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(C) Let the quadratic curve be $f(x) = A(x+1)(x-k)$. Since it touches $y=x$ at $(1,1)$,$f(1)=1$ and $f'(1)=1$.$f(1) = A(2)(1-k) = 1 \Rightarrow 2A(1-k

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$11$

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(C) Let the quadratic curve be $f(x) = A(x+1)(x-k)$. Since it touches $y=x$ at $(1,1)$,$f(1)=1$ and $f'(1)=1$.$f(1) = A(2)(1-k) = 1 \Rightarrow 2A(1-k) = 1$.$f'(x) = A(x-k) + A(x+1) = A(2x+1-k)$.$f'(1) = A(2+1-k) = A(3-k) = 1$.From $2A(1-k) = 1$ and $A(3-k) = 1$,we get $2A(1-k) = A(3-k) \Rightarrow 2-2k = 3-k \Rightarrow k = -1$.Then $A(3 - (-1)) = 1 \Rightarrow 4A = 1 \Rightarrow A = 1/4$.So,$f(x) = \frac{1}{4}(x+1)^2$.Given the point $(\alpha, \alpha+1)$ lies on the curve: $\alpha+1 = \frac{1}{4}(\alpha+1)^2$.Since $\alpha > -1$,$\alpha+1 = 4 \Rightarrow \alpha = 3$.The point is $(3, 4)$.$f'(x) = \frac{1}{2}(x+1)$,so $f'(3) = \frac{1}{2}(3+1) = 2$.The slope of the normal at $(3, 4)$ is $m_n = -1/2$.The equation of the normal is $y - 4 = -\frac{1}{2}(x - 3)$.For the $x$-intercept,set $y=0$: $-4 = -\frac{1}{2}(x - 3) \Rightarrow 8 = x - 3 \Rightarrow x = 11$.

Let the quadratic curve passing through the point $(-1, 0)$ and touching the line $y = x$ at $(1, 1)$ be $y = f(x)$. Then the $x$-intercept of the normal to the curve at the point $(\alpha, \alpha + 1)$ in the first quadrant is $..........$.

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