The problem gives two quadratic functions $f(x) = 2ax^2 + (2a+1)x + a$ and $g(x) = -ax^2 - x - 1$. (1) We need to find the expression for $f(x) - g(x)$ in the form $Jax^2 + K(a+L)x + a + M$ and determine the values of $J, K, L, M$. (2) Using the result from (1), we need to find the range of values of $a$ for which the two parabolas $y=f(x)$ and $y=g(x)$ have no intersection points. We are given options for the range of $a$.

1. Problem Description

The problem gives two quadratic functions $f(x) = 2ax^2 + (2a+1)x + a$ and $g(x) = -ax^2 - x - 1$.

(1) We need to find the expression for $f(x) - g(x)$ in the form $Jax^2 + K(a+L)x + a + M$ and determine the values of $J, K, L, M$.

(2) Using the result from (1), we need to find the range of values of $a$ for which the two parabolas $y=f(x)$ and $y=g(x)$ have no intersection points. We are given options for the range of $a$.

2. Solution Steps

(1) Calculate $f(x) - g(x)$:

$f(x) - g(x) = (2ax^2 + (2a+1)x + a) - (-ax^2 - x - 1)$

$f(x) - g(x) = 2ax^2 + (2a+1)x + a + ax^2 + x + 1$

$f(x) - g(x) = (2a+a)x^2 + (2a+1+1)x + (a+1)$

$f(x) - g(x) = 3ax^2 + (2a+2)x + a+1$

$f(x) - g(x) = 3ax^2 + 2(a+1)x + a+1$

Comparing this with the given form $Jax^2 + K(a+L)x + a + M$, we have:

$J=3$, $K=2$, $L=1$, $M=1$.

So, $f(x) - g(x) = 3ax^2 + 2(a+1)x + a+1$

(2) The two parabolas $y=f(x)$ and $y=g(x)$ have no intersection points if and only if the equation $f(x) = g(x)$ has no real solutions. This is equivalent to $f(x) - g(x) = 0$ having no real solutions.

Therefore, we need to find the values of $a$ for which the quadratic equation $3ax^2 + 2(a+1)x + (a+1) = 0$ has no real solutions.

If $a=0$, the equation becomes $2x+1=0$, which has a solution $x=-1/2$. So $a$ cannot be

0. If $a \ne 0$, the discriminant of the quadratic equation must be negative:

$D = b^2 - 4ac < 0$

$D = (2(a+1))^2 - 4(3a)(a+1) < 0$

$4(a+1)^2 - 12a(a+1) < 0$

$4(a+1)(a+1 - 3a) < 0$

$4(a+1)(1-2a) < 0$

$(a+1)(1-2a) < 0$

This inequality holds if either

(i) $a+1>0$ and $1-2a<0$, which means $a>-1$ and $2a>1$, so $a>-1$ and $a>1/2$. Thus $a > 1/2$.

(ii) $a+1<0$ and $1-2a>0$, which means $a<-1$ and $2a<1$, so $a<-1$ and $a<1/2$. Thus $a < -1$.

Therefore, the range of values of $a$ is $a<-1$ or $a>1/2$.

3. Final Answer

The range of $a$ is $a<-1$ or $a>1/2$, which corresponds to option (2).

$J = 3$

$K = 2$

$L = 1$

$M = 1$

Final Answer: $a<-1, 1/2<a$

Option (2) is the correct one.