Consider the identity function `I_N : N->N` defined as, `I_N(x)=x` for all `x in N` . Show that although `I_N` is onto but `I_N+I_N : N->N` defined as `(I_N+I_N)(x)=I_N(x)+I_N(x)=x+x=2x` is not onto.

Consider the identity function, I_N: N rarr N defined as I_n(x)=x forall x in N . Show that although I_N is onto but I_N+I_N: N rarr N defined as. (I_N+I_N)(x)=I_N(x)+I_N(x)=x+x=2 x is not onto

Consider the identity function I_N : N rarr N defined as I_N (x) = x, forall x in N Show that although I_N is onto but I_N + I_N : N rarr N defined as : (I_N + I_N) (x) = I_N (x) + I_N (x) = x + x = 2x is not onto.

Consider the identity function I_(N):NrarrN defined as : I_(N)(x)=xAAx inN . Show that although I_(N) is onto but I_(N)+I_(N):NrarrN defined as : (I_(N)+I_(N))(x)=I_(N)(x)+I_(N)(x)=x+x=2x is not onto.

Consider the identity function I_(N) : N rarr N defined as I_(N) (x) = x,AAx in N . Show that although I_(N) is onto but I_N + I_(N) :N rarr N defined as (I_(N) +I_(N))(x) = I_(N)(x) +I_(N) (x) = x+x=2x is not onto.

Consider the identity function I_(N):N to N defined as I _(N) (x)=x AA x in N. Show that although I _(N) is onto but I _(N) + I _(N) : N to N defined as (I _(N) + I _(N)) (x) = I _(N) (x) + I _(N) (x) = x +x = 2x is not onto.

Consider the identity function I_(N):N to N defined as I _(N) (x)=x AA x in N. Show that although I _(N) is onto but I _(N) + I _(N) : N to N defined as (I _(N) + I _(N)) (x) = I _(N) (x) + I _(N) (x) x +x = 2x is not onto.

If n in N and I_(n) = int (log x)^(n) dx , then I_(n) + n I_(n - 1) =