Case Based Questions (MCQ)

Chapter 7 Class 12 Integrals
Serial order wise

Β

## (d) e x /x + 1 + cΒ

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Question 2 β«1βγπ^π₯ [π(π₯)+π^β² (π₯)] γ ππ₯ = β«1βγπ^π π(π)ππγ+β«1βγπ^π₯ πβ²(π₯)ππ₯γ Using integration by parts = π(π) β«1βπ^π ππββ«1βγπ^β² (π) π^π γ ππ + β«1βγπ^β² (π₯) π^π₯ γ ππ₯ = π(π₯) π^π₯ββ«1βγπ^β² (π₯) π^π₯ γ ππ₯ + β«1βγπ^β² (π₯) π^π₯ γ ππ₯ = π^π₯ π(π₯)+πΆ Based on the above information, answer any four of the following questions. Question 1 β«1βγπ^π₯ (sinβ‘γπ₯ γ+cosβ‘π₯)γ ππ₯=______________. (a) π^π₯ cosβ‘π₯+π (b) π^π₯ sinβ‘π₯+π (c) π^π₯+π (d) π^π₯ (βcosβ‘π₯+sinβ‘π₯ )+c β«1βγπ^π₯ (sinβ‘γπ₯ γ+cosβ‘π₯)γ ππ₯ Putting π(π)=πππ π" " β΄ π^β² (π)=cosβ‘π₯ Thus, β«1βγπ^π₯ (sinβ‘γπ₯ γ+cosβ‘π₯)γ ππ₯=π^π πππ π+πͺ So, the correct answer is (b) Question 2 β«1βγπ^π₯ ((π₯ β 1)/π₯^2 ) γ ππ₯=______________. (a) π^π₯+π (b) π^π₯/π₯+π (c) π^π₯/π₯^2 +π (d) γβπγ^π₯/π₯^2 +π β«1βγπ^π₯ ((π₯ β 1)/π₯^2 ) γ ππ₯= β«1βγππ₯" " (π₯/π₯^2 β 1/π₯2) ππ₯γ = β«1βγππ" " (π/π β π/ππ) ππγ Putting π(π)=π/π β΄ π^β² (π)=(β1)/π₯^2 Thus, β«1βγππ₯" " (1/π₯ β 1/π₯2) ππ₯γ = π^π/π+πͺ So, the correct answer is (b) Question 3 β«1βγπ^π₯ (1+π₯) γ ππ₯=______________. (a) π₯π^π₯+π (b) π^π₯+π (c) π^(βπ₯)+π (d) none of these β«1βγπ^π₯ (1+π₯)γ ππ₯=β«1βγπ^π (π+π)γ ππ Putting π(π)=π β΄ π^β² (π)=1 Thus, β«1βγπ^π₯ (π₯+1)γ ππ₯=π^π Γ π+πͺ So, the correct answer is (a) Question 4 β«1_0^πβπ^π₯ (tanβ‘π₯+sec^2β‘π₯) ππ₯ = _________. (a) 0 (b) 1 (c) β1 (d) βπ^π β«1_0^πβπ^π₯ (tanβ‘π₯+sec^2β‘π₯) ππ₯ Putting π(π)=πππ π" " β΄ π^β² (π)=γπ ππγ^2β‘π₯ Thus, β«1_0^πβπ^π₯ (tanβ‘π₯+sec^2β‘π₯) ππ₯ = γ[π^π πππ π]γ_0^π =[π^π πππ πβπ^π πππ π] =[π^π πππ πβπ^π πππ π] =[π^π Γ πβπ^π Γ π] =0β0 =0 So, the correct answer is (a) Question 5 β«1βγγπ₯πγ^π₯/(1 + π₯)^2 γ ππ₯=______________. (a) π₯π^π₯+π (b) π^π₯/(π₯ + 1)^2 +π (c) (π₯ π^π₯)/(π₯ + 1)+π (d) π^π₯/(π₯ + 1)+π β«1βγπ^π₯ ((π₯ )/(1 + π₯)^2 ) γ ππ₯= β«1βγππ₯" " ((1 + π₯ β 1 )/(1 + π₯)^2 ) ππ₯γ =" " β«1βγππ₯" " ((1 + π₯ )/(1 + π₯)^2 β1/(1 + π₯)^2 ) ππ₯γ =" " β«1βγππ" " (π/((π + π))βπ/(π + π)^π ) ππγ It is of form β«1βγπ^π (π(π)+π^β² (π)) γ ππ=π^π₯ π(π₯)+πΆ Putting π(π)=π/((π + π)) β΄ π^β² (π)=(β1)/(π + π)^2 Thus, β«1βγππ₯" " (1/((1 + π₯))β1/(1 + π₯)^2 ) ππ₯γ = π^π/((π + π))+πͺ So, the correct answer is (d)