Complete set of values of x, satisfying the in equality x2+x2(x+1)2<54, is
A
(−12,1)
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B
(−52,12)
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C
(1,52)
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D
(−52,−12)
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Solution
The correct option is A(−12,1) x2+x2(x+1)2<54 ⇒(x−xx+1)2+2x2x+1<54 ⇒(x2x+1)2+(2x2x+1)<54 Let x2x+1=t ∴t2+2t−54<0 ⇒4t2+8t−5<0 ⇒4t2+10t−2t−5<0 ⇒2t(2t+5)−1(2t+5)<0 ⇒(2t−1)(2t+5)<0 ⇒−52<t<12
So x2x+1<12 and x2x+1>−52 ⇒2x2−x−1(x+1)<0,2x2+5x+52(x+1)>0 ⇒(2x+1)(x−1)x+1<0,2x2+5x+52(x+1)>0 So x∈(−12,1)