Q. 8

# <span lang="EN-US

y =loge (x + e) and x = loge look like – The curves intersect at (0, 1)

(Putting x = 0 in the 2 curves, y = loge(e) = 1 and 0 = loge(1/y),

i.e., y = 1/e0 = 1)

So, bounds are x = 1 – e to x = 0 for the first curve and then x = 0 to apparently x = ∞ for the second curve.

Therefore,   (Ans)

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