#### Question

If the line y = x touches the curve y = x^{2} + bx + c at a point (1, 1) then

(a) b = 1, c = 2

(b) b = −1, c = 1

(c) b = 2, c = 1

(d) b = −2, c = 1

#### Solution

(b) 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,

\[y = x^2 + bx + c . . . \left( 1 \right)\]

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

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) =2+b\]

\[\text { Given}:\]

\[\text { Slope of the tangent } = 1\]

\[ \Rightarrow 2 + b = 1\]

\[ \Rightarrow b = - 1\]

\[\text { On substituting b= - 1, x=1 and y=1 in (1), we get}\]

\[ \Rightarrow 1 = 1 - 1 + c\]

\[ \Rightarrow c = 1\]

Is there an error in this question or solution?

Solution If the Line Y = X Touches the Curve Y = X2 + Bx + C at a Point (1, 1) Then (A) B = 1, C = 2 (B) B = −1, C = 1 (C) B = 2, C = 1 (D) B = −2, C = 1 Concept: Tangents and Normals.