MCQ
The function $f(x) = {x^2}\,\,\sin \frac{1}{x},\,x \ne \,0,\,\,f(0)\, = 0$ at $x = 0$
- AIs continuous but not differentiable
- BIs discontinuous
- CIs having continuous derivative
- ✓Is continuous and differentiable
but $ - 1 \le \sin \left( {\frac{1}{x}} \right) \le 1$ and $x \to 0$
$\therefore $ $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = 0 = \mathop {\lim }\limits_{x \to {0^ - }} f(x) = f(0)$
Therefore $f(x)$ is continuous at $x = 0$.
Also, the function $f(x) = {x^2}\sin \frac{1}{x}$ is differentiable because
$Rf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\sin \frac{1}{h} - 0}}{h} = 0$,
$Lf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\sin (1/ - h)}}{{ - h}} = 0$.
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If matrix A = [aij]2×2, where aij = 1, if $\text{i}\neq\text{j}$ and 0 if i = j then A2 equal to: