Question:
Let \(f:\mathbf{R}\to\mathbf{R}\) be a function such that \(|f(x)|\le x^2\), for all \(x\in\mathbf{R}\). Then, at \(x=0\), \(f\) is :
Let \(f:\mathbf{R}\to\mathbf{R}\) be a function such that \(|f(x)|\le x^2\), for all \(x\in\mathbf{R}\). Then, at \(x=0\), \(f\) is :
Show Hint
Use the squeeze theorem on \(f(x)\) and on \(\frac{f(x)-f(0)}{x}\).
Updated On: Nov 11, 2026
- continuous but not differentiable
- continuous as well as differentiable
- neither continuous nor differentiable
- differentiable but not continuous.
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The Correct Option is B
Solution and Explanation
At \(x=0\): \(|f(0)|\le 0\Rightarrow f(0)=0\).
\(-x^2\le f(x)\le x^2\Rightarrow \lim_{x\to0}f(x)=0=f(0)\), so \(f\) is continuous at 0.
Also \(\left|\frac{f(x)-f(0)}{x}\right|=\frac{|f(x)|}{|x|}\le|x|\to0\), so \(f'(0)=0\) exists. Hence \(f\) is differentiable at 0.
Answer: (2) continuous as well as differentiable
\(-x^2\le f(x)\le x^2\Rightarrow \lim_{x\to0}f(x)=0=f(0)\), so \(f\) is continuous at 0.
Also \(\left|\frac{f(x)-f(0)}{x}\right|=\frac{|f(x)|}{|x|}\le|x|\to0\), so \(f'(0)=0\) exists. Hence \(f\) is differentiable at 0.
Answer: (2) continuous as well as differentiable
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