The values of k, for which `|k" "veca|ltveca| andk" "veca+(1)/(2)veca` is parallel to `veca` holds true are……..

Explanation

We have, `|k" "veca|lt|veca| andkveca+(1)/(2)veca`
`|kveca|lt|veca|implies|k||veca|lt|veca|`
`implies|k|lt1implies-1ltklt1`
Also, since `kveca+(1)/(2)veca` is parallel to `veca`, then we see that at `k=(-1)/(2),kveca+(1)/(2)veca` becomes a null vector and then it will not be parallel to `veca`.
So, `kveca+(1)/(2)veca` is parallel to `veca` holds true when `kin]-1,1[kne(-1)/(2)`,