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Question
If the derivative of tan−1 (a + bx) takes the value 1 at x = 0, prove that 1 + a2 = b ?
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Solution
\[\text{Here}, \frac{d}{dx}\left[ \tan^{- 1} \left( a + bx \right) \right] = 1 \text{ at }x = 0\]
\[ \Rightarrow \left[ \left\{ \frac{1}{1 + \left( a + bx \right)^2} \right\}\frac{d}{dx}\left( a + bx \right) \right]_{x = 0} = 1\]
\[ \Rightarrow \left[ \frac{1}{1 + \left( a + bx \right)^2} \times \left( b \right) \right]_{x = 0} = 1\]
\[ \Rightarrow \frac{b}{1 + \left( a + 0 \right)^2} = 1\]
\[ \Rightarrow b = 1 + a^2 \]
\[ \therefore 1 + a^2 = b\]
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