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The values of k that satisfies the inequation (k-12) ≥ k2-9k + 12 are in the range [a, b]. Then find the values of a and b,

Question

The values of k that satisfies the inequation

(k12)k29k+12(k-12) \geq k^2 - 9k + 12
are in the range [a,b][a, b]. Then find the values of a and b.

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Solution

To solve this inequality, we first need to rearrange the terms to one side of the inequality. This gives us:

k^2 - 10k + 12 ≤ 0

This is a quadratic inequality. To solve it, we first need to find the roots of the corresponding quadratic equation, which are the values of k for which:

k^2 - 10k + 12 = 0

This can be factored to:

(k - 2)(k - 6) = 0

Setting each factor equal to zero gives the roots k = 2 and k = 6.

The quadratic inequality will change sign at each of these roots. To determine the sign in each interval, we can test a value in each interval in the inequality.

For k < 2, we can test k = 0. Substituting k = 0 into the inequality gives 12 ≤ 0, which is false. So the inequality is not satisfied for k < 2.

For 2 ≤ k ≤ 6, we can test k = 4. Substituting k = 4 into the inequality gives 4 ≤ 0, which is also false. So the inequality is not satisfied for 2 ≤ k ≤ 6.

For k > 6, we can test k = 7. Substituting k = 7 into the inequality gives -1 ≤ 0, which is true. So the inequality is satisfied for k > 6.

Therefore, the values of k that satisfy the inequality are in the range (-∞, 2) and (6, ∞). So, a = -∞ and b = ∞.

This problem has been solved

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