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If the maximum and minimum values of y=(...

If the maximum and minimum values of `y=(x^2-3x+c)/(x^2+3x+c)` are 7 and `1/7` respectively then the value of c is equal to

A

3

B

4

C

5

D

6

Text Solution

Verified by Experts

The correct Answer is:
4
`:'y=(x^(2)-3x+c)/(x^(2)+3x+c)`
`impliesx^(2)(y-1)+3x(y+1)+c(y-1)=0`
`:' x epsilon R`
`:.9(y+1)^(2)-4c(y-1)^(2) ge 0`
`(2sqrt(c)y-2sqrt(c))^(2)-(3y+3)^(2)le0`
`implies{(2sqrt(c)+3)y-(2sqrt(c)-3)}{(2sqrt(c)-3)y-(2sqrt(c)+3)}le0`
or `(2sqrt(c)-3)/(2sqrt(c)+3)leyle(2sqrt(c)+3)/(2sqrt(c)-3)`
But given `(2sqrt(c)+3)/(2sqrt(c)-3)=7`
`implies2sqrt(c)+3=14sqrt(c)-21`
or `12sqrt(c)=24` or `sqrt(c)=2`
`:.c=4`
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