If the curve $y=x^2+bx+c$ touches the line $y=x$ at the point (1,1), then the value of x for which the curve has a negative gradient are

We have, $y=x^2+bx+c$

$\Rightarrow \frac{dy}{dx}=2x+b$

Since the curve touches the line y = x at the point (1,1)

$\therefore {(2x+b)]}_{(1,1)}=1$

i.e. $2+b=1\Rightarrow b=-1$

Also, the curve passes through the point (1,1)

$\therefore 1=1+b+c$ i.e., c=-b=1

$\therefore y=x^2-x+1\Rightarrow \frac{dy}{dx}=2x-1$

Now, $\frac{dy}{dx}<0\Rightarrow 2x-1<0\Rightarrow x<\frac{1}{2}$