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 √3 cm 2     (d) 5 cm 2

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

### Transcript

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