Question

The derivative of \[\sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { w . r . t }. \sqrt{1 + 3 x} \text { at } x = - 1/3\]

विकल्प

• does not exist

• 0

• 1/2

• 1/3

Answer 1

does not exist

\[\text { We know that } \sec^{- 1} \alpha\text { is not defined for }\alpha \in \left( - 1, 1 \right)\]
\[\text { Here for } x = \frac{- 1}{3}, \frac{1}{2 x^2 + 1} = \frac{9}{11} \in \left( - 1, 1 \right)\]
\[ \therefore \sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right)\text { is not defined at } x = \frac{- 1}{3}\]
\[ \therefore \text { Derivative of } \sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { does not exist at } x = \frac{- 1}{3}\]