If the line y = x touches the curve y = x2 + bx + c at a point (1, 1) then _____________ .

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b = −1, c= 1 We can find the slope of the line by differentiating w.r.t. x. Slope of the given line = 1 Now, y = x^2 + bx + c . . . \left( 1 ight) \Rightarrow dy/dx = 2x + b 	ext { Slope of the tangent }= \left( dy/dx ight)_\left( 1, 1 ight) =2+b 	ext { Given}: 	ext { Slope of the tangent } = 1 \Rightarrow 2 + b = 1 \Rightarrow b = - 1 	ext { On substituting b= - 1, x=1 and y=1 in (1), we get} \Rightarrow 1 = 1 - 1 + c \Rightarrow c = 1

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