Let f(x), x ≥ 0, be a non-negative...

MathematicsLinear Differential EquationsJEE Advanced 2001Moderate

Let $f (x), x \geq 0$ , be a non-negative continuous function, and let $F (x) = \int_{0}^{x} f (t) d t, x \geq 0$ . If for some $c > 0, f (x) \leq c F (x)$ for all $x \geq 0$ , then show that $f (x) = 0$ for all $x \geq 0$ .

Visualized Solution (English)

Defining the Relationship

Given: $f (x) \geq 0$ for $x \geq 0$
Defined: $F (x) = \int_{0}^{x} f (t) d t$
Constraint: $f (x) \leq c F (x)$ for some $c > 0$

The Fundamental Connection

By Fundamental Theorem of Calculus:
$\frac{d}{d x} F (x) = F^{'} (x) = f (x)$

Forming the Differential Inequality

Substitute $f (x) = F^{'} (x)$ into $f (x) \leq c F (x)$ :
F'(x) \leq c F(x)
Rearranging gives:
F'(x) - c F(x) \leq 0

The Integrating Factor

Integrating Factor ( $I . F .$ ): $e^{\int - c d x} = e^{- c x}$
Multiply the inequality by $e^{- c x}$ :
e^{-cx} F'(x) - c e^{-cx} F(x) \leq 0

Applying the Product Rule

Recognize the LHS as a derivative:
\frac{d}{dx} (e^{-cx} F(x)) \leq 0
Integrate from $0$ to $x$ :
\int_0^x \frac{d}{dt} (e^{-ct} F(t)) dt \leq \int_0^x 0 dt

Evaluating the Integral

Fundamental Theorem of Calculus (Part 2):
[e^{-ct} F(t)]_0^x \leq 0
e^{-cx} F(x) - e^0 F(0) \leq 0
Since $F (0) = \int_{0}^{0} f (t) d t = 0$ :
e^{-cx} F(x) \leq 0

The Sandwich Logic

From $e^{- c x} F (x) \leq 0$ and $e^{- c x} > 0$ :
F(x) \leq 0
Since $f (t) \geq 0$ for all $t$ :
F(x) = \int_0^x f(t) dt \geq 0

Final Conclusion

Since $0 \leq F (x) \leq 0$ :
F(x) = 0 \text{ for all } x \geq 0
Differentiating both sides:
F'(x) = \frac{d}{dx}(0) = 0
Therefore:
f(x) = 0 \text{ for all } x \geq 0

00:00 / 00:00

Conceptually Similar Problems

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(A)

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Let $g (x) = \int_{0}^{x} f (t) d t$ , where $f$ is such that $\frac{1}{2} \leq f (t) \leq 1$ for $t \in [0, 1]$ and $0 \leq f (t) \leq \frac{1}{2}$ for $t \in [1, 2]$ . Then $g (2)$ satisfies the inequality

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(D)

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\int_{a}^{a + t} f (x) d x

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a

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View in:English Hindi

Let $f (x), x \geq 0$ , be a non-negative continuous function, and let $F (x) = \int_{0}^{x} f (t) d t, x \geq 0$ . If for some $c > 0, f (x) \leq c F (x)$ for all $x \geq 0$ , then show that $f (x) = 0$ for all $x \geq 0$ .

Visualized Solution (English)

Defining the Relationship

The Fundamental Connection

Forming the Differential Inequality

The Integrating Factor

Applying the Product Rule

Evaluating the Integral

The Sandwich Logic

Final Conclusion

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x),x≥0, be a non-negative continuous function, and let F(x)=∫0x​f(t)dt,x≥0. If for some c>0,f(x)≤cF(x) for all x≥0, then show that f(x)=0 for all x≥0.

Visualized Solution (English)

Defining the Relationship

The Fundamental Connection

Forming the Differential Inequality

The Integrating Factor

Applying the Product Rule

Evaluating the Integral

The Sandwich Logic

Final Conclusion

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