Question

If $\frac{(1-3x{)}^{\frac{1}{2}}+(1-x{)}^{\frac{5}{3}}}{2}$ is approximately equal to a + bx for small values of x, then (a, b) =

• A. $(2,\frac{35}{12})$
• B. $(1,-\frac{35}{24})$
• C. $(2,-\frac{35}{12})$
• D. $(2,-\frac{35}{12})$

Answer 1

The correct option is B $(1,-\frac{35}{24})$

$\frac{(1-3x{)}^{\frac{1}{2}}+(1-x{)}^{\frac{5}{3}}}{2{[1-\frac{x}{4}]}^{\frac{1}{2}}}$
$=\frac{[1+\frac{1}{2}(-3x)+\frac{1}{2}(-\frac{1}{2})\frac{1}{2}(-3x{)}^2+\dots ]+[1+\frac{5}{8}(-x)+\frac{5}{3}\frac{2}{3}\frac{1}{2}(-x{)}^2+\dots ]}{2[1+\frac{1}{2}(-\frac{x}{4})+\frac{1}{2}(-\frac{1}{2})\frac{1}{2}{(-\frac{x}{2})}^2+\dots ]}$
$=\frac{[1+\frac{19}{12}x+\frac{53}{144}{x}^{2-\dots }]}{[1+\frac{x}{2}-\frac{1}{8}{x}^2-\dots ]}=1-\frac{35}{24}x+\dots$
Neglecting higher powers of x, then
$a+bx=1-\frac{35}{24}\times \Rightarrow a=1,b=-\frac{35}{24}$