Let $f$ be a non-negative function defined on the interval $[0,1]$. If $$ \int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} d t=\int_{0}^{x} f(t) d t, \quad 0 \leq x \leq 1 $$ and $f(0)=0$, then

(C) $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

(A) $f\left(\frac{1}{2}\right) \frac{1}{3}$

(B) $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$

(D) $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Differential Equations

Let

f

be a non-negative function defined on the interval

[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

(A) $f (\frac{1}{2}) < \frac{1}{2}$ and $f (\frac{1}{3}) > \frac{1}{3}$
(B) $f (\frac{1}{2}) > \frac{1}{2}$ and $f (\frac{1}{3}) > \frac{1}{3}$
(C) $f (\frac{1}{2}) < \frac{1}{2}$ and $f (\frac{1}{3}) < \frac{1}{3}$ ✓ Correct
(D) $f (\frac{1}{2}) > \frac{1}{2}$ and $f (\frac{1}{3}) < \frac{1}{3}$

Answer: (C)

Differentiating both sides with respect to

x

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

Squaring:

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

, so

(f^{'} (x))^{2} + (f (x))^{2} = 1

.

This suggests

f (x) = sin x

(since

f (0) = 0

and

f

is non-negative). We can verify:

f^{'} (x) = cos x

, and

(sin x)^{2} + (cos x)^{2} = 1

✓

Since

sin x < x

for all

x > 0

, we have:

f (\frac{1}{2}) = sin (\frac{1}{2}) < \frac{1}{2}

f (\frac{1}{3}) = sin (\frac{1}{3}) < \frac{1}{3}

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option