f ( x ) = x 2 − 3 x + 4 x 2 + 3 x + 4 the range of f(x) is Options: A . &nbsp [ 0 , 1 7 ] B . &nbsp ( − ∞ , 1 7 ) ∪ ( 7 , ∞ ) C . &nbsp ( − ∞ , 7 ) D . &nbsp [ 1 7 , 7 ]

Answer: Option D : D y = 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 ≥ 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 ) ( 7 y − 1 ) ≥ 0 ( y − 7 ) ( y − 1 7 ) ≤ 0 1 7 ≤ y ≤ 7

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

Discussion Forum : Relations And Functions Ii

Question -

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

Options:

A .

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

B .

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

C .

(- \infty, 7)

D .

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

Answer: Option D
:
D

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

\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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Discussion Forum : Relations And Functions Ii

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