Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 11 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 10 Maths Textbook Solution.

Answers (1)

Answer:\text { The required points are }\left(\frac{5}{3}, \frac{4}{3}\right) \text { or }\left(\frac{4}{3}, \frac{4}{3}\right)

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

Given:\text { The curve is } y=3 x^{2}-9 x+8

Solution: y=3 x^{2}-9 x+8

\text { Differentiating the above with respect to } x

\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}=3(2) x^{2-1}-9

\frac{d y}{d x}=6 x-9 \quad \rightarrow(1)

\frac{d y}{d x}=\text { The slope of tangent }=\tan \theta

\text { The tangent is equally inclined with the axis } \theta=\frac{\pi}{4} \text { or } \frac{-\pi}{4}

\frac{d y}{d x}=\tan \left(\frac{\pi}{4}\right) \text { or } \tan \left(-\frac{\pi}{4}\right)

=1 \text { or }-1 \rightarrow(2)

From(1)&(2) we get

6 x-9=1 \text { or } 6 x-9=-1

6 x=1+9 \text { or } 6 x=-1+9

6 x=10 \quad \text { or } 6 x=8

x=\frac{10}{6} \quad \text { or } \quad x=\frac{8}{6}

x=\frac{5}{3} \quad \text { or } \quad x=\frac{4}{3}

Thus the requried points are\left(\frac{5}{3}, \frac{4}{3}\right) \text { or }\left(\frac{4}{3}, \frac{4}{3}\right)

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Result 2025 Live: UK Board 10th, 12th results tomorrow; AP SSC, TS Inter, UP, CBSE updates
anam-khan

When will CBSE Class 12 results 2025 be declared? Last two years' trends
anam-khan

CBSE Class 10, 12 board exams 2025 from April 7 to 11 for sports students

Latest Question