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 April 16, 2024 by Teachoo

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

Question 34 The area of a trapezium is defined by function š and given by š(š„) = (10 + š„) ā("100 ā š„2" ) , then the area when it is maximised is: (a) 75 cm2 (b) 7 ā3 cm2 (c) 75 ā3 cm2 (d) 5 cm2
š(š„) = (š+šš) (ā(šššāšš))
Since A has a square root
It will be difficult to differentiate
Let Z = [š(š„)]2
= (š„+10)^2 (100āš„2)
Where f'(x) = 0, there Zā(x) = 0
Differentiating Z
Z =(š„+10)^2 " " (100āš„2)
Differentiating w.r.t. x
Zā = š((š„ + 10)^2 " " (100 ā š„2))/šš
Zā = [(š„ + 10)^2 ]^ā² (100 ā š„^2 )+(š„ + 10)^2 " " (100 ā š„^2 )^ā²
Zā = 2(š„ + 10)(100 ā š„^2 )ā2š„(š„ + 10)^2
Zā = 2(š„ + 10)[100 ā š„^2āš„(š„+10)]
Zā = 2(š„ + 10)[100 ā š„^2āš„^2ā10š„]
Zā = 2(š„ + 10)[ā2š„^2ā10š„+100]
Zā = āš(š + šš)[š^š+šš+šš]
Putting š š/š š=š
ā4(š„ + 10)[š„^2+5š„+50] =0
(š„ + 10)[š„^2+5š„+50] =0
(š„ + 10) [š„2+10š„ā5š„ā50]=0
(š„ + 10) [š„(š„+10)ā5(š„+10)]=0
(š + šš)(šāš)(š+šš)=š
So, š„=š & š=āšš
Since x is length, it cannot be negative
ā“ x = 5
Finding maximum area of trapezium
A = (š„+10) ā(100āš„2)
= (5+10) ā(100ā(5)2)
= (15) ā(100ā25)
= 15 ā75
= 15 ā(25 Ć 3)
= 15 Ć āšš Ć āš
= 15 Ć 5 Ć ā3
= 75āš 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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