1. Given functions

We have $f(x)=e^x-x, \qquad g(x)=x^2-x$ and $h(x)=(f\circ g)(x)=f(g(x))=e^{x^2-x}-(x^2-x).$

We need the set of all real $x$ for which $h(x)$ is increasing.

2. Differentiate using chain rule

Since $h(x)=f(g(x)),$ we get $h'(x)=f'(g(x))\cdot g'(x).$

Now, $f'(x)=e^x-1,$ so $f'(g(x))=e^{g(x)}-1=e^{x^2-x}-1.$

Also, $g'(x)=2x-1.$

Hence, $h'(x)=(e^{x^2-x}-1)(2x-1).$

3. Find where $h'(x)\ge 0$

For $h(x)$ to be increasing, we need $h'(x)\ge 0.$ So we analyze the sign of $(e^{x^2-x}-1)(2x-1).$

4. Sign of $e^{x^2-x}-1$

Recall:

• $e^t-1>0$ if $t>0$
• $e^t-1=0$ if $t=0$
• $e^t-1<0$ if $t<0$

So the sign depends on $x^2-x=x(x-1).$

Now, $x^2-x\ge 0 \iff x\in(-\infty,0]\cup[1,\infty),$ $x^2-x\le 0 \iff x\in[0,1].$

Therefore:

• for $x<0$ or $x>1$, $e^{x^2-x}-1>0$
• at $x=0,1$, $e^{x^2-x}-1=0$
• for $0<x<1$, $e^{x^2-x}-1<0$

5. Sign of $2x-1$

$2x-1\ge 0 \iff x\ge \frac12,$ $2x-1<0 \iff x<\frac12.$

6. Combine signs interval-wise

Critical points are $0,\frac12,1$.

We check intervals:

(i) $x<0$

• $e^{x^2-x}-1>0$
• $2x-1<0$

So $h'(x)<0.$ Not increasing.

(ii) $0<x<\frac12$

• $e^{x^2-x}-1<0$
• $2x-1<0$

So $h'(x)>0.$ Increasing.

(iii) $\frac12<x<1$

• $e^{x^2-x}-1<0$
• $2x-1>0$

So $h'(x)<0.$ Not increasing.

(iv) $x>1$

• $e^{x^2-x}-1>0$
• $2x-1>0$

So $h'(x)>0.$ Increasing.

At boundary points:

• $x=0$: $h'(0)=0$
• $x=\frac12$: $h'\left(\frac12\right)=0$
• $x=1$: $h'(1)=0$

Thus the function is increasing on $\left[0,\frac12\right]\cup[1,\infty).$

7. Match with options

This corresponds to:

$\boxed{\left[0,\frac12\right]\cup[1,\infty)}$

So the correct option is D.

8. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They match.