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

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

Answers (1)

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)

 

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