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```				   Q:-Let 0∫1 f(x)dx=1 , 0∫1 xf(x)dx=2 and 0∫1 x2f(x)dx=4 , then the number of functions y=f(x) for all x belongs to [0,1],such that f(x)>0 and continuous for all x belongs to [0,1] is
A. 0
B. 1
C. 2
D. 8
```

6 years ago

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```										  it' s answer is A i.e 0 becouse if we integrate xf(x) from 0 to 1 we get (X-1) which is given to be 2 hence x=3 and if integrate x^2f(x) from 0 to 1 we get 2x-2 which is given to 4 hence x=3 and putting the value of X=3 in second and third integrals.

int. of f(X) from 0 to 1=1 in first case
int. of f(X) from 0 to 1=2/3 in second case
int. of f(X) from 0 to 1=4/9 in third case

How can same function have different area from 0 to 1 . Therefore answers is 0
```
6 years ago

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