tgt

## Sunday, 3 August 2014

### CHAPTER 23- Miscellaneous Examples

 Example: 1
 If $g\left( x \right) = f\left( x \right) + f\left( {1 - x} \right)$ and $f''\left( x \right) < 0\,{\rm{for\, all}}\,x \in \left[ {0\,,1} \right]$, prove that $g(x)$ is increasing in $[0,1/2)$ and decreasing in $(1/2,1]$.
 Solution: 1
Our requirement is to somehow show that $g'(x) > 0\,{\rm{for\,all}}\, x \in [0,1/2)$and $g'(x) < 0\,{\rm{for\,all}}\, x \in (1/2,1]$.
From the given functional relation between $f(x)$ and $g(x)$:
 $g'\left( x \right) = f'\left( x \right) - f'\left( {1 - x} \right)$
Therefore, we must show that:
 $f'(x) > f'(1 - x)\forall x \in [0,1/2)$ $\ldots(i)$
and
 $f'\left( x \right) < f'\left( {1 - x} \right)\,\,\,\,\,\,\,\,\,\forall x \in (1/2,\,1]\,\,$ $\ldots(ii)$
Since $f''\left( x \right) < 0\,\,\,\forall x \in [0,\,1],\,\,f'\left( x \right)$ is decreasing on $[0,1]$. This means that if we take any $x$ value in $[0,1/2),(1 - x)$ will be greater than $x$ so that $f'\left( {1 - x} \right)$ will be less than $f'\left( x \right)$. In other words, ($i$) is satisfied by virtue of the fact that $f'(x)$ is decreasing.
On similar lines, when we assume any $x$ value in $(1/2,1]$, we will see that ($ii$) is also satisfied for the same reason (that $f'(x)$ is decreasing).
$\Rightarrow \,\,\,\,g(x)$ satisfies the stated assertion
 Example: 2
 Let $f\left( x \right) = \dfrac{{\ln \left( {\pi + x} \right)}}{{\ln \left( {e + x} \right)}}$. Prove that $f(x)$ is decreasing on $\left[ {0,\,\infty } \right)$
 Solution: 2
 $= \dfrac{{\left( {e + x} \right)\ln \left( {e + x} \right) - \left( {\pi + x} \right)\ln \left( {\pi + x} \right)}}{{\left( {e + x} \right)\left( {\pi + x} \right){{\left( {\ln \left( {e + x} \right)} \right)}^2}}}$ $= \dfrac{{\left( {e + x} \right)\ln \left( {e + x} \right) - \left( {\pi + x} \right)\ln \left( {\pi + x} \right)}}{{\left( {e + x} \right)\left( {\pi + x} \right){{\left( {\ln \left( {e + x} \right)} \right)}^2}}}$ $= \dfrac{{g\left( x \right)}}{{h\left( x \right)}} \,\,\,\,\,\,\,{\rm{{This\, substitution\, was\, done\, for\, convenience}}}$
To determine the sign of $f'\left( x \right)$ in $\left[ {0,\,\infty } \right)$, we first note that $h\left( x \right) > 0\,\,\forall \,x \in \left[ {0,\infty } \right)$, so that we need to only worry about the sign of $g(x)$. The form of $g(x)$ suggests that we can construct a new$G\left( x \right) = x\ln x$ function to determine the sign of $g(x)$ as follows:
$G\left( x \right) = x\ln x$
$\Rightarrow \,\,\,\, G'\left( x \right) = 1 + \ln x$
$\Rightarrow \,\,\,\, G'\left( x \right) > 0\,\,\forall x \in \left( {\dfrac{1}{e},\infty } \right)$
and $\,\,\,\,$ $G'\left( x \right) < 0\,\,\forall \,x \in \left( {0,\dfrac{1}{e}} \right)$
$\Rightarrow \,\,\,\, G\left( x \right)\,{\rm{is}}\,{\rm{increasing}}\,{\rm{on}}\,\left( {\dfrac{1}{e},\infty } \right)$
$\Rightarrow \,\,\,\, x\ln x\,\,{\rm{increases}}\,{\rm{on}}\,\left( {\dfrac{1}{e},\infty } \right)$
$\left( {\pi + x} \right)\ln \left( {\pi + x} \right) > \left( {e + x} \right)\ln \left( {e + x} \right)\,\,\,\,\,\,\forall x \in \left[ {0,\infty } \right) \,\,\,\,\,\,\,\,\left\{ \begin{array}{l} \,{\rm{since}}\,\left( {\pi + x} \right) > \left( {e + x} \right)\\ > \dfrac{1}{e}\forall x \in \left[ {0,\infty } \right) \end{array} \right\}$
$\Rightarrow \,\,\,\, g\left( x \right) < 0\;\,\,\,\,\,\,\forall \,x \in \left[ {0,\infty } \right)$
$\Rightarrow \,\,\,\, f'\left( x \right) < 0\;\,\,\,\,\,\forall \,x \in \left[ {0,\infty } \right)\,$
$\Rightarrow \,\,\,\, f\left( x \right)$ is decreasing on $\left[ {0,\infty } \right)$

 Example: 3
 Let $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} { - {x^3} + \dfrac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}},}\,\,\,\,{0 \le x < 1}\\ {2x - 3,}\,\,\,\,{1 \le x \le 3} \end{array}} \right\}$ Find all possible real values of $b$ such that $f(x)$ has the smallest value at $x = 1$.
 Solution: 3
Notice that $f\left( 1 \right) = - 1$ (from the lower definition $f(x)$)
Also, $f(x)$ is monotonically decreasing on $[0,1)$ and monotonically increasing on $[1,3)$.
Therefore, all we require for $f(x)$ to have its minimum at $x = 1$ is:
$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ge f\left( 1 \right)$ {i.e., the minimum of the left side function must not be less than $f(1)$}
 $\Rightarrow \,\,\,\, - 1 + \dfrac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}} \ge - 1$ $\Rightarrow \,\,\,\, \dfrac{{{b^3} - {b^2} + b - 1}}{{{b^2} + 3b + 2}} \ge 0$ $\Rightarrow \,\,\,\, \dfrac{{\left( {b - 1} \right)\left( {{b^2} + 1} \right)}}{{\left( {b + 1} \right)\left( {b + 2} \right)}} \ge 0$
Upon solving, this yields:
 $b \in \left( { - 2, - 1} \right) \cup \left[ {1,\infty } \right)$
 Example: 4
 Using the relation $2\left( {1 - \cos x} \right) < {x^2},x \ne 0$ or otherwise, prove that $\sin \left( {\tan x} \right) \ge x$ for all $x \in \left[ {0,\pi /4} \right]$
 Solution: 4
Notice that and ‘$x$‘ have equal values at $x = 0$. If we consider the function
 $f\left( x \right) = \sin \left( {\tan x} \right) - x$
and try to show that it is increasing, we would obtain
 $f\left( x \right) > f\left( 0 \right)$ or $\,\,\,\,$ $\sin \left( {\tan x} \right) - x \ge 0$
Hence, our task could be accomplished by showing that $f(x)$ is increasing.
 $f'\left( x \right) = \cos \left( {\tan x} \right){\sec ^2}x - 1$ $= \cos \left( {\tan x} \right)\left( {1 + {{\tan }^2}x} \right) - 1$ $= {\tan ^2}x\cos \left( {\tan x} \right) - \left( {1 - \cos \left( {\tan x} \right)} \right)$ $> {\tan ^2}x\cos \left( {\tan x} \right) - \dfrac{{{{\tan }^2}x}}{2}\;\;\;\; \left( {{\rm{using\, the\, given \,inequality}}} \right)$ $= \dfrac{1}{2}{\tan ^2}x\left( {2\cos \left( {\tan x} \right) - 1} \right)$ $= \dfrac{1}{2}{\tan ^2}x\left\{ {2\left( {\cos \left( {\tan x} \right) - 1} \right) + 1} \right\}$ $> \dfrac{1}{2}{\tan ^2}x\left( {1 - {{\tan }^2}x} \right)\,\,\,\,\,\,{\rm{(again\,using\, the\, given\,inequality)}}$
For, $x \in \left[ {0,\pi /4} \right],\,\,\tan x \in \left[ {0,1} \right]$ so that $\left( {1 - {{\tan }^2}x} \right) \ge 0$
 $\Rightarrow \,\,\,\, f'\left( x \right) > \dfrac{1}{2}{\tan ^2}x\left( {1 - {{\tan }^2}x} \right) \ge 0$ $\Rightarrow \,\,\,\, f'\left( x \right) > 0$ $\Rightarrow \,\,\,\, f\left( x \right)$ is increasing on $\left[ {0,\pi /4} \right]$ $\Rightarrow \,\,\, \,\sin \left( {\tan x} \right) \ge \,x\,\,\forall x \in \left[ {0,\pi /4} \right]$
 Example: 5
 Show that $\cos \left( {\sin x} \right) > \sin \left( {\cos x} \right)\,\,\,\forall \,x\,\, \in \left( {0,\pi /2} \right)$
 Solution: 5
The approach we have followed in the previous questions could be applied here to prove that $f\left( x \right) = \cos \left( {\sin x} \right) - \sin \left( {\cos x} \right)$ is increasing. However, $f'\left( x \right)$ becomes complicated and proving that it is positive is not straightforward like in the previous cases (you are urged to try this out).
Instead of considering the expressions $\cos \left( {\sin x} \right)$ and $\sin \left( {\cos x} \right)$, we can consider $\sin \left( {\dfrac{\pi }{2} - \sin x} \right)\,\,{\rm{and}}\,\,\,\sin \left( {\cos x} \right)$. This is because ‘$sin$’ is a monotonically increasing function in $\left( {0,\,\,\dfrac{\pi }{2}} \right)$, so that to determine the larger of the two values above, we just need to compare their arguments, i.e, $\left( {\dfrac{\pi }{2} - \sin x} \right)$ and $\cos x$
For $\left( {0,\,\,\dfrac{\pi }{2}} \right)$
 $\sin x + \cos x\,\, < \,\dfrac{\pi }{2}$ $\Rightarrow \,\, \,\,\cos x < \dfrac{\pi }{2}\,\,\, - \,\,\,\sin x\,\,\,\,\,\,\,\forall \,x \in \left( {0,\dfrac{\pi }{2}} \right)$ $\Rightarrow \,\, \,\,\sin \left( {\cos x} \right) < \,\,\sin \left( {\dfrac{\pi }{2} - \sin x} \right)\,\,\,\,\,\,\, \forall \,x \in \left( {0,\dfrac{\pi }{2}} \right)$ $\Rightarrow \,\, \,\,\sin \left( {\cos x} \right) < \,\,\cos \left( {\sin x} \right)\,\,\,\,\,\,\,\forall \,x \in \left( {0,\dfrac{\pi }{2}} \right)$
 Example: 6
 Find the values of $a$ for which the function $f\left( x \right) = \sin x - a\sin 2x - \dfrac{1}{3}\sin 3x + 2ax\,\,{\rm{increases}}\,\,{\rm{on}}\,\,\mathbb{R}$.
 Solution: 6
We want $f'\left( x \right) \ge \,0\,\forall x \in\mathbb{R}$
 $f'\left( x \right) = \cos x - 2a\cos 2x - \cos 3x + 2a$ $= \cos x - 2a\left( {2{{\cos }^2}x - 1} \right) - \left( {4{{\cos }^3}x - 3\cos x} \right) + 2a$ $= 4a + 4\cos x - 4a{\cos ^2}x - 4{\cos ^3}x$ $= 4a{\sin ^2}x + 4\cos x{\sin ^2}x$ $= 4{\sin ^2}x\left( {a + \cos x} \right)$
This is always non-negative if (since the minimum value of $\cos x$ is $- 1$).
Therefore, the required values of $a$ are:
 $a \in \left[ {1,\infty } \right)$