Question 2 - Case Based Questions (MCQ) - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
∫ e
x
[f(x) + f′(x)] dx
=
∫
e
x
f(x)dx
+ ∫ e
x
f′(x) dx
Using integration by parts
=
f(x) ∫ e
x
dx - ∫ f′ (x) e
x
dx
+ ∫ f′ (x) e
x
dx
= f(x) e
x
∫ f′ (x) e
x
dx + ∫ f′ (x) e
x
dx
= e
x
f(x) + C
Based on the above information, answer any four of the following questions.
Question 1
∫e
x
(sin x + cos x) dx = ______________.
(a) e
x
cosx + c
(b) e
x
sinx + c
(c) e
x
+ c
(d) e
x
(-cosx + sinx ) + c
Question 2
∫e
x
(x - 1)/x
2
) dx =______________.
(a) e
x
+ c
(b) e
x
/x + c
(c) e
x
/x
2
+ c
(d) -e
x
/x
2
+ c
Question 3
∫e
x
(1 + x) dx =______________.
(a) xe
x
+ c
(b) e
x
+ c
(c) e
-x
+ c
(d) none of these
Question 4
∫
π
0
e
x
(tanx + sec
2
x) ππ₯ = _________.
(a) 0
(b) 1
(c) -1
(d) -e
π
Question 5
∫xe
x
/(1 + x)
2
dx =______________.
(a) xe
x
+ c
(b) e
x
/(x + 1)
2
+ c
(c) x e
x
/x + 1 + c
(d) e
x
/x + 1 + c
Transcript
Question 2 β«1βγπ^π₯ [π(π₯)+π^β² (π₯)] γ ππ₯ = β«1βγπ^π π(π)π
πγ+β«1βγπ^π₯ πβ²(π₯)ππ₯γ Using integration by parts = π(π) β«1βπ^π π
πββ«1βγπ^β² (π) π^π γ π
π + β«1βγπ^β² (π₯) π^π₯ γ ππ₯ = π(π₯) π^π₯ββ«1βγπ^β² (π₯) π^π₯ γ ππ₯ + β«1βγπ^β² (π₯) π^π₯ γ ππ₯ = π^π₯ π(π₯)+πΆ Based on the above information, answer any four of the following questions.
Question 1 β«1βγπ^π₯ (sinβ‘γπ₯ γ+cosβ‘π₯)γ ππ₯=______________. (a) π^π₯ cosβ‘π₯+π (b) π^π₯ sinβ‘π₯+π (c) π^π₯+π (d) π^π₯ (βcosβ‘π₯+sinβ‘π₯ )+c
β«1βγπ^π₯ (sinβ‘γπ₯ γ+cosβ‘π₯)γ ππ₯
Putting π(π)=πππ π" "
β΄ π^β² (π)=cosβ‘π₯
Thus,
β«1βγπ^π₯ (sinβ‘γπ₯ γ+cosβ‘π₯)γ ππ₯=π^π πππ π+πͺ
So, the correct answer is (b)
Question 2 β«1βγπ^π₯ ((π₯ β 1)/π₯^2 ) γ ππ₯=______________. (a) π^π₯+π (b) π^π₯/π₯+π (c) π^π₯/π₯^2 +π (d) γβπγ^π₯/π₯^2 +π
β«1βγπ^π₯ ((π₯ β 1)/π₯^2 ) γ ππ₯= β«1βγππ₯" " (π₯/π₯^2 β 1/π₯2) ππ₯γ
= β«1βγππ" " (π/π β π/ππ) π
πγ
Putting π(π)=π/π
β΄ π^β² (π)=(β1)/π₯^2
Thus,
β«1βγππ₯" " (1/π₯ β 1/π₯2) ππ₯γ = π^π/π+πͺ
So, the correct answer is (b)
Question 3 β«1βγπ^π₯ (1+π₯) γ ππ₯=______________. (a) π₯π^π₯+π (b) π^π₯+π (c) π^(βπ₯)+π (d) none of these
β«1βγπ^π₯ (1+π₯)γ ππ₯=β«1βγπ^π (π+π)γ π
π
Putting π(π)=π
β΄ π^β² (π)=1
Thus,
β«1βγπ^π₯ (π₯+1)γ ππ₯=π^π Γ π+πͺ
So, the correct answer is (a)
Question 4 β«1_0^πβπ^π₯ (tanβ‘π₯+sec^2β‘π₯) ππ₯ = _________. (a) 0 (b) 1 (c) β1 (d) βπ^π
β«1_0^πβπ^π₯ (tanβ‘π₯+sec^2β‘π₯) ππ₯
Putting π(π)=πππ π" "
β΄ π^β² (π)=γπ ππγ^2β‘π₯
Thus,
β«1_0^πβπ^π₯ (tanβ‘π₯+sec^2β‘π₯) ππ₯ = γ[π^π πππ π]γ_0^π
=[π^π πππ πβπ^π πππ π]
=[π^π πππ πβπ^π πππ π]
=[π^π Γ πβπ^π Γ π]
=0β0
=0
So, the correct answer is (a)
Question 5 β«1βγγπ₯πγ^π₯/(1 + π₯)^2 γ ππ₯=______________. (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«1βγπ^π₯ ((π₯ )/(1 + π₯)^2 ) γ ππ₯= β«1βγππ₯" " ((1 + π₯ β 1 )/(1 + π₯)^2 ) ππ₯γ
=" " β«1βγππ₯" " ((1 + π₯ )/(1 + π₯)^2 β1/(1 + π₯)^2 ) ππ₯γ
=" " β«1βγππ" " (π/((π + π))βπ/(π + π)^π ) π
πγ
It is of form
β«1βγπ^π (π(π)+π^β² (π)) γ π
π=π^π₯ π(π₯)+πΆ
Putting π(π)=π/((π + π))
β΄ π^β² (π)=(β1)/(π + π)^2
Thus,
β«1βγππ₯" " (1/((1 + π₯))β1/(1 + π₯)^2 ) ππ₯γ = π^π/((π + π))+πͺ
So, the correct answer is (d)
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