Case Based Questions (MCQ)

Chapter 7 Class 12 Integrals
Serial order wise

## (d) None of these

This question is inspired from Ex 7.2, 5 - Chapter 7 Class 12 - Integrals

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## (d) None of these

This question is inspired from Ex 7.2, 3 - Chapter 7 Class 12 - Integrals

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Question 3 The given integral β«1βγπ(π₯)γ ππ₯ can be transformed into another form by changing the independent variable π₯ to π‘ by substituting π₯=π(π‘) Consider I = β«1βγπ(π₯)γ ππ₯ Put π₯ = π(π‘) so that ππ₯/ππ‘=πβ²(π‘) We write ππ₯=π^β² (π‘)ππ‘ Thus I = β«1βγπ(π₯)γ ππ₯= β«1βγπ(π(π‘)) π^β² (π‘) ππ‘γ This change of variable formula is one of the important tools available to us in the name of integration by substitution. For example: β«1βγ2π₯ sinβ‘γ(π₯^2+1)γ ππ₯γ Put π₯^2+1=π‘ 2π₯ππ₯=ππ‘ Thus, β«1βγsinβ‘γ(π‘) ππ‘γ=βcosβ‘(π‘)+πΆγ = βcosβ‘γ(π₯^2+1)+πΆγ Based on the above information, answer any four of the following questions. Question 1 β«1βγsinβ‘γ(tan^(β1)β‘π₯)γ/(1 + π₯^2 ) ππ₯γ is equal to: (a) βsinβ‘(tan^(β1)β‘π₯ )+πΆ (b) βcosβ‘(tan^(β1)β‘π₯ )+πΆ (c) tanβ‘π₯+πΆ (d) None of these β«1βγsinβ‘γ(tan^(β1)β‘π₯)γ/(1 + π₯^2 ) ππ₯γ Let γπ­ππ§γ^(βπ)β‘π=π Differentiating both sides π€.π.π‘.π₯ π/(π + π^π )= ππ/ππ ππ₯/(1 + π₯^2 )=ππ‘ Now, Integrating the function β«1βγπππβ‘γ(γπππγ^(βπ)β‘π)γ/(π + π^π ) ππγ Putting tan^(β1)β‘π₯=π‘ & ππ₯/(1 + π₯^2 )=ππ‘ = β«1βγπ¬π’π§β‘π . ππγ = βcosβ‘π‘+πΆ Putting back t = tan^(β1)β‘π₯ = βππ¨π¬β‘γ(πππ^(βπ) π)γ+πͺ So, the correct answer is (b) Question 2 β«1βγtanβ‘π₯ ππ₯γ is equal to: (a) secβ‘π₯+πΆ (b) cotβ‘π₯+πΆ (c) logβ‘γ|π₯|γ+πΆ (d) None of these β«1βγtanβ‘π₯ ππ₯γ=β«1βγsinβ‘π₯/cosβ‘π₯ ππ₯γ Let πππβ‘π=π Differentiating both sides π€.π.π‘.π₯ βπππ π= ππ/ππ sin x ππ₯=βππ‘ Now, β«1β(πππ π)/(πππ π) ππ Putting πππ β‘π₯=π‘ & sinβ‘π₯ ππ₯=ππ‘ = β«1βγπ/π . (βππ)γ = ββ«1βππ/π = βlogβ‘γ|π‘|γ+πΆ Putting back t = cosβ‘π₯ = βπ₯π¨π |πππ π|+πͺ = loπ |πππ  π₯|^(β1)+πΆ = loπ 1/(|πππ  π₯|)+πΆ = π₯π¨π |πππ π|+πͺ So, the correct answer is (d) Question 3 β«1βγ2π₯/(1 + π₯^2 ) ππ₯γ is equal to: (a) 1+π₯^2+πΆ (b) logβ‘γ|1+π₯^2 |+πΆγ (c) logβ‘|2/(1 + π₯^2 )|+πΆ (d) None of these β«1βγ2π₯/(1 + π₯^2 ) ππ₯γ Let π+π^π=π Differentiating both sides π€.π.π‘.π₯ ππ= ππ/ππ 2π₯ ππ₯=ππ‘ Now, Integrating the function β«1βγππ/(π + π^π ) ππγ Putting 1+π₯^2=π‘ & 2π₯ππ₯=ππ‘ = β«1βγπ/π . ππγ = logβ‘γ|π‘|γ+πΆ Putting back t = 1+π₯^2 = π₯ππ|π+π^π |+πͺ So, the correct answer is (b) Question 4 β«1βγsinβ‘γ(ππ₯+π)γ cosβ‘γ(ππ₯+π)γ ππ₯γ is equal to: (a) cos^2 π₯ (ππ₯+π)+πΆ (b) sin^2β‘γ(ππ₯+π)γ (c) (β1)/4π cosβ‘2 (ππ₯+π)+πΆ (d) None of these Taking the given function πππβ‘(ππ + π) πππβ‘(ππ + π) = 1/2 sinβ‘(2(ππ₯+π)) = π/π πππβ‘(πππ+ππ) Let (πππ+ππ)=π Differentiating both sides π€.π.π‘.π₯ 2π+0=ππ‘/ππ₯ 2π=ππ‘/ππ₯ 2π.ππ₯=ππ‘ ππ=ππ/ππ Integrating the function β«1βγ" " sinβ‘(ππ₯+π) cosβ‘(ππ₯+π)" " γ.ππ₯ = 1/2 β«1βγsinβ‘(2ππ₯+2π)γ .ππ₯ Putting π‘=2ππ₯+2π & ππ₯=ππ‘/2π = π/π β«1βγπππ πγ .ππ/ππ = 1/4π β«1βγsinβ‘(π‘)γ .ππ‘ = 1/4π [βcosβ‘π‘+πΆ1] = β 1/4π . cosβ‘π‘ + πΆ1/4π = β π/ππ . πππβ‘π + πͺ Putting back π‘=2ππ₯+2π = β 1/4π . cosβ‘γ(2ππ₯+2π)γ + πΆ = β π/ππ . πππβ‘γπ(ππ+π)γ+πͺ So, the correct answer is (c) Question 5 β«1βγ1/(π₯ + π₯ logβ‘π₯ ) ππ₯γ is equal to: (a) |1+logβ‘π₯ |+πΆ (b) logβ‘γ|1+logβ‘π₯ |+πΆγ (c) logβ‘π₯+πΆ (d) None of these β«1βγ1/(π₯ + π₯ logβ‘π₯ ) ππ₯γ = β«1βγ1/(π₯(1 + logβ‘π₯)) ππ₯γ Let 1 + log π= π Differentiate π€.π.π‘.π₯ 0 + ππ‘/ππ₯= 1/π₯ 0 + ππ‘/ππ₯= 1/π₯ ππ.π=ππ Now, β«1β1/(π₯ + π₯ logβ‘π₯ ) .ππ₯ =β«1β1/(π₯ (1 + logβ‘π₯ ) ) .ππ₯" " Putting 1+logβ‘π₯ & ππ₯=π₯ ππ‘ = β«1β1/(π₯(π‘)) ππ‘.π₯ = β«1β1/π‘ ππ‘ = πππ|π‘|+πΆ Putting back π‘ =1+πππβ‘π₯ = πππ|π+π₯π¨π β‘π |+πͺ So, the correct answer is (b)

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.