यदि \(f:\mathbb{R}\to\mathbb{R}\) और (f(x)=\cos x) है, तो (f) एकैकी नहीं है क्योंकि

If \(f:\mathbb{R}\to\mathbb{R}\) and (f(x)=\cos x), (f) is not one-one because

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Correct Answer

A. (f(0)=f\(2\pi\)) और \(0\neq 2\pi\)(f(0)=f\(2\pi\)) and \(0\neq 2\pi\)

Step 1

Concept

\(\cos x\) is a periodic function.

Step 2

Why this answer is correct

\(\cos 0=1\) and \(\cos 2\pi=1\), while \(0\neq 2\pi\).

Step 3

Exam Tip

Periodic functions on all real numbers are generally not one-one. चरण 1: \(\cos x\) एक आवर्ती फलन है। चरण 2: \(\cos 0=1\) और \(\cos 2\pi=1\), जबकि \(0\neq 2\pi\)। चरण 3: आवर्ती फलन पूरे \(\mathbb{R}\) पर सामान्यतः एकैकी नहीं होते।

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Mathematics Answer, Explanation and Revision Hints

यदि \(f:\mathbb{R}\to\mathbb{R}\) और (f(x)=\cos x) है, तो (f) एकैकी नहीं है क्योंकि / If \(f:\mathbb{R}\to\mathbb{R}\) and (f(x)=\cos x), (f) is not one-one because

Correct Answer: A. (f(0)=f\(2\pi\)) और \(0\neq 2\pi\) / (f(0)=f\(2\pi\)) and \(0\neq 2\pi\). Explanation: चरण 1: \(\cos x\) एक आवर्ती फलन है। चरण 2: \(\cos 0=1\) और \(\cos 2\pi=1\), जबकि \(0\neq 2\pi\)। चरण 3: आवर्ती फलन पूरे \(\mathbb{R}\) पर सामान्यतः एकैकी नहीं होते। / Step 1: \(\cos x\) is a periodic function. Step 2: \(\cos 0=1\) and \(\cos 2\pi=1\), while \(0\neq 2\pi\). Step 3: Periodic functions on all real numbers are generally not one-one.

Which concept should I revise for this Mathematics MCQ?

\(\cos x\) is a periodic function.

What exam hint can help solve this Mathematics question?

Periodic functions on all real numbers are generally not one-one. चरण 1: \(\cos x\) एक आवर्ती फलन है। चरण 2: \(\cos 0=1\) और \(\cos 2\pi=1\), जबकि \(0\neq 2\pi\)। चरण 3: आवर्ती फलन पूरे \(\mathbb{R}\) पर सामान्यतः एकैकी नहीं होते।