The set of values of p for which x2−px+sin−1(sin4)>0 for all real x is given by :
Options:
A .  
(–4,4)
B .  
(−∞,−4)∪(4,∞)
C .  
ϕ
D .  
None of these
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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