R a n g e o f t h e f u n c t i o n f ( x ) = x 2 + x + 2 x 2 + x + 1 ; x ϵ R i s Options: A . &nbsp ( 1 , ∞ ) B . &nbsp ( 1 , 11 7 ] C . &nbsp ( 1 , 7 3 ] D . &nbsp ( 1 , 7 5 ]

Answer: Option C : C W e h a v e , f ( x ) = x 2 + x + 2 x 2 + x + 1 = ( x 2 + x + 1 ) x 2 + x + 1 = 1 + 1 ( x + 1 2 ) 2 + 3 4 W e c a n s e e h e r e t h a t a s x → ∞ , f ( x ) → 1 w h i c h i s t h e m i n v a l u e o f f ( x ) . A l s o f ( x ) i s m a x w h e n ( x + 1 2 ) 2 + 3 4 i s m i n w h i c h i s s o w h e n x = − 1 2 a n d t h e n 3 4 . ∴ f m a x = 1 + 1 3 4 = 7 3 ∴ R i = ( 1 , 7 3 ]

Range of the function f(x)=x2+x+2x2+x+1;xϵR is

Question

$R a n g e o f t h e f u n c t i o n f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}; x ϵ R i s$

Options:

A .

(1, \infty

)

B .

(1, \frac{11}{7}]

C .

(1, \frac{7}{3}]

D .

(1, \frac{7}{5}]

Answer: Option C
:
C

$W e h a v e, f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1} = \frac{(x^{2} + x + 1)}{x^{2} + x + 1} = 1 + \frac{1}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} W e c a n s e e h e r e t h a t a s x \to \infty, f (x) \to 1 w h i c h i s t h e m i n v a l u e o f f (x) . A l s o f (x) i s m a x w h e n {(x + \frac{1}{2})}^{2} + \frac{3}{4} i s m i n w h i c h i s s o w h e n x = - \frac{1}{2} a n d t h e n \frac{3}{4} . ∴ f_{m a x} = 1 + \frac{1}{\frac{3}{4}} = \frac{7}{3} ∴ R_{i} = (1, \frac{7}{3}]$