Functions Question - None

Question

Range of

f(x) = \frac{\sec x + an x - 1}{ an x - \sec x + 1}; x \in \left(0, \frac{\pi}{2} ight)

is

Accepted Answer

Step 1: Simplify using trigonometric identities

Let s = sec x and t = tan x. We know that s² - t² = 1 (fundamental identity).

The function becomes: f(x) = (s + t - 1)/(t - s + 1)

Step 2: Factor using the identity s² - t² = 1

Note that s² - t² = (s - t)(s + t) = 1, so s + t = 1/(s - t)

Let u = s - t. Then s + t = 1/u

Numerator: s + t - 1 = 1/u - 1 = (1 - u)/u

Denominator: t - s + 1 = -u + 1 = (1 - u)

Step 3: Simplify the fraction

f(x) = [(1 - u)/u]/[(1 - u)] = 1/u = 1/(s - t) = 1/(sec x - tan x)

Step 4: Use another identity

Recall that (sec x - tan x)(sec x + tan x) = sec² x - tan² x = 1

Therefore: sec x - tan x = 1/(sec x + tan x)

So: f(x) = sec x + tan x

Step 5: Analyze the range for x ∈ (0, π/2)

As x → 0⁺: sec x → 1 and tan x → 0, so f(x) → 1

As x → (π/2)⁻: both sec x and tan x → ∞, so f(x) → ∞

Since f(x) = sec x + tan x is continuous and strictly increasing on (0, π/2), the range is (1, ∞).

∴ Answer: B

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