## Task B-4.6.

For what value of the real number $x$ is the sum of the third and fifth term in the expansion of $\left(\sqrt{2^{x}}+\frac{1}{\sqrt{2^{x-1}}}\right)^{m}$ less than or equal to 135 if the sum of the binomial coefficients of the last three terms is 22?
## Solution.

From the fact that the sum of the binomial coefficients of the last three terms is 22, we obtain the equation

$$
\binom{m}{m-2}+\binom{m}{m-1}+\binom{m}{m}=22
$$

from which, by applying the symmetry property of binomial coefficients, we conclude that

$$
\binom{m}{2}+\binom{m}{1}+\binom{m}{0}=22
$$

Furthermore, from the definition of binomial coefficients, we obtain the equation $\frac{m(m-1)}{2}+m+1=22$, which simplifies to $m^{2}+m-42=0$.

The solutions to this equation are $m_{1}=6, m_{2}=-7$. From the definition of binomial coefficients, it follows that $m \geqslant 2$, so we conclude that $m=6$.

From $\left(\sqrt{2^{x}}+\frac{1}{\sqrt{2^{x-1}}}\right)^{m}=\sum_{k=0}^{m}\binom{m}{k}\left(\sqrt{2^{x}}\right)^{m-k}\left(\frac{1}{\sqrt{2^{x-1}}}\right)^{k}$ and the fact that the sum of the third and fifth terms is less than or equal to 135, we obtain the following inequality

$$
\binom{6}{2}\left(\sqrt{2^{x}}\right)^{4}\left(\frac{1}{\sqrt{2^{x-1}}}\right)^{2}+\binom{6}{4}\left(\sqrt{2^{x}}\right)^{2}\left(\frac{1}{\sqrt{2^{x-1}}}\right)^{4} \leqslant 135
$$

Solving this equation step by step, we get:

$$
\begin{aligned}
& 15 \cdot 2^{x+1}+\frac{15}{2^{x-2}} \leqslant 135 \\
& 2^{x+1}+\frac{1}{2^{x-2}} \leqslant 9 \\
& 2 \cdot 2^{2 x}-9 \cdot 2^{x}+4 \leqslant 0
\end{aligned}
$$

By substituting $2^{x}=t$, we obtain the quadratic inequality $2 t^{2}-9 t+4 \leqslant 0$ whose solution is $t \in\left[\frac{1}{2}, 4\right]$.

Finally, from $\frac{1}{2} \leqslant 2^{x} \leqslant 4$ we get $x \in[-1,2]$.