Last updated at March 16, 2023 by Teachoo
This question is similar to Question 20 - CBSE Class 12 - Sample Paper for 2019 Board
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Question 7 β«1_(β1)^1βγ (π₯^3 + |π₯| + 1)/(π₯^2 + 2|π₯| + 1)γ ππ₯ is equal to (A) logβ‘2 (B) 2 logβ‘2 (C) 1/2 logβ‘2 (D) 4 logβ‘2 We know that |π₯|={β(β&π₯, π₯<[email protected]&π₯, π₯β₯0)β€ So, for β1 to 0, |x| = βx and for 0 to 1 |x| = x So, our integral becomes β«1_(β1)^1β(π₯+|π₯|+1)/(π₯^2+2|π₯|+1) β«1_(β1)^0β(π₯+|π₯|+1)/(π₯^2+2|π₯|+1) β«1_0^1β(π₯+|π₯|+1)/(π₯^2+2|π₯|+1) β«1_(βπ)^πβ(πβπ+π)/(π^πβππ+π) β«1_π^πβ(π+π+π)/(π^π+ππ+π) β«1_(β1)^0β1/(π₯^2β2π₯+1) β«1_0^1β(2π₯+1)/(π₯^2+2π₯+1) β«1_(βπ)^πβπ/(πβπ)^π β«1_π^πβ(ππ+π)/(π+π)^π β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2π₯+1+1β1)/(π₯+1)^2 β«1_0^1β(2π₯+1+1β1)/(π₯+1)^2 β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2π₯+2β1)/(π₯+1)^2 β«1_(βπ)^πβπ/(πβπ)^π β«1_π^πβ(π(π+π)βπ)/(π+π)^π β«1_(β1)^0β1/(π₯β1)^2 β«1_0^1β(2(π₯+1) )/(π₯+1)^2 β«1_0^1β1/(π₯+1)^2 β«1_(β1)^0β1/(π₯β1)^2 β«1_π^πβ(π )/((π+π) ) β«1_π^πβπ/(π+π)^π β«1_(β1)^0β1/(π₯β1)^2 2 β«1_0^1β"dx " /((π₯+1) ) β«1_0^1β1/(π₯+1)^2 [(π₯β1)^(β1)/(β1)]_(β1)^0 2[logβ‘γ|π₯+1|γ ]_0^1 [(π₯+1)^(β1)/(β1)]_0^1 1/2+2 logβ‘2β1/2 2 π₯π¨π β‘π So, the correct answer is (b)
