The number of the real roots of the equation 2cos(x(x + 1)) = 2x + 2–x is
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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