फलन (f(x)=\frac{1}{x-2-4x+5}) का परिसर क्या है?

What is the range of (f(x)=\frac{1}{x-2-4x+5})?

Explanation opens after your attempt
Correct Answer

A. ((0,1])

Step 1

Concept

The denominator (x-2-4x+5=(x-2)2+1\ge 1). Hence (0<f(x)\le 1).

Step 2

Why this answer is correct

The correct answer is A. ((0,1]). The denominator (x-2-4x+5=(x-2)2+1\ge 1). Hence (0<f(x)\le 1).

Step 3

Exam Tip

हर (x-2-4x+5=(x-2)2+1\ge 1)। इसलिए (0<f(x)\le 1)।

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फलन (f(x)=\frac{1}{x-2-4x+5}) का परिसर क्या है? / What is the range of (f(x)=\frac{1}{x-2-4x+5})?

Correct Answer: A. ((0,1]). Explanation: हर (x-2-4x+5=(x-2)2+1\ge 1)। इसलिए (0<f(x)\le 1)। / The denominator (x-2-4x+5=(x-2)2+1\ge 1). Hence (0<f(x)\le 1).

Which concept should I revise for this Mathematics MCQ?

The denominator (x-2-4x+5=(x-2)2+1\ge 1). Hence (0<f(x)\le 1).

What exam hint can help solve this Mathematics question?

हर (x-2-4x+5=(x-2)2+1\ge 1)। इसलिए (0<f(x)\le 1)।