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For x>1, y= log x satisfy the inequality...

For x>1, y= log x satisfy the inequality

Text Solution

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Here, `y = logx => dy/dx = 1/x`
(a) Let `y_1 = x-1`
`=>dy_1/dx = 1`
As, `x gt 1`, so `dy_1/dx gt dy/dx`
`:. x-1 gt y ` is true.

(b)Let `y_2 = x^2-1`
`=>dy_2/dx = 2x`
...
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