The range of f x = 4 sin - 1 x 2 x 2 + 1 is

Given, f x = 4 sin - 1 x 2 x 2 + 1 Now taking, x 2 1 + x 2 = 1 - 1 1 + x 2 < 1 ⇒ 0 ≤ x 2 1 + x 2 < 1 Now taking sin - 1 we get, ⇒ 0 ≤ sin - 1 x 2 1 + x 2 < π 2 ⇒ 0 ≤ 4 sin - 1 x 2 1 + x 2 < 2 π Hence, the range of 4 sin - 1 x 2 x 2 + 1 is [ 0 , 2 π )

Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsDomain-RangeEasy2 minPYQ_2023

MathematicsEasysingle choice

The range of $f (x) = 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$ is

Options:

Answer:

Solution:

Given,
$f (x) = 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$
Now taking, $\frac{x^{2}}{1 + x^{2}} = 1 - \frac{1}{1 + x^{2}} < 1$
$\Rightarrow 0 \leq \frac{x^{2}}{1 + x^{2}} < 1$
Now taking $\sin^{- 1}$ we get,
$\Rightarrow 0 \leq \sin^{- 1} (\frac{x^{2}}{1 + x^{2}}) < \frac{π}{2}$
$\Rightarrow 0 \leq 4 \sin^{- 1} (\frac{x^{2}}{1 + x^{2}}) < 2 π$
Hence, the range of $4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$ is $[0, 2 π)$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Domain-Range

⏱ 2mℹ️ Source: PYQ_2023

Doubts & Discussion

Loading discussions...