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Question

The least positive value of a for which 4xa2xa+30 is satisfied by atleast one real value of x is

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Solution

Let f(x)=4xa2xa+3 and x1,x2 be the roots of f(x)=0
Let t=2x,t(0,) xR
g(t)=t2ata+3
Let t1,t2 be the roots of g(t)=0

For f(x)0 to have atleast one real x
g(t)0 should have one real t, where t(0,)

Required cases:
Case I: Both roots are positive


So,
(1) D0a2+4a120
(a+6)(a2)0
a(,6][2,)

(2) B2A>0a2>0a>0

(3) g(0)>0a+3>0a<3 a[2,3) ...(i)

Case II: One root is positive other root is negative


0 lies in between the roots
So, g(0)<0a>3 ...(ii)

Case III: One root is 0 and other root is greater than 0


g(0)=0a=3
B2A>0a2>0a>0
a=3 ...(iii)

From (i),(ii) and (iii)
a[2,)

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