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Maths-

General

Easy

Question

Let $f left parenthesis x right parenthesis$ and $g left parenthesis x right parenthesis$ are defined and differentiable for $x greater or equal than x subscript 0 end subscript$ and $f open parentheses x subscript 0 end subscript close parentheses equals g open parentheses x subscript 0 end subscript close parentheses comma f to the power of ´ end exponent open parentheses x close parentheses greater than g ´ left parenthesis x right parenthesis$ for $x greater than x subscript 0 end subscript comma$ then

$f (x) < g (x)$ for some $x > x_{0}$
$f (x) = g (x)$ for some $x > x_{0}$
$f (x) > g (x)$ for all $x > x_{0}$
None of these

The correct answer is: $f open parentheses x close parentheses greater than g left parenthesis x right parenthesis$ for all $x greater than x subscript 0 end subscript$

Consider the function $ϕ open parentheses x close parentheses equals f open parentheses x close parentheses minus g left parenthesis x right parenthesis$ on the interval $open square brackets x subscript 0 end subscript comma x close square brackets.$ Clearly, $ϕ left parenthesis x right parenthesis$ satisfies conditions of Lagrange’s theorem on $open square brackets x subscript 0 end subscript comma x close square brackets.$ Therefore, there exists $c element of left parenthesis x subscript 0 end subscript comma x right parenthesis$ such that
$ϕ open parentheses x close parentheses minus ϕ open parentheses x subscript 0 end subscript close parentheses equals ϕ to the power of ´ end exponent open parentheses c close parentheses open parentheses x minus x subscript 0 end subscript close parentheses$

$rightwards double arrow ϕ open parentheses x close parentheses equals ϕ ´ left parenthesis c right parenthesis left parenthesis x minus x subscript 0 end subscript right parenthesis$ …(i)

Now,

$ϕ open parentheses x close parentheses equals f open parentheses x close parentheses minus g left parenthesis x right parenthesis$

$rightwards double arrow ϕ to the power of ´ end exponent open parentheses x close parentheses equals f to the power of ´ end exponent open parentheses x close parentheses minus g ´ left parenthesis x right parenthesis$

$rightwards double arrow ϕ to the power of ´ end exponent open parentheses c close parentheses equals f to the power of ´ end exponent open parentheses c close parentheses minus g to the power of ´ end exponent open parentheses c close parentheses greater than 0 open square brackets because f to the power of ´ end exponent open parentheses x close parentheses greater than g to the power of ´ end exponent open parentheses x close parentheses f o r blank x greater than x subscript 0 end subscript close square brackets$

From (i), we have

$ϕ open parentheses x close parentheses greater than 0$ for all $x greater than x subscript 0 end subscript open square brackets because ϕ to the power of ´ end exponent open parentheses c close parentheses greater than 0 blank a n d blank x minus x subscript 0 end subscript greater than 0 close square brackets$

$rightwards double arrow f open parentheses x close parentheses minus g open parentheses x close parentheses greater than 0$ for all $x greater than x subscript 0 end subscript$

$rightwards double arrow f open parentheses x close parentheses greater than g left parenthesis x right parenthesis$ for all $x greater than x subscript 0 end subscript$

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