Solve all your doubts with Teachoo Black (new monthly pack available now!)

Miscellaneous

Misc 1 (a)
Deleted for CBSE Board 2023 Exams

Misc 1 (b) Important Deleted for CBSE Board 2023 Exams

Misc 2 Important

Misc 3 Important

Misc 4 Deleted for CBSE Board 2023 Exams

Misc 5 Important Deleted for CBSE Board 2023 Exams

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc. 19 (MCQ)

Misc 20 (MCQ) Important Deleted for CBSE Board 2023 Exams You are here

Misc 21 (MCQ) Important

Misc 22 (MCQ)

Misc. 23 (MCQ) Important

Misc 24 (MCQ) Important

Chapter 6 Class 12 Application of Derivatives

Serial order wise

Last updated at Aug. 9, 2021 by Teachoo

Misc 20 The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2,– 1) is (A) 22/7 (B) 6/7 (C) 7/6 (D) (− 6)/7We need to find slope of tangent at (2, −1) We know that slope of tangent is 𝑑𝑦/𝑑𝑥 𝒅𝒚/𝒅𝒙= (𝒅𝒚/𝒅𝒕)/(𝒅𝒙/𝒅𝒕) Finding 𝒅𝒙/𝒅𝒕 Given 𝑥 = t2 + 3t – 8 Differentiating w.r.t t 𝑑𝑥/𝑑𝑡= (𝑑(𝑡^2 + 3𝑡 −8))/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 2t + 3 Finding 𝒅𝒚/𝒅𝒕 Given 𝑦 = 2t2 − 2t − 5 Differentiating w.r.t t 𝑑𝑦/𝑑𝑡= (𝑑 (2𝑡2 − 2𝑡 − 5))/𝑑𝑡 𝑑𝑦/𝑑𝑡=4t −2 Thus, 𝑑𝑦/𝑑𝑥= (𝑑𝑦∕𝑑𝑡)/(𝑑𝑥∕𝑑𝑡) 𝒅𝒚/𝒅𝒙= (𝟒𝒕 − 𝟐)/(𝟐𝒕 + 𝟑) Now we need to find value of t, Given point is (2, –1) Putting 𝑥 = 2 & 𝑦 = –1 in the curve x = t2 + 3t – 8 2 = t2 + 3t − 8 t2 + 3t – 8 – 2 =0 t2 + 3t − 10 = 0 t2 + 5t – 2t −10 = 0 t (t + 5) – 2 (t − 5) = 0 (t −2) (t + 5) = 0 So, t = 2 & t = − 5 y = 2t2 – 2t – 5 – 1 = 2t2 – 2t – 5 2t2 – 2t – 5 + 1 = 0 2t2 – 2t – 4 = 0 2(t2 – t – 2 ) = 0 t2 – t – 2 = 0 t2 – 2t + t – 2 = 0 t (t − 2) + 1(t − 2) = 0 (t + 1) (t – 2) = 0 So, t = −1 & t = 2 Since t = 2 is common in both parts So, we will calculate 𝑑𝑦/𝑑𝑥= (4𝑡 −2)/(2𝑡 −3) at t = 2 At t = 2 𝑑𝑦/𝑑𝑥= (4𝑡 −2)/(2𝑡 + 3) 𝑑𝑦/𝑑𝑥= (4 (2) − 2)/(2 (2) + 3) = (8 − 2)/(4 +3) = 𝟔/𝟕 Hence the correct answer is (B)