फलन (f(x)=\frac{1}{x-2+4}) का परिसर ज्ञात कीजिए।

Find the range of (f(x)=\frac{1}{x-2+4}).

Explanation opens after your attempt
Correct Answer

A. (\(0,\frac{1}{4}]\)

Step 1

Concept

Since \(x^2+4\ge 4\), (0<f(x)\le \frac{1}{4}). The value (0) is never attained.

Step 2

Why this answer is correct

The correct answer is A. (\(0,\frac{1}{4}]\). Since \(x^2+4\ge 4\), (0<f(x)\le \frac{1}{4}). The value (0) is never attained.

Step 3

Exam Tip

क्योंकि \(x^2+4\ge 4\), इसलिए (0<f(x)\le \frac{1}{4})। शून्य प्राप्त नहीं होता।

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Mathematics Answer, Explanation and Revision Hints

फलन (f(x)=\frac{1}{x-2+4}) का परिसर ज्ञात कीजिए। / Find the range of (f(x)=\frac{1}{x-2+4}).

Correct Answer: A. (\(0,\frac{1}{4}]\). Explanation: क्योंकि \(x^2+4\ge 4\), इसलिए (0<f(x)\le \frac{1}{4})। शून्य प्राप्त नहीं होता। / Since \(x^2+4\ge 4\), (0<f(x)\le \frac{1}{4}). The value (0) is never attained.

Which concept should I revise for this Mathematics MCQ?

Since \(x^2+4\ge 4\), (0<f(x)\le \frac{1}{4}). The value (0) is never attained.

What exam hint can help solve this Mathematics question?

क्योंकि \(x^2+4\ge 4\), इसलिए (0<f(x)\le \frac{1}{4})। शून्य प्राप्त नहीं होता।