# Tutor profile: Ayushraj M.

## Questions

### Subject: Pre-Calculus

Question: Find the values of a, b and c if the following function is differentiable at x=5 and continuous everywhere f(x) = -x +5 ( -infi < x < 0) ax^3 + b ( 0<= x < 5) cx^2 + 5 (5<= x < 10) 105 ( x >= 10)

Answer: Since the function is continuous everywhere, it must be so at points where it changes definitions too, i.e at x={0,5,10}. At x=0, Left hand limit = (-0)+5 = 5 Right hand function value = a(0^3) +b =b These values must be equal for continuity, hence b = 5 At x=5 Left hand limit = 125a + 5 Right hand function value = 25c + 5 Again establishing equality of the left and right sides => 125a + 5 = 25 c+5 => c=5a At x=10, Left hand limit = 25c+5 Right hand function value = 105 Hence, 25c+5=105. This implies, 25c=100 => c=4 Since c=5a, a=4/5 = 0.8 Hence the solution (a,b,c) is (5,4,0.8)

### Subject: Basic Math

Question: Person A is twice as productive as person B, who in turn is thrice as productive as person C. C takes 30 days to a job. How long will it take A and B to complete the job together?

Answer: The productivity to do a job is inversely proportional to the time taken. For instance if I take a larger number of days to do my homework as compared to my friend, my friend will be more productive. Its given that for productivity, A > B > C. Hence it should take A the least number of days to finish the job and C should take the maximum. Since productivity is inversely proportional to the time taken, we will assume the proportionality constant as K, i.e D = K/P, where P is the productivity and D is the number of days taken. Notice how as P grows larger, D reduces and vice versa! Assume that Da means the number of days taken for A to complete the job, Db means the number of days taken for B to complete the job, Dc means the number of days taken for C to complete the job. Same goes for Pa, Pb and Pc. So its given that Dc=30. This means Pc=K/30. This implies that since Pb = 3Pc, Pb = 3K/30 = K/10 This implies that since Pa = 2Pb, Pa = 2K/10 = K/5. Now when A and B work together, their productivities add up. Hence P = Pa + Pb = (K/5) + (K/10) = 3K/10. So D=K/P => number of days taken by them to complete the job = K/(3K/10) = (10/3) =3.333. Hence it takes (10/3) or 3.33 days for A and B to complete the job together.

### Subject: Calculus

A typical question on calculus: Calculate the range of the following function f(x) = -xlogx - (1-x)log(1-x) for 0<x<1, where the base of the log function is 2, i.e. log 2 will be 1, log 4 will be 2, log 8 will be 3 and so on.

We first observe that the domain of the variable x doesn't violate anything. For a log function, the acceptable range of the argument is (0, infinity). Since in the question, the domain of x is (0,1), in both log x and log (1-x), the arguments of the log function lie in (0,1). Proceeding to the next part of the question, we analyze the boundary points of the domain. For x tending to 0 (x->0), f(x)->0. This is because the (1-x)log(1-x) part will go to (1)log(1) = 0. And the xlogx part will also go to 0. This can be affirmed by finding the limit to the following function [Lim x->0 ](log x)/(1/x), which by LHospital's rule will get reduced to [Lim x->0]-x = 0. In a similar fashion, for x -> 1, f(x) ->0. This is because the xlogx term goes to log(1) = 0. And [Lim x->1] (1-x)log(1-x) is same as [Lim y->0] ylogy = 0, where y =1-x. Hence we observed that at boundary points the function goes to 0. Note that in (0,1) f(x) is differentiable, hence we can analyze its monotonicity. Also since f(0) = f(1), there must exist a pt. c in (0,1) such that derivative of f(x) = f'(x) becomes 0 at c, i.e. f'(c) = 0. f'(x) for any general x = -logx -1 +1 + log(1-x) = log[(1-x)/x]. For f'(c) = 0 => (1-c)/c = 1 => c=1/2. Hence since f'(x) becomes 0 at only one point in (0,1), this point must correspond to a global extrema. f(1/2) = -(1/2)log(1/2) - (1/2)log(1/2) = -log(1/2) = log 2 =1. The double derivative f''(x) = [x/(1-x)](-1/x^2) => f''(1/2)=-4 = negative. This means that x=1/2 is a point of global maxima. Since the derivative goes to 0 only once, the boundary points must be the global minima. Hence global maxima =1, and global minima =0 (but remember f(x) =0 is only reached in limit). This means the range of f(x) for 0<x<1 is (0,1] (notice the usage of the square and simple brackets here).

## Contact tutor

needs and Ayushraj will reply soon.