Ex 6.3, 27 (MCQ) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Minima/ maxima (statement questions) - Geometry questions
Ex 6.3, 27 (MCQ) You are here
Example 23 Important
Example 29 Important
Example 24
Misc 9 Important
Example 25 Important
Misc 6 Important
Ex 6.3,21
Ex 6.3, 20 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3,22 Important
Misc 7
Misc 8 Important
Ex 6.3,17
Ex 6.3,18 Important
Example 36 Important
Ex 6.3,19 Important
Misc 5 Important
Ex 6.3,23 Important
Misc 12 Important
Misc 14 Important
Example 26 Important
Misc 15 Important
Minima/ maxima (statement questions) - Geometry questions
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 27 (Method 1) The point on the curve š„2=2š¦ which is nearest to the point (0, 5) is (A) (2 ā2 ,4) (B) (2 ā2,0) (C) (0, 0) (D) (2, 2) Let (ā , š) be the point on the curve š„2 = 2š¦ Where is nearest to the point (0, 5) Since (ā, š) lie on the curve š„2= 2š¦ ā (ā š) will satisfy the equation of curve š„2=2š¦ ā Putting š„=ā & y=š in equation ā^2=2š We need to minimize the distance of a point (ā ,š) from(0, 5) Let D be the distant between (ā,š) & (0,5) D = ā((0āā)^2+(5āš)^2 ) D = ā(ā^2+(5āš^2 ) ) From (1) ā^2=2š D = ā(2š+(5āš)^2 ) Diff w.r.t š šš·/šš=š(ā(2š + (5 ā š)^2 ))/šš =1/(2ā(2š + (5 ā š)^2 )) Ćš(2š + (5 ā š)^2 )/šš¾ =1/(2ā(2š + (5 ā š)^2 )) Ć [2+2(5āš).š(5 ā š)/šš¾] Let D be the distant between (ā,š) & (0,5) D = ā((0āā)^2+(5āš)^2 ) D = ā(ā^2+(5āš^2 ) ) From (1) ā^2=2š D = ā(2š+(5āš)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2š+(5āš)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2š+(5āš)^2 Diff w.r.t. k Zā = š(2š + (5 ā š)^2 )/šš Zā = 2 + 2 (5 ā k) Ć (5 ā k)ā Zā = 2 + 2 (5 ā k) Ć (0 ā 1) Zā = 2 ā 2 (5 ā k) Zā = 2 ā 10 + 2k Zā = ā8 + 2k Putting Zā = 0 ā8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (š^2 š)/(šā^2 ) " " šš/šš=ā8+2š Differentiating again w.r.t k (š^2 š)/(šā^2 ) = 0+2 (š^2 š)/(šā^2 ) = 2 ā“ (š^2 š)/(šā^2 ) > 0 for k = 4 ā“ Z is minimum when k = 4 Thus, D is Minimum at š= 4 Finding h From (1) h^2=2š h^2=2(4) h=ā8 h=2ā2 Hence, Required Point is (ā,š)=(2ā(2 ,) 4) Correct answer is A Ex 6.3, 27 (Method 2) The point on the curve š„2= 2š¦ which is nearest to the point (0, 5) is (A) (2 ā2,4) (B) (2 ā2,0) (C) (0, 0) (D) (2, 2)Since points given lie on the curve, it will satisfy equation of curve Option 1 Point is (2ā(2 ,) 4) Putting š„=2ā2 , & š¦=4 in š„2=2š¦ ā (2ā2)^2=2(4) ā 4 Ć 2 = 8 Which is true Thus, (2ā2,4) lie on the curve Now, finding distance between (2ā(2 ,) 4" " ) & (0 ,5) D = ā((0ā2ā(2 ))^2+(5ā4)^2 ) = ā(8+1) = ā9 = 3 Option 2 Point (2ā(2 ,) 0) Putting š„=2ā2 & š¦=0 in š„2=2š¦ (2ā(2 ))^2=2(0) (4 Ć2)=0 8 = 0 Since 8 ā 0 ā (2ā(2 ,) 0) is not the required point Option 3 Point (0, 0) Putting š„=0 & š¦=0 in š„2=2š¦ (0)^2=2(0) 0=0 ā“ (0 , 0) lie on the curve Now, Finding distance between (0, 0) ššš (0 , 5) D = ā((0ā0)^2+(5ā0)^2 ) = ā(0+5^2 ) = 5 Option 4 Point (2 ,2) Putting š„=2 & š¦=2 in š„2=2š¦ (2)^2=2(2) 4=4 ā“ (2, 2) lie on the curve Now, Finding distance between (2, 2) ššš (0 , 5) D = ā((0ā2)^2+(5ā2)^2 ) = ā((ā2)^2+(3)^2 ) = ā(4+9) = ā13 Thus, Point (2ā2,4) is on the curve š„2=2š¦ & nearest to the point (0, 5) Hence correct answer is A