Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

Last updated at April 15, 2021 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

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Ex 6.5, 27 (Method 1) The point on the curve ๐ฅ2=2๐ฆ which is nearest to the point (0, 5) is (A) (2 โ2 ,4) (B) (2 โ2,0) (C) (0, 0) (D) (2, 2) Let (โ , ๐) be the point on the curve ๐ฅ2 = 2๐ฆ Where is nearest to the point (0, 5) Since (โ, ๐) lie on the curve ๐ฅ2= 2๐ฆ โ (โ ๐) will satisfy the equation of curve ๐ฅ2=2๐ฆ โ Putting ๐ฅ=โ & y=๐ in equation โ^2=2๐ We need to minimize the distance of a point (โ ,๐) from(0, 5) Let D be the distant between (โ,๐) & (0,5) D = โ((0โโ)^2+(5โ๐)^2 ) D = โ(โ^2+(5โ๐^2 ) ) From (1) โ^2=2๐ D = โ(2๐+(5โ๐)^2 ) Diff w.r.t ๐ ๐๐ท/๐๐=๐(โ(2๐ + (5 โ ๐)^2 ))/๐๐ =1/(2โ(2๐ + (5 โ ๐)^2 )) ร๐(2๐ + (5 โ ๐)^2 )/๐๐พ =1/(2โ(2๐ + (5 โ ๐)^2 )) ร [2+2(5โ๐).๐(5 โ ๐)/๐๐พ] Let D be the distant between (โ,๐) & (0,5) D = โ((0โโ)^2+(5โ๐)^2 ) D = โ(โ^2+(5โ๐^2 ) ) From (1) โ^2=2๐ D = โ(2๐+(5โ๐)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2๐+(5โ๐)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2๐+(5โ๐)^2 Diff w.r.t. k Zโ = ๐(2๐ + (5 โ ๐)^2 )/๐๐ Zโ = 2 + 2 (5 โ k) ร (5 โ k)โ Zโ = 2 + 2 (5 โ k) ร (0 โ 1) Zโ = 2 โ 2 (5 โ k) Zโ = 2 โ 10 + 2k Zโ = โ8 + 2k Putting Zโ = 0 โ8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (๐^2 ๐)/(๐โ^2 ) " " ๐๐/๐๐=โ8+2๐ Differentiating again w.r.t k (๐^2 ๐)/(๐โ^2 ) = 0+2 (๐^2 ๐)/(๐โ^2 ) = 2 โด (๐^2 ๐)/(๐โ^2 ) > 0 for k = 4 โด Z is minimum when k = 4 Thus, D is Minimum at ๐= 4 Finding h From (1) h^2=2๐ h^2=2(4) h=โ8 h=2โ2 Hence, Required Point is (โ,๐)=(2โ(2 ,) 4) Correct answer is A Ex 6.5, 27 (Method 2) The point on the curve ๐ฅ2= 2๐ฆ which is nearest to the point (0, 5) is (A) (2 โ2,4) (B) (2 โ2,0) (C) (0, 0) (D) (2, 2)Since points given lie on the curve, it will satisfy equation of curve Option 1 Point is (2โ(2 ,) 4) Putting ๐ฅ=2โ2 , & ๐ฆ=4 in ๐ฅ2=2๐ฆ โ (2โ2)^2=2(4) โ 4 ร 2 = 8 Which is true Thus, (2โ2,4) lie on the curve Now, finding distance between (2โ(2 ,) 4" " ) & (0 ,5) D = โ((0โ2โ(2 ))^2+(5โ4)^2 ) = โ(8+1) = โ9 = 3 Option 2 Point (2โ(2 ,) 0) Putting ๐ฅ=2โ2 & ๐ฆ=0 in ๐ฅ2=2๐ฆ (2โ(2 ))^2=2(0) (4 ร2)=0 8 = 0 Since 8 โ 0 โ (2โ(2 ,) 0) is not the required point Option 3 Point (0, 0) Putting ๐ฅ=0 & ๐ฆ=0 in ๐ฅ2=2๐ฆ (0)^2=2(0) 0=0 โด (0 , 0) lie on the curve Now, Finding distance between (0, 0) ๐๐๐ (0 , 5) D = โ((0โ0)^2+(5โ0)^2 ) = โ(0+5^2 ) = 5 Option 4 Point (2 ,2) Putting ๐ฅ=2 & ๐ฆ=2 in ๐ฅ2=2๐ฆ (2)^2=2(2) 4=4 โด (2, 2) lie on the curve Now, Finding distance between (2, 2) ๐๐๐ (0 , 5) D = โ((0โ2)^2+(5โ2)^2 ) = โ((โ2)^2+(3)^2 ) = โ(4+9) = โ13 Thus, Point (2โ2,4) is on the curve ๐ฅ2=2๐ฆ & nearest to the point (0, 5) Hence correct answer is A

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.