If m and n respectively are the number of local maximum and local minimum points of the function
\(f(x)=∫_0 ^{x^2} \frac{t^2–5t+4}{2+et}dt\), then the ordered pair (m, n) is equal to
\(f(x)=∫_0 ^{x^2} \frac{t^2–5t+4}{2+et}dt\), then the ordered pair (m, n) is equal to
- (3, 2)
- (2, 3)
- (2, 2)
- (3, 4)
The Correct Option is B
Solution and Explanation
\(f(x)=∫_0 ^{x^2} \frac{t^2–5t+4}{2+et}dt\)
\(f'(x)=2x\bigg(\frac{x^4–5x^2+4}{2+ex^2}\bigg)=0\)
\(x=0\), or \((x^2–4)(x^2–1)=0\)
\(x = 0, x = ±2, ±1\)
Now,
\(f'(x)=\frac{2x(x+1)(x-1)(x+2)(x-2)}{(ex^2+2)}\)
changes sign from positive to negative at
\(x = –1\), \(1\) So, number of local maximum points = \(2\)
changes sign from negative to positive at
\(x = –2, 0, 2\)
Hence, number of local minimum points = \(3\)
\(∴ m = 2, n = 3\)
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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