Question

The number of solutions of the inequality $2^{\frac{1}{{\sin }^2{x}_2}}.3^{\frac{1}{{\sin }^2{x}_3}}.....n^{\frac{1}{{\sin }^2{x}_n}}\le n!$, where $x_i\in (0,4\pi )$ for $i\equiv 1,2,3,....,n$, is:

• A. $1$
• B. $4^{n-1}$
• C. $n^n$
• D. Infinite

Answer 1

Answer: (A)

$1$

Since ${\sin }^2{x}_i\le 1.$ Thus $\frac{1}{{\sin }^2{x}_i}\ge 1$

Thus LHS is always greater than RHS. So the inequality can only be satisfied when equality occurs. I.e. when

$2^{\frac{1}{{\sin }^2{x}_2}}\times 3^{\frac{1}{{\sin }^2{x}_3}}\times ...n^{\frac{1}{{\sin }^2{x}_n}}=2\times 3\times ...n$

Thus $\frac{1}{{\sin }^2{x}_i}=1$

$\Rightarrow {\sin }^2{x}_i=1$

Thus $x_i=2\pi$

Hence, only one solution exists between $(0,4\pi )$