Question

Range of f(x) = cos⁻¹$\left( \frac{x^{2}}{\sqrt{1 + x^{2}}} \right)$

• A. (0, π/2)
• B. (0, π/4)
• C. (0, π/2]
• D. [0, π/4)

Answer 1

Answer: (C)

(0, π/2]

Answer 2

For x = 0, $\frac { x ^ { 2 } } { 1 + x ^ { 2 } }$ = 0

For x ≠ 0, $\frac { x ^ { 2 } } { 1 + x ^ { 2 } } = \frac { 1 } { 1 + \frac { 1 } { x ^ { 2 } } }$

Since for x ≠ 0 ⇒ 0 < x² < ∞

⇒ ∞ > $\frac { 1 } { x ^ { 2 } }$ > 0

⇒ $\infty > 1 + \frac { 1 } { x ^ { 2 } } > 1$

⇒ $0 < \frac { 1 } { 1 + \frac { 1 } { x ^ { 2 } } } < 1$

⇒ 0 ≤ $\frac { x ^ { 2 } } { 1 + x ^ { 2 } }$ < 1 ∀ x ∈ R

Thus range is $\left( 0 , \frac { \pi } { 2 } \right]$