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