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Question : 32 of 160
Marks:
+1,
-0
Solution:
Given,
=x2+x+3 =x−4+2 a . b > 6 ⇒(x2+x+3)(x−4+2)>6 ⇒x2.x+(x)(−4)+3(2>6 [a.b=a1a2+b1b2+c1c2, where
a=a1+b1j+c1k and
b=a2+b2j+c2]
⇒x3−4x+6>6 ⇒x3−4x+6−6>0 ⇒x3−4x>0 ⇒x(x2−4)>0 ⇒x(x−2)(x+2)>0 Using wavy curve method critical points are
x=0,2,−2
∴ Solution is
x∈(−2,0)∪(2,∞).
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