Question

If $\sum _{r=1}^k{\cos }^{-1}{\beta }_r=\frac{k\pi }{2}$ for any $k\ge 1$ and $A=\sum _{r=1}^k({\beta }_r{)}^r$ then $\underset{x\to A}{lim}\frac{(1+x{)}^{1/3}-(1-2x{)}^{1/4}}{x+x^2}$ is equal to

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{\pi }{2}$
• D. $\frac{5}{6}$

Answer 1

The correct option is D $\frac{5}{6}$

$\sum _{r=1}^k{\cos }^{-1}\left({\beta }_r\right)=k.\frac{\pi }{2}$

${\cos }^{-1}{\beta }_1+{\cos }^{-1}{\beta }_2.....{\cos }^{-1}{\beta }_k=k,\frac{T}{2}$

only possible, when each ${\cos }^{-1}{\beta }_r=\frac{\pi }{2}\Rightarrow {\beta }_r=0$

$A=\sum _{r=1}^k(0{)}^r=0$

$A=0$

$\underset{x\to 0}{lim}\frac{(1+x{)}^{1/3}-(1-2x{)}^{1/4}}{x+x^2}$

$\frac{0}{0}$ Form, using Binomial approx. $(1+x{)}^r\simeq 1+nx$

$\underset{x\to 0}{lim}\frac{1+\frac{x}{3}-(1-\frac{2x}{4})}{x+x^2}$

$\underset{x\to 0}{lim}\frac{5x}{6(x+x^2)}=\underset{x\to 0}{lim}\frac{5}{6(1+x)}=\frac{5}{6}$