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If f is a real-valued differentiable function satisfying |f(x) – f(y)| ≤ (x – y)2, x, y ∈ R and f(0) = 0, then f(1) equals
- a)1
- b)2
- c)0
- d)–1
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If f is a real-valued differentiable function satisfying |f(x) –...
The given inequality |f(x) < f'(x)|="" holds="" for="" all="" real="" numbers="" x="" if="" and="" only="" if="" f(x)="" />< f'(x)="" for="" all="" x="" in="" the="" domain="" of="" />
To prove this, we can consider two cases:
Case 1: f'(x) > 0 for all x in the domain of f.
In this case, since f is differentiable, f'(x) > 0 implies that f is strictly increasing. Therefore, if f(x) < f'(x)="" for="" all="" x,="" then="" f(x)="" />< f'(x)="" />< f'(x+h)="" for="" all="" x="" and="" for="" all="" positive="" h.="" this="" implies="" that="" |f(x+h)="" -="" f(x)|="" />< |f'(x+h)="" -="" f(x)|="" for="" all="" x="" and="" for="" all="" positive="" h.="" taking="" the="" limit="" as="" h="" approaches="" 0,="" we="" have="" |f'(x)="" -="" 0|="" ≤="" 0,="" which="" is="" a="" contradiction.="" therefore,="" it="" is="" not="" possible="" for="" f'(x)="" /> 0 for all x in the domain of f.
Case 2: f'(x) < 0="" for="" all="" x="" in="" the="" domain="" of="" />
In this case, since f is differentiable, f'(x) < 0="" implies="" that="" f="" is="" strictly="" decreasing.="" therefore,="" if="" f(x)="" />< f'(x)="" for="" all="" x,="" then="" f(x)="" />< f'(x)="" />< f'(x+h)="" for="" all="" x="" and="" for="" all="" negative="" h.="" this="" implies="" that="" |f(x+h)="" -="" f(x)|="" />< |f'(x+h)="" -="" f(x)|="" for="" all="" x="" and="" for="" all="" negative="" h.="" taking="" the="" limit="" as="" h="" approaches="" 0,="" we="" have="" |f'(x)="" -="" 0|="" ≤="" 0,="" which="" is="" true.="" therefore,="" it="" is="" possible="" for="" f'(x)="" />< 0="" for="" all="" x="" in="" the="" domain="" of="" />
In conclusion, the given inequality |f(x) < f'(x)|="" holds="" for="" all="" real="" numbers="" x="" if="" and="" only="" if="" f'(x)="" />< 0="" for="" all="" x="" in="" the="" domain="" of="" f.="" 0="" for="" all="" x="" in="" the="" domain="" of="" />
To prove this, we can consider two cases:
Case 1: f'(x) > 0 for all x in the domain of f.
In this case, since f is differentiable, f'(x) > 0 implies that f is strictly increasing. Therefore, if f(x) < f'(x)="" for="" all="" x,="" then="" f(x)="" />< f'(x)="" />< f'(x+h)="" for="" all="" x="" and="" for="" all="" positive="" h.="" this="" implies="" that="" |f(x+h)="" -="" f(x)|="" />< |f'(x+h)="" -="" f(x)|="" for="" all="" x="" and="" for="" all="" positive="" h.="" taking="" the="" limit="" as="" h="" approaches="" 0,="" we="" have="" |f'(x)="" -="" 0|="" ≤="" 0,="" which="" is="" a="" contradiction.="" therefore,="" it="" is="" not="" possible="" for="" f'(x)="" /> 0 for all x in the domain of f.
Case 2: f'(x) < 0="" for="" all="" x="" in="" the="" domain="" of="" />
In this case, since f is differentiable, f'(x) < 0="" implies="" that="" f="" is="" strictly="" decreasing.="" therefore,="" if="" f(x)="" />< f'(x)="" for="" all="" x,="" then="" f(x)="" />< f'(x)="" />< f'(x+h)="" for="" all="" x="" and="" for="" all="" negative="" h.="" this="" implies="" that="" |f(x+h)="" -="" f(x)|="" />< |f'(x+h)="" -="" f(x)|="" for="" all="" x="" and="" for="" all="" negative="" h.="" taking="" the="" limit="" as="" h="" approaches="" 0,="" we="" have="" |f'(x)="" -="" 0|="" ≤="" 0,="" which="" is="" true.="" therefore,="" it="" is="" possible="" for="" f'(x)="" />< 0="" for="" all="" x="" in="" the="" domain="" of="" />
In conclusion, the given inequality |f(x) < f'(x)|="" holds="" for="" all="" real="" numbers="" x="" if="" and="" only="" if="" f'(x)="" />< 0="" for="" all="" x="" in="" the="" domain="" of="" f.="" 0="" for="" all="" x="" in="" the="" domain="" of="" />
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If f is a real-valued differentiable function satisfying |f(x) –f(y)| ≤(x –y)2, x, y ∈ R and f(0) = 0, then f(1) equalsa)1b)2c)0d)–1Correct answer is option 'C'. Can you explain this answer?
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If f is a real-valued differentiable function satisfying |f(x) –f(y)| ≤(x –y)2, x, y ∈ R and f(0) = 0, then f(1) equalsa)1b)2c)0d)–1Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If f is a real-valued differentiable function satisfying |f(x) –f(y)| ≤(x –y)2, x, y ∈ R and f(0) = 0, then f(1) equalsa)1b)2c)0d)–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If f is a real-valued differentiable function satisfying |f(x) –f(y)| ≤(x –y)2, x, y ∈ R and f(0) = 0, then f(1) equalsa)1b)2c)0d)–1Correct answer is option 'C'. Can you explain this answer?.
If f is a real-valued differentiable function satisfying |f(x) –f(y)| ≤(x –y)2, x, y ∈ R and f(0) = 0, then f(1) equalsa)1b)2c)0d)–1Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If f is a real-valued differentiable function satisfying |f(x) –f(y)| ≤(x –y)2, x, y ∈ R and f(0) = 0, then f(1) equalsa)1b)2c)0d)–1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If f is a real-valued differentiable function satisfying |f(x) –f(y)| ≤(x –y)2, x, y ∈ R and f(0) = 0, then f(1) equalsa)1b)2c)0d)–1Correct answer is option 'C'. Can you explain this answer?.
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