CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

Question 34 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Sept. 4, 2021 by

The area of a trapezium is defined by function π and given by π(π₯) = (10 + π₯) β(100 - x
^{
2
}
) , then the area when it is maximised is:

Question 34 The area of a trapezium is defined by function π and given by π(π₯) = (10 + π₯) β("100 β π₯2" ) , then the area when it is maximised is: (a) 75 cm2 (b) 7 β3 cm2 (c) 75 β3 cm2 (d) 5 cm2
π(π₯) = (π+ππ) (β(πππβππ))
Since A has a square root
It will be difficult to differentiate
Let Z = [π(π₯)]2
= (π₯+10)^2 (100βπ₯2)
Where f'(x) = 0, there Zβ(x) = 0
Differentiating Z
Z =(π₯+10)^2 " " (100βπ₯2)
Differentiating w.r.t. x
Zβ = π((π₯ + 10)^2 " " (100 β π₯2))/ππ
Zβ = [(π₯ + 10)^2 ]^β² (100 β π₯^2 )+(π₯ + 10)^2 " " (100 β π₯^2 )^β²
Zβ = 2(π₯ + 10)(100 β π₯^2 )β2π₯(π₯ + 10)^2
Zβ = 2(π₯ + 10)[100 β π₯^2βπ₯(π₯+10)]
Zβ = 2(π₯ + 10)[100 β π₯^2βπ₯^2β10π₯]
Zβ = 2(π₯ + 10)[β2π₯^2β10π₯+100]
Zβ = βπ(π + ππ)[π^π+ππ+ππ]
Putting π π/π π=π
β4(π₯ + 10)[π₯^2+5π₯+50] =0
(π₯ + 10)[π₯^2+5π₯+50] =0
(π₯ + 10) [π₯2+10π₯β5π₯β50]=0
(π₯ + 10) [π₯(π₯+10)β5(π₯+10)]=0
(π + ππ)(πβπ)(π+ππ)=π
So, π₯=π & π=βππ
Since x is length, it cannot be negative
β΄ x = 5
Finding maximum area of trapezium
A = (π₯+10) β(100βπ₯2)
= (5+10) β(100β(5)2)
= (15) β(100β25)
= 15 β75
= 15 β(25 Γ 3)
= 15 Γ βππ Γ βπ
= 15 Γ 5 Γ β3
= 75βπ cm2
So, the correct answer is (C)

CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

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.