Let f be a non-negative function defined on the interval [0,1] . If ∫_0^x 1-\ f^ (t)\ ^2 d t=∫_0^x f(t) d t, 0 ≤ x ≤ 1 and f(0)=0 , then

Correct answer: C. f((1)/(2)) < (1)/(2) and f((1)/(3)) < (1)/(3) Given ∫_0^x 1-(f^ (t))^2 d t=∫_0^x f(t) d t , 0 ≤ x ≤ 1 Applying Leibnitz theorem, we get array rlrl & & 1-(f^ (x))^2 & =f(x) & & 1-(f^ (x))^2 & =f²(x) & & (f^ (x))^2 & =1-f²(x) & & f^ (x) & =± 1-f^2(x) & & (d y)/(d x)& =± 1-y^2 & where y=f(x) & d y 1-y^2 = x array On integrating both sides, we get aligned & ^ -1 (y)= +C & f(0)=0 C=0 y=± & y= x=f(x) given f(x) ≥ 0 for & x ∈[0,1] aligned It is known that x x, x ∈ R^ + ((1)/(2)) (1)/(2) f((1)/(2)) (1)/(2) and ((1)/(3)) (1)/(3) f((1)/(3)) (1)/(3)

JEE Advanced 2009Paper 1

Exam: JEE Advanced
Year: 2009
Paper: Paper 1
Subject: Mathematics
Chapter: Definite Integration

JEE Advanced 2009 · Paper 1 · Mathematics · Definite Integration

Let f be a non-negative function defined on the interval [0,1] . If...

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 \text{\text{ and } }

f(0)=0

\text{,\text{ then } }

Options

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

Correct answer

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

Step-by-step solution

Given

\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t

, 0 \leq x \leq 1 Applying Leibnitz theorem, we get

\begin{array}{rlrl}& & \sqrt{1-\left(f^{\prime}(x)\right)^2}& =f(x) \\ \Rightarrow & & 1-\left(f^{\prime}(x)\right)^2 & =f²(x) \\ \Rightarrow & & \left(f^{\prime}(x)\right)^2 & =1-f²(x) \\ \Rightarrow & & f^{\prime}(x) & =\pm \sqrt{1-f^2(x)}\\ \Rightarrow & & \quad \frac{d y}{d x}& =\pm \sqrt{1-y^2}\\ & \text{ where }y=f(x) \Rightarrow & \frac{d y}{\sqrt{1-y^2}}=\pmd x \end{array}

On integrating both sides, we get

\begin{aligned}& \sin ^{-1}(y)=\pmx+C \\ & \because \quad f(0)=0 \Rightarrow C=0 \Rightarrow y=\pm \sinx \\ & y=\sin x=f(x) \text{ given }f(x) \geq 0 \text{ for }\\ & x \in[0,1] \end{aligned}

It is known that

\sin x \lt x, \forall x \in R^{+}\therefore \quad \sin \left(\frac{1}{2}\right) \lt \frac{1}{2} \Rightarrow f\left(\frac{1}{2}\right) \lt \frac{1}{2}

and

\sin \left(\frac{1}{3}\right) \lt \frac{1}{3} \Rightarrow f\left(\frac{1}{3}\right) \lt \frac{1}{3}

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