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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 newG\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)
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