If f(x) = cos–1 \left(\right. \frac{1}{\sqrt{x}} - 1 \left.\right) + \left(tan\right)^{- 1} \left(\right. \frac{2 - \sqrt{\frac{1}{\sqrt{x}}}}{\sqrt{\frac{1}{\sqrt{x}}}} \left.\right) then identify which of the following statement(s) is(are) correct?

Engineering

Mathematics

Basic of Inverse Trigonometric Functions

Question

If f(x) = cos^–1 $(\frac{1}{\sqrt{x}} - 1) + \tan^{- 1} (\frac{2 - \sqrt{\frac{1}{\sqrt{x}}}}{\sqrt{\frac{1}{\sqrt{x}}}})$ then identify which of the following statement(s) is(are) correct?

The equation f(x) = 1 has atleast one solution.

Domain of the function f (x) is $[\frac{1}{4}, \infty)$ .

The largest integer in the range of the function f (x) is 5.

Number of integers in the range of f(x) is 4.

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For the domain of f(x)
$- 1 \leq \frac{1}{\sqrt{x}} - 1 \leq 1 \Rightarrow 0 \leq \frac{1}{\sqrt{x}} \leq 2 \Rightarrow x \geq \frac{1}{4} \Rightarrow (C)$
f (x) = cos^–1 $(\frac{1}{\sqrt{x}} - 1) + \tan^{- 1} (2 \sqrt{\sqrt{x}} - 1)$
f (x)|_min = f $(\frac{1}{4})$ = cos^–1(1) + tan^–1 $(\sqrt{2} - 1) = 0 + \frac{π}{8} = \frac{π}{8}$
f (x)|_max = f( $\infty$ ) = cos^–1 (–1) + tan^–1 $(\infty) = π + \frac{π}{2} = \frac{3 π}{2}$
$∴$ Range of f(x) is $[\frac{π}{8}, \frac{3 π}{2})$
⇒ Integers in the range of f(x) are 1, 2, 3, 4 ⇒ (A)
⇒ From I.V.T equation f(x) = 1 has atleast one solution. ⇒ (B) ]