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Question

# Let g (x), x≥0, be a non-negative continuous function and let F(x)=∫x0f(t)dt, x≥0. If for some c > 0, f(x) ≤ c F(x) for all x≥0, then

A
f(x)=0,x1
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B
f(x)=0,x1
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C
f(x)=1,x1
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D
f(x)=1,x1
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Solution

## The correct option is C f(x)=0,∀x≤1f(x)≥0 for x≥0F(x)=∫x0f(t)dt for all x≥0As c<0,f(x)≤cF(x) for x≥0f(x)=0⇒F′(x)=f(x) for x≥0Since f(x)≤cF(x) since F′(x)=f(x)⇒e−cxF′(x)−ce−cxF(x)≤0 by multiplying by e−cx where e−cx>0When x→∞,e−cx→0⇒ddx(e−cxF(x))≤0 for all x≥0⇒e−cxF(x) is a decreasing function for x≥0Let g(x)=e−cxF(x) be a decreasing function.⇒g(0)≥g(x) for x≥0⇒1×F(0)≥e−cxF(x)Given:F(0)=∫00F(t)dt=0⇒e−cxF(x)≤0⇒F(x)≤0 and e−cx≥0⇒c>0⇒cF(x)≤0⇒f(x)≤cF(x)≤0Since f(x) is non-negative f(x)≤0⇒f(x)=0

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