## Based on the above information answer the following:

This question is inspired from Example 37 - Chapter 6 Class 12 (AOD) - Maths

## (D) 75 √7  cm 2

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

Transcript

Question There is a bridge whose length of three sides of a trapezium other than base are equal to 10 cm. Based on the above information answer the following: Question 1 What is the value of DP? (A) โ(100โ๐ฅ^2 ) (B) โ(๐ฅ^2โ100) (C) 100โ๐ฅ^2 (D)ใ ๐ฅใ^2 โ 100 In ฮ ADP By Pythagoras theorem DP2 + AP2 = AD2 DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 โ ๐ฅ2 DP = โ(๐๐๐ โ๐๐) So, the correct answer is (A) Question 2 What is the area of trapezium A(x)? (A) (๐ฅ โ10)โ(100โ๐ฅ^2 ) (B) (๐ฅ+10)โ(100โ๐ฅ^2 ) (C) (๐ฅโ10)(100โ๐ฅ^2) (D) (๐ฅ+10)(100โ๐ฅ^2 ") " Let A be the area of trapezium ABCD A = 1/2 (Sum of parallel sides) ร (Height) A = ๐/๐ (DC + AB) ร DP A = 1/2 (10+2๐ฅ+10) (โ(100โ๐ฅ2)) A = 1/2 (2๐ฅ+20) (โ(100โ๐ฅ2)) A = (๐+๐๐) (โ(๐๐๐โ๐๐)) So, the correct answer is (B) Question 3 If A'(x) = 0, then what are the values of x? (A) 5,โ10 (B) โ5, 10 (C) โ5,โ10 (D) 5,10 A = (๐+๐๐) (โ(๐๐๐โ๐๐)) Since A has a square root It will be difficult to differentiate Let Z = A2 = (๐ฅ+10)^2 (100โ๐ฅ2) Where A'(x) = 0, there Zโ(x) = 0 So, the correct answer is (A) Differentiating Z Z =(๐ฅ+10)^2 " " (100โ๐ฅ2) Differentiating w.r.t. x Zโ = ๐((๐ฅ + 10)^2 " " (100 โ ๐ฅ2))/๐๐ Using product rule As (๐ข๐ฃ)โฒ = uโv + vโu Zโ = [(๐ฅ + 10)^2 ]^โฒ (100 โ ๐ฅ^2 )+(๐ฅ + 10)^2 " " (100 โ ๐ฅ^2 )^โฒ Zโ = 2(๐ฅ + 10)(100 โ ๐ฅ^2 )โ2๐ฅ(๐ฅ + 10)^2 Zโ = 2(๐ฅ + 10)[100 โ ๐ฅ^2โ๐ฅ(๐ฅ+10)] Zโ = 2(๐ฅ + 10)[100 โ ๐ฅ^2โ๐ฅ^2โ10๐ฅ] Zโ = 2(๐ฅ + 10)[โ2๐ฅ^2โ10๐ฅ+100] Zโ = โ๐(๐ + ๐๐)[๐^๐+๐๐+๐๐] Putting ๐๐/๐๐=๐ โ4(๐ฅ + 10)[๐ฅ^2+5๐ฅ+50] =0 (๐ฅ + 10)[๐ฅ^2+5๐ฅ+50] =0 (๐ฅ + 10) [๐ฅ2+10๐ฅโ5๐ฅโ50]=0 (๐ฅ + 10) [๐ฅ(๐ฅ+10)โ5(๐ฅ+10)]=0 (๐ + ๐๐)(๐โ๐)(๐+๐๐)=๐ So, ๐ฅ=๐ & ๐=โ๐๐ So, the correct answer is (A) Question 4 What is the value of maximum Area? (A) 75 โ2 cm^2 (B) 75 โ3 cm^2 (C) 75 โ5 cm^2 (D) 75 โ7 cm^2 We know that Zโ(x) = 0 at x = 5, โ10 Since x is length, it cannot be negative โด x = 5 Finding sign of Zโโ for x = 5 Now, Zโ = โ4(๐ฅ + 10)[๐ฅ^2+5๐ฅ+50] Zโ = โ4[๐ฅ(๐ฅ^2+5๐ฅ+50)+10(๐ฅ^2+5๐ฅ+50)] Zโ = โ4[๐ฅ^3+5๐ฅ^2+50๐ฅ+10๐ฅ^2+50๐ฅ+500] Zโ = โ๐[๐^๐+๐๐๐^๐+๐๐๐๐+๐๐๐] Differentiating w.r.t x Zโโ = ๐(โ4[๐^๐ + ๐๐๐^๐ + ๐๐๐๐ + ๐๐๐])/๐๐ Zโโ =โ4[3๐ฅ^2+15 ร 2๐ฅ+100] Zโโ =โ4[3๐ฅ^2+30๐ฅ+100] Putting x = 5 Zโโ (5) = โ4[3(5^2) +30(5) +100] = โ4 ร 375 = โ1500 < 0 Hence, ๐ฅ = 5 is point of Maxima โด Z is Maximum at ๐ฅ = 5 That means, Area A is maximum when x = 5 Finding maximum area of trapezium A = (๐ฅ+10) โ(100โ๐ฅ2) = (5+10) โ(100โ(5)2) = (15) โ(100โ25) = 15 โ75 = 75โ๐ cm2 So, the correct answer is (C)