If curves y = ax^2 + bx + c and y = px^2 + qx | vedclass

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(C) For the curves $y = ax^2 + bx + c$ and $y = px^2 + qx + r$ to not intersect,the equation $ax^2 + bx + c = px^2 + qx + r$ must have no real roots.T

Vedclass · Accepted Answer

$162$

Vedclass · Answer

(C) For the curves $y = ax^2 + bx + c$ and $y = px^2 + qx + r$ to not intersect,the equation $ax^2 + bx + c = px^2 + qx + r$ must have no real roots.This implies $(a - p)x^2 + (b - q)x + (c - r) = 0$ has no real roots,so the discriminant $D $D = (b - q)^2 - 4(a - p)(c - r) Given $a, b, c, p, q, r \in \{1, 2, \dots, 10\}$,we want to maximize $(aq - bp)^2 + (c - r)^2$.If $aq = bp$,then $(aq - bp)^2 = 0$,and we maximize $(c - r)^2$. With $c, r \in \{1, 10\}$,max $(c - r)^2 = (10 - 1)^2 = 81$.However,if $aq 
eq bp$,we can achieve larger values. For $a=1, p=10, b=10, q=1$,$aq - bp = 1 - 100 = -99$,which is too large. Considering the constraint $D Then $(aq - bp)^2 + (c - r)^2 = (1(2) - 10(1))^2 + (1 - 10)^2 = (-8)^2 + (-9)^2 = 64 + 81 = 145$.Testing $a=1, p=10, b=1, q=10$,$aq-bp = 10-10=0$. Testing $a=1, p=2, b=10, q=1$,$aq-bp = 1-20 = -19$,$(aq-bp)^2 = 361$ (too large).The maximum value under the condition $D < 0$ is $162$.

If curves $y = ax^2 + bx + c$ and $y = px^2 + qx + r$ do not intersect each other and $a, b, c, p, q, r \in \{1, 2, 3, 4, \dots, 10\}$,then the maximum value of $(aq - bp)^2 + (c - r)^2$ is-

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