If f:R−{−1,K}→R−{α,β} is a bijective function defined by f(x)=(2x−1)(2x2−4px+p3)(x+1)(x2−p2x+p2), where P≥0, then which of the following is/are TRUE?

I f K ∈ ( 1 , 3 ) then α + β = 6 , I f K ∈ ( 1 , 3 ) then α + β = 2

If f:R−{−1,K}→R−{α,β} is a bijective function defined by f(x)=(2x−1)(2x2−4px+p3)(x+1)(x2−p2x+p2), where P≥0, then which of the following is/are TRUE?

If x + 1 s a factor of 2 x 2 − 4 p x + p 3 ⇒ p 3 4 p + 2 = 0 Then 2 x − 1 is a factor of x 2 − p 2 x + p 2 ⇒ 2 p 2 + 1 = 0 = 0 P ∉ R So this case is not true Now, 2 x 2 − 4 p x + p 3 x 2 − p 2 x + p 2 will be some constant for bijective of f ( x ) . So 2 1 = − 4 p − p 2 = p 3 p 2 ⇒ p = 2 ⇒ f ( x ) = 4 x − 2 x + 1 , x ≠ 2 Domain of f ( x ) is R − { − 1 , 2 } , Range of f ( x ) is R − { 2 , 4 } K = 2 , α + β = 6

I f K ∈ ( 1 , 3 ) then α + β = 6

If K ∈ ( − 1 , 1 ) then α + β = 6

If K ∈ ( − 1 , 1 ) then α + β = 2

I f K ∈ ( 1 , 3 ) then α + β = 2

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If $f : R - {- 1, K} \to R - {α, β}$ is a bijective function defined by $f (x) = \frac{(2 x - 1) (2 x^{2} - 4 p x + p^{3})}{(x + 1) (x^{2} - p^{2} x + p^{2})}$ , where $P \geq 0$ , then which of the following is/are TRUE?

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$I f K \in (1, 3) then α + β = 6$

If $K \in (- 1, 1) then α + β = 6$

If $K \in (- 1, 1)$ then $α + β = 2$

$I f K \in (1, 3) then α + β = 2$

answer is A, D.

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Detailed Solution

If $x + 1$ s a factor of $2 x^{2} - 4 p x + p^{3} \Rightarrow p^{3} 4 p + 2 = 0$
Then $2 x - 1$ is a factor of $x^{2} - p^{2} x + p^{2} \Rightarrow 2 p^{2} + 1 = 0 = 0 P \notin R$
So this case is not true
Now, $\frac{2 x^{2} - 4 p x + p^{3}}{x^{2} - p^{2} x + p^{2}}$ will be some constant for bijective of $f (x)$ . So
$\frac{2}{1} = \frac{- 4 p}{- p^{2}} = \frac{p^{3}}{p^{2}} \Rightarrow p = 2 \Rightarrow f (x) = \frac{4 x - 2}{x + 1}, x \neq 2$
Domain of $f (x)$ is $R - {- 1, 2}$ , Range of $f (x)$ is $R - {2, 4}$
$K = 2, α + β = 6$