Practicing Success
The solution set of the inequation $\frac{1}{\log_2x}<\frac{1}{\log_2\sqrt{x+2}}$, is |
$(0,1)$ $(2, 0)$ $(0,1) ∪ (2,∞)$ none of these |
$(0,1) ∪ (2,∞)$ |
Two sides of the given inequation are defined for $x>0$ and $x ≠ 1$. Now, two cases arise. CASE I When $x>1$ In this case, we have $\log_2x>0$ and $\log_2\sqrt{x+2}>0$ $∴\frac{1}{\log_2x}<\frac{1}{\log_2\sqrt{x+2}}$ $⇒\log_2\sqrt{x+2}<\log_2x$ $⇒\sqrt{x+2}<x$ $⇒x+2<x^2$ $⇒x^2-x-2>0$ $⇒(x-2)(x+1)>0$ $⇒x-2>0$ $[∵x>1⇒x+1>0]$ $⇒x>2⇒x∈(2,∞)$ CASE II When $0<x<1$ In this case, we have $\log_2x<0$ and $\log_2\sqrt{x+2}>0$ So, the inequation $\frac{1}{\log_2x}<\frac{1}{\log_2\sqrt{x+2}}$ holds for all $x ∈(0, 1)$. Hence, the solution set of the given inequation is $(0,1) ∪ (2,∞)$. |