Let $$f$$ be a non-negative function in $$[0,1]$$ and twice differentiable in $$(0,1)$$. If $$\int_{0}^{x} \sqrt{1-(f(t))^{2}} d t=\int_{0}^{x} f(t) d t, 0 \leq x \leq 1$$ and $$f(0)=0$$, then $$\lim _{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} f(t) d t$$ :

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Definite Integrals

Let

f

be a non-negative function in

[0, 1]

and twice differentiable in

(0, 1)

. If

\int_{0}^{x} 1 - (f (t))^{2} d t = \int_{0}^{x} f (t) d t, 0 \leq x \leq 1

and

f (0) = 0

, then

x \to 0 lim \frac{1}{x ^{2}} \int_{0}^{x} f (t) d t

Answer: (C)

Differentiating the given condition:

1 - f^{2} (x) = f (x)

. Squaring:

1 - f^{2} (x) = f^{2} (x)

, so

f (x) = \frac{1}{2}

for

x > 0

. But

f (0) = 0

, so

f

has a jump. Using L'Hôpital's rule twice on

x \to 0 lim \frac{\int _{0}^{x} f ( t ) d t}{x ^{2}}

: first application gives

x \to 0 lim \frac{f ( x )}{2 x}

, second gives

x \to 0 lim \frac{f ^{'} ( x )}{2}

. From the functional equation,

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

near

x = 0

. Thus the limit equals

\frac{1}{2}

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option