Question

The expression $\frac{1}{\sqrt{\left(3x+1\right)}}\left[{\left(\frac{1+\sqrt{3x+1}}{2}\right)}^7-{\left(\frac{1-\sqrt{3x+1}}{2}\right)}^7\right]$ is a polynomial in $x$ of degree

• A. 7
• B. 5
• C. 4
• D. 3

Answer 1

Answer: (D)

$\frac{1}{\sqrt{\left(3x+1\right)}}\left\{{\left(\frac{1+\sqrt{3x+1}}{2}\right)}^7-{\left(\frac{1-\sqrt{3x+1}}{2}\right)}^7\right\}$
$=\frac{1}{2^7\sqrt{\left(3x+1\right)}}\left\{{\left(1+\sqrt{3x+1}\right)}^7-{\left(1-\sqrt{3x+1}\right)}^7\right\}$ ···(i)
Now,
${\left(1+\sqrt{3x+1}\right)}^7-{\left(1-\sqrt{3x+1}\right)}^7$
$=\left[{}^7{C}_0+{}^7{C}_1\sqrt{\left(3x+1\right)}+{}^7{C}_2{\left(\sqrt{3x+1}\right)}^2+...+{}^7{C}_7{\left(\sqrt{3x+1}\right)}^7\right]$
$-\left[{}^7{C}_0-{}^7{C}_1\sqrt{\left(3x+1\right)}+{}^7{C}_2{\left(\sqrt{3x+1}\right)}^2.....-{}^7{C}_7{\left(\sqrt{3x+1}\right)}^7\right]$
$=2\left[{}^7{C}_1\sqrt{\left(3x+1\right)}+{}^7{C}_3{\left(\sqrt{3x+1}\right)}^3+{}^7{C}_5{\left(\sqrt{3x+1}\right)}^5+{}^7{C}_7{\left(\sqrt{3x+1}\right)}^7\right]$
$=2\sqrt{3x+1}\times \left[7+35(3x+1)+21(3x+1{)}^2+(3x+1{)}^3\right]$
Now, putting above value in (i), so the given expression becomes
$\frac{1}{2^6}[42+105x+21(3x+1{)}^2+(3x+1{)}^3]$
$\therefore$ Degree of a polynomial is the highest power of x. So, degree of given expression is 3.