Identify the correct statements

A. As a nearly equal to b so $\frac{a}{b}=1$(approx.) 

$\therefore \frac{a}{b}=1+x$ where x is very small

$\Rightarrow \frac{b+2a}{a+2b}\Rightarrow \frac{1+\frac{2a}{b}}{2+\frac{a}{1b}}\Rightarrow (1+\frac{2a}{b}){(2+\frac{a}{b})}^{-1}\Rightarrow (3+2x){(3+x)}^{-1}\Rightarrow \frac{1}{3}(3+2x)(1-\frac{x}{3})$

$\Rightarrow \frac{1}{3}(3+2x-x)=1+\frac{x}{3}\Rightarrow {(1+x)}^{\frac{1}{3}}={\left(\frac{a}{b}\right)}^{\frac{1}{3}}$

$\therefore K=\frac{1}{3}$

B. Coefficient of $x^{100}=3´coefficientof{x}^{100}in{(1–x)}^{-2}–5´coefficientof{x}^{99}in{(1–x)}^{-2}$

$=3\times {}^{100+2-1}{C}_{100}-5.{}^{99+2-1}{C}_{99}=3\times {}^{101}{C}_{100}-5.{}^{100}{C}_{99}=3\times 101-5\times 100=-197$

C. Since, $(r+1)th$ term in the expansion of ${(1+x)}^{27/5}=\frac{\frac{27}{5}(\frac{27}{5}-1)\mathrm{.....}(\frac{27}{5}-r+1)}{r!}{x}^r$

Now, this term will be negative, if the last factor in numerator is the only one negative factor.

$\Rightarrow \frac{27}{5}-r+1<0\Rightarrow \frac{32}{5}<r\Rightarrow 6.4<r\Rightarrow$Least value of r is 7.
Thus, first negative term will be $8^{th}.$

D. ${(1+2x+3x^2+\mathrm{....})}^{-3/2}={\left[{(1-x)}^{-2}\right]}^{-3/2}={(1-x)}^3=1-3x+3x^2-x^3$

Therefore, coefficient of $x^5$ is 0.