Is \(f(x)=\cos(x+2)\) Invertible?

📝 Question

Let:

\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=\cos(x+2) \]

Is \(f\) invertible? Justify your answer.

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✅ Solution

🔹 Step 1: Check whether \(f\) is one-one

Assume:

\[ f(x_1)=f(x_2) \]

\[ \cos(x_1+2)=\cos(x_2+2) \]

Using identity:

\[ \cos A=\cos B \Rightarrow A=B \ \text{or} \ A=2\pi-B \]

So,

\[ x_1+2=x_2+2 \quad \text{or} \quad x_1+2=2\pi-(x_2+2) \]

\[ x_1=x_2 \quad \text{or} \quad x_1=2\pi-x_2-4 \]

This shows different inputs can give the same output.

Hence, \(f\) is not one-one.

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🔹 Step 2: Check whether \(f\) is onto

Range of cosine function is:

\[ [-1,1] \]

But codomain is \(\mathbb{R}\).

So, \(f\) is not onto.

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🔹 Step 3: Conclusion

Since \(f\) is neither one-one nor onto, it is not bijective.

\[ \boxed{f(x)=\cos(x+2)\text{ is not invertible on }\mathbb{R}} \]

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🎯 Final Answer

\[ \boxed{f \text{ is not invertible}} \]

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🚀 Exam Shortcut

• Cosine is periodic ⇒ not one-one
• Range is \([-1,1]\) ⇒ not onto \(\mathbb{R}\)
• Not bijective ⇒ no inverse exists