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
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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Answer:
p cannot be 10.
Step-by-step explanation: