NIMCET Continuity Previous Year Questions (PYQs) – Page 1 of 2
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Test the continuity of the function at x = 2
$f(x)= \begin{cases} \frac{5}{2}-x & \text{ if } x<2 \\ 1 & \text{ if } x=2 \\ x-\frac{3}{2}& \text{ if } x>2 \end{cases}$
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Solution
LHL ≠ f(2)
Let $f(x) = \lfloor x^2 - 3 \rfloor$ where $\lfloor \cdot \rfloor$ is the greatest integer function.
Number of points in $(1,2)$ where $f$ is discontinuous:
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Solution
Discontinuity occurs when
$x^2 - 3$ is an integer
Let
$x^2 - 3 = k \Rightarrow x = \sqrt{k+3}$
We need $1 < x < 2$ → square both sides:
$1 < \sqrt{k+3} < 2$
$\Rightarrow 1 < k+3 < 4$
$\Rightarrow -2 < k < 1$
Possible integer values: $k = -1, 0$
So number of discontinuities = $2$
Let \(f(x) = \begin{cases} \cos [x], & x\geq 0\\ |x| + a, & x< 0 \end{cases}\) where \([x]\) denotes the greatest integer \(\leq x\). If f should be continuous at \(x = 0\) then a must be:
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Solution
If $f\colon R\rightarrow R$ is defined by $f(x)=\begin{cases}{\frac{x+2}{{x}^2+3x+2}} & {,\, if\, x\, \in R-\{-1,-2\}} \\ {-1} & {,if\, x=-2} \\ {0} & {,if\, x=-1}\end{cases}$ , then f(x) is continuous on the set
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Solution
If
$
f(x)=
\begin{cases}
x \sin\left(\frac{1}{x}\right), & x \ne 0 \\
0, & x = 0
\end{cases}
$
then
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Solution
Solution:
lim_{x→0} x·sin(1/x) = 0
⇒ f is continuous at 0.
f'(0) = lim_{x→0} [x sin(1/x)]/x = sin(1/x)
But sin(1/x) has no limit as x→0⁺ or x→0⁻.
⇒ Both f'(0+) and f'(0-) do not exist.
Let $g:\mathbb{R}\rightarrow \mathbb{R}$ and $h:\mathbb{R}\rightarrow \mathbb{R}$, be two functions such that $h(x) = sgn(g(x))$. Then select
which of the following is not true?( $\mathbb{R}$ denotes the set of all real numbers, sgn stands for
signum function)
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Solution
Let $g:\mathbb{R}\to\mathbb{R}$ and $h:\mathbb{R}\to\mathbb{R}$ be such that
$h(x) = \operatorname{sgn}(g(x))$.
Recall:
$\operatorname{sgn}(t) =
\begin{cases}
1, & t>0\\
0, & t=0\\
-1, & t<0
\end{cases}$
Check each statement:
1) "The domain of $h(x)$ is the same as the domain of $g(x)$."
$\Rightarrow$ True, because $\operatorname{sgn}(g(x))$ is defined for every $x$ where $g(x)$ is defined.
2) "The domain of continuity of $h(x)$ equals the domain of continuity of $g(x) - \{x\in\mathbb{R} : g(x)=0\}$."
At points where $g(x)\neq 0$, $h(x)$ is locally constant ($1$ or $-1$), hence continuous there (provided $g$ itself is continuous).
At points where $g(x)=0$, $h(x)$ jumps from $-1$ to $1$, so it is discontinuous.
$\Rightarrow$
This statement is true.
3) "The domain of $h(x)$ is different from the domain of $g(x)$ at the same point."
Since for every $x$ in the domain of $g$, $h(x)=\operatorname{sgn}(g(x))$ is defined, the domains are exactly the same; they never differ.
$\Rightarrow$ This statement is false.
4) " $h(x)$ is discontinuous at $g(x)=0$."
At any $x_0$ where $g(x_0)=0$, the left and right limits of $h(x)$ are $-1$ and $1$, not equal to $h(x_0)=0$.
$\Rightarrow$ $h$ is discontinuous there, so this statement is true.
Therefore, the statement which is **not true** is:
$\boxed{\text{Option 3}}$
Consider the function $f$ defined by
$f(x)=\begin{cases} x^2-1, & x<3 \ 2ax, & x\geq 3 \end{cases}$
for all real number $x$. If $f$ is continuous at $x=3$, then value of $a$ is
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Solution
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