Question

Find all values of 'k' for which the inequality $k. 4^x + (k-1)2^{x+2} + k > 1$ is satisfied for all $x \in R$.

Answer 1

[1, ∞)

Answer 2

Let the given inequality be $k. 4^x + (k-1)2^{x+2} + k > 1$

Let $y = 2^x$. Since $x \in R$, $y = 2^x$ can take any positive real value, i.e., $y \in (0, \infty)$. The inequality can be rewritten in terms of $y$: $k \cdot (2^x)^2 + (k-1) \cdot 2^x \cdot 2^2 + k > 1$ $k y^2 + 4(k-1)y + k > 1$ $k y^2 + 4(k-1)y + k - 1 > 0$

Let $f(y) = k y^2 + 4(k-1)y + k - 1$. We need to find the values of $k$ for which $f(y) > 0$ for all $y \in (0, \infty)$.

Case 1: $k = 0$. The inequality becomes $0 \cdot y^2 + 4(0-1)y + 0 - 1 > 0$, which simplifies to $-4y - 1 > 0$. $-4y > 1 \implies y < -1/4$. This inequality is not satisfied for any $y \in (0, \infty)$ since $y$ must be positive. So, $k=0$ is not a solution.

Case 2: $k \neq 0$. $f(y)$ is a quadratic function of $y$.

Subcase 2.1: $k > 0$. The parabola $f(y) = k y^2 + 4(k-1)y + k - 1$ opens upwards. For $f(y) > 0$ for all $y \in (0, \infty)$, we consider two possibilities for the roots of $f(y) = 0$.

Possibility A: The quadratic has no real roots. This occurs when the discriminant $\Delta < 0$. $\Delta = (4(k-1))^2 - 4k(k-1) = 16(k-1)^2 - 4k(k-1) = 4(k-1)[4(k-1) - k] = 4(k-1)(4k - 4 - k) = 4(k-1)(3k-4)$. $\Delta < 0 \implies 4(k-1)(3k-4) < 0$. Since $k > 0$, this inequality holds when $(k-1)(3k-4) < 0$. The roots of $(k-1)(3k-4) = 0$ are $k=1$ and $k=4/3$. The inequality $(k-1)(3k-4) < 0$ holds for $1 < k < 4/3$. If $1 < k < 4/3$, then $\Delta < 0$ and $k > 0$, so the parabola opens upwards and is entirely above the y-axis. Thus $f(y) > 0$ for all $y \in R$. Since $(0, \infty) \subset R$, $f(y) > 0$ for all $y \in (0, \infty)$. So, the interval $(1, 4/3)$ is part of the solution.

Possibility B: The quadratic has real roots, but both roots are less than or equal to 0. This occurs when $\Delta \ge 0$ and the roots $y_1, y_2$ satisfy $y_1 \le 0$ and $y_2 \le 0$. For real roots, $\Delta = 4(k-1)(3k-4) \ge 0$. Since $k > 0$, this holds for $0 < k \le 1$ or $k \ge 4/3$. The sum of the roots is $y_1 + y_2 = -\frac{4(k-1)}{k}$. For $y_1 \le 0, y_2 \le 0$, we need $y_1 + y_2 \le 0$. $-\frac{4(k-1)}{k} \le 0$. Since $k > 0$, $-4(k-1) \le 0 \implies k-1 \ge 0 \implies k \ge 1$. The product of the roots is $y_1 y_2 = \frac{k-1}{k}$. For $y_1 \le 0, y_2 \le 0$, we need $y_1 y_2 \ge 0$. $\frac{k-1}{k} \ge 0$. Since $k > 0$, $k-1 \ge 0 \implies k \ge 1$. Combining the conditions for Possibility B ($k > 0, \Delta \ge 0, y_1+y_2 \le 0, y_1 y_2 \ge 0$): $k > 0$ and ($0 < k \le 1$ or $k \ge 4/3$) and ($k \ge 1$) and ($k \ge 1$). The intersection of these conditions is $k=1$ or $k \ge 4/3$. If $k=1$, $f(y) = y^2$. $y^2 > 0$ for all $y \in (0, \infty)$. So $k=1$ is a solution. If $k \ge 4/3$, both roots are negative or zero. Since the parabola opens upwards, $f(y) > 0$ for all $y$ greater than the maximum root. Since the maximum root is $\le 0$, $f(y) > 0$ for all $y > 0$. So $k \ge 4/3$ is part of the solution.

Combining the results from Subcase 2.1 ($k > 0$): The solution is the union of $(1, 4/3)$, $\{1\}$, and $[4/3, \infty)$, which is $[1, \infty)$.

Subcase 2.2: $k < 0$. The parabola $f(y) = k y^2 + 4(k-1)y + k - 1$ opens downwards. As $y \to \infty$, $f(y) = k y^2 + 4(k-1)y + k - 1 \approx k y^2$. Since $k < 0$, $k y^2 \to -\infty$. Thus, for sufficiently large $y$, $f(y)$ will be negative. It is impossible for $f(y) > 0$ for all $y \in (0, \infty)$ if $k < 0$. So, there are no solutions for $k < 0$.

Combining all cases, the values of $k$ for which the inequality is satisfied for all $x \in R$ are $k \in [1, \infty)$.