The complete set of values of x satisfying x2+x2(x+1)2<54 is
A
(−12,1)
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B
(−52,−12)
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C
(−52,12)
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D
(1,52)
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Solution
The correct option is A(−12,1) Given: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
Case I:t<12 ⇒x2x+1<12
⇒2x2−x−1(x+1)<0
⇒(2x+1)(x−1)x+1<0
⇒x∈(−∞,−1)∪(−12,1)⋯(i)
Case II.t>52 x2x+1>−52
⇒2x2+5x+52(x+1)>0 ⇒2x2+5x+5(x+1)>0⋯(A)
Now, 2x2+5x+5>0∀x∈R ⇒2x2+5x+5(x+1)>0
Thus, (A) becomes: ⇒1(x+1)>0 ⇒x∈(−1,∞)⋯(ii)
Thus, from both the solution set (i)&(ii), we get: x∈(−12,1)