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$y = {Tan}^{- 1} \sqrt{\frac{1 - x}{1 + x}} then \frac{d y}{d (\cos^{- 1} x)} =$

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a

1

b

12

c

2

d

11−x2

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detailed solution

Correct option is B

x=cos⁡2θ⇒θ=12cos−1⁡x(−1

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Let $f$ be a twice differentiable function defined on R such that $f (0) = 1, f^{'} (0) = 2$ and $f^{1} (x) \neq 0$ for all $x \in R$ . If $|\begin{matrix} f (x) & f^{'} (x) \\ f^{'} (x) & f^{"} (x) \end{matrix}| = 0$ for all $x \in R$ , then the value of $f (1)$ lies in the interval

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