## (D) r 1 /2 = r/3

1. Chapter 6 Class 12 Application of Derivatives (Term 1)
2. Serial order wise
3. Case Based Questions (MCQ)

Transcript

Question A right circular cylinder is inscribed in a cone. S = Curved Surface Area of Cylinder. Based on the above information answer the following questions: Question 1 (๐ )/๐1 = ? (A) (โ โโ1)/โ1 (B) (โ1โโ)/โ1 (C) (โ โโ1)/โ (D) (โ + โ1)/โ1 In ฮ OAD tan ๐ถ = ๐ด๐ท/๐๐ด = ๐/(๐_๐โ๐) In ฮ OBC tan ๐ถ = ๐ต๐ถ/๐๐ต = ๐_๐/๐_๐ Comparing (1) and (2) ๐/(โ_1 โ โ) = ๐_1/โ_1 ๐/๐_๐ = (๐_๐ โ ๐)/๐_๐ So, the correct answer is (B) Question 2 Find the value of โSโ? (A) 2๐๐/โ (โ_1โโ)โ (B) 2๐๐/โ_1 (โ_1โโ)โ (C) (2๐๐_1)/โ_1 (โ_1โโ)โ (D) (2๐๐_1)/โ_1 (โ_1+โ)โ Now, S = Curved Surface Area of Cylinder = 2๐๐โ We know that ๐/๐_๐ = (๐_๐ โ ๐)/๐_๐ ๐ = ((๐_๐ โ ๐)/๐_๐ ) ร ๐_๐ = 2๐((๐_๐ โ ๐)/๐_๐ ) ร ๐_๐ ร โ =(2๐๐_1)/โ_1 (โ_1โโ)โ So, the correct answer is (C) Question 3 Find the value of ๐๐/๐โ ? (A) (2๐๐_1)/โ (โ_1โ2โ) (B) (2๐๐_1)/โ_1 (โโ2โ_1) (C) 2๐๐/โ (โ_1โ2โ) (D) 2๐๐1/โ_1 (โ_1โ2โ) Now, ๐=(2๐๐_1)/โ_1 (โ_1โโ)โ ๐=(2๐๐_1)/โ_1 (โ_1 ร โโโ^2 ) Differentiating w.r.t h ๐๐/๐โ=(๐๐๐_๐)/๐_๐ (๐_๐ โ๐๐) So, the correct answer is (D) Question 4 Find the value of (๐^2 ๐)/(๐โ^2 ) (A) โ (4๐๐_1)/โ_1 (B) โ 4๐๐/โ (C) โ (4๐๐_1)/โ (D) (4๐๐_1)/โ Now, ๐๐/๐โ=(2๐๐_1)/โ_1 (โ_1 โ2โ)" " Differentiating w.r.t h (๐^2 ๐)/(๐โ^2 )=(2๐๐_1)/โ_1 (0โ2) (๐^2 ๐)/(๐โ^2 )=(โ๐๐๐_๐)/๐_๐ So, the correct answer is (A) Question 5 What is the relation between ๐_1 and ๐? (A) ๐_1=๐/๐ (B) 2๐_1=3๐ (C) ๐_1=2๐ (D) ๐_1/2=๐/3 Putting ๐๐บ/๐๐=๐ (2๐๐_1)/โ_1 (โ_1 โ2โ)"= 0 " โ_1=2โ And, (๐^2 ๐)/(๐โ^2 )=(โ๐๐๐_๐)/๐_๐ โด (๐^2 ๐)/(๐โ^2 ) < 0 for โ_1=2โ So, Surface area is maximum for โ_1=2โ Now, from Question 1 ๐/๐_๐ = (๐_๐ โ ๐)/๐_๐ Putting โ_1=2โ ๐/๐_1 = (2โ โ โ)/2โ ๐/๐_1 = โ/2โ ๐/๐_1 = 1/2 2๐=๐_1 ๐_๐=๐๐ So, the correct answer is (C)