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Application of Derivatives question

2019 · 11 Jan · Shift 2 · Q25
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  5. /2019 · 11 Jan · Shift 2 · Q25

Application of Derivatives question

2019 · 11 Jan · Shift 2 · Q25

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1
Let f(x) = xa2+x2−d−xb2+(d−x)2,  {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\,a2+x2​x​−b2+(d−x)2​d−x​, x  ∈\, \in∈ R, where a, b and d are non-zero real constants. Then :
  1. A
    f is an increasing function of x
  2. B
    f is neither increasing nor decreasing function of x
  3. C
    f ' is not a continuous function of x
  4. D
    f is a decreasing function of x
View written solutionFree

Correct answer: A

  1. Given function

We have

f(x)=xa2+x2−d−xb2+(d−x)2,x∈R,f(x)=\frac{x}{\sqrt{a^2+x^2}}-\frac{d-x}{\sqrt{b^2+(d-x)^2}}, \qquad x\in\mathbb R,f(x)=a2+x2​x​−b2+(d−x)2​d−x​,x∈R,

where a,b,da,b,da,b,d are non-zero real constants.

We need to determine whether fff is increasing/decreasing and whether f′f'f′ is continuous.

  1. Differentiate the first term

Let

g(x)=xa2+x2=x(a2+x2)−1/2.g(x)=\frac{x}{\sqrt{a^2+x^2}}=x(a^2+x^2)^{-1/2}.g(x)=a2+x2​x​=x(a2+x2)−1/2.

Then

g′(x)=(a2+x2)−1/2+x(−12)(a2+x2)−3/2(2x).g'(x)=(a^2+x^2)^{-1/2}+x\left(-\frac12\right)(a^2+x^2)^{-3/2}(2x).g′(x)=(a2+x2)−1/2+x(−21​)(a2+x2)−3/2(2x).

So,

g′(x)=1a2+x2−x2(a2+x2)3/2=a2(a2+x2)3/2.g'(x)=\frac{1}{\sqrt{a^2+x^2}}-\frac{x^2}{(a^2+x^2)^{3/2}} =\frac{a^2}{(a^2+x^2)^{3/2}}.g′(x)=a2+x2​1​−(a2+x2)3/2x2​=(a2+x2)3/2a2​.

Since a≠0a\neq 0a=0, we have a2>0a^2>0a2>0, and also (a2+x2)3/2>0(a^2+x^2)^{3/2}>0(a2+x2)3/2>0 for all xxx. Hence,

g′(x)>0for all x∈R.g'(x)>0 \quad \text{for all } x\in\mathbb R.g′(x)>0for all x∈R.
  1. Differentiate the second term

Let

h(x)=d−xb2+(d−x)2.h(x)=\frac{d-x}{\sqrt{b^2+(d-x)^2}}.h(x)=b2+(d−x)2​d−x​.

Put u=d−xu=d-xu=d−x. Then

h(x)=ub2+u2,dudx=−1.h(x)=\frac{u}{\sqrt{b^2+u^2}}, \qquad \frac{du}{dx}=-1.h(x)=b2+u2​u​,dxdu​=−1.

Now for

ϕ(u)=ub2+u2,\phi(u)=\frac{u}{\sqrt{b^2+u^2}},ϕ(u)=b2+u2​u​,

we have

ϕ′(u)=b2(b2+u2)3/2.\phi'(u)=\frac{b^2}{(b^2+u^2)^{3/2}}.ϕ′(u)=(b2+u2)3/2b2​.

Therefore, by chain rule,

h′(x)=ϕ′(u)⋅dudx=−b2(b2+(d−x)2)3/2.h'(x)=\phi'(u)\cdot \frac{du}{dx} =-\frac{b^2}{(b^2+(d-x)^2)^{3/2}}.h′(x)=ϕ′(u)⋅dxdu​=−(b2+(d−x)2)3/2b2​.
  1. Differentiate f(x)f(x)f(x)

Since

f(x)=g(x)−h(x),f(x)=g(x)-h(x),f(x)=g(x)−h(x),

we get

f′(x)=g′(x)−h′(x).f'(x)=g'(x)-h'(x).f′(x)=g′(x)−h′(x).

Thus,

f′(x)=a2(a2+x2)3/2+b2(b2+(d−x)2)3/2.f'(x)=\frac{a^2}{(a^2+x^2)^{3/2}}+\frac{b^2}{(b^2+(d-x)^2)^{3/2}}.f′(x)=(a2+x2)3/2a2​+(b2+(d−x)2)3/2b2​.

Each term is strictly positive for every real xxx, because a,b≠0a,b\neq 0a,b=0. Hence,

f′(x)>0for all x∈R.f'(x)>0 \quad \text{for all } x\in\mathbb R.f′(x)>0for all x∈R.

Therefore, fff is a strictly increasing function on R\mathbb RR.

So:

  • Option A is true.
  • Option D is false.
  • Option B is false.
  1. Check continuity of f′f'f′

We have

f′(x)=a2(a2+x2)3/2+b2(b2+(d−x)2)3/2.f'(x)=\frac{a^2}{(a^2+x^2)^{3/2}}+\frac{b^2}{(b^2+(d-x)^2)^{3/2}}.f′(x)=(a2+x2)3/2a2​+(b2+(d−x)2)3/2b2​.

Since a≠0a\neq 0a=0 and b≠0b\neq 0b=0, the denominators are never zero:

a2+x2>0,b2+(d−x)2>0∀x.a^2+x^2>0, \qquad b^2+(d-x)^2>0 \quad \forall x.a2+x2>0,b2+(d−x)2>0∀x.

Thus f′(x)f'(x)f′(x) is continuous for all real xxx.

So option C is false.

  1. Conclusion

The only correct option is A\boxed{\text{A}}A​

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer is also A, so they agree.

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