NCERT Exemplar MCQ

Chapter 7 Class 12 Integrals
Serial order wise

## (D) 4 log⁡2

This question is similar to Question 20 - CBSE Class 12 - Sample Paper for 2019 Board

### Transcript

Question 7 โซ1_(โ1)^1โใ (๐ฅ^3 + |๐ฅ| + 1)/(๐ฅ^2 + 2|๐ฅ| + 1)ใ ๐๐ฅ is equal to (A) logโก2 (B) 2 logโก2 (C) 1/2 logโก2 (D) 4 logโก2 We know that |๐ฅ|={โ(โ&๐ฅ, ๐ฅ<0@&๐ฅ, ๐ฅโฅ0)โค So, for โ1 to 0, |x| = โx and for 0 to 1 |x| = x So, our integral becomes โซ1_(โ1)^1โ(๐ฅ+|๐ฅ|+1)/(๐ฅ^2+2|๐ฅ|+1) โซ1_(โ1)^0โ(๐ฅ+|๐ฅ|+1)/(๐ฅ^2+2|๐ฅ|+1) โซ1_0^1โ(๐ฅ+|๐ฅ|+1)/(๐ฅ^2+2|๐ฅ|+1) โซ1_(โ๐)^๐โ(๐โ๐+๐)/(๐^๐โ๐๐+๐) โซ1_๐^๐โ(๐+๐+๐)/(๐^๐+๐๐+๐) โซ1_(โ1)^0โ1/(๐ฅ^2โ2๐ฅ+1) โซ1_0^1โ(2๐ฅ+1)/(๐ฅ^2+2๐ฅ+1) โซ1_(โ๐)^๐โ๐/(๐โ๐)^๐ โซ1_๐^๐โ(๐๐+๐)/(๐+๐)^๐ โซ1_(โ1)^0โ1/(๐ฅโ1)^2 โซ1_0^1โ(2๐ฅ+1+1โ1)/(๐ฅ+1)^2 โซ1_0^1โ(2๐ฅ+1+1โ1)/(๐ฅ+1)^2 โซ1_(โ1)^0โ1/(๐ฅโ1)^2 โซ1_0^1โ(2๐ฅ+2โ1)/(๐ฅ+1)^2 โซ1_(โ๐)^๐โ๐/(๐โ๐)^๐ โซ1_๐^๐โ(๐(๐+๐)โ๐)/(๐+๐)^๐ โซ1_(โ1)^0โ1/(๐ฅโ1)^2 โซ1_0^1โ(2(๐ฅ+1) )/(๐ฅ+1)^2 โซ1_0^1โ1/(๐ฅ+1)^2 โซ1_(โ1)^0โ1/(๐ฅโ1)^2 โซ1_๐^๐โ(๐ )/((๐+๐) ) โซ1_๐^๐โ๐/(๐+๐)^๐ โซ1_(โ1)^0โ1/(๐ฅโ1)^2 2 โซ1_0^1โ"dx " /((๐ฅ+1) ) โซ1_0^1โ1/(๐ฅ+1)^2 [(๐ฅโ1)^(โ1)/(โ1)]_(โ1)^0 2[logโกใ|๐ฅ+1|ใ ]_0^1 [(๐ฅ+1)^(โ1)/(โ1)]_0^1 1/2+2 logโก2โ1/2 2 ๐ฅ๐จ๐ โก๐ So, the correct answer is (b)