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Explanation:

−2 ≤ 2cos(x(x + 1)) ≤ 2 

∴ −2 ≤ 2x + 2–x ≤ 2 

Let 2x be a, so 2–x is 1/a. 

So, −2 ≤ a + (1/a) ≤ 2     

∴ −2 ≤ (a2 + 1)/a ≤ 2 

∴ −2a ≤ (a2 + 1) ≤ 2a 

∴ (a2 + 1 + 2a) ≥ 0 ⇒ (a + 1)2 ≥ 0, so a ∈ R. 

Also, a2 + 1 − 2a  ≤ 0  ⇒ (a − 1)2 ≤ 0, so a = 1. 

Hence a = 1. 

So, 2x = 1. 

∴ x = 0.  

So, there is only one real root. 

Hence, option (d).

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