Of all triangles with given perimeter, find the triangle with the maximum area. Justify your answer

A 40 feet high screen is put on a vertical wall 10 feet above your eye-level. How far should you stand to maximize the angle subtended by the screen (from top to bottom) at your eye?

Study the derivatives of the function

and sketch its graph on the real line.

Suppose P and Q are the centres of two disjoint circles and respectively, such that P lies outside and Q lies outside . Two tangents are drawn from the point P to the circle , which intersect the circle at point A and B. Similarly, two tangents are drawn from the point Q to the circle , which intersect the circle at points M and N. Show that AB=MN

Suppose ABC is a triangle with inradius r. The incircle touches the sides BC, CA, and AB at D,E and F respectively. If BD=x, CE=y and AF=z, then show that

Evaluate:

Consider the equation . Show that
(a) the equation has only one real root;
(b) this root lies between 1 and 2;
(c) this root must be irrational.

In how many ways can you divide the set of eight numbers into 4 pairs such that no pair of numbers has equal to 2?

Suppose S is the set of all positive integers. For , define

For example 8*12=6.
Show that exactly two of the following three properties are satisfied:
(i) If , then .
(ii) for all .
(iii) There exists an element such that for all

.Two subsets A and B of the (x,y)-plane are said to be equivalent if there exists a function f: Ato B which is both one-to-one and onto.
(i) Show that any two line segments in the plane are equivalent.
(ii) Show that any two circles in the plane are equivalent.

Solution to 6: Let y be the given limit. Then e^y = (2nCn)^(1/2n). Now 2nCn = (2n)!/(n!)^2 . We divide both Nr and Dr by (2n)^2n. Hence e^y = [ {(2n)!/(2n)^2n}^(1/2n)]/[{n!/n^n}^2]^(1/2n)*2. Now limit n tends to infinity (n!/n^n)^(1/2n) is e^-1. Hence on simplification limit n tends to infinity e^y =2 and finally result is log 2. Do it yourself to understand the solution.

How to solve problem number 5 and 6?

Solution to 6: Let y be the given limit. Then e^y = (2nCn)^(1/2n). Now 2nCn = (2n)!/(n!)^2 . We divide both Nr and Dr by (2n)^2n. Hence e^y = [ {(2n)!/(2n)^2n}^(1/2n)]/[{n!/n^n}^2]^(1/2n)*2. Now limit n tends to infinity (n!/n^n)^(1/2n) is e^-1. Hence on simplification limit n tends to infinity e^y =2 and finally result is log 2. Do it yourself to understand the solution.