Question

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thanxx ​

Answer 1

Given

$y = \frac{x}{2} where \: x > 0 \\ point \: p(a \: {a}^{2}) \\ so \\ y \: - \frac{x}{2} > 0 \\ {a}^{2} - \frac{a}{2} > 0 equation \: 1\\ second \: equation \\ y = 3x \\ y - 3x > 0 \\ {a}^{2} - 3 a > 0 equation \: 2 \\ on \: solving \: above \:two \: equation \: \\a/3 < a^2 <3a\\ </p><p>so \\ \frac{1}{2} < a < 3\\(1/3,3)$

Now

a

Answer 2

Answer:

a ∈ (1/2, 3)

Step-by-step explanation:

Given lines are y = x/2 and y = 3x.

a² - (a/2) > 0  

=> a(a - 1/2) > 0

=> a ∈ (-∞,0) ∪ (1/2, ∞) ----- (1)

Now,

a² - 3a < 0  

=> a(a - 3) < 0

=> a ∈ (0,3)  ------- (2)

Solve (1) and (2), we will get

a ∈ (1/2, 3)

Hope this helps!