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

Solution

✅ Correct Option: 2

For a function to be increasing on an interval [1,2][1,2], the derivative must be non-negative throughout that interval.

Given: f(x)=2x2−kx+7f(x) = 2x^2 - kx + 7

Taking the derivative:

f′(x)=4x−kf'(x) = 4x - k


For f(x)f(x) to be increasing on [1,2][1,2]:

f′(x)≥0f'(x) \geq 0 for all x∈[1,2]x \in [1,2]

4x−k≥04x - k \geq 0

k≤4xk \leq 4x


Since k≤4xk \leq 4x must hold for all values of xx in [1,2][1,2], kk must be less than or equal to the minimum value of 4x4x on this interval.

The function 4x4x is increasing, so its minimum on [1,2][1,2] occurs at x=1x = 1:

4(1)=44(1) = 4

Therefore:

k≤4k \leq 4


The interval for kk is (−∞,4](-\infty, 4].

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