Calculus

Let's simplify the numerator and denominator separately.

Numerator (N): . Using the integral approximation for large n:

Denominator (D): Let's simplify the two parts.

Part 1: . For large n, this is approximately .

Part 2: The sum .

This is an arithmetic progression with n terms. The sum is . For large n, the constants are negligible, so . So, the denominator is approximately . Now let's evaluate the limit of the fraction : We are given that this limit is equal to . We can solve this quadratic equation.

Let's test the options.

(A) a=5:

(B) a=8:

(C) a=-15/2:

(D) a=-17/2: . This value satisfies the equation.

Let the acceleration be and velocity be . We are given . We know that acceleration can be written as . So, we have the differential equation: Separate the variables and integrate: The particle starts from rest at the origin, which means when , . Substituting these initial conditions to find C: So, the relationship between velocity and position is: The particle next comes to rest when again, for some . Since we are looking for a distance , we can divide by : The particle next comes to rest after covering a distance of .

We can write the function as a piecewise function by considering the definition of .

So,

The function is composed of two linear pieces, and . Linear functions are continuous everywhere. The only point where continuity might be an issue is at the "join" point, . To check for continuity at , we check the left-hand limit, the right-hand limit, and the function value. Since LHL = RHL = , the function is continuous at . Because it is continuous for , continuous for , and continuous at , the function is continuous for all real numbers, i.e., .

Given the parametric equations of the curve: The slope of the tangent is denoted by . We need to find the equation of the tangent at any point on this curve. To do so, we first compute the derivative (which represents the slope of the tangent).

Step 1: Find and - The derivative of with respect to is: - The derivative of with respect to is:

Step 2: Find To find the slope of the tangent, we use the formula: Substituting the values: Since , we can simplify the expression as:

Step 3: Find the equation of the tangent The general form of the equation of the tangent to the curve is: Here, is the point on the curve at any given value of , and is the slope of the tangent at that point. Substitute and into the equation. After some simplification, we get the equation of the tangent as: Thus, the equation of the tangent is . This corresponds to option .

We are given the two equations: 1. 2.

Step 1: Behavior of The equation implies that the product of (an exponentially increasing function) and (a periodic function oscillating between -1 and 1) equals 1. Since is always positive, we can infer that this equation will have real roots at certain intervals where reaches values such that the product equals 1.

Step 2: Behavior of Similarly, the equation suggests that for certain values of , the function and (which oscillates between -1 and 1) will multiply to give -1. For real solutions to exist, needs to be large enough and the cosine function will take negative values (between -1 and 0).

Step 3: Intermediate Roots By the Intermediate Value Theorem, since has real roots, and the function is continuous, there must be at least one root of between any two consecutive real roots of , as the functions and oscillate between positive and negative values. Thus, between any two real roots of , there is at least one root of .

Hence, the correct answer is .

We are given the function:

Step 1: Understand the behavior of The function involves , which is the absolute value of . The function is continuous and differentiable everywhere, except at points where , because the absolute value function introduces a corner at those points, where the derivative is undefined.

Step 2: Identify points where The function at within the interval . At these points, will have a sharp corner, and the function will not be differentiable. Thus, the points where is not differentiable are at .

Step 3: Conclusion The function is not differentiable at two points: and .

Therefore, the number of points at which is not differentiable is 2. Thus, the correct answer is .