Question

The range of cosh(x/2)​

Answer 1

Range of cosh(x/2) ∈ [1, ∞) .

we have to find range of hyperbolic of cosine ; cosh(x/2).

we know, cosh(x) = (e^x + e^-x)/2

so, cosh(x/2) = {e^(x/2) + e^(-x/2)}/2

let y = cosh(x/2)

so, y = {e^(x/2) + e^(-x/2)}/2

⇒2y = e^(x/2) + 1/e^(x/2)

⇒e^2(x/2) - 2ye^(x/2) + 1 = 0

let e^(x/2) = p

then, p² - 2yp + 1 = 0

for real value of p, D = (-2y)² - 4 × 1 ≥ 0

⇒4y² - 4 ≥ 0

⇒(y - 1)(y + 1) ≥ 0

y ≥ 1 or, y ≤ -1

but y = cosh(x/2) = {e^(x/2)+e^(-x/2)}/2 > 0 for all value of x

hence, y ≤ -1 is false statement.

hence, range of cosh(x/2) , y ≥ 1

or, range ∈ [1, ∞ )

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