्यूनतम \(\sqrt{3}\) मिलेगा।

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फलन (f(x)=\sqrt{x-2+4x+7}) का परिसर क्या है? / What is the range of (f(x)=\sqrt{x-2+4x+7})?

Correct Answer: A. \( [\sqrt{3},\infty\) ). Explanation: (x-2+4x+7=(x+2)2+3), इसलिए अंदर का न्यूनतम (3) है। वर्गमूल लेने पर न्यूनतम \(\sqrt{3}\) मिलेगा। / Since (x-2+4x+7=(x+2)2+3), the inside minimum is (3). Taking the square root gives the minimum \(\sqrt{3}\).

Which concept should I revise for this Mathematics MCQ?

Since (x-2+4x+7=(x+2)2+3), the inside minimum is (3). Taking the square root gives the minimum \(\sqrt{3}\).

What exam hint can help solve this Mathematics question?

(x-2+4x+7=(x+2)2+3), इसलिए अंदर का न्यूनतम (3) है। वर्गमूल लेने पर न्यूनतम \(\sqrt{3}\) मिलेगा।