kyotsu-test 2018 QCourse1-II-Q2
Intersection/tangency conditions between two curves
Let us consider the three quadratic functions
$$f ( x ) = - x ^ { 2 } - 2 x + 1 , \quad g ( x ) = - x ^ { 2 } + 4 x , \quad h ( x ) = 2 x ^ { 2 } + a x + b$$
(1) When we denote the discriminant of the quadratic equation $h ( x ) - f ( x ) = 0$ by $D _ { 1 }$ and the discriminant of the quadratic equation $h ( x ) - g ( x ) = 0$ by $D _ { 2 }$, we have
$$D _ { 1 } = \mathbf { N } , \quad D _ { 2 } = \mathbf { O }$$
(for N and O, choose the correct answers from among choices (0) $\sim$ (5) below). (0) $a ^ { 2 } + 4 a - 3 b + 7$
(1) $a ^ { 2 } - 8 a - 12 b + 16$
(2) $a ^ { 2 } + 4 a - 12 b + 16$
(3) $a ^ { 2 } + 8 a + 12 b + 16$
(4) $a ^ { 2 } - 4 a + 12 b + 16$
(5) $a ^ { 2 } - 8 a - 3 b + 7$
(2) The values of $a$ and $b$ such that both of the two equations $f ( x ) = h ( x )$ and $g ( x ) = h ( x )$ have only one real solution are
$$a = \mathbf { P } , \quad b = \frac { \mathbf { Q } } { \mathbf{4} } .$$
In this case, the solution of $f ( x ) = h ( x )$ is $x = - \frac { \mathbf { S } } { \mathbf{T} }$ and the solution of $g ( x ) = h ( x )$ is $x = \frac { \mathbf { U } } { \mathbf{4} }$.
(3) Let $b = 3$. Then the range of the values of $a$ such that both $f ( x ) < h ( x )$ and $g ( x ) < h ( x )$ hold for any $x$ is $\square$ W (for $\square$ W, choose the correct answer from among choices (0) $\sim$ (5) below). (0) $- 2 - 2 \sqrt { 6 } < a < 10$
(1) $a < - 2 - 2 \sqrt { 6 } , 10 < a$
(2) $a < - 1 - \sqrt { 6 } , 10 < a$
(3) $- 2 < a < - 1 + \sqrt { 6 }$
(4) $- 2 < a < - 2 + 2 \sqrt { 6 }$
(5) $- 1 - \sqrt { 6 } < a < 10$