INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 
Bose Olympiad Project Round is Live now. Learn More 

February 16, 2021

Test of Mathematics Solution Subjective 75 - Continuity and Roots of a Polynomial

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 75 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Show that there is at least one value of $ {x}$ for which ${\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} = 1}$


Solution


$ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} = 1}$
$ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1}$ is a continuous function. Now if we can chose $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1}$ takes both negative & positive numbers then the $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1 = 0}$ have a solution.

For $ {\displaystyle{x = {\frac{1}{64}}}}$, $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1 < 0}$ & for $ {x = 64}$, $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1 > 0}$.
So by intermediate value theorem we can say that $ {\displaystyle{\sqrt[3]{x}} + {\sqrt{x}} - 1 = 0}$ has a solution for some real $ {x}$.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
enter