We are given a random variable X which has an exponential distribution of parameter λ=1.
$$X\sim\exp(λ=1)$$
We know that $$E[X]=\frac{1}{λ}$$
Hence for us $E[X]=1$. By virtue of the memoryless property of exponential distribution, I have understood that:
$$E[X\mid X>x]=x+E[X]=x+\frac{1}{λ}=x+1$$
but what about the either case:
$$E[X\mid X\le x]= ??$$
I suppose something like:
$$E[X\mid X\le x]=x-E[X]=x-1$$
but then if $x$ is smaller than $E[X]$ (in this case $E[X]=1$) it would result in a negative expectation. Is that correct or such either case is not expressed like I did?
We have $$E(X)=E(X|X\le k)\Pr(X\le k)+E(X|X\gt k)\Pr(X\gt k).\tag{1}$$ You know $E(X)$, $E(X|X\gt k)$ and the probabilities mentioned in (1). Now $E(X|X\le k)$ is determined.
There are other ways to solve the problem, for example by finding the conditional distribution of $X$ given $X\le k$.