Question

Higest degree in expansion of ${[x+(x^3-1{)}^{1/2}]}^5+{[x-(x^3-1{)}^{1/2}]}^5$ is

• A. $8$
• B. $7$
• C. $6$
• D. $5$

Answer 1

Answer: (B)

$7$

${(x+\sqrt{x^3-1})}^5+{(x-\sqrt{x^3-1})}^5$

Take $a=x$ and $b=\sqrt{x^3-1}$

$={(a+b)}^5+{(a-b)}^5$

${=}^5{C}_0{a}^5{+}^5{C}_1{a}^{4}b{+}^5{C}_2{a}^3{b}^2{+}^5{C}_3{a}^2{b}^3{+}^5{C}_{4}a{b}^4{+}^5{C}_5{b}^5{+}^5{C}_0{a}^5{-}^5{C}_1{a}^{4}b{+}^5{C}_2{a}^3{b}^2{-}^5{C}_3{a}^2{b}^3{}^{}{+}^5{C}_{4}a{b}^4{-}^5{C}_5{b}^5$

$=2\left[{}^5{C}_0{a}^5{+}^5{C}_2{a}^3{b}^2{+}^5{C}_{4}a{b}^4\right]$

We have ${}^5{C}_0=1$,${}^5{C}_2=\frac{5!}{3!2!}=\frac{5\times 4\times 3!}{3!2!}=10$

${}^5{C}_4=\frac{5\times 4!}{4!}=5$

$=2[a^5+10a^3{b}^2+5a{b}^4]$

$=2[x^5+10x^3{\left(\sqrt{x^3-1}\right)}^2+5x{\left(\sqrt{x^3-1}\right)}^4]$

$=2[x^5+10x^3(x^3-1)+5x{(x^3-1)}^2]$

$=2[x^5+10x^3(x^3-1)+5x(x^6-2x^3+1)]$

$=2x^5+20x^6-20x^3+10x^7-20x^4+10x$

$=10x^7+20x^6+2x^5-20x^4-20x^3+10x$

The highest power$=7$

$\therefore$ degree $=7$