Let S=x∈R:0<x<1 and 2tan^(-1)(1-x)/(1+x)=cos^(-1)(1-x^2)/(1+x^2). If n(S) denotes the number of elements in S then :

Correct answer: (C) n(S)=1 and the element in S is less then 1/2. Given, 0<x<1 2tan^(-1)(1-x)/(1+x)=cos^(-1)(1-x^2)/(1+x^2) Put x=tanθ ; 0<θ<(π)/4 2tan^(-1)(1-tan θ)/(1+tan θ)=cos^(-1)(1-tan^2θ)/(1+tan^2θ) ⇒2tan^(-1)tan(π)/4-θ=cos^(-1)(cos2θ) 2(π)/4-θ=2θ⇒θ=(π)/8 Put θ=tan^(-1)x we get, x=tan(π)/8∴x=√2-1≃0.414<1/2

Q33 / 90 · JEE Main 2023 (01 Feb Shift 2) · Mathematics

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 :

A
$n (S) = 2$ and only one element in $S$ is less then $\frac{1}{2}$
B
$n (S) = 1$ and the element in $S$ is more than $\frac{1}{2}$
Correct answer
C
$n (S) = 1$ and the element in $S$ is less then $\frac{1}{2}$
D
$n (S) = 0$

Solution

Given, $0 < x < 1$

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

Put $x = \tan (θ); 0 < θ < \frac{π}{4}$

$2 \tan^{- 1} (\frac{1 - \tan θ}{1 + \tan θ}) = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ})$

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

$2 (\frac{π}{4} - θ) = 2 θ \Rightarrow θ = \frac{π}{8}$

Put $θ = \tan^{- 1} x$ we get,

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

Hence, the correct answer is option: 'C'.

Let S=x∈R:0<x<1 and 2tan-11-x1+x=cos-11-x21+x2. If n(S) denotes the number of elements in S then :

Solution

Relevant Questions

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 :