Question

Find the number of integral values of a parameter for which the naquality 1 + log 2x square + 2x + 7 / 2 to the base 2 is greater than log ax square + a the base 2

Answer 1

7

Answer 2

The given inequality is $1 + \log_2\left(2x^2 + 2x + \frac{7}{2}\right) > \log_2(ax^2 + a)$.

1. Determine the domain of the logarithmic expressions: For $\log_2\left(2x^2 + 2x + \frac{7}{2}\right)$ to be defined, $2x^2 + 2x + \frac{7}{2} > 0$. The discriminant of $2x^2 + 2x + \frac{7}{2}$ is $\Delta = 2^2 - 4(2)\left(\frac{7}{2}\right) = 4 - 28 = -24$. Since $\Delta < 0$ and the leading coefficient $2 > 0$, the quadratic $2x^2 + 2x + \frac{7}{2}$ is always positive for all real $x$.

For $\log_2(ax^2 + a)$ to be defined, $ax^2 + a > 0$. This can be factored as $a(x^2 + 1) > 0$. Since $x^2 + 1$ is always positive (as $x^2 \ge 0 \implies x^2+1 \ge 1$), for $a(x^2 + 1)$ to be positive, we must have $a > 0$. This is a necessary condition for the parameter $a$.

2. Simplify the inequality: The inequality can be rewritten using logarithm properties: $1 + \log_2\left(2x^2 + 2x + \frac{7}{2}\right) > \log_2(ax^2 + a)$ Since $1 = \log_2(2)$: $\log_2(2) + \log_2\left(2x^2 + 2x + \frac{7}{2}\right) > \log_2(a(x^2 + 1))$ $\log_2\left(2\left(2x^2 + 2x + \frac{7}{2}\right)\right) > \log_2(ax^2 + a)$ $\log_2(4x^2 + 4x + 7) > \log_2(ax^2 + a)$

Since the base of the logarithm is $2$, which is greater than $1$, the $\log_2(t)$ function is strictly increasing. Thus, we can remove the logarithm while preserving the inequality direction: $4x^2 + 4x + 7 > ax^2 + a$ Rearrange the terms to form a quadratic inequality: $(4 - a)x^2 + 4x + (7 - a) > 0$

Let $f(x) = (4 - a)x^2 + 4x + (7 - a)$. We need to find the integral values of $a$ (with $a > 0$) for which $f(x) > 0$ has at least one real solution for $x$.

3. Analyze the quadratic inequality based on the coefficient of $x^2$:

Case 1: The coefficient of $x^2$ is zero. $4 - a = 0 \implies a = 4$. Substitute $a=4$ into the inequality: $0 \cdot x^2 + 4x + (7 - 4) > 0$ $4x + 3 > 0$ $4x > -3 \implies x > -3/4$. This linear inequality has infinitely many solutions for $x$. Since $a=4$ satisfies $a>0$, $a=4$ is a valid integral value.

Case 2: The coefficient of $x^2$ is non-zero. $4 - a \ne 0$. Let $A = 4-a$, $B = 4$, $C = 7-a$. The discriminant of the quadratic $f(x)$ is $\Delta = B^2 - 4AC = 4^2 - 4(4-a)(7-a)$. $\Delta = 16 - 4(28 - 4a - 7a + a^2)$ $\Delta = 16 - 4(a^2 - 11a + 28)$ $\Delta = 16 - 4a^2 + 44a - 112$ $\Delta = -4a^2 + 44a - 96$.

Subcase 2a: The parabola opens upwards ($A > 0$). $4 - a > 0 \implies a < 4$. For $f(x) > 0$ to have at least one solution when $A > 0$: If $\Delta \ge 0$: The parabola intersects or touches the x-axis. Since it opens upwards, $f(x)$ will be positive for $x$ values outside the roots (if $\Delta > 0$) or for all $x \ne -B/(2A)$ (if $\Delta = 0$). Thus, $f(x) > 0$ has solutions. If $\Delta < 0$: The parabola is entirely above the x-axis. Thus, $f(x) > 0$ for all real $x$. This means $f(x) > 0$ has solutions. Therefore, if $A > 0$, the inequality $f(x) > 0$ always has solutions. Combining this with the condition $a > 0$ (from domain) and $a < 4$, we get $0 < a < 4$. The integral values of $a$ in this range are $1, 2, 3$.

Subcase 2b: The parabola opens downwards ($A < 0$). $4 - a < 0 \implies a > 4$. For $f(x) > 0$ to have at least one solution when $A < 0$: The parabola opens downwards. For $f(x)$ to be positive for some $x$, the parabola must intersect the x-axis. This requires the discriminant to be strictly positive ($\Delta > 0$). If $\Delta = 0$, $f(x)$ touches the x-axis at one point, $f(x_0)=0$, and $f(x) < 0$ for all other $x$. So $f(x) > 0$ has no solution. If $\Delta < 0$, $f(x)$ is entirely below the x-axis. So $f(x) < 0$ for all $x$, and $f(x) > 0$ has no solution. So, we need $\Delta > 0$: $-4a^2 + 44a - 96 > 0$ Divide by $-4$ and reverse the inequality sign: $a^2 - 11a + 24 < 0$ Factor the quadratic: $(a - 3)(a - 8) < 0$ This inequality holds when $3 < a < 8$. Combining this with the conditions $a > 0$ (from domain) and $a > 4$, we get $4 < a < 8$. The integral values of $a$ in this range are $5, 6, 7$.

4. Collect all valid integral values of $a$: From Case 1: $a = 4$. From Subcase 2a: $a \in \{1, 2, 3\}$. From Subcase 2b: $a \in \{5, 6, 7\}$. Combining all these values, the set of integral values of $a$ for which the inequality has at least one root is $\{1, 2, 3, 4, 5, 6, 7\}$. The number of integral values is 7.

The final answer is $\boxed{7}$.

Explanation of the solution:

1. Domain: Ensure logarithmic arguments are positive. $2x^2+2x+7/2 > 0$ for all $x$ (discriminant negative, leading coefficient positive). $ax^2+a > 0 \implies a(x^2+1) > 0 \implies a > 0$.
2. Simplify Inequality: Use $\log_2(2)=1$ and $\log_b(M)+\log_b(N)=\log_b(MN)$ to combine terms. Then, since base $2 > 1$, remove logarithms while preserving the inequality sign: $4x^2+4x+7 > ax^2+a$.
3. Form Quadratic Inequality: Rearrange to $(4-a)x^2+4x+(7-a) > 0$. Let $f(x) = (4-a)x^2+4x+(7-a)$. We need $f(x)>0$ for at least one $x$.
4. Case Analysis:
   • If $4-a=0 \implies a=4$: Inequality becomes $4x+3 > 0 \implies x > -3/4$. This has solutions, so $a=4$ is valid.
   • If $4-a>0 \implies a<4$: Parabola opens upwards. $f(x)>0$ always has solutions (either always positive or positive outside roots). Combined with $a>0$, this gives $0 < a < 4$. Integral values: $1, 2, 3$.
   • If $4-a<0 \implies a>4$: Parabola opens downwards. For $f(x)>0$ to have solutions, the parabola must intersect the x-axis, meaning the discriminant $\Delta$ must be positive. $\Delta = 4^2 - 4(4-a)(7-a) = -4a^2+44a-96$. Set $\Delta > 0 \implies -4a^2+44a-96 > 0 \implies a^2-11a+24 < 0 \implies (a-3)(a-8) < 0 \implies 3 < a < 8$. Combined with $a>4$, this gives $4 < a < 8$. Integral values: $5, 6, 7$.
5. Count Values: Sum all integral values: $1, 2, 3, 4, 5, 6, 7$. Total 7 values.

Answer: 7