Discontinuity Problem | ISI B.Stat Objective | TOMATO 734
Try this beautiful problem based on Discontinuity, useful for ISI B.Stat Entrance.
Discontinuity Problem | ISI B.Stat TOMATO 734
$\quad$ Let $f(x)=|| x-1|-1|$ if $x<1$ and $f(x)=[x]$ if $x \geq 1,$ where, for any real number $x,[x]$ denotes the largest integer $\leq x$ and $|y|$ denotes absolute value of $y .$ Then, the set of discontinuity-points of the function f consists of
all integers $\geq 0$
all integers $\geq 1$
all integers $> 1$
the integer 1
Key Concepts
Limit
Calculas
Continuous
Check the Answer
Answer: \(C\)
TOMATO, Problem 734
Challenges and Thrills in Pre College Mathematics
Try with Hints
Let us first check at $x=1$ $\lim f(x)$ as $x\to1-=\lim || x-1|-1|$ as $x\to 1-=1$ $\lim f(x)$ as $x\to1+=\lim [x]$ as $x\to1+=1$ So, continuous at $x=1$ Let us now check at $x=0$ $\lim f(x)$ as $x\to 0-=\lim || x-1|-1|$ as $x\to 0-=0$ $\lim f(x)$ as $x\to 0+=\lim || x-1|-1|$ as $x\to 0+=0$ So continuous at $x=0$
Can you now finish the problem ..........
Let us now check at $x=2$ $\lim f(x)$ as $x\to 2-=\lim [x]$ as $x\to 2-=1$ $\lim f(x)$ as $x\to 2+=\lim [x]$ as $x \rightarrow 2+=2$ Discontinuous at $x=2$ Option (c) is correct.
Real valued function | ISI B.Stat Objective | TOMATO 690
Try this beautiful problem based on Real valued function, useful for ISI B.Stat Entrance
Real valued functions | ISI B.Stat TOMATO 690
Let \(f(x)\) be a real-valued function defined for all real numbers x such that \(|f(x) – f(y)|≤(1/2)|x – y|\) for all x, y. Then the number of points of intersection of the graph of \(y = f(x)\) and the line \(y = x\) is
0
1
2
none of these
Key Concepts
Limit
Calculas
Real valued function
Check the Answer
Answer: \(1\)
TOMATO, Problem 690
Challenges and Thrills in Pre College Mathematics
Try with Hints
Now,
\(|f(x) – f(y)| ≤ (1/2)|x – y|\)
\(\Rightarrow lim |{f(x) – f(y)}/(x – y)|\)( as x -> y ≤ lim (1/2)) as x - > y
\(\Rightarrow |f‟(y)| ≤ ½\)
\(\Rightarrow -1/2 ≤ f‟(y) ≤1/2\)
\(\Rightarrow -y/2 ≤ f(y) ≤ y/2\) (integrating)
\(\Rightarrow -x/2 ≤ f(x) ≤ x/2\)
Can you now finish the problem ..........
Therefore from the picture we can say that intersection point is \(1\)
