Jamia Millia Islamia MCA Continuity Previous Year Questions (PYQs)
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If $f(x)=x^{2}\sin\frac{1}{x}$ for $x\neq0$, then the value of the function $f$ at $x=0$,
so that $f$ is continuous at $x=0$, is
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Solution
Which statement is false?
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Solution
Continuity of $f(x)$ at $x=a$ does **not imply** that its inverse $f^{-1}(x)$ is continuous there (the inverse may not even exist).
$\boxed{(B)}$
The function $f$ is defined in $[-5,5]$ as
$
f(x)=
\begin{cases}
x, & \text{if }x\text{ is rational}\\
-x, & \text{if }x\text{ is irrational}
\end{cases}
$
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Solution
For rationals $f(x)=x$, for irrationals $f(x)=-x$.
At $x=0$, both give $0$, so it’s continuous there.
At any $x\neq0$, limits from rationals and irrationals differ.
A function is continuous at x = a if:
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Solution
Continuity condition
$\lim_{x \to a} f(x) = f(a)$
Conditions:
$f(a)$ exists
$\lim_{x \to a} f(x)$ exists
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