IGDTUW 2026 — Mathematics PYQ

To find the intervals where the function is increasing or decreasing, we need to find its first derivative, $f^{\prime }(x)$, and analyze its sign.

Step 1: Differentiate the function

The given function is:

$$f(x)=x{e}^{x-x^2}$$

Using the Product Rule of differentiation $\left(\frac{d}{dx}[uv]=u^{\prime }v+u{v}^{\prime }\right)$ and the Chain Rule:

$$f^{\prime }(x)=\frac{d}{dx}(x)\cdot e^{x-x^2}+x\cdot \frac{d}{dx}\left(e^{x-x^2}\right)$$

$$f^{\prime }(x)=1\cdot e^{x-x^2}+x\cdot \left[e^{x-x^2}\cdot \frac{d}{dx}(x-x^2)\right]$$

$$f^{\prime }(x)=e^{x-x^2}+x\cdot e^{x-x^2}\cdot (1-2x)$$

Step 2: Simplify the derivative

Factor out the common term $e^{x-x^2}$:

$$f^{\prime }(x)=e^{x-x^2}\left[1+x(1-2x)\right]$$

$$f^{\prime }(x)=e^{x-x^2}\left(1+x-2x^2\right)$$

Rearranging the quadratic expression inside the bracket:

$$f^{\prime }(x)=-e^{x-x^2}\left(2x^2-x-1\right)$$

Step 3: Factorize the quadratic equation

Let's split the middle term for $2x^2-x-1$:

$$2x^2-2x+x-1=2x(x-1)+1(x-1)=(2x+1)(x-1)$$

Substitute this back into $f^{\prime }(x)$:

$$f^{\prime }(x)=-e^{x-x^2}(2x+1)(x-1)$$

Alternatively, incorporating the negative sign inside the factors:

$$f^{\prime }(x)=e^{x-x^2}(2x+1)(1-x)$$

Step 4: Determine the sign of $f^{\prime }(x)$

Since the exponential term $e^{x-x^2}$ is always positive for all real values of $x$, the sign of $f^{\prime }(x)$ depends entirely on the product $(2x+1)(1-x)$.

Let's find the critical points by setting $f^{\prime }(x)=0$:

$$2x+1=0⟹x=-\frac{1}{2}$$

$$1-x=0⟹x=1$$

Now, analyze the intervals on a number line using the wavy curve method:

• For $x\in \left(-\infty ,-\frac{1}{2}\right)$, f'(x) < 0 (Decreasing)

• For $x\in \left[-\frac{1}{2},1\right]$, $f^{\prime }(x)\ge 0$ (Increasing)

• For $x\in (1,\infty )$, f'(x) < 0 (Decreasing)

Since $f^{\prime }(x)\ge 0$ for all $x\in \left[-\frac{1}{2},1\right]$, the function $f(x)$ is increasing on this interval.

Correct Answer:

(a) increasing on $\left[-\frac{1}{2},1\right]$