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Let f:RtoR be a differentiable function such that f(x)=x^(2)+int_(0)^(x)e^(-t)f(x-t)dt. y=f(x) is |
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Answer» injective but not surjective `x^(2)+int_(0)^(x)e^(-x-t)f(x-(x-t))dt` [Using `int_(a)^(b)f(x)DX=int_(a)^(b)f(a+b-x)dx`] `=x^(2)+e^(-x)int_(0)^(x)e^(t)f(t)dt`……………..2 Differentiating w.R.t.`x` we get `f'(x)=2x-e^(-x)int_(0)^(x)e^(t)f(t)dt+e^(-x)e^(x)f(x)` `=2x-e^(-x)int_(0)^(x)e^(t)f(t)dt+f(x)` `=2x+x^(2)` [using equation 2] `:. f(x)=(x^(3))/3+x^(2)+c` Also `f(0)=0` [from equation 1] or `f(x)=(x^(3))/3+x^(2)` or `f'(x)=x^(2)+2x` Thus `f'(x)=` has real roots. Hence `f(x)` is non monotonic. Hence `f(x)` is many one but range is `R` and hence, is surjective `int_(0)^(1)f(x)dx=int_(0)^(1)((x^(3))/3+x^(2))dx` `=[(x^(4))/12+(x^(3))/3]_(0)^(1)` `=1/12+1/3` `=5/12` |
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