Question

QuestioN!!
It is given that x > 3. Then, evaluate the value of
$\sf\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}}$

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\sf\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}}$

Let assume that

$\rm :\longmapsto\:f(x) = \sf\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}}$

can be rewritten as

$\rm :\longmapsto\:f(x) = \sf\dfrac{1}{\dfrac{x + 2 + x + 3}{(x + 3)(x+2)}}$

$\rm :\longmapsto\:f(x) =\dfrac{(x + 3)(x + 2)}{2x + 5}$

$\rm :\longmapsto\:f(x) =\dfrac{ {x}^{2} + 3x + 2x + 6 }{2x + 5}$

$\rm :\longmapsto\:f(x) =\dfrac{ {x}^{2} +5x + 6 }{2x + 5}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} f(x) =\dfrac{d}{dx} \: \dfrac{ {x}^{2} +5x + 6 }{2x + 5}$

$\rm :\longmapsto\:f'(x) = \dfrac{(2x + 5)\dfrac{d}{dx}( {x}^{2} + 5x + 6) - ( {x}^{2} + 5x + 6)\dfrac{d}{dx}(2x + 5)}{ {(2x + 5)}^{2} }$

$\rm :\longmapsto\:f'(x) = \dfrac{(2x + 5)(2x + 5)- ( {x}^{2} + 5x + 6)(2)}{ {(2x + 5)}^{2} }$

$\rm :\longmapsto\:f'(x) = \dfrac{ {4x}^{2} + 25 + 20x - 2{x}^{2} - 10x - 12}{ {(2x + 5)}^{2} }$

$\rm :\longmapsto\:f'(x) = \dfrac{ {2x}^{2} + 10x + 13 }{ {(2x + 5)}^{2} }$

Now, Let us consider,

$\rm :\longmapsto\: {2x}^{2} + 10x + 13$

Its a quadratic expression with a = 2 > 0 and its Discriminant is

$\rm :\longmapsto\:Discriminant, \: D = {10}^{2} - 4 \times 2 \times 13$

$\rm :\longmapsto\:Discriminant, \: D = 100 - 104$

$\rm :\longmapsto\:Discriminant, \: D \: = \: - \: 4 < 0$

Since,

$\rm :\longmapsto\:a > 0 \: \: and \: Discriminant \: < 0$

$\bf\implies \: {2x}^{2} + 10x + 13 > 0$

So, using this, we get

$\rm :\longmapsto\:f'(x) = \dfrac{ {2x}^{2} + 10x + 13 }{ {(2x + 5)}^{2} } > 0$

$\bf\implies \:f(x) \: is \: increasing \: function$

As it is given that,

$\rm :\longmapsto\:x > 3$

and

$\rm :\longmapsto\: \:f(x) \: is \: increasing \: function$

$\bf\implies \:f(x) > f(3)$

$\rm \implies\:\sf\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}} > \sf\dfrac{1}{\dfrac{1}{3+2}+\dfrac{1}{3+3}}$

$\rm \implies\:\sf\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}} > \sf\dfrac{1}{\dfrac{1}{5}+\dfrac{1}{6}}$

$\rm \implies\:\sf\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}} \: > \: \sf\dfrac{1}{\dfrac{6 + 5}{5 \times 6}}$

$\rm \implies\:\sf\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}} \: > \: \sf\dfrac{1}{\dfrac{11}{30}}$

$\rm \implies\:\sf\dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}} \: > \: \dfrac{30}{11}$

Hence,

$\red{\rm \implies\:\boxed{ \tt{ \: \dfrac{1}{\dfrac{1}{x+2}+\dfrac{1}{x+3}} \: > \: \dfrac{30}{11} \: }}}$

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