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No. It shows the weak output with only a 0.5V voltage drop.
With a 10V supply, its output sinking voltage is +0.5V and its current is typically 2.6ma.
When its output sourcing voltage is +9.5V then its current is also typically 2.6mA.
If you have a load like a 2V LED that causes more voltage drop in the output transistors then the current is typically 14ma sinking and 18mA sourcing.
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