$ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{r}{ \sqrt{ n^2 + r^2}} $ equals
- 1 + $ \sqrt 5 $
- 1 - $ \sqrt 5 $
- - 1 + $ \sqrt 2 $
- 1 + $ \sqrt 2 $
The Correct Option is B
Solution and Explanation
= $ lim_{ n \to \infty} \frac{1}{n} \displaystyle \sum_{r = 1}^ {2n} \frac{ r / n}{ \sqrt {1 + ( r / n)^2}} $
= $ \int \limits_0^2 \frac{x}{\sqrt{ 1 + x^2}} dx = [ \sqrt{ 1 + x^2 }]_0^2 = \sqrt 5 - 1 $
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Questions Asked in JEE Advanced exam
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
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Concepts Used:
Limits
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).