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

Answer: $y-6=0$and 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   $y-6=0$