Question

Column-I	Column-II
(A) $\frac{3}{x^4}^{(\log_3 x)^2 + \log_3 x} = \sqrt{3}$	(p) number of solutions = 2
(B) $\log_{0.2} \frac{x+2}{x} \le 1$	(q) number of solutions = 3
(C) $\log_2 (x+5) = 6-x$	(r) number of solution = 1
(s) number of solution = 0
(t) infinite number of solutions

Answer 1

(A)-(r), (B)-(t), (C)-(r)

Answer 2

For (A): Let $y = \log_3 x$. The equation becomes $\frac{3}{x^4}^{y^2+y} = \sqrt{3}$. Taking $\log_3$ on both sides: $(\log_3 3 - \log_3 x^4)(y^2+y) = \log_3 \sqrt{3}$ $(1 - 4\log_3 x)(y^2+y) = \frac{1}{2}$ $(1 - 4y)(y^2+y) = \frac{1}{2}$ $y^2+y - 4y^3 - 4y^2 = \frac{1}{2}$ $-4y^3 - 3y^2 + y = \frac{1}{2}$ $8y^3 + 6y^2 - 2y + 1 = 0$. Let $f(y) = 8y^3 + 6y^2 - 2y + 1$. $f'(y) = 24y^2 + 12y - 2$. The roots of $f'(y)=0$ are $y = \frac{-12 \pm \sqrt{144 - 4(24)(-2)}}{48} = \frac{-12 \pm \sqrt{144 + 192}}{48} = \frac{-12 \pm \sqrt{336}}{48}$. Since $f'(y)$ has two real roots, $f(y)$ has local extrema. $f(y) \to \infty$ as $y \to \infty$ and $f(y) \to -\infty$ as $y \to -\infty$. The discriminant of the cubic is positive, indicating one real root. Thus, there is one real value for $y$, which means one real value for $x$. So (A) matches (r).

For (B): Domain is $\frac{x+2}{x} > 0$, so $x \in (-\infty, -2) \cup (0, \infty)$. $\log_{0.2} \frac{x+2}{x} \le 1$. Since base is $0.2 < 1$, we reverse the inequality: $\frac{x+2}{x} \ge (0.2)^1 = \frac{1}{5}$ $\frac{x+2}{x} - \frac{1}{5} \ge 0$ $\frac{5(x+2) - x}{5x} \ge 0 \implies \frac{4x+10}{5x} \ge 0 \implies \frac{2(2x+5)}{5x} \ge 0$. Critical points are $x = -5/2$ and $x=0$. The inequality holds for $x \in (-\infty, -5/2] \cup (0, \infty)$. Intersecting with the domain $(-\infty, -2) \cup (0, \infty)$, we get $(-\infty, -5/2] \cup (0, \infty)$. This set has infinite solutions. So (B) matches (t).

For (C): Domain is $x+5 > 0 \implies x > -5$. Consider $f(x) = \log_2(x+5)$ (increasing) and $g(x) = 6-x$ (decreasing). They can intersect at most once. By inspection, $x=3$: $\log_2(3+5) = \log_2 8 = 3$ and $6-3 = 3$. So $x=3$ is the unique solution. Thus (C) matches (r).