Question 4 - Case Based Questions (MCQ) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at July 30, 2021 by Teachoo

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.

Question 1

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

(a) ± 1/4

(b) ± 1/2

(c) ± 1

(d) None

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

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

Question 4

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

(a) 598

(b) 1176

(c) 3588

(d) 3312

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)

Question The shape of a toy is given as π(π₯)=6(2π₯^4βπ₯^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 πβ²(π)=π
π(π₯)=6(2π₯^4βπ₯^2 )
Finding πβ²(π)
π^β² (π₯)=π(6(2π₯^4βπ₯^2 ) )/ππ₯
=6(2 Γ 4π₯^3β2π₯)
=6(8π₯^3β2π₯)
=6 Γ 2π₯(4π₯^2β1)
=πππ(ππ^πβπ)
Putting π^β² (π)= 0
12π₯(4π₯^2β1)=0
12π₯ = 0
π = 0
4π₯^2β1= 0
4π₯^2=1
π₯^2=1/4
π = Β± π/π
β΄ x = 0, Β± π/π
So, the correct answer is (b)
Question 2 Find the slope of the normal based on the position of the stick. (a) 360 (b) β360 (c) 1/360 (d) (β1)/360
We need to find Slope of Normal at (2, 3)
First, letβs find Slope of Tangent at (2, 3)
Slope of tangent = π^β² (π₯)
= 12π₯(4π₯^2β1)
Putting x = 2
= 12 Γ 2 Γ (4(2)2 β 1)
= 24(4 Γ 4 β 1)
= 24(16 β 1)
= 24 Γ 15
= 360
Thus,
Slope of Normal = (β1)/(πππππ ππ πππππππ‘)
= (βπ)/πππ
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 β 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 β y1 = Slope Γ (x β x1)
y β 3 = 360 (x β 2)
y β 3 = 360x β 720
y = 360x β 720 + 3
y = 360x β 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
π(π₯)=6(2π₯^4βπ₯^2 )
And,
π^β² (π) = 12π₯(4π₯^2β1) = πππ^πβπππ
Now,
π^β²β² (π₯)=(π(48π₯^3 β 12π₯))/ππ₯
= 48 Γ 3π₯^2β12
= ππππ^πβππ
Putting x = 5
= 144 Γ 5^2β12
= 144 Γ 25β12
= 3600β12
= ππππ
So, the correct answer is (c)
Question 5 At which of the following intervals will f(x) be increasing? (a) (ββ", " (β1)/2) βͺ (1/2 ", " β) (b) (1/2,0) βͺ (1/2 ", " β) (c) (0", " 1/2) βͺ (1/2 ", " β) (d) (ββ", " (β1)/2) βͺ (0", " 1/2)
π(π₯) is increasing where π^β² (π)>π
Putting fβ(x) = 0
12π₯(4π₯^2β1)=0
12π₯((2π₯)^2β1^2 )=0
12π₯(2π₯β1)(2π₯+1)=0
So, x = 0, π/π , (βπ)/π
Hence,
β΄ f is strictly increasing in ((βπ)/π,π) and (π/π ", " β)
So, the correct answer is (b)

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.