The shape of a toy is given as f(x)=6(2x 4 - x 2 ). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy.

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Question 1

Which value from the following may be abscissa of critical point?

(a) ± 1/4

(b) ± 1/2

(c) ± 1

(d) None

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Question 2

Find the slope of the normal based on the position of the stick.

(a) 360

(b) –360

(c) 1/360

(d) (-1)/360

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Question 3

What will be the equation of the tangent at the critical point if it passes through (2,3)?

(a) x + 360y = 1082

(b) y = 360x – 717

(c) x = 717y + 360

(d) none

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Question 4

Find the second order derivative of the function at x = 5.

(a) 598

(b) 1176

(c) 3588

(d) 3312

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Question 5
At which of the following intervals will f(x) be increasing?
(a) (-∞", "  (-1)/2) ∪ (1/2 ", " ∞)
(b) (1/2,0) ∪ (1/2 ", " ∞)
(c) (0", "  1/2) ∪ (1/2 ", " ∞)
(d) (-∞", "  (-1)/2) ∪ (0", "  1/2)

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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Question The shape of a toy is given as 𝑓(π‘₯)=6(2π‘₯^4βˆ’π‘₯^2). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy. Question 1 Which value from the following may be abscissa of critical point? (a) Β± 1/4 (b) Β± 1/2 (c) Β± 1 (d) None Critical point is point where 𝒇′(𝒙)=𝟎 𝑓(π‘₯)=6(2π‘₯^4βˆ’π‘₯^2 ) Finding 𝒇′(𝒙) 𝑓^β€² (π‘₯)=𝑑(6(2π‘₯^4βˆ’π‘₯^2 ) )/𝑑π‘₯ =6(2 Γ— 4π‘₯^3βˆ’2π‘₯) =6(8π‘₯^3βˆ’2π‘₯) =6 Γ— 2π‘₯(4π‘₯^2βˆ’1) =πŸπŸπ’™(πŸ’π’™^πŸβˆ’πŸ) Putting 𝒇^β€² (𝒙)= 0 12π‘₯(4π‘₯^2βˆ’1)=0 12π‘₯ = 0 𝒙 = 0 4π‘₯^2βˆ’1= 0 4π‘₯^2=1 π‘₯^2=1/4 𝒙 = Β± 𝟏/𝟐 ∴ x = 0, Β± 𝟏/𝟐 So, the correct answer is (b) Question 2 Find the slope of the normal based on the position of the stick. (a) 360 (b) –360 (c) 1/360 (d) (βˆ’1)/360 We need to find Slope of Normal at (2, 3) First, let’s find Slope of Tangent at (2, 3) Slope of tangent = 𝑓^β€² (π‘₯) = 12π‘₯(4π‘₯^2βˆ’1) Putting x = 2 = 12 Γ— 2 Γ— (4(2)2 βˆ’ 1) = 24(4 Γ— 4 βˆ’ 1) = 24(16 βˆ’ 1) = 24 Γ— 15 = 360 Thus, Slope of Normal = (βˆ’1)/(π‘†π‘™π‘œπ‘π‘’ π‘œπ‘“ π‘‡π‘Žπ‘›π‘”π‘’π‘›π‘‘) = (βˆ’πŸ)/πŸ‘πŸ”πŸŽ So, the correct answer is (d) Question 3 What will be the equation of the tangent at the critical point if it passes through (2, 3)? (a) x + 360y = 1082 (b) y = 360x – 717 (c) x = 717y + 360 (d) none We found the slope of tangent at (2, 3) Slope of tangent at (2, 3) = 360 Finding equation of line passing through (2, 3) with slope 360 y βˆ’ y1 = Slope Γ— (x βˆ’ x1) y βˆ’ 3 = 360 (x βˆ’ 2) y βˆ’ 3 = 360x βˆ’ 720 y = 360x βˆ’ 720 + 3 y = 360x βˆ’ 717 So, the correct answer is (b) Question 4 Find the second order derivative of the function at x = 5. (a) 598 (b) 1176 (c) 3588 (d) 3312 We know that 𝑓(π‘₯)=6(2π‘₯^4βˆ’π‘₯^2 ) And, 𝒇^β€² (𝒙) = 12π‘₯(4π‘₯^2βˆ’1) = πŸ’πŸ–π’™^πŸ‘βˆ’πŸπŸπ’™ Now, 𝑓^β€²β€² (π‘₯)=(𝑑(48π‘₯^3 βˆ’ 12π‘₯))/𝑑π‘₯ = 48 Γ— 3π‘₯^2βˆ’12 = πŸπŸ’πŸ’π’™^πŸβˆ’πŸπŸ Putting x = 5 = 144 Γ— 5^2βˆ’12 = 144 Γ— 25βˆ’12 = 3600βˆ’12 = πŸ‘πŸ“πŸ–πŸ– So, the correct answer is (c) Question 5 At which of the following intervals will f(x) be increasing? (a) (βˆ’βˆž", " (βˆ’1)/2) βˆͺ (1/2 ", " ∞) (b) (1/2,0) βˆͺ (1/2 ", " ∞) (c) (0", " 1/2) βˆͺ (1/2 ", " ∞) (d) (βˆ’βˆž", " (βˆ’1)/2) βˆͺ (0", " 1/2) 𝑓(π‘₯) is increasing where 𝒇^β€² (𝒙)>𝟎 Putting f’(x) = 0 12π‘₯(4π‘₯^2βˆ’1)=0 12π‘₯((2π‘₯)^2βˆ’1^2 )=0 12π‘₯(2π‘₯βˆ’1)(2π‘₯+1)=0 So, x = 0, 𝟏/𝟐 , (βˆ’πŸ)/𝟐 Hence, ∴ f is strictly increasing in ((βˆ’πŸ)/𝟐,𝟎) and (𝟏/𝟐 ", " ∞) So, the correct answer is (b)

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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.