# The interval in which y = x^{2}e^{- }^{x }is increasing is

(A) (- ∞, ∞) (B) (- 2, 0) (C) (2, ∞) (D) (0, 2)

**Solution:**

Increasing functions are those functions that increase monotonically within a particular domain,

and decreasing functions are those which decrease monotonically within a particular domain.

We have,

y = x^{2}e^{- }^{x}

Therefore,

dy/dx = 2xe^{- }^{x} - x^{2}e^{- x}

= xe^{- x} (2 - x)

Now,

dy/dx = 0

Hence, x = 0 and x = 2

The points x = 0 and x = 2 divide the real line into three disjoint intervals

i.e., (- ∞, 0) , (0, 2) and (2, ∞)

In intervals (- ∞, 0) and (2, ∞), f' (x) < 0 as e^{- }^{x} is always positive.

Hence, f is decreasing on (- ∞, 0) and (2, ∞)

In interval (0, 2) ,

f' (x) > 0

Hence, f is strictly increasing in (0, 2)

Thus, the correct option is D

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.2 Question 19

## The interval in which y = x^{2}e^{- }^{x }is increasing is (A) (- ∞, ∞) (B) (- 2, 0) (C) (2, ∞) (D) (0, 2)

**Summary:**

For the function y = x^{2}e^{- }^{x}, f is strictly increasing in (0, 2) Thus, the correct option is D. The points x = 0 and x = 2 divide the real line into three disjoint intervals

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