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

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## (c) 75 β3 cm 2 Β  Β  (d) 5 cm 2

Β

This question is inspired from Example 37 - Chapter 6 Class 12 Β  - Application of Derivatives

This video is only available for Teachoo black users

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

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 β3 cm2 (c) 75 β3 cm2 (d) 5 cm2 π(π₯) = (π+ππ) (β(πππβππ)) Since A has a square root It will be difficult to differentiate Let Z = [π(π₯)]2 = (π₯+10)^2 (100βπ₯2) Where f'(x) = 0, there Zβ(x) = 0 Differentiating Z Z =(π₯+10)^2 " " (100βπ₯2) Differentiating w.r.t. x Zβ = π((π₯ + 10)^2 " " (100 β π₯2))/ππ Zβ = [(π₯ + 10)^2 ]^β² (100 β π₯^2 )+(π₯ + 10)^2 " " (100 β π₯^2 )^β² Zβ = 2(π₯ + 10)(100 β π₯^2 )β2π₯(π₯ + 10)^2 Zβ = 2(π₯ + 10)[100 β π₯^2βπ₯(π₯+10)] Zβ = 2(π₯ + 10)[100 β π₯^2βπ₯^2β10π₯] Zβ = 2(π₯ + 10)[β2π₯^2β10π₯+100] Zβ = βπ(π + ππ)[π^π+ππ+ππ] Putting ππ/ππ=π β4(π₯ + 10)[π₯^2+5π₯+50] =0 (π₯ + 10)[π₯^2+5π₯+50] =0 (π₯ + 10) [π₯2+10π₯β5π₯β50]=0 (π₯ + 10) [π₯(π₯+10)β5(π₯+10)]=0 (π + ππ)(πβπ)(π+ππ)=π So, π₯=π & π=βππ Since x is length, it cannot be negative β΄ x = 5 Finding maximum area of trapezium A = (π₯+10) β(100βπ₯2) = (5+10) β(100β(5)2) = (15) β(100β25) = 15 β75 = 15 β(25 Γ 3) = 15 Γ βππ Γ βπ = 15 Γ 5 Γ β3 = 75βπ cm2 So, the correct answer is (C)