If `2^2pi//sin^((-1)x)-2(a+2)^pi//sin^((-1)x)+8a<0` for at least one real `x ,` then `1/8lt=a<2` (b) `a<2` `a in R-{2}` (d) `a in [0,1/8]uu(2,oo)`
If `2^2pi//sin^((-1)x)-2(a+2)^pi//sin^((-1)x)+8a<0`
for at least one real `x ,`
then
`1/8lt=a<2`
(b) `a<2`
`a in R-{2}`
(d) `a in [0,1/8]uu(2,oo)`
A. `(1)/(8) le a le 2`
B. `a le 2`
C. `a in R - {2}`
D. `a in [0, (1)/(8)) uu (2, oo)`
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Correct Answer - D
`2^(2pi//sin^(-1)x) - 2 (a + 2) 2^(pi//sin^(-1) x) + 8a lt 0`
`(2^(pi//sin^(-1)x) - 4) (2^(pi//sin^(-1)x) - 2a) lt 0`
Now `2^(pi//sin^(-1) x) in (0, (1)/(4)] uu [4, oo)`
For `2^(pi//sin^(-1)x) in (0, (1)/(4)]`. We have
`(2^(pi//sin^(-1)x) -4) lt 0`
`:. 2^(pi//sin^(-1)x) -2a gt 0`
or `2a lt (1)/(4)`
or `0 le a (1)/(8)`
Similarly, for `2^(pi//sin^(-1) x) in [4, oo), a gt 2`, we get
So, `a in [0, (1)/(8)) uu (2, oo)`
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