Question

Q 53/75 Let P(x)=x2-(2-P)x+P-2. If
P(x) assumes both positive and negative
values for xe R, then the range of Pis:

Answer 1

Given :  P(x)   = x²  - (2 -P)x  + P - 2 P(x) assumes both positive and negative values for x ∈ R

To find : Range of  P

Solution:

P(x)   = x²  - (2 -P)x  + P - 2

let sat P - 2  = y

P=> (x)  = x²  + yx  + y

=>P(x) = (x  + y /2)² - y²/4  + y

(x  + y /2)²  is always ≥ 0

so P(x)to have - ve  vales  - y²/4  + y  must be - ve

=> y - y²/4   <  0

=> 4y  - y²  < 0

=> y(4 - y)  < 0

=>   y < 0    or  y > 4  

=> P - 2 < 0    P - 2 > 4

=> P < 2   or  P > 6

Range of  P   = (-∞ , 2 ) ,  (6 , ∞ )

or P ∈ R  - [2 , 6 ]

Range of  P   = (-∞ , 2 ) ,  (6 , ∞ )

or P ∈ R  - [2 , 6 ]

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