CSIR NET MATHEMATICS | Real Analysis | Continuous & One-One Functions & Solutions June 2019
Booklet C Section B Example 21 June 2019
Real Analysis, Continuous function, One-One function, Onto function, Bounded Function
CSIR NET Answer Key June 2019, Mathematical Science
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Video Summary
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This video provides a step-by-step solution to a real analysis problem from the CSIR NET June 2019 exam. The goal is to determine which property must hold for a continuous and one-to-one function $f: \\mathbb{R} \\to \\mathbb{R}$. Through the use of counterexamples, the instructor eliminates incorrect options to conclude that such a function must be strictly monotonic.
In this video, we are going to discuss the problem
of real analysis from CSIR NET June 2019.
Example number 21 from Booklet C, Marks 3.
Let f defined from real number to real number be a continuous
and one-one function.
Which of the following is true?
f is onto, f is either strictly decreasing or strictly increasing.
Third option, there exists x belongs to R such that f is equal to 1
and f is unbounded.
We'll solve this problem using counter example.
So, first example.
f of x is equal to e power x
so f of x is equal to e power x is continuous as well as 1 1 but it is
not onto why because f of x is equal to 0
is not possible for any real number f of x is equal to
e power x and it is 0 when x has value minus infinity
and then minus infinity is not in real
Therefore, f of x never be zero.
Therefore, we cannot get any.
Inverse image of zero therefore f is not onto.
Now the second example f of x is equal to 1 plus e power x.
So f of x is continuous as well as 1 1.
Now f of x never be 1 because
e power x never be 0 which
is discussed in the example number 1.
So the similar way f of x never be 1.
So, the third option that exists x belongs to R says that f of x
is equal to 1 for all continuous and 1, 1 function on real number
to real number which is not possible because f of x is equal to 1 plus
e power x does not have value f of x is equal to 1 for real number.
So, third option is also wrong.
Now, the example number 3 f of x is equal to 1 upon 1 plus
e power x. It is also continuous as well as 1,
1 since 0 less than e power x less than infinity.
If we add 1, then the inequality will be 1 less than 1 plus
e power x less than infinity.
Now, if we take the reciprocal, then 1 greater than 1 upon 1 plus
e power x greater than 1 upon infinity, which is 0.
Now, rewrite 0 less than 1 upon 1 plus e power x less than 1,
which shows that f of x is bounded.
So, f is bounded when f is continuous and 1, 1.
So, third option that f is unbounded if it is continuous
and 1,1 on real values which is wrong.
So, there is only one correct option
which is second one that f is either strictly decreasing or increasing.
Thanks for watching.
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