Question

If $\sum _{r=1}^{k}co{s}^{-1}{\alpha }_r=\frac{k\pi }{2}$ for all $k\ge 1$ and $A=\sum _{r=1}^k{\left({\alpha }_r\right)}^r$, then $\underset{x\to A}{lim}\frac{{(1+x^2)}^{1/3}-{(1-2x)}^{1/4}}{x+x^2}=$

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

Answer 1

The correct option is A $\frac{1}{2}$

Since maximum value of ${\cos }^{-1}x=\frac{\pi }{2}$
${\sum }_{r=1}^k{\cos }^{-1}\alpha r=\frac{k\pi }{2}$ is possible if and only if each ${\cos }^{-1}\alpha r=\frac{\pi }{2}\Rightarrow \alpha r=0$
Therefore $\theta ={\sum }_{r=1}^k{\left(\alpha r\right)}^r=0$
Hence
${lim}_{x\to 0}\frac{{(1+x^2)}^{1/3}-{(1-2x)}^{1/4}}{x+x^2}={lim}_{x\to 0}\frac{{(1+x^2)}^{1/3}-{(1-2x)}^{1/4}}{x+x^2}$
$={lim}_{x\to 0}\frac{(1+\frac{1}{3}{x}^2+O\left(x^4\right))-(1-\frac{x}{2}+O\left(x^2\right))}{x+x^2}$ (Where $O\left(x^2\right)$ means terms containing higher power)
$={lim}_{x\to 0}\frac{\frac{1}{2}+\frac{x}{3}+O\left(x^2\right)}{1+x}=\frac{1}{2}$
Hence, option 'A' is correct.