160 - Questions Answers

two intersecting points means two solutions

multiply with the conjugate in numerator again and again you will get 

lim_x->∞ ((x+1)(x+2)(x+3)(x+4) - x⁴)/((x+1)(x+2)(x+3)(x+4))^1/4 + x )*((x+1)(x+2)(x+3)(x+4))^1/2 + x² )

after solving the numerator get a 3 power expression of x then take 3 power of x common, similarly take x and 2 power of x from the first and second denominator terms

its I. E. Maron 

There should be any variabe in the question 

calculate dx/dθ and dy/dθ

then dy/dx = [2cosθ-2cos2θ]/[-2sinθ+2sin2θ]

now solve

use cosC-cosD = 2sin[(C+D)/2]sin[(D-C)/2]

and sinC+sinD = 2sin[(C+D)/2]cos[(C-D)/2]

sinA + sinB + sinC = 2sin[(A+B)/2]cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2sin[(π-C)/2]cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2cosC/2cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2cosC/2* [cos[(A-B)/2]+[cos(A+B)/2]                     here sinC/2 = cos(A+B)/2

now use cosC + cosD = 2 cos[(C+D)/2]cos[(C-D)/2]

You should remember that the given formula is based on a triangle. 

limn->∞ 4ⁿ/ n!

= lim_n->∞ [4/1][4/2][4/3][4/4][4/5][4/6][4/7]................................[4/n]

besides first four bracket, all brackets are real numbers less than 1 so their product will be zero for n tends to infinite

since first one is a straight line so at every point only one tangent is possible so it is differentiable

but second one which is modulus has a sharp turn at x = -5/2 so two tangents at x = -5/2 are possible so is not differentiable.

perimeter = 4 so 2Y + X + πX/2 = 4

and area = XY + [π(X/2)²]/2

put Y from 1st equation to 2nd we get A = X[4-X{1+(π/2)}]/2 + πX²/8

now calculate dA/dX and then compare it to zero

X will be 8/(4+π)

for n = 4, we can form a square with integral coordinates of all the vertices.