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Definite Integration question

2019 · 9 Jan · Shift 2 · Q31
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Definite Integration question

2019 · 9 Jan · Shift 2 · Q31

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f be a differentiable function from R to R such that ∣f(x)−f(y)∣≤2∣x−y∣32,\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},∣f(x)−f(y)∣≤2∣x−y∣23​, for all x,y∈x,y \inx,y∈ R. If f(0)=1f\left( 0 \right) = 1f(0)=1 then ∫01f2(x)dx\int\limits_0^1 {{f^2}} \left( x \right)dx0∫1​f2(x)dx is equal to :
  1. A
    1
  2. B
    2
  3. C
    12{1 \over 2}21​
  4. D
    0
View written solutionFree

Correct answer: A

  1. Given condition

We have a differentiable function f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R such that

∣f(x)−f(y)∣≤2∣x−y∣3/2∀x,y∈R|f(x)-f(y)|\le 2|x-y|^{3/2}\qquad \forall x,y\in\mathbb{R}∣f(x)−f(y)∣≤2∣x−y∣3/2∀x,y∈R

and

f(0)=1.f(0)=1.f(0)=1.

We need to find

∫01f2(x) dx.\int_0^1 f^2(x)\,dx.∫01​f2(x)dx.
  1. Use differentiability to extract information about f′(x)f'(x)f′(x)

Fix any x∈Rx\in\mathbb{R}x∈R. Since fff is differentiable at xxx,

f′(x)=lim⁡h→0f(x+h)−f(x)h.f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.f′(x)=h→0lim​hf(x+h)−f(x)​.

Now use the given inequality with y=xy=xy=x and xxx replaced by x+hx+hx+h:

∣f(x+h)−f(x)∣≤2∣h∣3/2.|f(x+h)-f(x)|\le 2|h|^{3/2}.∣f(x+h)−f(x)∣≤2∣h∣3/2.

So,

∣f(x+h)−f(x)h∣≤2∣h∣1/2.\left|\frac{f(x+h)-f(x)}{h}\right| \le 2|h|^{1/2}.​hf(x+h)−f(x)​​≤2∣h∣1/2.

As h→0h\to 0h→0,

2∣h∣1/2→0.2|h|^{1/2}\to 0.2∣h∣1/2→0.

Hence,

∣f′(x)∣=∣lim⁡h→0f(x+h)−f(x)h∣≤0,|f'(x)|=\left|\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\right|\le 0,∣f′(x)∣=​h→0lim​hf(x+h)−f(x)​​≤0,

which gives

f′(x)=0∀x∈R.f'(x)=0\qquad \forall x\in\mathbb{R}.f′(x)=0∀x∈R.
  1. Conclude that fff is constant

Since f′(x)=0f'(x)=0f′(x)=0 for all xxx, the function is constant on R\mathbb{R}R. Therefore,

f(x)=C.f(x)=C.f(x)=C.

Using f(0)=1f(0)=1f(0)=1, we get

C=1.C=1.C=1.

Thus,

f(x)=1∀x∈R.f(x)=1\qquad \forall x\in\mathbb{R}.f(x)=1∀x∈R.
  1. Compute the integral

Then

f2(x)=1,f^2(x)=1,f2(x)=1,

so

∫01f2(x) dx=∫011 dx=1.\int_0^1 f^2(x)\,dx=\int_0^1 1\,dx=1.∫01​f2(x)dx=∫01​1dx=1.
  1. Check options
  • A: 111 ✅
  • B: 222 ❌
  • C: 12\tfrac1221​ ❌
  • D: 000 ❌

Therefore, the correct answer is A.

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