Fx-x2-3 x-4-x2-3 x-4 the range of fx is

The correct option is B [ 1/7,7 ]y=x^2 3x+4/x^2+3x+4 yx^2+3xy+4y=x^2 3x+4 x^2(y 1)+3x(y+1)+4(y 1)=0 D ≥ 0⇒ 9(y+1)^2 4.4(y 1)^2 ≥ 0 (3(y+1) 4(y 1)) (3(y+1)+4(y ...

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Byju's Answer

Standard XII

Mathematics

Expressing x in Terms of y, to Find the Range of a Function

fx=x2-3 x+4/x...

Question

$f (x) = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ the range of f(x) is

[0, \frac{1}{7}]

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(- \infty, \frac{1}{7}) \cup (7, \infty)

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(- \infty, 7)

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[\frac{1}{7}, 7]

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Solution

The correct option is D $[\frac{1}{7}, 7]$
$y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$
$y x^{2} + 3 x y + 4 y = x^{2} - 3 x + 4$
$x^{2} (y - 1) + 3 x (y + 1) + 4 (y - 1) = 0$
D $\geq 0 \Rightarrow 9 (y + 1)^{2} - 4.4 (y - 1)^{2} \geq 0$
$(3 (y + 1) - 4 (y - 1))$ $(3 (y + 1) + 4 (y - 1)) \geq 0$
$(- y + 7) (7 y - 1) \geq 0$
$(y - 7) (y - \frac{1}{7}) \leq 0$
$\frac{1}{7} \leq y \leq 7$

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f(x)=x2−3x+4x2+3x+4 the range of f(x) is

The correct option is D [17,7]y=x2−3x+4x2+3x+4 yx2+3xy+4y=x2−3x+4 x2(y−1)+3x(y+1)+4(y−1)=0 D ≥0⇒9(y+1)2−4.4(y−1)2≥0 (3(y+1)−4(y−1)) (3(y+1)+4(y−1))≥0 (−y+7)(7y−1)≥0 (y−7)(y−17)≤0 17≤y≤7

$f (x) = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ the range of f(x) is