Question
The set of values of p for which x2−px+sin−1(sin4)>0 for all real x is given by :
Answer: Option C
:
C
x2−px+sin−1(sin4)>0 for all real x.
⇒x2−px+sin−1(sin4)>0
⇒x2−px+(π−4)>0∀xϵR
⇒D=p2−4(π−4)<0 ⇒p2+16−4π<0
Since 16−4π>0,p2+16−4π cannot be negative for any value of pϵR.
∴ Set of values of p=ϕ
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:
C
x2−px+sin−1(sin4)>0 for all real x.
⇒x2−px+sin−1(sin4)>0
⇒x2−px+(π−4)>0∀xϵR
⇒D=p2−4(π−4)<0 ⇒p2+16−4π<0
Since 16−4π>0,p2+16−4π cannot be negative for any value of pϵR.
∴ Set of values of p=ϕ
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