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Differentiation question

2024 · 30 Jan · Shift 2 · Q36
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  5. /2024 · 30 Jan · Shift 2 · Q36

Differentiation question

2024 · 30 Jan · Shift 2 · Q36

JEE MainMathematicsDifferentiationMCQ+4 / −1
Let f:R−{0}→Rf: \mathbb{R}-\{0\} \rightarrow \mathbb{R}f:R−{0}→R be a function satisfying f(xy)=f(x)f(y)f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}f(yx​)=f(y)f(x)​ for all x,y,f(y)eq0x, y, f(y) eq 0x,y,f(y)eq0. If f′(1)=2024f^{\prime}(1)=2024f′(1)=2024, then
  1. A
    xf′(x)+2024f(x)=0x f^{\prime}(x)+2024 f(x)=0xf′(x)+2024f(x)=0
  2. B
    xf′(x)−2023f(x)=0x f^{\prime}(x)-2023 f(x)=0xf′(x)−2023f(x)=0
  3. C
    xf′(x)−2024f(x)=0x f^{\prime}(x)-2024 f(x)=0xf′(x)−2024f(x)=0
  4. D
    xf′(x)+f(x)=2024x f^{\prime}(x)+f(x)=2024xf′(x)+f(x)=2024
View written solutionFree

Correct answer: C

  1. Given functional equation

We have

f(xy)=f(x)f(y)f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}f(yx​)=f(y)f(x)​

for all admissible x,yx,yx,y with f(y)≠0f(y)\neq 0f(y)=0.

We are also given

f′(1)=2024.f'(1)=2024.f′(1)=2024.

We need to determine which differential equation must be satisfied.

  1. First find f(1)f(1)f(1)

Put x=y≠0x=y\neq 0x=y=0. Then

f(1)=f(xx)=f(x)f(x)=1f(1)=f\left(\frac{x}{x}\right)=\frac{f(x)}{f(x)}=1f(1)=f(xx​)=f(x)f(x)​=1

(whenever f(x)≠0f(x)\neq 0f(x)=0; the relation itself indicates the function behaves multiplicatively and in particular f(1)=1f(1)=1f(1)=1).

So,

f(1)=1.f(1)=1.f(1)=1.
  1. Rewrite the relation in a more useful form

Let

xy=u⇒x=uy.\frac{x}{y}=u \quad \Rightarrow \quad x=uy.yx​=u⇒x=uy.

Then the given equation becomes

f(u)=f(uy)f(y).f(u)=\frac{f(uy)}{f(y)}.f(u)=f(y)f(uy)​.

Hence,

f(uy)=f(u)f(y).f(uy)=f(u)f(y).f(uy)=f(u)f(y).

So for all nonzero real numbers,

f(xy)=f(x)f(y).f(xy)=f(x)f(y).f(xy)=f(x)f(y).

This is the standard multiplicative functional equation.

  1. Differentiate with respect to one variable

From

f(xy)=f(x)f(y),f(xy)=f(x)f(y),f(xy)=f(x)f(y),

fix xxx and differentiate with respect to yyy:

ddyf(xy)=ddy[f(x)f(y)].\frac{d}{dy}f(xy)=\frac{d}{dy}[f(x)f(y)].dyd​f(xy)=dyd​[f(x)f(y)].

Using chain rule,

xf′(xy)=f(x)f′(y).xf'(xy)=f(x)f'(y).xf′(xy)=f(x)f′(y).

Now put y=1y=1y=1. Since f(1)=1f(1)=1f(1)=1, this gives

xf′(x)=f(x)f′(1).xf'(x)=f(x)f'(1).xf′(x)=f(x)f′(1).

Given f′(1)=2024f'(1)=2024f′(1)=2024, we get

xf′(x)=2024f(x).xf'(x)=2024f(x).xf′(x)=2024f(x).

Therefore,

xf′(x)−2024f(x)=0.xf'(x)-2024f(x)=0.xf′(x)−2024f(x)=0.
  1. Match with the options

The obtained equation is

xf′(x)−2024f(x)=0,xf'(x)-2024f(x)=0,xf′(x)−2024f(x)=0,

which is exactly Option C.

  1. Verification with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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