JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Inverse Trigonometric Functions

Considering only the principal values of the inverse trigonometric functions, the domain of the function

f (x) = cos^{- 1} (\frac{x ^{2} - 4 x + 2}{x ^{2} + 3})

is :

Answer: (B)

Solution

For

cos^{- 1}

to be defined, we need

\frac{x ^{2} - 4 x + 2}{x ^{2} + 3} \leq 1

. This is equivalent to

(x^{2} - 4 x + 2)^{2} \leq (x^{2} + 3)^{2}

, which simplifies to

(x^{2} - 4 x + 2)^{2} - (x^{2} + 3)^{2} \leq 0

. Factoring:

(2 x^{2} - 4 x + 5) (- 4 x - 1) \leq 0

. Since

2 x^{2} - 4 x + 5 = 2 (x - 1)^{2} + 3 > 0

for all

x

, we need

- 4 x - 1 \leq 0

, giving

x \geq - \frac{1}{4}

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