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Question

Let f(x) be one-one and onto function, such that |f(x)f1(x)|>0 for x ϵ (0,2)(2,4) given f(2)=2,f(4)=0 and f(0)=4. Let g(x)=16x2. Given f(x)<g(x)  x ϵ (0,4) and graph of y=f(x) and y=f1(x) are symmetrical about the line x+y=4.

If f(x)f1(x)<0 for x ϵ (0,2) and 20f(x)dx=5, then 42f1(x)dx is
 
  1. 2
  2. 1
  3. 5
  4. 7


Solution

The correct option is B 1

20f(x)dx=5= shaded region PSTR

42f1(x)dx= area of TRM which is equal to the area of PQR.

Area of PQR = area of PSTR – area QSTR
=5(2×2)=54=1

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