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Joint Entrance Examination
Mathematics
Application of Derivatives
Question 167
Question 167 of 212
Medium
MCQ
Let f be any function defined on R and let it satisfy the condition : \[|f(x) - f(y)|\, \le \,|{(x - y)^2}|,\forall (x,y) \in R\] If f(0) = 1, then :
A
f(x) can take any value in R
B
\[f(x) < 0,\forall x \in R\]
C
\[f(x) > 0,\forall x \in R\]
D
\[f(x) = 0,\forall x \in R\]
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Official solution
\[|f(x) - f(y)|\, \le \,|{(x - y)^2}|\] \[ \Rightarrow \left| {{{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|x - y|\] \[ \Rightarrow \left| {\mathop {\lim }\limits_{x \to y} {{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|\mathop {\lim }\limits_{x \to y} (x - y)|\] \[ \Rightarrow |f'(x)|\, \le 0\] \[ \Rightarrow f'(x) = 0\] \[ \Rightarrow f(x)\] is constant function. \[ \because \]
f
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0
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f(0)
f
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0
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= 1 then f(x) = 1
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