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Q:-Let 0 ∫ 1 f(x)dx=1 , 0 ∫ 1 xf(x)dx=2 and 0 ∫ 1 x 2 f(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

Q:-Let 01 f(x)dx=1 , 01 xf(x)dx=2 and 01 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

Grade:12

1 Answers

Harish kumar
18 Points
11 years ago

  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

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