logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Prove the following inequalities: $\cos x> 1-\large\frac{x^{2}}{2}, $$x>0$

Note: This is part 1st of a 4 part question, split as 4 separate questions here.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
  • (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$
  • A function $f(x)$ is said to be increasing on an interval I $f(x_1) \leq f(x_2)$ whenever $x_1 < x_2$
  • It is strictly increasing if $ x_1 < x_2 =>f(x_1) < f(x_2)$
  • A function $f$ is said to be decreasing on an interval I if $f(x_1) \geq f(x_2)$ whenever $x_1 < x_2$
  • It is strictly decreasing if $ x_1 < x_2 =>f(x_1) > f(x_2)$
$\cos x > 1- \large\frac{x^2}{2}$$, x >0$
Step 1:
Let $f(x)=\cos x -\bigg(1-\large\frac{x^2}{2}\bigg), $$x >0$
$f'(x)=\sin x +x$
Step 2:
$f''(x)=1- \cos x= 2 \sin ^2 \large\frac{x}{2}$$ >0$ for all x
$\therefore \;f'(x)$ is a strictly increasing function for $ x >0$
$\therefore f'(x) > f'(0) \;for\; x>0$
$=> x -\sin x >0$
Step 3:
We have $f'(x)=x -\sin x>0$ for $x>0$
$\therefore f(x)$ is a strictly increasing function
Step 4:
=>$ f(x) >f(0) \;for\; x >0$
$\cos x -\bigg(1-\large\frac{x^2}{2}\bigg)$$ >1-(1-0)$
$\cos x -\bigg(1-\large\frac{x^2}{2}\bigg)$$ >0$
$\therefore \cos x > 1-\large\frac{x^2}{2}$ for $x>0$
answered Jul 30, 2013 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...