Question 1 - Case Based Questions (MCQ) - Chapter 7 Class 12 Integrals
Last updated at May 29, 2023 by Teachoo
Odd functionΒ f(x) = -f(x)Β
Even functionΒ g(x) = g(x)Β
Based on the above information, answer any fourΒ of the following questions.
Question 1
β«
1
(-1)Β
x
99
Β Β dx=______________.
(a) 0 Β (b) 1Β (c) β1Β (d) 2
Question 2
β«
Ο
-Ο
x cos x dx =______________.
(a) 1 Β (b) 0Β Β Β (c) β1Β Β Β (d) Ο/2
Question 3
β«
Ο/2
-Ο/2
Β sin
3
π₯ ππ₯ = _________.
(a) 1Β Β (b) 0Β Β Β (c) β1Β Β Β (d) Ο
Β
Question 4
β«
Ο
-Ο
Β π₯ sin π₯ ππ₯ = _________.
(a) ΟΒ Β Β (b) 0Β Β Β Β Β Β (c) 2ΟΒ Β Β Β Β Β (d) Ο/2
Question 5
β«
Ο
-Ο
Β tanπ₯ sec
2
π₯ ππ₯ = _________.
(a) 1Β Β Β Β Β Β (b) β1Β Β Β Β Β (c) 0Β Β Β Β Β Β (d) 2
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Transcript
Question 1 Based on the above information, answer any four of the following questions.
Question 1 β«1_(β1)^1βπ₯^99 ππ₯=______________. (a) 0 (b) 1 (c) β1 (d) 2
This is of form β«_(βπ)^πβπ(π)π
π
π(π)=π₯^99
π(βπ)=(βπ₯)^99=βπ₯^99
Thus, π(βπ) =βπβ‘(π)
β«1_(β1)^1βπ₯^99 ππ= 0
So, the correct answer is (a)
Question 2 β«1_(βπ)^πβγ π₯ cosβ‘π₯ γ ππ₯=______________. (a) 1 (b) 0 (c) β1 (d) π/2
This is of form β«_(βπ)^πβπ(π)π
π
π(π)=π₯ πππ π₯
π(βπ)=(βπ₯) cosβ‘γ(βπ₯)γ=βπ₯ cosβ‘π₯
Thus, π(βπ) =βπβ‘(π)
β«1_(βπ)^πβγ π₯ cosβ‘π₯ γ ππ₯= 0
So, the correct answer is (b)
Question 3 β«1_((βπ)/2)^(π/2 )βγ sinγ^3β‘π₯ ππ₯ = _________. (a) 1 (b) 0 (c) β1 (d) π
This is of form β«_(βπ)^πβπ(π)π
π
π(π)=sin^3β‘π₯
π(βπ)=sin^3β‘γ(βπ₯)γ=(βsinβ‘π₯ )^3=βsin^3β‘π₯
Thus, π(βπ) =βπβ‘(π)
β«1_((βπ)/2)^(π/2 )βγ sinγ^3β‘π₯ ππ₯ = 0
So, the correct answer is (b)
Question 4 β«1_(βπ)^πβγπ₯ π ππ γβ‘π₯ ππ₯ = _________. (a) π (b) 0 (c) 2π (d) π/2
This is of form β«_(βπ)^πβπ(π)π
π
π(π)=π₯ π ππ π₯
π(βπ)=(βπ₯) γπ ππ γβ‘γ(βπ₯)γ=βπ₯ Γ βsinβ‘π₯=π₯ sinβ‘γπ₯ γ
Thus, π(βπ) =πβ‘(π)
Therefore,
β«1_(βπ)^πβγπ₯ π ππ γβ‘π₯ ππ₯ = 2β«1_0^πβγπ₯ π ππ γβ‘π₯ ππ₯
Let I = πβ«1_π^π
βγπ πππ γβ‘π ππ₯
I = 2β«1_0^πβγ(π
βπ) π ππ γβ‘γ(π βπ₯) γ ππ₯
I = 2β«1_0^πβγ(π βπ₯) π ππ γβ‘γπ₯ γ ππ₯
I = πβ«1_π^π
βγπ πππ γβ‘π ππ₯ β πβ«1_π^π
βγπ πππ γβ‘π ππ₯
Adding (1) and (2)
I + I = 2β«1_0^πβγπ₯ π ππ γβ‘π₯ ππ₯ + 2β«1_0^πβγπ π ππ γβ‘π₯ ππ₯ β 2β«1_0^πβγπ₯ π ππ γβ‘π₯ ππ₯
2I = 2β«1_0^πβγπ π ππ γβ‘π₯ ππ₯
I = β«1_π^π
βγπ
πππ γβ‘π ππ₯
I = πβ«1_0^πβγπ ππ γβ‘π₯ ππ₯
I = π
γ[βππ¨π¬β‘π]γ_π^π
I = π[βcosβ‘γπ β(βcosβ‘0)γ]
I = π[βcosβ‘π+cosβ‘0]
I = π[β(β1)+1]
I = π [1+1]
I = ππ
So, the correct answer is (c)
Question 5 β«1_(βπ)^πβγtanβ‘π₯ sec^2β‘π₯ γ ππ₯ = _________. (a) 1 (b) β1 (c) 0 (d) 2
This is of form β«_(βπ)^πβπ(π)π
π
π(π)=tanβ‘π₯ sec^2β‘π₯
π(βπ)=tanβ‘γ(βπ₯)γ sec^2β‘γ(βπ₯)γ=βtanβ‘π₯ sec^2β‘γπ₯ γ
Thus, π(βπ) =βπβ‘(π)
β«1_(βπ)^πβγtanβ‘π₯ sec^2β‘π₯ γ " ππ₯"= 0
So, the correct answer is (c)
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