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 + 105 x + 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.