2 graphs of functions and intersect at points A and B. At both A and B, x is not equal to 0. At on the second graph, abscissa is p. Find the perimeter of triangle ABC if the point (p,0) is considered as point C. (Till 2 decimal points)
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Taking logarithm and equating both of functions:
ln x 2 = ln x x 1 ( 2 − x 1 ) ln x = 0 ⟹ x = 1 and x = 2 1
So, we have A ( 1 , 1 ) and B ( 2 1 , 4 1 ) .
Now, taking the derivative of the second function,
ln y = x 1 ln x y y ′ = − x 2 1 ln x + x 1 × x 1 = x 2 1 ( 1 − ln x ) x 2 x x ( 1 − ln x ) = 0 ⟹ x = e = p
and we have C ( e , 0 ) . Now, the distance of point A to B , A to C , and B to C are 0 . 9 0 1 , 1 . 9 8 8 , and 2 . 2 3 2 respectively, and, finally the perimeter of triangle ABC is 5 . 1 2