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Question: Let \(f(x)=\dfrac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}}+\dfrac{d-x}{\sqrt{{{b}^{2}}+{{\left( d-x \right)}^{...

Let f(x)=xx2+a2+d−xb2+(d−x)2, x∈Rf(x)=\dfrac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}}+\dfrac{d-x}{\sqrt{{{b}^{2}}+{{\left( d-x \right)}^{2}}}},\,x\in R. Where a, b, d are non-zero real constants. Then:
(a) f is decreasing function of x
(b) f is neither increasing nor decreasing function of x
(c) f is not continuous function of x
(d) f is increasing function of x

Explanation

Solution

To solve the above question, we will differentiate the above equation and then we will check the value of f’(x). While differentiating we will be applying the chain rule. In chain rule we differentiate as follows: suppose our function is f(g(x)) then after differentiating we get f′(g(x))×g′(x)f'\left( g\left( x \right) \right)\times g'\left( x \right). A function is called increasing if f’(x) > 0 at each point in an interval and a function is called decreasing if f’(x) < 0 at each point in an interval.

Complete step by step answer:
Now, differentiate the equation with respect to ‘x’,
=ddx[xx2+a2+d−xb2+(d−x)2] =ddx[x(x2+a2)12+d−x(b2+(d−x)2)12] =ddx(x(x2+a2)−12)+ddx((d−x)(b2+(d−x)2)−12) \begin{aligned} & =\dfrac{d}{dx}\left[ \dfrac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}}+\dfrac{d-x}{\sqrt{{{b}^{2}}+{{\left( d-x \right)}^{2}}}} \right] \\\ & =\dfrac{d}{dx}\left[ \dfrac{x}{{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{\dfrac{1}{2}}}}+\dfrac{d-x}{{{\left( {{b}^{2}}+{{\left( d-x \right)}^{2}} \right)}^{\dfrac{1}{2}}}} \right] \\\ & =\dfrac{d}{dx}\left( x{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{\dfrac{-1}{2}}} \right)+\dfrac{d}{dx}\left( (d-x){{\left( {{b}^{2}}+{{(d-x)}^{2}} \right)}^{\dfrac{-1}{2}}} \right) \\\ \end{aligned}
Here, we will apply chain rule to differentiate x(x2+a2)−12x{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{\dfrac{-1}{2}}}and (d−x)(b2+(d−x)2)−12(d-x){{\left( {{b}^{2}}+{{\left( d-x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}.
In chain rule, suppose our function is f(g(x)) then after differentiating we get f′(g(x))×g′(x)f'\left( g\left( x \right) \right)\times g'\left( x \right).

We will start by differentiating x(x2+a2)−12x{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{\dfrac{-1}{2}}}.
Here, our first function i.e. f(x) will be ‘x’ and second i.e. g(x) will be (x2+a2)−12{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{\dfrac{-1}{2}}}
Therefore, differentiation of the term will be
=[x(−12(2x)−1−12)+1(x2+a2)−12] =[x(−12(2x)−32)+1(x2+a2)−12]......(1) \begin{aligned} & =\left[ x\left( \dfrac{-1}{2}{{\left( 2x \right)}^{-1-\dfrac{1}{2}}} \right)+1{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{\dfrac{-1}{2}}} \right] \\\ & =\left[ x\left( \dfrac{-1}{2}{{\left( 2x \right)}^{\dfrac{-3}{2}}} \right)+1{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{\dfrac{-1}{2}}} \right]......(1) \\\ \end{aligned}
Now we differentiate the other equation (d−x)(b2+(d−x)2)−12(d-x){{\left( {{b}^{2}}+{{\left( d-x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}
Here, our first function i.e. f(x) will be ‘x’ and second i.e. g(x) will be (b2+(d−x)2)−12{{\left( {{b}^{2}}+{{\left( d-x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}
Therefore, differentiation of the term will be
=1[b2+(d−x)2]−12−[(d−x)×−12(b2+(d−x)2)−1−12×2(d−x)×−1] =1[b2+(d−x)2]−12−[(d−x)×−12(b2+(d−x)2)−32×2(d−x)×−1]......(2) \begin{aligned} & =1{{\left[ {{b}^{2}}+{{\left( d-x \right)}^{2}} \right]}^{\dfrac{-1}{2}}}-\left[ \left( d-x \right)\times \dfrac{-1}{2}{{\left( {{b}^{2}}+{{(d-x)}^{2}} \right)}^{-1-\dfrac{1}{2}}}\times 2\left( d-x \right)\times -1 \right] \\\ & =1{{\left[ {{b}^{2}}+{{\left( d-x \right)}^{2}} \right]}^{\dfrac{-1}{2}}}-\left[ \left( d-x \right)\times \dfrac{-1}{2}{{\left( {{b}^{2}}+{{(d-x)}^{2}} \right)}^{\dfrac{-3}{2}}}\times 2\left( d-x \right)\times -1 \right]......(2) \\\ \end{aligned}
Now add (1) and (2)
=[x(−12(2x)−32)+1(x2+a2)−12]+1[b2+(d−x)2]−12−[(d−x)×−12(b2+(d−x)2)−32×2(d−x)×−1] =1(a2+x2)12−x2(a2+x2)32+1[b2+(d−x)2]12−(d−x)2[b2+(d−x)2]32 =(a2+x2−x2)(a2+x2)32+b2+(d−x)2−(d−x)2[b2+(d−x)2]32 =a2(a2+x2)32+b2[b2+(d−x)2]32  \begin{aligned} & =\left[ x\left( \dfrac{-1}{2}{{\left( 2x \right)}^{\dfrac{-3}{2}}} \right)+1{{\left( {{x}^{2}}+{{a}^{2}} \right)}^{\dfrac{-1}{2}}} \right]+1{{\left[ {{b}^{2}}+{{\left( d-x \right)}^{2}} \right]}^{\dfrac{-1}{2}}}-\left[ \left( d-x \right)\times \dfrac{-1}{2}{{\left( {{b}^{2}}+{{(d-x)}^{2}} \right)}^{\dfrac{-3}{2}}}\times 2\left( d-x \right)\times -1 \right] \\\ & =\dfrac{1}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{\dfrac{1}{2}}}}-\dfrac{{{x}^{2}}}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{\dfrac{3}{2}}}}+\dfrac{1}{{{\left[ {{b}^{2}}+{{(d-x)}^{2}} \right]}^{\dfrac{1}{2}}}}-\dfrac{{{(d-x)}^{2}}}{{{\left[ {{b}^{2}}+{{(d-x)}^{2}} \right]}^{\dfrac{3}{2}}}} \\\ & =\dfrac{\left( {{a}^{2}}+{{x}^{2}}-{{x}^{2}} \right)}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{\dfrac{3}{2}}}}+\dfrac{{{b}^{2}}+{{(d-x)}^{2}}-{{(d-x)}^{2}}}{{{[{{b}^{2}}+{{(d-x)}^{2}}]}^{\dfrac{3}{2}}}} \\\ & =\dfrac{{{a}^{2}}}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{\dfrac{3}{2}}}}+\dfrac{{{b}^{2}}}{{{[{{b}^{2}}+{{(d-x)}^{2}}]}^{\dfrac{3}{2}}}} \\\ & \\\ \end{aligned}
In the final expression all the terms are positive with positive powers. Therefore, our function will not become 0 or undefined at any point.
As, f’(x) is positive,
We can write f’(x) > 0
i.e. f’(x) is an increasing function.

So, the correct answer is “Option A”.

Note: You can most probably make mistake while differentiating x2+a2\sqrt{{{x}^{2}}+{{a}^{2}}} and b2+(d−x)2\sqrt{{{b}^{2}}+{{(d-x)}^{2}}}. Also, remember differentiation of xn{{x}^{n}} is xn−1{{x}^{n-1}} .Use brackets everywhere possible to avoid confusion. Try to avoid mistakes in positive negative signs. Remember the condition for a function to be increasing i.e. f’(x) > 0.