Let S = { ? x R : 0 x 1 and 2 tan - 1 ( 1 - x 1 + x ) = cos - 1 ( 1 - x 2 1 + x 2 ) } . If n ( S ) denotes the number of elements in S, then [2023]

Question

Let $S = {x \in R : 0 < x < 1 and 2 \tan^{- 1} (\frac{1 - x}{1 + x}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})} .$ If $n (S)$ denotes the number of elements in S, then [2023]

Smart Achievers · Accepted Answer

(2)

$0 < x < 1$

$2 \tan^{- 1} (\frac{1 - x}{1 + x}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$ ...(i)

Let $\tan^{- 1} x = θ \in (0, \frac{π}{4}) \Rightarrow x = \tan θ$

$\Rightarrow 2 \tan^{- 1} (\tan (\frac{π}{4} - θ)) = \cos^{- 1} (\cos 2 θ)$

$\Rightarrow 2 (\frac{π}{4} - θ) = 2 θ$

$\Rightarrow 4 θ = \frac{π}{2}$ $∴$ $θ = \frac{π}{8}$ $x = \tan \frac{π}{8} \Rightarrow x = \sqrt{2} - 1 ≃ 0.414$

Solution

Q.
Let $S = {x \in R : 0 < x < 1 and 2 \tan^{- 1} (\frac{1 - x}{1 + x}) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})} .$ If $n (S)$ denotes the number of elements in S, then [2023]

1 $n (S) = 2 and only one element in S is less than \frac{1}{2}$

2 $n (S) = 1 and the element in S is less than \frac{1}{2}$

3 $n (S) = 1 and the element in S is more than \frac{1}{2}$

4 $n (S) = 0$

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Solution

Q. Let S={ x∈R:0<x<1 and 2tan-1(1-x1+x)=cos-1(1-x21+x2)}. If n(S) denotes the number of elements in S, then [2023]

1 n(S)=2 and only one element in S is less than 12

2 n(S)=1 and the element in S is less than 12

3 n(S)=1 and the element in S is more than 12

4 n(S)=0