If f(x) = $\sin^{-1}$ $\left(\frac{2x}{1+x^{2}}\right)$, then f' $(\sqrt{3})$ is
- $-\frac {1}{2}$
- $\frac {1}{2}$
- $\frac {1}{\sqrt{3}}$
- $-\frac {1}{\sqrt{3}}$
The Correct Option is B
Solution and Explanation
Put $x=\tan\, \theta$, where $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$f(x)=\sin ^{-1}\left(\frac{2 \tan \,\theta}{1+\tan ^{2} \theta}\right)$
$\Rightarrow f(x)=\sin ^{-1}(\sin 2 \theta)$
$\Rightarrow f(x)=2 \,\theta=2 \tan ^{-1} x \,\,\left(\because \theta=\tan ^{-1} x\right)$
$f'(x)=\frac{2}{1+x^{2}}$
$\therefore f'(\sqrt{3})=\frac{2}{1+3}=\frac{1}{2}$
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions