Binomial Theorem And Its Simple Applications Question 21
Question: The expression $ \frac{1}{\sqrt{3x+1}}[ {{( \frac{1+\sqrt{3x+1}}{2} )}^{7}}-{{( \frac{1-\sqrt{3x+1}}{2} )}^{7}} ] $ is a polynomial in x of degree equal to
Options:
A) 3
B) 4
C) 2
D) 5
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ [ {{( \frac{1-\sqrt{3x+1}}{2} )}^{4}}-{{( \frac{1-\sqrt{3x+1}}{2} )}^{7}} ] $
$ =,\frac{1}{\sqrt{3x+1}},[ \frac{{{( 1+\sqrt{3x+1} )}^{7}}-{{( 1-\sqrt{3x+1} )}^{7}}}{2^{7}} ] $
$ =\frac{1}{\sqrt{3x+1}}[ \frac{2{ {{,}^{7}}C_1(\sqrt{3x+1})+{{,}^{7}}C_3{{(\sqrt{3x+1})}^{3}}+{{,}^{7}}C_5{{(\sqrt{3x+1})}^{5}}+{{,}^{7}}C_7{{(\sqrt{3x+1})}^{7}} }}{2^{7}} ] $
$ =\frac{1}{2^{6}}[ ^{7}C_1+{{,}^{7}}C_3(3x+1)+{{,}^{7}}C_5{{(3x+1)}^{2}}+{{,}^{7}}C_7{{(3x+1)}^{3}} ] $ Clearly above is a polynomial of degree 3 in x.
BETA
