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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

If both the roots of the quadratic equation $x^2 - 2kx + k^2 + k - 5 = 0$ are less than $5$,then $k$ lies in the interval:

For the quadratic equation $x^2 - (K + 1)x + (K^2 + K - 8) = 0$,if one root is greater than $2$ and the other root is less than $2$,then $K$ lies in which interval?

If $y = f(x) = ax^2 + 2bx + c = 0$ has imaginary roots and $4a + 4b + c < 0$,then :-

The real number $k$ for which the equation $2x^2 + 3x + k = 0$ has two distinct real roots in the interval $[0, 1]$.

The value of $p$ for which both the roots of the equation $4x^2 - 20px + (25p^2 + 15p - 66) = 0$ are less than $2$ lies in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module