The point at which the tangent to the curve $y=2 x^2-x+1$ is parallel to $y=3 x+9$ will be

(A) $(2,1)$

(B) $(1,2)$

(C) $(3,9)$

(D) $(-2,1)$

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Best answer

**SOLUTION —**

On differentiating, we get

$\frac{d y}{d x}=4 x-1$

Since, this is parallel to the given curve $y=3 x+9$ $\therefore$ These slopes are equal.

$\begin{array}{lrlrl}\Rightarrow & 4 x-1 & =3 \\\Rightarrow & x & =1 \\& \text { At } x=1 & y & =2(1)^2-1+1 \Rightarrow y=2\end{array}$

Thus, the point is $(1,2)$

So, The correct option will be **(B).**

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