Case Based Questions (MCQ)

Chapter 7 Class 12 Integrals
Serial order wise

Β

## (a) 1Β  Β  Β  Β  Β  Β  (b) β1Β  Β  Β  Β  Β  (c) 0Β  Β  Β  Β  Β  Β  (d) 2

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Question 1 Based on the above information, answer any four of the following questions. Question 1 β«1_(β1)^1βπ₯^99 ππ₯=______________. (a) 0 (b) 1 (c) β1 (d) 2 This is of form β«_(βπ)^πβπ(π)ππ π(π)=π₯^99 π(βπ)=(βπ₯)^99=βπ₯^99 Thus, π(βπ) =βπβ‘(π) β«1_(β1)^1βπ₯^99 ππ= 0 So, the correct answer is (a) Question 2 β«1_(βπ)^πβγ π₯ cosβ‘π₯ γ ππ₯=______________. (a) 1 (b) 0 (c) β1 (d) π/2 This is of form β«_(βπ)^πβπ(π)ππ π(π)=π₯ πππ  π₯ π(βπ)=(βπ₯) cosβ‘γ(βπ₯)γ=βπ₯ cosβ‘π₯ Thus, π(βπ) =βπβ‘(π) β«1_(βπ)^πβγ π₯ cosβ‘π₯ γ ππ₯= 0 So, the correct answer is (b) Question 3 β«1_((βπ)/2)^(π/2 )βγ sinγ^3β‘π₯ ππ₯ = _________. (a) 1 (b) 0 (c) β1 (d) π This is of form β«_(βπ)^πβπ(π)ππ π(π)=sin^3β‘π₯ π(βπ)=sin^3β‘γ(βπ₯)γ=(βsinβ‘π₯ )^3=βsin^3β‘π₯ Thus, π(βπ) =βπβ‘(π) β«1_((βπ)/2)^(π/2 )βγ sinγ^3β‘π₯ ππ₯ = 0 So, the correct answer is (b) Question 4 β«1_(βπ)^πβγπ₯ π ππ γβ‘π₯ ππ₯ = _________. (a) π (b) 0 (c) 2π (d) π/2 This is of form β«_(βπ)^πβπ(π)ππ π(π)=π₯ π ππ π₯ π(βπ)=(βπ₯) γπ ππ γβ‘γ(βπ₯)γ=βπ₯ Γ βsinβ‘π₯=π₯ sinβ‘γπ₯ γ Thus, π(βπ) =πβ‘(π) Therefore, β«1_(βπ)^πβγπ₯ π ππ γβ‘π₯ ππ₯ = 2β«1_0^πβγπ₯ π ππ γβ‘π₯ ππ₯ Let I = πβ«1_π^πβγπ πππ γβ‘π ππ₯ I = 2β«1_0^πβγ(π βπ) π ππ γβ‘γ(π βπ₯) γ ππ₯ I = 2β«1_0^πβγ(π βπ₯) π ππ γβ‘γπ₯ γ ππ₯ I = πβ«1_π^πβγπ πππ γβ‘π ππ₯ β πβ«1_π^πβγπ πππ γβ‘π ππ₯ Adding (1) and (2) I + I = 2β«1_0^πβγπ₯ π ππ γβ‘π₯ ππ₯ + 2β«1_0^πβγπ π ππ γβ‘π₯ ππ₯ β 2β«1_0^πβγπ₯ π ππ γβ‘π₯ ππ₯ 2I = 2β«1_0^πβγπ π ππ γβ‘π₯ ππ₯ I = β«1_π^πβγπ πππ γβ‘π ππ₯ I = πβ«1_0^πβγπ ππ γβ‘π₯ ππ₯ I = πγ[βππ¨π¬β‘π]γ_π^π I = π[βcosβ‘γπ β(βcosβ‘0)γ] I = π[βcosβ‘π+cosβ‘0] I = π[β(β1)+1] I = π [1+1] I = ππ So, the correct answer is (c) Question 5 β«1_(βπ)^πβγtanβ‘π₯ sec^2β‘π₯ γ ππ₯ = _________. (a) 1 (b) β1 (c) 0 (d) 2 This is of form β«_(βπ)^πβπ(π)ππ π(π)=tanβ‘π₯ sec^2β‘π₯ π(βπ)=tanβ‘γ(βπ₯)γ sec^2β‘γ(βπ₯)γ=βtanβ‘π₯ sec^2β‘γπ₯ γ Thus, π(βπ) =βπβ‘(π) β«1_(βπ)^πβγtanβ‘π₯ sec^2β‘π₯ γ " ππ₯"= 0 So, the correct answer is (c)