Question

the number of solutions of the equation a^f(x)+g(x)=0 where a>0,g(x) is not equals to 0 and g(x)has minimum value1/2 is
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Answer 1

Given :   $a^{f(x)} + g(x) = 0$   a>0 g(x)has minimum value = 1/2

To find : possible number of solutions of the equation

Solution:

$a^{f(x)} + g(x) = 0$

g(x) minimum value  = 1/2

=> g(x) ≥ 1/2

=>   $a^{f(x)} \leq \frac{-1}{2}$

Lets consider three cases  f(x) < 0   , f(x) = 0  , f(x) > 0

f(x) = - n  , f(x) = 0  , f(x) = n  where n > 0

 a⁻ⁿ  =  1/aⁿ     > 0    as   a  > 0  

0  > -1/2   hence not possible

 a⁰  =  1    

1  > -1/2   hence not possible

=> aⁿ    > 0    as   a  > 0  

0  > -1/2   hence not possible

so no possible Solution

Hence number of Solution = Zero

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