Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8 Important

Example 9 Important

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20 Important

Example 21 Important

Example 22

Example 23 Important

Example 24

Example 25 Important

Example 26 Important

Example 27

Example 28 Important

Example 29 Important

Example 30 Important

Example 31 Important

Example 32 Important

Example 33 Important

Example 34 Important

Example 35

Example 36 Important

Example 37

Question 1

Question 2

Question 3 You are here

Question 4 Important

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13 Important

Question 14 Important

Last updated at April 16, 2024 by Teachoo

Question 3 Find the equation of all lines having slope 2 and being tangent Equation of curve is y + 2/(𝑥 − 3) = 0 Differentiating both sides w.r.t x (𝑑𝑦 )/𝑑𝑥 + (𝑑 )/𝑑𝑥 (2/(𝑥 − 3))=0 (𝑑𝑦 )/𝑑𝑥 =−(𝑑 )/𝑑𝑥 (2/(𝑥 − 3)) (𝑑𝑦 )/𝑑𝑥 =−2 (𝑑 )/𝑑𝑥 (𝑥−3)^(−1) (𝑑𝑦 )/𝑑𝑥 =−2〖 × −(𝑥−3)〗^(−1−1) (𝑑𝑦 )/𝑑𝑥 =2(𝑥−3)^(−2) (𝒅𝒚 )/𝒅𝒙 =𝟐/(𝒙 − 𝟑)^𝟐 Given that slope = 2 𝑑𝑦/𝑑𝑥 = 2 2/(𝑥 − 3)^2 = 2 1/(𝑥 − 3)^2 = 1 (𝑥−3)^2 = 1 𝑥−3 = ±1 x − 3 = 1 x = 4 x − 3 = − 1 x = 2 x − 3 = − 1 x = 2 So, x = 4 & x = 2 Finding value of y If x = 2 y = (−2)/(𝑥 − 3) y = (−2)/(2 − 3) 𝑦=2 Thus, point is (2, 2) If x = 4 y = (−2)/(𝑥 − 3) y = (−2)/(4 − 3) 𝑦=−2 Thus, point is (4, –2) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 2) and (4, − 2) We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent through (2, 2) is 𝑦 − 2 = 2 (𝑥 −2) 𝑦 − 2 = 2𝑥 − 4 𝒚−𝟐𝒙+𝟐 = 𝟎 Equation of tangent through (4, −2) is 𝑦 −(−2) = 2 (𝑥 −4) 𝑦 + 2 = 2𝑥 − 8 𝒚 − 𝟐𝒙 + 𝟏𝟎 = 𝟎 to the curve y + 2/(𝑥 − 3) = 0