Let fx=x^2+2px+2q^2 and gx=-x^2-2qx+p^2 (where q≠0). If x∈R and the the minimum value of fx is equal to the maximum value of gx, then the value of…

Question

Let fx=x^2+2px+2q^2 and gx=-x^2-2qx+p^2 (where q≠0). If x∈R and the the minimum value of fx is equal to the maximum value of gx, then the value of (p^2)/(q^2) is equal to

MARKS Content Team · Accepted Answer

Correct answer: 0.5. Since the minimum/maximum value of a quadratic expression ax^2+bx+c is given by (4bc-b^2)/4a So, f_min=2q^2-p^2, g_min=p^2+q^2 Now, 2q^2-p^2=p^2+q^2⇒q^2=2p^2 Hence, p^2q^2=1/2=0.5

Let $f (x) = x^{2} + 2 p x + 2 q^{2}$ and $g (x) = - x^{2} - 2 q x + p^{2}$ (where $q \neq 0$ ). If $x \in R$ and the the minimum value of $f (x)$ is equal to the maximum value of $g (x)$ , then the value of $\frac{p^{2}}{q^{2}}$ is equal to

Solution

More Quadratic Equation Questions

Let fx=x2+2px+2q2 and gx=-x2-2qx+p2 (where q≠0). If x∈R and the the minimum value of fx is equal to the maximum value of gx, then the value of p2q2 is equal to

Solution

More Quadratic Equation Questions

Let $f (x) = x^{2} + 2 p x + 2 q^{2}$ and $g (x) = - x^{2} - 2 q x + p^{2}$ (where $q \neq 0$ ). If $x \in R$ and the the minimum value of $f (x)$ is equal to the maximum value of $g (x)$ , then the value of $\frac{p^{2}}{q^{2}}$ is equal to