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Q.

Number of integers satisfying logx⁡4x+56−5x<−1 is

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a

0

b

1

c

2

d

3

answer is A.

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Detailed Solution

logx⁡4x+56−5x<−1 We must have 4x+56−5x>0⇒ 4x+55x−6<0⇒ x∈−54,65 Also x>0 and x≠1∴ x∈0,65−{1}-----(1) Case I: 01x⇒ 4x+56−5x−1x>0⇒ 4x2+5x+5x−6x(6−5x)>0⇒ 4x2+10x−6x(5x−6)<0⇒ 2(x+3)(2x−1)x(5x−6)<0⇒ x∈(−3,0)∪12,65-----(3) From (i), (ii) and (iii), we get ∴ x∈12,1 Case II: x>1----(4)logx⁡4x+56−5x<−1⇒ 4x+56−5x<1x⇒ x∈(−∞,−3)∪0,12∪65,∞-----(5) From (i), (iv) and (v), we get x∈ϕ Thus, x∈12,1
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