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If f(tanx)=sin2x: x ne (2n+1)(pi)/(2), n...

If `f(tanx)=sin2x: x ne (2n+1)(pi)/(2), n in I` then which of the following is an incorrect statement?

A

Domain of `f(x)` is `r-(2n+1)(pi)/(2), n in I`

B

Range of `f(x)` is `[-1, 1]`

C

f(x) is odd function

D

`f(x) is many - one function

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The correct Answer is:
To solve the given problem, we need to analyze the function \( f(\tan x) = \sin 2x \) and determine which of the provided statements about the function is incorrect. ### Step-by-Step Solution: 1. **Understanding the Function**: We start with the equation: \[ f(\tan x) = \sin 2x \] We know from trigonometric identities that: \[ \sin 2x = \frac{2 \tan x}{1 + \tan^2 x} \] This allows us to express \( f(t) \) in terms of \( t \) where \( t = \tan x \). 2. **Rewriting the Function**: By substituting \( t = \tan x \), we can rewrite the equation as: \[ f(t) = \frac{2t}{1 + t^2} \] 3. **Finding the Domain of \( f(t) \)**: The function \( \tan x \) is undefined at odd multiples of \( \frac{\pi}{2} \) (i.e., \( x = \frac{(2n+1)\pi}{2} \) for \( n \in \mathbb{Z} \)). Therefore, the domain of \( f(t) \) is all real numbers except those points where \( \tan x \) is undefined. Thus, the domain of \( f(t) \) is: \[ \text{Domain of } f(t) = \mathbb{R} \setminus \left\{ \frac{(2n+1)\pi}{2} \, | \, n \in \mathbb{Z} \right\} \] 4. **Finding the Range of \( f(t) \)**: To find the range, we analyze the function: \[ y = \frac{2t}{1 + t^2} \] Rearranging gives: \[ yt^2 - 2t + y = 0 \] The discriminant \( D \) of this quadratic must be non-negative for \( y \) to have real values: \[ D = (-2)^2 - 4y^2 = 4 - 4y^2 \] Setting \( D \geq 0 \): \[ 4 - 4y^2 \geq 0 \implies 1 - y^2 \geq 0 \implies -1 \leq y \leq 1 \] Thus, the range of \( f(t) \) is: \[ \text{Range of } f(t) = [-1, 1] \] 5. **Checking if \( f(t) \) is an Odd Function**: To check if \( f(t) \) is odd, we compute \( f(-t) \): \[ f(-t) = \frac{2(-t)}{1 + (-t)^2} = -\frac{2t}{1 + t^2} = -f(t) \] Since \( f(-t) = -f(t) \), \( f(t) \) is indeed an odd function. 6. **Determining if \( f(t) \) is Many-One**: A function is many-one if different inputs can produce the same output. In this case, we can find that: \[ f(2) = f\left(\frac{1}{2}\right) = \frac{4}{5} \] This shows that \( f(t) \) is many-one. ### Conclusion: From the analysis, we can summarize the statements: - **Statement 1**: Domain of \( f(x) \) is \( \mathbb{R} \setminus \left\{ \frac{(2n+1)\pi}{2} \, | \, n \in \mathbb{Z} \right\} \) (correct). - **Statement 2**: Range of \( f(x) \) is \( [-1, 1] \) (correct). - **Statement 3**: \( f(x) \) is an odd function (correct). - **Statement 4**: \( f(x) \) is a many-one function (correct). Thus, the **incorrect statement** is: - **Statement 1**: The domain of \( f(x) \) is incorrectly stated as \( \mathbb{R} \setminus \left\{ \frac{(2n+1)\pi}{2} \, | \, n \in \mathbb{Z} \right\} \).
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