Math, asked by vini785, 1 year ago

If p(x) = -x^2 + px - p - 8 and p(x) <0 for all the real values of x , then the value of p cannot be

Options
1. 7
2. -3
3. 0
4. 10


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Answers

Answered by amitnrw
7

Given :   p(x)  =  -x² + px - p - 8 p(x) <0 for all the real values of x ,

To find : the value of p cannot be

Solution:

p(x)  =  -x² + px - p - 8

p(x)  < 0

=>   -x² + px - p - 8 < 0

=> x² - px + p +  8 >  0

=> (x -  p/2)² - p²/4  + p +  8  > 0

=>  (x -  p/2)² > p²/4 - p  - 8

=> (x -  p/2)² >  ( p/2 - 1)² - 1 - 8

=> (x -  p/2)² >  ( p/2 - 1)² - 9

=> ( p/2 - 1)² <   9

=>  -3    < ( p/2 - 1) <  3    

=>    -2 <  p/2 <  4

=>   - 4 <  p <  8

-3 , 0 & 7 lies in between but 10 Does not

hence p can not be 10

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Answered by lovish165
1

Answer:

p cannot be 10.

Step-by-step explanation:

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