Question

If the $6^{th}$ term in the expansion of the binomial ${\left(\sqrt{2^{\log \left(10-3^x\right)}}+\sqrt[5]{2^{\left(x-2\right)\log 3}}\right)}^m=21$ and known that the binomial coefficient of $2^{nd}$, $3^{rd}$ and $4^{th}$ terms in the expansion represent respectively the $1^{st}$, $3^{rd}$ and $5^{th}$ of an AP, then $x=$

• A. $0$
• B. $1$
• C. $-2$
• D. $3$

Answer 1

The correct option is A

$0$

Explnation for Correct Option

Step 1: Solving the given equation

Given ${\left(\sqrt{2^{\log \left(10-3^x\right)}}+\sqrt[5]{2^{\left(x-2\right)\log 3}}\right)}^m$ and $\Rightarrow {\mathrm{T}}_6=21$

Since $2^{\mathrm{nd}},3^{\mathrm{rd}}\mathrm{and}{4}^{\mathrm{th}}$ term is in A.P.

So

$\Rightarrow 2{}^{m}C_2={}^{m}C_1+{}^{m}C_3$

,$$\begin{gathered} \Rightarrow 2\times \frac{m(m-1)}{2}=m+\frac{m(m-1)(m-2)}{6} \\ \Rightarrow m^2-m=m+\frac{m(m-1)(m-2)}{6} \\ \Rightarrow 6m-6=6+m^2-3m+2 \\ \Rightarrow {\mathrm{m}}^2-9\mathrm{m}+14=0 \\ \Rightarrow (\mathrm{m}-7)(\mathrm{m}-2)=0 \\ \Rightarrow \mathrm{m}=2\mathrm{or}\mathrm{m}=7 \end{gathered}$$

Step 2:Putting the value of m in the given equation

$$\begin{gathered} T_6={}^{7}C_5\times 2^{\log \left(10-3^x\right)}\times 2^{\left(x-2\right)\log 3}=21 \\ \Rightarrow \frac{7\times 6}{2}\times 2^{\left[\log \left(10-3^x\right)+\left(x-2\right)\log 3\right]}=21 \\ \Rightarrow 2^{\left[\log \left(10-3^x\right)+\left(x-2\right)\log 3\right]}=\frac{21}{21} \\ \Rightarrow 2^{\left[\log \left(10-3^x\right)+\left(x-2\right)\log 3\right]}=\frac{\sout{21}}{\sout{21}} \\ \Rightarrow 2^{\left[\log \left(10-3^x\right)+\left(x-2\right)\log 3\right]}=1 \\ \Rightarrow 2^{\left[\log \left(10-3^x\right)+\left(x-2\right)\log 3\right]}=2^0 \end{gathered}$$

Step 3: Compare the power $2$

$$\begin{gathered} \Rightarrow \left[\log \left(10-3^x\right)+\left(x-2\right)\log 3\right]=0 \\ \Rightarrow \log \left(10-3^x\right)=\log 3^{-\left(x-2\right)} \\ \Rightarrow 10-3^x=3^{-\left(x-2\right)} \\ \Rightarrow \left(10-3^x\right)\left(3^{\left(x-2\right)}\right)=1 \\ \Rightarrow 3^{2x}-9·3^x-3^x+9=0 \\ \Rightarrow \left(3^x-1\right)\left(3^x-9\right)=0 \end{gathered}$$

Step 4: Compare the power $3$

$$\begin{gathered} \Rightarrow \left(3^x-1\right)=0\mathrm{or}\left(3^{\mathrm{x}}-9\right)=0 \\ \Rightarrow 3^{\mathrm{x}}=1\mathrm{or}{3}^{\mathrm{x}}=9 \\ \Rightarrow 3^{\mathrm{x}}=3^0\mathrm{or}{3}^{\mathrm{x}}=3^2 \\ \Rightarrow x=0\mathrm{or}x=2 \end{gathered}$$

So,$\mathrm{x}=0$ or $\mathrm{x}=2$ is the correct answer

Hence Option(1) is correct