SATHEE: Limit Continuity and Differentiability 3 Question 1

Limit Continuity and Differentiability 3 Question 1

1. If $α$ and $β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$ , then $lim_{n \to \infty} \sum_{r = 1}^{n} α^{r} + lim_{n \to \infty} \sum_{r = 1}^{n} β^{r}$ is equal to

(2019 Main, 12 April I)

(a) $\frac{21}{346}$

(b) $\frac{29}{358}$

(d) $\frac{7}{116}$

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Answer:

Correct Answer: 1. (c)

Solution:

Given function is

$f (x) = \begin{array}{cc} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} & , x \neq \frac{π}{4} \\ k & , x = \frac{π}{4} \end{array}$

$∵$ Function $f (x)$ is continuous, so it is continuous at $x = \frac{π}{4}$ .

$\begin{aligned} ∴ & f \frac{π}{4} & = lim_{x \to \frac{π}{4}} f (x) \\ \Rightarrow & k & = lim_{x \to \frac{π}{4}} \frac{\sqrt{2} \cos x - 1}{\cot x - 1} \end{aligned}$

Put $x = \frac{π}{4} + h$ , when $x \to \frac{π}{4}$ , then $h \to 0$

$\begin{aligned} k & = lim_{h \to 0} \frac{\sqrt{2} \cos \frac{π}{4} + h - 1}{\cot \frac{π}{4} + h - 1} \\ = lim_{h \to 0} \frac{\sqrt{2} \frac{1}{\sqrt{2}} \cos h - \frac{1}{\sqrt{2}} \sin h - 1}{\frac{\cot h - 1}{\cot h + 1} - 1} \end{aligned}$

$[∵ \cos (x + y) = \cos x \cos y - \sin x \sin y$ and

$\cot (x + y) = \frac{\cot x \cot y - 1}{\cot y + \cot x}]$

$\begin{aligned} = lim_{h \to 0} \frac{\cos h - \sin h - 1}{\frac{- 2}{1 + \cot h}} \\ = \frac{lim_{h \to 0}}{\frac{(1 - \cos h) + \sin h}{2 \sin h} (\sin h + \cos h)} \\ = lim_{h \to 0} \frac{2 \sin^{2} \frac{h}{2} + 2 \sin \frac{h}{2} \cos \frac{h}{2}}{4 \sin \frac{h}{2} \cos \frac{h}{2}} (\sin h + \cos h) \\ = lim_{h \to 0} \frac{\sin \frac{h}{2} + \cos \frac{h}{2}}{2 \cos \frac{h}{2}} \times (\sin h + \cos h) \Rightarrow k = \frac{1}{2} \end{aligned}$

Limit Continuity and Differentiability 3 Question 1

1. If $α$ and $β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$ , then $lim_{n \to \infty} \sum_{r = 1}^{n} α^{r} + lim_{n \to \infty} \sum_{r = 1}^{n} β^{r}$ is equal to

Answer:

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Limit Continuity and Differentiability 3 Question 1

1. If α and β are the roots of the equation 375x2−25x−2=0, then limn→∞∑r=1nαr+limn→∞∑r=1nβr is equal to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. If $α$ and $β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$ , then $lim_{n \to \infty} \sum_{r = 1}^{n} α^{r} + lim_{n \to \infty} \sum_{r = 1}^{n} β^{r}$ is equal to