Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If \int_0^x \sqrt 1 - (f'(t))^2 dt = \int_0^x f(t)dt , 0 \le x \le 1 and f(0) = 0, then \mathop \lim \limits_x \to 0 1 \over x^2\int_0^x f(t)dt :

Definite Integration PYQ — JEE Main 2021 · 31 Aug · Shift 1 · Q23 with Solution

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Correct answer: D

Given condition

We have, for $0 \le x \le 1$ ,

\int_0^x \sqrt{1-(f'(t))^2}\,dt=\int_0^x f(t)\,dt

and $f(0)=0$ , with $f$ non-negative on $[0,1]$ and twice differentiable on $(0,1)$ .

We need to find

\lim_{x\to 0}\frac{1}{x^2}\int_0^x f(t)\,dt.

Differentiate the given integral identity

Since both sides are integrals from $0$ to $x$ , by the Fundamental Theorem of Calculus,

\sqrt{1-(f'(x))^2}=f(x), \qquad 0<x<1.

Because $f(x)\ge 0$ , this is consistent with the square root.

Now square both sides:

f(x)^2=1-(f'(x))^2.

So,

f(x)^2+(f'(x))^2=1.

Differentiate again

Differentiate

f^2+(f')^2=1

with respect to $x$ :

2ff'+2f'f''=0.

Thus,

2f'(f+f'')=0.

So for each $x$ ,

f'(x)=0 \quad \text{or} \quad f''(x)+f(x)=0.

Instead of splitting cases, use the relation near $x=0$ more directly.

Find the behavior at $x=0$

From

\sqrt{1-(f'(x))^2}=f(x),

and since $f(0)=0$ , taking $x\to 0$ gives

0=\sqrt{1-(f'(0))^2}

(if $f'$ extends continuously to $0$ ; equivalently the limiting relation forces this), hence

(f'(0))^2=1.

f'(0)=\pm 1.

But $f(x)\ge 0$ for $x\in[0,1]$ and $f(0)=0$ . If $f'(0)=-1$ , then for small positive $x$ , $f(x)<0$ , impossible. Therefore

f'(0)=1.

Evaluate the required limit

Since $f$ is differentiable at $0$ with $f(0)=0$ and $f'(0)=1$ , we have

f(t)=t+o(t) \qquad (t\to 0).

Therefore,

\int_0^x f(t)\,dt=\int_0^x (t+o(t))\,dt =\frac{x^2}{2}+o(x^2).

Hence

\lim_{x\to 0}\frac{1}{x^2}\int_0^x f(t)\,dt =\frac{1}{2}.

Check options

A: $0$ — incorrect
B: $1$ — incorrect
C: does not exist — incorrect
D: $\dfrac12$ — correct

So the correct option is D.

More from Definite Integration

View written solutionFree

Correct answer: D

Given condition

We have, for $0 \le x \le 1$ ,

\int_0^x \sqrt{1-(f'(t))^2}\,dt=\int_0^x f(t)\,dt

and $f(0)=0$ , with $f$ non-negative on $[0,1]$ and twice differentiable on $(0,1)$ .

We need to find

\lim_{x\to 0}\frac{1}{x^2}\int_0^x f(t)\,dt.

Differentiate the given integral identity

Since both sides are integrals from $0$ to $x$ , by the Fundamental Theorem of Calculus,

\sqrt{1-(f'(x))^2}=f(x), \qquad 0<x<1.

Because $f(x)\ge 0$ , this is consistent with the square root.

Now square both sides:

f(x)^2=1-(f'(x))^2.

So,

f(x)^2+(f'(x))^2=1.

Differentiate again

Differentiate

f^2+(f')^2=1

with respect to $x$ :

2ff'+2f'f''=0.

Thus,

2f'(f+f'')=0.

So for each $x$ ,

f'(x)=0 \quad \text{or} \quad f''(x)+f(x)=0.

Instead of splitting cases, use the relation near $x=0$ more directly.

Find the behavior at $x=0$

From

\sqrt{1-(f'(x))^2}=f(x),

and since $f(0)=0$ , taking $x\to 0$ gives

0=\sqrt{1-(f'(0))^2}

(if $f'$ extends continuously to $0$ ; equivalently the limiting relation forces this), hence

(f'(0))^2=1.

f'(0)=\pm 1.

But $f(x)\ge 0$ for $x\in[0,1]$ and $f(0)=0$ . If $f'(0)=-1$ , then for small positive $x$ , $f(x)<0$ , impossible. Therefore

f'(0)=1.

Evaluate the required limit

Since $f$ is differentiable at $0$ with $f(0)=0$ and $f'(0)=1$ , we have

f(t)=t+o(t) \qquad (t\to 0).

Therefore,

\int_0^x f(t)\,dt=\int_0^x (t+o(t))\,dt =\frac{x^2}{2}+o(x^2).

Hence

\lim_{x\to 0}\frac{1}{x^2}\int_0^x f(t)\,dt =\frac{1}{2}.

Check options

A: $0$ — incorrect
B: $1$ — incorrect
C: does not exist — incorrect
D: $\dfrac12$ — correct

So the correct option is D.

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