#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 12 Maths Textbook Solution.

Answer:$\text { The required point is }(1,2)$

Hint: $\text { The slope of the tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }$

Given:

$y=2 x^{2}-x+1 \quad \rightarrow(1)$

$y=3 x+4 \quad \rightarrow(2)$

Solution: $\text { Differentiating eqn(1) with respect to } x$

\begin{aligned} &\therefore \frac{d}{d x}\left(\mathrm{x}^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0 \\ &\frac{d y}{d x}=2 \times 2(x)^{2-1}-1+0 \\ &\frac{d y}{d x}=4 x-1 \quad \rightarrow(3) \end{aligned}

$\text { Again differentiating eqn }(2) \text { with respect to } x$

$\frac{d y}{d x}=3 \quad \rightarrow(3)$

ATQ

$\text { The curve } y=2 x^{2}-x+1 \text { is the tangent parallel to the line } y=3 x+4$

\begin{aligned} &4 x-1=3 \\ &4 x=3+1 \\ &4 x=4 \\ &x=\frac{4}{4}=1 \end{aligned}

Thus from(1)

$y=2$

Hence the point is (1,2)