Fx-x2-3 x-4-x2-3 x-4 then range of fx is A- -0- 1-7-B- - 1-7 -7- -1-7- 7-

The correct option is C 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 Dis 1 ≥ 0 ⇒ 9 (y+1)^2 4 × 4(y 1)^2 ≥ 0 (3(y+1) 4(y ...

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Byju's Answer

Standard XII

Mathematics

Range

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

Question

$f (x) = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4} t h e n r a n g e o f f (x) i s$

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Solution

The correct option is C
$y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4} y x^{2} + 3 x y + 4 y = x^{2} - 3 x + 4 x^{2} (y - 1) + 3 x (y + 1) + 4 (y - 1) = 0 D i s 1 \geq 0 \Rightarrow 9 (y + 1)^{2} - 4 \times 4 (y - 1)^{2} \geq 0 (3 (y + 1) - 4 (y - 1) 3 (y + 1) 4 - 4 (y - 1) \geq 0) (- y + 7) (7 y - 1) \geq 0 (y - 7) (y - \frac{1}{7}) \leq 0 \frac{1}{7} \leq y \leq 7$

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

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

The correct option is C y=x2−3x+4x2+3x+4yx2+3xy+4y=x2−3x+4x2(y−1)+3x(y+1)+4(y−1)=0Dis 1≥0⇒9(y+1)2−4×4(y−1)2≥0(3(y+1)−4(y−1)3(y+1)4−4(y−1)≥0)(−y+7)(7y−1)≥0(y−7)(y−17)≤017≤y≤7

$f (x) = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4} t h e n r a n g e o f f (x) i s$