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Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 14 Maths Textbook Solution.

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Equation of tangent is  y=3


Tangent is parallel to X-axis so slope becomes 0


Given curve   y=x+\frac{4}{x^{2}}

To find:

We have to find the equation of the tangent to the given curve that is parallel to X-axis


Given tangent is parallel to X-axis so slope is 0

                \Rightarrow \frac{d y}{d x}=0                                                                                                                                          … (i)

                Here   y=x+\frac{4}{x^{2}}                                                                                                                             … (ii)

On differentiate both side with respect to  x  we get

                \frac{d y}{d x}=1-\frac{8}{x^{3}}

Substituting the value of  \frac{dy}{dx}  in equation (i), we get

               \begin{aligned} &1-\frac{8}{x^{3}}=0 \\\\ &x^{3}=8 \\\\ &x=2 \end{aligned}

\because From (ii) we get  y=2+\frac{4}{4}=3

Hence equation of tangent is y= 3                                         [ \because equation of tangent y-3=0(x-2) ]

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