If f(x) = x^2 - 2(4K - 1)x + g(K) 0 for all | vedclass

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(D) For $f(x) = x^2 - 2(4K - 1)x + g(K) > 0$ to hold for all $x \in R$,the discriminant $D$ must be less than $0$.$D = [-2(4K - 1)]^2 - 4(1)(g(K)) < 0

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$g(K)$ attains no maximum and no minimum in $(a, b)$

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(D) For $f(x) = x^2 - 2(4K - 1)x + g(K) > 0$ to hold for all $x \in R$,the discriminant $D$ must be less than $0$.$D = [-2(4K - 1)]^2 - 4(1)(g(K)) $4(16K^2 - 8K + 1) - 4(15K^2 - 2K - 7) $16K^2 - 8K + 1 - 15K^2 + 2K + 7 $K^2 - 6K + 8 $(K - 2)(K - 4) Thus,$K \in (2, 4)$,so $a = 2$ and $b = 4$.The function $g(K) = 15K^2 - 2K - 7$ is an upward-opening parabola with its vertex at $K = -(-2) / (2 	imes 15) = 1/15$.Since $1/15 
otin (2, 4)$,the function $g(K)$ is strictly increasing on the interval $(2, 4)$.Therefore,$g(K)$ does not attain a maximum or a minimum value within the open interval $(2, 4)$.

If $f(x) = x^2 - 2(4K - 1)x + g(K) > 0$ for all $x \in R$ and for $K \in (a, b)$. If $g(K) = 15K^2 - 2K - 7$,then:

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