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If the function f(x)=x2−2kx+7f(x) = x^2-2kx+7 is increasing on [1, 5] then kk belongs to

Solution

✅ Correct Option: 2

f′(x)=2x−2k≥0f'(x) = 2x-2k \geq 0 for all x∈[1,5]x\in[1,5] requires k≤xk \leq x throughout, so kk must not exceed the least value of xx.

Hence k≤1k \leq 1, i.e. k∈(−∞,1]k \in (-\infty, 1].

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