The function f ( x ) = x 2 + 2 x - 15 x 2 - 4 x + 9 , x ? R is [2024]

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Solution

Q.
The function $f (x) = \frac{x^{2} + 2 x - 15}{x^{2} - 4 x + 9}, x \in R$ is [2024]

1 both one-one and onto.

2 neither one-one nor onto.

3 onto but not one-one.

4 one-one but not onto.

Ans.
(2)

   $We have, f (x) = \frac{x^{2} + 2 x - 15}{x^{2} - 4 x + 9}$

   $f (x) = \frac{(x + 5) (x - 3)}{x^{2} - 4 x + 9}$

   $f (x) has two roots as x = - 5, 3$

   $So f (- 5) = 0 and f (3) = 0$

   $So f cannot be one - one .$

   $Now consider, x^{2} - 4 x + 9$

   $D = 16 - 36 = - 20 < 0$

   $Now, let y = \frac{x^{2} + 2 x - 15}{x^{2} - 4 x + 9}$

   $\Rightarrow y x^{2} - 4 x y + 9 y = x^{2} + 2 x - 15$

   $\Rightarrow x^{2} (y - 1) - 2 x (2 y + 1) + (9 y + 15) = 0$

$D = 4 {(2 y + 1)}^{2} - 4 (y - 1) (9 y + 15) \geq 0$    $[∵ x \in R]$

$\Rightarrow 4 (4 y^{2} + 1 + 4 y) - (36 y^{2} - 36 y + 60 y - 60) \geq 0$

$\Rightarrow 16 y^{2} + 4 + 16 y - 36 y^{2} - 24 y + 60 \geq 0$

$\Rightarrow - 20 y^{2} - 8 y + 64 \geq 0$

$\Rightarrow 5 y^{2} + 2 y - 16 \leq 0$

$\Rightarrow (5 y - 8) (y + 2) \leq 0$

$\Rightarrow y \in [- 2, \frac{8}{5}] which is the range of function f$

$So, f is not onto if f : R \to R .$

$Here in given question co - domain is not defined .$

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