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 March 22, 2023 by Teachoo

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)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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