Let y = f(x) be the equation of a parabola which is touched by the line y = x at the point where x = 1. Then,

2f(0) = 1 - f'(0)

Let $y=a x^2+b x+c$ be the given parabola. Then,

$f(x)=a x^2+b x+c$

Clearly, $\frac{d y}{d x}=2 a x+b$

It is given that y = x touches the parabola at x = 1.

∴  $\left(\frac{d y}{d x}\right)_{x=1}$ = (Slope of the line y = x)

$\Rightarrow 2 a+b=1$           .......(i)

Putting x = 1 in y = x, we get y = 1

So, the line y = x touches the parabola $y=a x^2+b x+c$ at (1, 1).

$a+b+c=1$             .......(ii)

Now,

$f(x) =a x^2+b x+c \Rightarrow f'(x)=2 a x+b$ and $f"(x)=2 a$

∴  $f(0) =c, f'(0)=b, f"(0)=2 a$ and $f'(1)=2 a+b$

From (ii), we have

$a+b+c=1$

$\Rightarrow 2 a+2 b+2 c=2$

$\Rightarrow 2 a+b+(b+2 c)=2$

$\Rightarrow 1+(b+2 c)=2$             [∵ 2a + b = 1 from (i)]

$\Rightarrow b+2 c=1 \Rightarrow 2 c=1-b \Rightarrow 2 f(0)=1-f'(0)$

Also, $f'(1)=2 a+b=1$                   [Using (i)]