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For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

This problem is a simple application of the sequential definition of continuity from ISI MStat 2015 PSB Problem 1 based on Telescopic Continuity.

Let \(f: R \rightarrow R\) be a function which is continuous at 0 and \(f(0)=1\)

Also assume that \(f\) satisfies the following relation for all \(x\) :

$$

f(x)-f(\frac{x}{2})=\frac{3 x^{2}}{4}+x

$$ Find \(f(3)\).

$$

f(3)-f(\frac{3}{2})=\frac{3 \times 3^{2}}{4}+3

$$

$$

f(\frac{3}{2})-f(\frac{\frac{3}{2}}{2})=\frac{3 \times \frac{3}{2}^{2}}{4}+\frac{3}{2}

$$

$$

f(\frac{3}{2^2})-f(\frac{\frac{3}{2^2}}{2})=\frac{3 \times \frac{3}{2^2}^{2}}{4}+\frac{3}{2^2}

$$

\( \cdots \)

$$

f(\frac{3}{2^n})-f(\frac{\frac{3}{2^n}}{2})=\frac{3 \times \frac{3}{2^n}^{2}}{4}+\frac{3}{2^n}

$$

Add them all up. That's the telescopic elegance.

$$

f(3)-f(\frac{3}{2^{n+1}})= \frac{3 \times 3^{2}}{4} \times \sum_{k = 0}^{n} \frac{1}{2^{2k}} + 3 \times \sum_{k = 0}^{n} \frac{1}{2^k} \rightarrow [*]

$$

Observe that \( a_n \to 0 \Rightarrow f(a_n) \to f(0) = 1\) since, \(f(x)\) is continuous at \(x=0\).

Hence take limit \( n \to \infty \) on \([*]\), and we get \( f(3) - f(0) = \frac{3 \times 3^{2}}{4} \times \frac{4}{3} + 3 \times 2 = 15 \).

- Find the general function from the given condition.
- $$f(x)-f(\frac{x}{2})=g(x)$$ and g(x) is continuous, then prove that \(g(0) =0\).
- What if \(g(x)\) is not continuous?

What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

- What are some of the best colleges for Mathematics that you can aim to apply for after high school?
- How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
- What are the best universities for MS, MMath, and Ph.D. Programs in India?
- What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
- How can you pursue a Ph.D. in Mathematics outside India?
- What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

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