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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 20 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 20 Maths Textbook Solution.

Answers (1)

Answer: y-6=0and y =7  at points (2,7),(3,6)

Hint: Differentiate both side with respect to x

Given : y=2x^{3}-15x^{2}+36x-21

Differentiating both sides w.r.t x, we get

\frac{dy}{dx}=6x^{2}-30x+36

For the points on the curve, where tangents are parallel to x-axis

\frac{dy}{dx}=0

\begin{aligned} &\Rightarrow 6 x^{2}-30 x+36=0 \Rightarrow x^{2}-5 x+6=0 \\ &\therefore(x-2)(x-3)=0 \Rightarrow x=2,3 \\ &\therefore \text { from }(1), y=2(2)^{3}-15(2)^{2}+36(2)-21,2(3)^{2}-15(3)^{2}+36(3)-21 \\ &=16-60+72-21,54-135+108-21=7,6 \\ &\therefore \text { points are }(2,7),(3,6) \end{aligned}

The equation of tangent at (2,7) parallel to x-axis is

y-7=0\left ( x=2 \right )                                                                         \left [ \because m=0 \right ]

The equation of tangent at (2,6) parallel to x-axis is

y-6=0\left ( x=3 \right )  or   y-6=0

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