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Question

If 22x+2a2x+2+54a0 has atleast one real solution, Then a ϵ


A

(,1]

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B

(3,1]

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C

[1,)

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D

(87,1]

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Solution

The correct option is A

(,1]


22x+2a2x+2+54a0Let t=2x 4t24at+54a04t2+54a+4at4t2+5t+14a 4a4t2+5t+1
t such that 4a4t2+5t+14amin4t2+5t+1
Let f(t)=4t2+5t+1f(t)=(t+1)(8t)(4t2+5)(t+1)2
f(t)=8t2+8t4t25(t+1)2 f(t)=0 4t2+8t5=0 t=12, 52 (rejected as t>0) f(12)=4×14+512+1=4Also as t0f(t)5 4a4=1 a1


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