If f - R -1- k - R - - is a bijective function defined by f x -2 x -12 x 2-4 px - p 3- x -1 x 2- p 2 x - p 2 for p - 0- then which of thefollowing statements is are CORRECT -A- If k -1-1- then -2B- If k -1-1- then -6C- If k -1-3- then -4D- If k -1-3- then -6

The correct options are A If k∈( 1,1), then α+β=2 D If k∈(1,3), then α+β=6f:ℝ { 1,k}→ℝ {α,β} f(x)=(2x 1)(2x^2 4px+p^3)(x+1)(x^2 p^2x+p^2) For the domain to be ...

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Byju's Answer

Standard XII

Mathematics

Range

If f : R --1,...

Question

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

k \in (- 1, 1),

then

α + β = 2

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k \in (- 1, 1),

then

α + β = 6

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k \in (1, 3),

then

α + β = 4

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k \in (1, 3),

then

α + β = 6

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Solution

The correct options are
A If $k \in (- 1, 1),$ then $α + β = 2$
D If $k \in (1, 3),$ then $α + β = 6$
$f : R - {- 1, k} \to R - {α, β}$
$f (x) = \frac{(2 x - 1) (2 x^{2} - 4 p x + p^{3})}{(x + 1) (x^{2} - p^{2} x + p^{2})}$

For the domain to be $R - {- 1, k}, k$ must be a repeated root of $x^{2} - p^{2} x + p^{2} = 0$
$∴ p^{4} - 4 p^{2} = 0 \Rightarrow p^{2} (p^{2} - 4) = 0$
$∴ p = 0, \pm 2,$ but $p \geq 0$
$\Rightarrow p = 0, 2$

For $p = 0, k = 0$
$f (x) = \frac{(2 x - 1) (2 x^{2})}{(x + 1) (x^{2})}$
$\Rightarrow D_{f} = R - {- 1, 0}, R_{f} = R - {4, - 2}$
$∴ α + β = 2$

For $p = 2, k = 2$
$f (x) = \frac{(2 x - 1) 2 (x - 2)^{2}}{(x + 1) (x - 2)^{2}}$
$\Rightarrow D_{f} = R - {- 1, 2}, R_{f} = R - {4, 2}$
$∴ α + β = 6$

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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) for p≥0, then which of the following statements is (are) CORRECT ?

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