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

  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