106|x|2 - |x| + m = 0
= > |x| = 1 ± √1 - 4m2
|x| is real and ≥ 0
so, 1 - 4m ≥ 0 or m ≤ 1/4 .... (1)
|1 - 4m | ≤ 1
=> 16m2 - 8m ≤ 0
=> 2m(m-1) ≤ 0
=> m ε [0,1] ... (2)
from (1) and (2) m ε [0,1/4]
\hspace{-16}$Find all Real Value of $\mathbf{m}$ for Which the Given Equation\\\\ $x^2-\mid x \mid+m=0 $ has Real solution
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1 Answers
106