If the curve $y=x^2+b x+c$ touches the line y = x at the point (1, 1), then the set of values of x for which the curve has a negative gradient is

$(-\infty, 1 / 2)$

We have,

$y=x^2+b x+c$          ..........(i)

This curve touches y = x at P(1, 1). Therefore, P lies on the curve and slope of the tangent to the curve at P is equal to the slope of the line y = x.

∴  $1=1+b+c$ and $\left(\frac{d y}{d x}\right)_P=1$

$\Rightarrow b+c=0$ and $2+b=1 \Rightarrow b=-1$ and c = 1

Substituting the values of b and c in (i), we have

$y=x^2-x+1 \Rightarrow \frac{d y}{d x}=2 x-1$

For negative gradient, we must have

$\frac{d y}{d x}<0 \Rightarrow 2 x-1<0 \Rightarrow x<\frac{1}{2} \Rightarrow x \in(-\infty, 1 / 2)$