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+x2 y=0whenx=−1If y≠0,yx2−x+ky−1=0;x∈R⇒D≥0⇒1−4y(ky−1)≥0⇒4ky2−4y−1≤0 ∀y∈[0,1] f(0)≤0andf(1)≤0⇒−1≤0and4k−4−1≤0⇒k≤54∴k∈(−∞,54]Also, 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