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Ex 6.3,2
Ex 6.3,3 Important
Ex 6.3,4
Ex 6.3, 5 Important
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10 You are here
Ex 6.3,11 Important
Ex 6.3,12
Ex 6.3,13
Ex 6.3, 14 (i)
Ex 6.3, 14 (ii) Important
Ex 6.3, 14 (iii)
Ex 6.3, 14 (iv) Important
Ex 6.3, 14 (v)
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19
Ex 6.3,20
Ex 6.3,21 Important
Ex 6.3,22
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25
Ex 6.3,26 (MCQ) Important
Ex 6.3,27 (MCQ)
Last updated at April 14, 2021 by Teachoo
Ex 6.3, 10 Find the equation of all lines having slope –1 that are tangents to the curve 𝑦=1/(𝑥 − 1) , 𝑥≠1.Equation of Curve is 𝑦=1/(𝑥 − 1) Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑(1/(𝑥 − 1))/𝑑𝑥 (𝑑𝑦 )/𝑑𝑥 =(𝑑 )/𝑑𝑥 (𝑥−1)^(−1) (𝑑𝑦 )/𝑑𝑥 =−1〖 × (𝑥−1)〗^(−1−1) (𝑑𝑦 )/𝑑𝑥 =−(𝑥−1)^(−2) 𝑑𝑦/𝑑𝑥=(− 1)/(𝑥 − 1)^2 Given that slope = −1 Hence, 𝑑𝑦/𝑑𝑥 = −1 ∴ (− 1)/(𝑥 − 1)^2 =−1 1/(𝑥 − 1)^2 =1 1=(𝑥 − 1)^2 (𝑥 − 1)^2=1 𝑥 − 1=±1 x − 1 = 1 x = 2 x − 1 = −1 x = 0 So, x = 2 & x = 0 Finding value of y If x = 2 y = 1/(𝑥 − 1) y = 1/(2 − 1) y = 1/1 y = 1 Thus, point is (2, 1) If x = 0 y = 1/(𝑥 − 1) y = 1/(0 − 1) y = 1/(−1) y = −1 Thus, point is (0, –1) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 1) and (0, −1) We know that Equation of line at (𝑥1 , 𝑦1)& having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent through (2, 1) is 𝑦 −1 =−1 (𝑥 −2) 𝑦 −1 =−𝑥+2 𝒚+𝒙−𝟑 = 𝟎 Equation of tangent through (0, −1) is 𝑦 −(−1)=−1 (𝑥 −0) 𝑦 +1 =−𝑥 𝒚+𝒙 + 𝟏 = 𝟎