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Bounded by the axes of reference and the normal

to y=logex(log base e) at the point (1, 0) is
(a)1unitsquare (b)2unitsquare (c)1/2unit square (d)
none of these
i need this solution for this problem..please, help me out..:)


First of all find the equation of normal to curve at (1, 0)

y = log_ex \\  \frac{dy}{dx} =  \frac{1}{x} \\  (\frac{dy}{dx})_(_x _= _1_, _y _= _0_) = 1

This is slope of tangent so slope of normal is -1.
So equation of line -> (y - 0) = (-1) * (x - 1) -> x + y = 1

This is the line intersection x-axis at (1, 0) and y-axis at (0, 1), forming a right angle triangle with height = 1 unit and base = 1 unit.

So area = \frac{1}{2} * 1 * 1

1 5 1
is it not the best answer?? if you know what i mean...
offcourse...but,it is right answer na..it's enough for me..:)
then why don't you pick this as best answer... it will help me na.. :P
i did it...i have given full stars for this answer...:)..a return thanks in the form which u would like it..:P
of course :P
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