Let e be the base of natural logarithm and let f : \1, 2, 3, 4\ \1, e, e2, e3 \ and g : \1, e, e2, e3 \ \1, 12, 13, 14 \ be two bijective functions such that f is strictly decreasing and g is strictly increasing. If (x)= [f-1 \g-1 ( 12 ) \ ]x, then the area of the region R= \(x, y) : x2 y (x), . . 0 x 1 \ is :
- 3- e(2)3 e(2)
- 13 e(2)
- 3+ e(2)
- 3+ e(2)2+ e(3)
Answer
(1) 3- e(2)3 e(2)
Solution
[figure]
f strictly decreasing : f(1)=e3, f(2)=e2, f(3)=e, f(4)=1
g strictly increasing, 14< 13< 12<1 : g(1)= 14, g(e)= 13, g (e2 )= 12, g (e3 )=1
g-1 ( 12 )=e2 and f-1 (e2 )=2
(x)=2x
For 0 x 1 : 2x 1 x2, so R lies between y=x2 (below) and y=2x (above)
Area = 01 (2x-x2 )dx= [ 2x e 2- x33 ]01
= 2-1 e 2- 13= 3- e(2)3 e(2)
Answer: (1)