If the range of function f(x)=x+1k+x2 contains the interval [0, 1], then value of k can be equal to

y∈[0,1]where y=x+1k+x2y=0 when x=−1If y≠0,yx2−x+ky−1=0;x∈R⇒D≥0⇒1−4y(ky−1)≥0⇒4ky2−4y−1≤0∀y∈[0,1]                      (Given)f(0)≤0 and f(1)≤0⇒−1≤0 and 4k−4−1≤0⇒k≤54∴ k∈−∞,54Also, k > 0, otherwise graph of 4ky2 - 4y - 1 is concave downward and is negative for unbounded interval of values of y.Hence, k∈0,54.