1) Let f(x) = x3+x2+100x+7sinx then show that the eqn
1 / [y-f(!)] + 2 / [y-f(2)] + 3 / [y-f(3)] = 0 has exactly one root lying in (f(1) , f(2)).
2) Let f(x) and g(x) be two functions which cuts each other orthogonally. At their common point of intersction (x1) , both f(x) and g(x) have equal to n, where n belongs to N, and n ≠1. Also if |f ' (x1)| = | g ' (x1)| at the common point of intersection. Then show that as
limit x approaches x1 [f(x).g(x)] is equals to n-1 , where [.] represents greatest integral functions.
11For the first one
let g(y) = 1 / [y-f(1)] + 2/[y-f(2)] +3/[y-f(3)] g' (y) = -(1/[y-f(1)]2 + 2/[ y-f(2)]2 + 3/ [y-f(3)]2) since g '(y) is negative summ of squares , its always negative so, g '(y) < 0 g(y) is strictly decreasing function here f(1) = 102 + 7 sin1 and f(2) = 212 + 7 sin2 clearly f(2) > f(1) g( f(1) + 0 ) ---> infinity g( f(2) - 0 ) ---> - infinity So, the mean value theorem between ( -infinity,infinity) as 0 will lie btn y=f(1) & y=f(2) hence proved
