Question

Let f(x) = arctan(x ^ 2 - 4x + 5) where x in R then range of f(x) is equal to

Answer 1

Correct option is (B) [π/4, π/2]

f(x) = tan^-1(x² - 4x + 5)

 = tan^-1((x - 2)² + 1)

\(\because\) (x - 2)² \(\geq\) 0

⇒ (x - 2)² + 1 \(\geq\) 1

⇒ tan^-1((x - 2)² + 1) \(\geq\) tan^-1(1)

(\(\because\) tan^-1 x is an increasing function)

⇒ f(x)  \(\geq\) \(\frac{\pi}4\)

But range of tan^-1x is (-π/2, π/2).

\(\therefore\) Range of tan^-1(x²- 4x + 5) is [π/4, π/2].