If \(\frac{1}{(20-a)(40-a)} + \frac{1}{(40-a)(60-a)} + \ldots + \frac{1}{(180-a)(200-a)} = \frac{1}{256}\), then the maximum value of a is :
- 198
- 202
- 212
- 218
The Correct Option is C
Solution and Explanation
\(\frac{1}{20}\left(\frac{1}{20-a} - \frac{1}{40-a} + \frac{1}{40-a} - \frac{1}{60-a} + \ldots + \frac{1}{180-a} - \frac{1}{200-a}\right) = \frac{1}{256}\)
\(⇒\) \(\frac{1}{20}\left(\frac{1}{20-a} - \frac{1}{200-a}\right) = \frac{1}{256}\)
\(⇒\) \(\frac{1}{20} \cdot \frac{180}{(20-a)(200-a)} = \frac{1}{256}\)
\(⇒\) \((20 - a)(200 - a) = 9.256\)
\(⇒\) \(a^2 - 220a + 1696 = 0\)
\(⇒\) \(a = 212, 8\)
Then the maxixum value of \(a = 212\)
So, the correct option is (C): 212
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Concepts Used:
Maxima and Minima
What are Maxima and Minima of a Function?
The extrema of a function are very well known as Maxima and minima. Maxima is the maximum and minima is the minimum value of a function within the given set of ranges.
There are two types of maxima and minima that exist in a function, such as:
- Local Maxima and Minima
- Absolute or Global Maxima and Minima