Question 37 of 82intermediate🔧 ApplyMCQ1 mark

The interval in which y = x² e⁻ˣ is increasing is

Exercise: EXERCISE 6.2 | Q: 19 | (Page 13)
For More Understanding

Explanation

To find where the function is increasing, we need to find where its derivative is positive. For y = x²e⁻ˣ, using the product rule: dy/dx = 2xe⁻ˣ - x²e⁻ˣ = e⁻ˣ(2x - x²) = xe⁻ˣ(2 - x). Since e⁻ˣ > 0 for all real x, the sign depends on x(2-x). Setting dy/dx = 0 gives x = 0 and x = 2. Testing intervals: for 0 < x < 2, both x and (2-x) are positive, so dy/dx > 0. Therefore, the function is increasing in (0, 2).

Solution Steps

  1. Step 1: Find derivative using product rule: dy/dx = d/dx(x²)·e⁻ˣ + x²·d/dx(e⁻ˣ) = 2xe⁻ˣ - x²e⁻ˣ

  2. Step 2: Simplify: dy/dx = e⁻ˣ(2x - x²) = xe⁻ˣ(2 - x)

  3. Step 3: Find critical points by setting dy/dx = 0: x = 0 or x = 2

  4. Step 4: Test sign in each interval. Since e⁻ˣ > 0 always, check sign of x(2-x)

  5. Step 5: For 0 < x < 2: x > 0 and (2-x) > 0, so dy/dx > 0. Function is increasing in (0, 2).

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