Limit Continuity And Differentiability Ques 86
- If $\alpha$ and $\beta$ are the roots of the equation $375 x^{2}-25 x-2=0$, then $\lim _{n \rightarrow \infty} \sum _{r=1}^{n} \alpha^{r}+\lim _{n \rightarrow \infty} \sum _{r=1}^{n} \beta^{r}$ is equal to
(2019 Main, 12 April)
(a) $\frac{21}{346}$
(b) $\frac{29}{358}$
(c) $\frac{1}{12}$
(d) $\frac{7}{116}$
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Answer:
Correct Answer: 86.(c)
Solution:
- Given function is defined
$$ f(x)=\begin{array}{c} \frac{\sqrt{2} \cos x-1}{\cot x-1} & , x \neq \frac{\pi}{4} \\ k &, x=\frac{\pi}{4} \end{array} $$
$\because$ Function $f(x)$ is continuous, so it is continuous at $x=\frac{\pi}{4}$.
$$ \begin{aligned} & \therefore & f\left(\frac{\pi}{4}\right) & =\lim _{x \rightarrow \frac{\pi}{4}} f(x) \\ \Rightarrow & & k & =\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1} \end{aligned} $$
Put $x=\frac{\pi}{4}+h$, when $x \rightarrow \frac{\pi}{4}$, then $h \rightarrow 0$
$$ \begin{aligned} k & =\lim _{h \rightarrow 0} \frac{\sqrt{2} \cos \left(\frac{\pi}{4}+h\right)-1}{\cot \left(\frac{\pi}{4}+h\right)-1} \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{2} \frac{1}{\sqrt{2}} \cos h-\frac{1}{\sqrt{2}} \sin h-1}{\frac{\cot h-1}{\cot h+1}-1} \end{aligned} $$
$[\because \cos (x+y)=\cos x \cos y-\sin x \sin y$ and
$$ \left.\cot (x+y)=\frac{\cot x \cot y-1}{\cot y+\cot x}\right] $$
$$ \begin{alignedat} & =\lim _{h \rightarrow 0} \frac{\cos h-\sin h-1}{\frac{-2}{1+\cot h}} \\ & =\frac{\lim _{h \rightarrow 0} \left( \frac{(1-\cos h)+\sin h}{2 \sin h} \right)(\sin h+\cos h)}{1} \\ & =\lim _{h \rightarrow 0} \frac{2 \sin ^{2} \frac{h}{2}+2 \sin \frac{h}{2} \cos \frac{h}{2}}{4 \sin \frac{h}{2} \cos \frac{h}{2}}(\sin h+\cos h) \\ & =\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}+\cos \frac{h}{2}}{2 \cos \frac{h}{2}} \times(\sin h+\cos h) \Rightarrow k=1 \end{aligned} $$
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