mathematics-2015 class-xii -...
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MATHEMATICS-2015 Class-XII
Section โ A
Q. 1. Write the value of โ= ๐ + ๐ ๐ + ๐ ๐ + ๐
๐ ๐ ๐โ๐ โ๐ โ๐
Sol.
โ= ๐ฅ + ๐ฆ ๐ฆ + ๐ง ๐ง + ๐ฅ
๐ง ๐ฅ ๐ฆโ3 โ3 โ3
= ๐ฅ + ๐ฆ + ๐ง ๐ฅ + ๐ฆ + ๐ง ๐ฅ + ๐ฆ + ๐ง
๐ง ๐ฅ ๐ฆโ3 โ3 โ3
๐๐ฉ๐ฉ๐ฅ๐ฒ๐ข๐ง๐ ๐๐ โ ๐๐ + ๐๐
Taking common ๐ฅ + ๐ฆ + ๐ง and โ 3 from R1 and R3 respectively
= โ3 ๐ฅ + ๐ฆ + ๐ง 1 1 1๐ง ๐ฅ ๐ฆ1 1 1
= 0
โต ๐๐๐๐ง๐ ๐๐ ๐๐ซ๐ ๐ข๐๐๐ง๐ญ๐ข๐๐๐ฅ
Q. 2. Write sum of the order and degree of the following differential equation :
๐
๐ ๐
๐ ๐
๐ ๐
๐ = ๐
Sol. Order = 2 Degree = 1 โด Sum = 2 + 1 = ๐
Q. 3. Write the integrating factor of the following differential equation :
๐ + ๐๐ + ๐๐๐ โ ๐๐จ๐ญ ๐ ๐ ๐
๐ ๐= ๐
Sol.
1 + ๐ฆ2 + 2๐ฅ๐ฆ โ cot ๐ฆ ๐๐ฆ
๐๐ฅ= 0
2๐ฅ๐ฆ โ cot ๐ฆ ๐๐ฆ
๐๐ฅ= โ 1 + ๐ฆ2
2๐ฅ๐ฆ โ cot ๐ฆ = โ 1 + ๐ฆ2 ๐๐ฅ
๐๐ฆ
2๐ฅ๐ฆ
โ 1+๐ฆ2 +
cot ๐ฆ
1+๐ฆ2=
๐๐ฅ
๐๐ฆ
๐๐ฅ
๐๐ฆ+
2๐ฅ๐ฆ
1+๐ฆ2=
cot ๐ฆ
1+๐ฆ2
Comparing with ๐๐ฅ
๐๐ฆ+ P๐ฅ = Q
P =2๐ฆ
1+๐ฆ2, Q =
โcot ๐ฆ
1+๐ฆ2
Integrating factor I. F. = ๐ P ๐๐ฆ
= ๐
2๐ฆ
1+๐ฆ2๐๐ฆ
= ๐ ๐A
A Let A = 1 + ๐ฆ2 , ๐A = 2๐ฆ ๐๐ฆ
= ๐ log A
= ๐ = ๐ + ๐๐
Q. 4. If ๐ , ๐ and ๐ are mutually perpendicular unit vectors, then find the value of
๐๐ + ๐ + ๐ .
Sol.
๐ , ๐ and ๐ are mutually perpendicular unit vectors,
๐ . ๐ = 0, ๐ = ๐ = ๐ = 1 โฆ (๐)
๐ . ๐ = 0, ๐ . ๐ = 0
2๐ + ๐ + ๐ 2
= 2๐ + ๐ + ๐ 2
= 2๐ + ๐ + ๐ . 2๐ + ๐ + ๐
= 4๐ 2 + 2๐ ๐ + 2๐ ๐ + 2๐ ๐ + ๐ 2 + ๐ ๐ + 2๐ ๐ + ๐ ๐ + ๐ 2
= 4 1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1 From (i)
and
๐ 2 = ๐ 2 = 12 = 1
๐ 2 = ๐ 2
= 1
๐ 2 = ๐ 2 = 1
= 6 2๐ + ๐ + ๐ = ๐ (squaring root on both sides)
Q. 5. Write a unit vector perpendicular to both the vectors ๐ = ๐ + ๐ + ๐ and ๐ = ๐ + ๐ . Sol.
๐ ร ๐ = ๐ ๐ ๐
1 1 11 1 0
= ๐ โ1 โ ๐ โ1 + ๐ 0 = โ๐ + ๐
๐ ร ๐ = โ1 2 + 1 2 = 2
A unit vector โฅ to both ๐ and ๐ =๐ ร๐
๐ ร๐
=โ๐ +๐
2=
โ๐
๐๐ +
๐
๐๐
Q. 6. The equation of a line are ๐๐ โ ๐ = ๐๐๐ + ๐ = ๐ โ ๐๐๐. Write the direction cosines
of the line. Sol. 5๐ฅ โ 3 = 15๐ฆ + 7 = 3 โ 10๐ง
5 ๐ฅ โ3
5 = 15 ๐ฆ +
7
15 = โ10 ๐ง โ
3
10 [LCM of 5,10,15 is 30
Dividing throughout by 30, we have
5 ๐ฅโ
3
5
306=
15 ๐ฆ+7
15
302=
โ10 ๐งโ3
10
303
Direction ratios of given line are 6,2, โ3 ๐ = 6, ๐ = 2, ๐ = โ3
๐2 + ๐2 + ๐2 = 62 + 22 + โ3 2 = 7 Direction cosines are
๐ =๐
๐2+๐2+๐2 ๐ =
๐
๐2+๐2+๐2 ๐ =
๐
๐2+๐2+๐2
=๐
๐ =
๐
๐ =
โ๐
๐
Section โ B Q. 7. To promote the making of toilets for women, an organisation tried to generate
awareness though (๐) House call, (๐๐) letters and (๐๐๐) announcements. The cost of each mode per attempt is given below : (๐) Rs. 50 (๐๐) Rs. 20 (๐๐๐) Rs. 40 The number of attempts made in tree villages ๐, ๐ and ๐ are given below :
(๐) (๐๐) (๐๐๐)
๐ ๐๐๐ ๐๐๐ ๐๐๐ ๐ ๐๐๐ ๐๐๐ ๐๐ ๐ ๐๐๐ ๐๐๐ ๐๐๐
Find the total cost incurred by the organisation for the three villages separately, using
matrices.
Write one value generated by the organisation in the society.
Sol.
The number of attempt made in 3 villages X, Y, Z is written in Row matrix
400 300 100300 250 75500 400 150
The cost for each mode per attempt (in column matrix)
502040
Total cost = 400 300 100300 250 75500 400 150
502040
= 20000 + 6000 + 400015000 + 5000 + 300025000 + 8000 + 6000
= 300002300039000
Total cost incurred by the organization for village X = Rs. 30,000
Total cost incurred by the organization for village Y = Rs. 23,000
Total cost incurred by the organization for village Z = Rs. 39,000
Value generated by the organisation :
1. Prevention is better than cure so cleanliness leads to better quality of life (healthy life) It will help to avoid various diseases.
2. Concern for safety of women.
Q. 8. Solve for ๐ โถ ๐ญ๐๐งโ๐ ๐ + ๐ + ๐ญ๐๐งโ๐ ๐ โ ๐ = ๐ญ๐๐งโ๐ ๐
๐๐
Sol.
tanโ1 ๐ฅ + 1 + tanโ1 ๐ฅ โ 1 = tanโ1 8
31
tanโ1 ๐ฅ+1 + ๐ฅโ1
1โ ๐ฅ+1 ๐ฅโ1 = tanโ1 8
31
โต ๐ญ๐๐งโ๐ ๐ + ๐ญ๐๐งโ๐ ๐ = ๐ญ๐๐งโ๐ ๐+๐
๐โ๐๐
2๐ฅ
1โ ๐ฅ2โ1 =
8
31
62๐ฅ = 8 2 โ ๐ฅ2 62๐ฅ = 16 โ 8๐ฅ2 8๐ฅ2 + 62๐ฅ โ 16 = 0 4๐ฅ2 + 31๐ฅ โ 8 = 0 (Dividing by 2) 4๐ฅ2 + 32๐ฅ โ ๐ฅ โ 8 = 0 4๐ฅ ๐ฅ + 8 โ 1 ๐ฅ + 8 = 0 ๐ฅ + 8 4๐ฅ โ 1 = 0 ๐ฅ + 8 = 0 or 4๐ฅ โ 1 = 0
๐ฅ = โ8 or ๐ฅ =1
4
Since ๐ฅ = โ8 does not satisfy the given equation
โด ๐ =๐
๐
Or Prove the following :
๐๐จ๐ญโ๐ ๐๐+๐
๐โ๐ + ๐๐จ๐ญโ๐
๐๐+๐
๐โ๐ + ๐๐จ๐ญโ๐
๐๐+๐
๐โ๐ = ๐ ๐ < ๐ฅ๐ฆ, ๐ฆ๐ฅ, ๐ง๐ฅ < 1
Sol. ๐. ๐. ๐. =
= cotโ1 ๐ฅ๐ฆ +1
๐ฅโ๐ฆ + cotโ1
๐ฆ๐ง +1
๐ฆโ๐ง + cotโ1
๐ง๐ฅ+1
๐งโ๐ฅ
= tanโ1 ๐ฅโ๐ฆ
1+๐ฅ๐ฆ + tanโ1
๐ฆโ๐ง
1+๐ฆ๐ง + tanโ1
๐งโ๐ฅ
1+๐ง๐ฅ
โต cotโ1 ๐
๐ = tanโ1
๐
๐
โต tanโ1 AโB
1+AB = tanโ1 A โ tanโ1 B
= tanโ1 ๐ฅ โ tanโ1 ๐ฆ + tanโ1 ๐ฆ โ tanโ1 ๐ง + tanโ1 ๐ง โ tanโ1 ๐ฅ = 0 = ๐. ๐. ๐.
Q. 9. Using properties of determinants, prove the following :
๐๐ ๐๐ ๐๐ + ๐๐
๐๐ + ๐๐ ๐๐ ๐๐๐๐ ๐๐ + ๐๐ ๐๐
= ๐๐๐๐๐๐๐.
Sol.
๐. ๐. ๐. = ๐2 ๐๐ ๐๐ + ๐2
๐2 + ๐๐ ๐2 ๐๐๐๐ ๐2 + ๐๐ ๐2
Taking common ๐, ๐, ๐ from C1, C2 and C3 respectively
= ๐๐๐ ๐ ๐ ๐ + ๐
๐ + ๐ ๐ ๐๐ ๐ + ๐ ๐
Applying C1 โ C1 โ C2 โ C3
= ๐๐๐ โ2๐ ๐ ๐ + ๐
0 ๐ ๐โ2๐ ๐ + ๐ ๐
Taking common โ2๐ from C1, we have
= โ2๐๐๐2 1 ๐ ๐ + ๐0 ๐ ๐1 ๐ + ๐ ๐
Applying R3 โ R3 โ R1
= โ2๐๐๐2 1 ๐ ๐ + ๐0 ๐ ๐0 ๐ โ๐
Expanding along C1, we have
= โ2๐๐๐2 . 1 โ๐๐ โ ๐๐
= โ2๐๐๐2 โ2๐๐
= 4๐2๐2๐2 = ๐. ๐. ๐.
Q. 10. Find the ad joint of the matrix ๐ = โ๐ โ๐ โ๐ ๐ ๐ โ๐ ๐ โ๐ ๐
and hence show that
๐. ๐๐ ๐ ๐ = ๐ ๐๐. Sol.
A = โ1 โ2 โ2 2 1 โ2 2 โ2 1
A11 = 1 + 4 = โ3, A12 = โ 2 + 4 = โ6 A13 = โ4 โ 2 = โ6, A21 = โ โ2 โ 4 = 6 A22 = 1 + 4 = 3, A23 = โ 2 + 4 = โ6 A31 = 4 + 2 = 6, A32 = โ 2 + 4 = โ6 A33 = โ1 + 4 = 3
๐๐๐ A = โ3 6 6โ6 3 โ6โ6 โ6 3
๐. ๐. ๐. = A. ๐๐๐ A
= โ1 โ2 โ2 2 1 โ2 2 โ2 1
โ3 6 6โ6 3 โ6โ6 โ6 3
= 3 + 12 + 12 โ6 โ 6 + 12 โ6 + 12 โ 6โ6 โ 6 + 12 12 + 3 + 12 12 โ 6 โ 6โ6 + 12 โ 6 12 โ 6 โ 6 12 + 12 + 3
= 27 0 00 27 00 0 27
= 27 1 0 00 1 00 0 1
โฆ (๐)
Expanding along R1of matrix A, A = โ1 1 โ 4 + 2 2 + 4 โ 2 โ4 โ 2 = 3 + 12 + 12 = 27 ๐. ๐. ๐. = A I3
= 27 1 0 00 1 00 0 1
โฆ (๐๐)
From โฆ (๐) and โฆ (๐๐) ๐. ๐. ๐. = ๐. ๐. ๐.
Q. 11. Show that the function ๐ ๐ = ๐ โ ๐ + ๐ + ๐ for all ๐ โ ๐, is not differentiable at
the points ๐ = โ๐ and ๐ = ๐.
Sol.
๐ฅ โ 1 = โ ๐ฅ โ 1 , ๐ฅ < 1+ ๐ฅ โ 1 , ๐ฅ โฅ 1
๐ฅ โ 1 = โ ๐ฅ โ 1 , ๐ฅ < โ1+ ๐ฅ โ 1 , ๐ฅ โฅ โ1
๐ ๐ฅ = โ ๐ฅ โ 1 โ ๐ฅ + 1 for ๐ฅ < โ1 โ ๐ฅ โ 1 + ๐ฅ + 1 for โ 1 โค ๐ฅ < 1 ๐ฅ โ 1 + ๐ฅ + 1 for ๐ฅ โฅ 1
๐ ๐ฅ = โ2๐ฅ, for ๐ฅ < โ1 2, for โ 1 โค ๐ฅ < 1 2๐ฅ for ๐ฅ โฅ 1
๐๐ญ ๐ = โ๐
LHD at ๐ฅ = โ1 RHD at ๐ฅ = โ1
= lim๐ฅโโ1โ1๐ ๐ฅ โ๐ โ1
๐ฅโ โ1 = lim๐ฅโโ1+
๐ ๐ฅ โ๐ โ1
๐ฅโ โ1
= lim๐ฅโโ1โโ2๐ฅโ2
๐ฅโ1 โต ๐ โ1 = 2 = lim๐ฅโโ1+
2โ2
๐ฅโ1
= lim๐ฅโโ1โ2 ๐ฅโ1
๐ฅโ1 = lim๐ฅโโ1+ 0
= โ2 = 0
LHD at ๐ฅ = โ1 โ RHD at ๐ฅ = โ1
So, ๐ ๐ is not differentiable at ๐ = โ๐
๐๐ญ ๐ = โ๐
LHD at ๐ฅ = 1 RHD at ๐ฅ = 1
= lim๐ฅโ1โ1๐ ๐ฅ โ๐ 1
๐ฅโ1 = lim๐ฅโ1+
๐ ๐ฅ โ๐ 1
๐ฅโ1
= lim๐ฅโ1โ2โ2
๐ฅโ1
at ๐ฅ = 1 ๐ ๐ฅ = 2๐ฅ
๐ 1 = 2
= lim๐ฅโ1+2๐ฅโ2
๐ฅโ1
= lim๐ฅโ1+ 0 = lim๐ฅโ1โ2 ๐ฅโ1
๐ฅโ1
= 0 = 2
LHD at ๐ฅ = 1 โ RHD at ๐ฅ = 1
So, ๐ ๐ is not differentiable at ๐ = ๐
Q. 12. If ๐ = ๐๐ ๐ฌ๐ข๐งโ๐ ๐, then show that ๐ โ ๐๐ ๐ ๐๐
๐ ๐๐ โ ๐๐ ๐
๐ ๐โ ๐๐๐ = ๐.
Sol.
๐ฆ = ๐๐ sin โ1 ๐ฅ โฆ (๐) Differentiating both sides w.r.t. ๐ฅ, we get
๐๐ฆ
๐๐ฅ= ๐๐ sin โ1 ๐ฅ .
๐
1โ๐ฅ2
๐๐ฆ
๐๐ฅ= ๐ฆ.
๐
1โ๐ฅ2 โฆ (๐๐)
1 โ ๐ฅ2 ๐๐ฆ
๐๐ฅ= ๐๐ฆ
Again differentiating both sides w.r.t. ๐ฅ, we get
1 โ ๐ฅ2 ๐2๐ฆ
๐๐ฅ 2+
๐๐ฆ
๐๐ฅ.
1
1โ๐ฅ22 . โ2๐ฅ = ๐๐๐ฆ
๐๐ฅ
1 โ ๐ฅ2 ๐2๐ฆ
๐๐ฅ 2โ
๐ฅ
1โ๐ฅ22
๐๐ฆ
๐๐ฅ= ๐
๐๐ฆ
๐๐ฅ
1โ๐ฅ2
๐2๐ฆ
๐๐ฅ 2โ๐ฅ๐๐ฆ
๐๐ฅ
1โ๐ฅ2= ๐
๐๐ฆ
๐๐ฅ
1 โ ๐ฅ2 ๐2๐ฆ
๐๐ฅ 2โ ๐ฅ
๐๐ฆ
๐๐ฅ= ๐ 1 โ ๐ฅ2
๐ฆ๐
1โ๐ฅ2 [From (ii)]
๐ โ ๐๐ ๐ ๐๐
๐ ๐๐ โ ๐๐ ๐
๐ ๐= ๐๐๐ = ๐
Q. 13. If ๐ ๐ = ๐๐ + ๐; ๐ ๐ =๐+๐
๐๐+๐ and ๐ ๐ = ๐๐ โ ๐, then find ๐โฒ ๐โฒ ๐ โฒ ๐ .
Sol.
g ๐ฅ =๐ฅ+1
๐ฅ2+1
Differentiating both sides w.r.t. ๐ฅ, we have
g ๐ฅ = ๐ฅ2+1 .
๐
๐๐ฅ ๐ฅ+1 โ ๐ฅ+1 .
๐
๐๐ฅ ๐ฅ2+1
๐ฅ2+1 2
= ๐ฅ2+1 .1โ ๐ฅ+1 .2๐ฅ
๐ฅ2+1 2
=๐ฅ2+1โ2๐ฅ2+2๐ฅ
๐ฅ2+1 2
g ๐ฅ =โ๐ฅ2โ2๐ฅ+1
๐ฅ2+1 2
โ ๐ฅ = 2๐ฅ โ 3 Differentiating both sides w.r.t. ๐ฅ, we have
โโฒ ๐ฅ = 2
โด โโฒ gโฒ ๐ฅ = 2 โฆ (๐) โต ๐๐๐ญ ๐โฒ ๐ = ๐ ๐ญ๐ก๐๐ง ๐โฒ ๐ = ๐ ๐๐ง๐ ๐โฒ โ๐๐ + ๐ = ๐
๐ ๐ฅ = ๐ฅ2 + 1 Differentiating both sides w.r.t. ๐ฅ, we have
๐ ๐ฅ =1
๐ฅ2+1ร 2๐ฅ =
๐ฅ
๐ฅ2+1
When ๐ฅ = 2, ๐ ๐ฅ =1
22+1=
2
5
๐โฒ โโฒ gโฒ ๐ฅ =2
5ร
5
5=
๐ ๐
๐ [From (i)
Q. 14. Evaluate : ๐ โ ๐๐ . ๐ + ๐ โ ๐๐ ๐ ๐ Sol.
3 โ 2๐ฅ . 2 + ๐ฅ โ ๐ฅ2 ๐๐ฅ
= 2 + 1 โ 2๐ฅ 2 + ๐ฅ โ ๐ฅ2 ๐๐ฅ
= 2 2 + ๐ฅ โ ๐ฅ2 ๐๐ฅ + 1 โ 2๐ฅ 2 + ๐ฅ โ ๐ฅ2 ๐๐ฅ [๐๐๐ญ ๐ = ๐๐ โ ๐๐, ๐ ๐ = ๐ โ ๐๐ ๐ ๐
= 2 โ ๐ฅ2 โ ๐ฅ โ 2 ๐๐ฅ + ๐ ๐๐
= 2 โ ๐ฅ2 โ ๐ฅ + 1
2
2โ 1
4โ 2 ๐๐ฅ + ๐
1
2 ๐๐
= 2 โ ๐ฅ โ1
2
2โ
9
4 ๐๐ฅ + ๐
1
2 ๐๐
= 2 3
2
2โ ๐ฅ โ
1
2
2๐๐ฅ + ๐
1
2 ๐๐
= 2 ร1
2 ๐ฅ โ
1
2
3
2
2โ ๐ฅ โ
1
2
2+
9
4sinโ1
๐ฅโ1
23
2
+2
3๐
3
2 + C
โต ๐๐ + ๐๐ ๐ ๐ =๐
๐ ๐ ๐๐ + ๐๐ + ๐๐ ๐ฌ๐ข๐งโ๐ ๐
๐ + ๐
= ๐๐โ๐
๐ ๐ + ๐ + ๐๐ +
๐
๐๐ฌ๐ข๐งโ๐
๐๐โ๐
๐ +
๐
๐ ๐ + ๐ + ๐๐
๐
๐ + ๐
Or
Evaluate : ๐๐+๐+๐
๐๐+๐ ๐+๐ ๐ ๐
Sol.
Let ๐ฅ2+๐ฅ+1
๐ฅ+2 ๐ฅ2+1 =
A
๐ฅ+2+
B๐ฅ+C
๐ฅ2+1 โฆ ๐
โ ๐ฅ2 + ๐ฅ + 1 = A ๐ฅ2 + 1 + B๐ฅ + C ๐ฅ + 2 โ ๐ฅ2 + ๐ฅ + 1 = A ๐ฅ2 + 1 + B๐ฅ ๐ฅ + 2 + C ๐ฅ + 2 Comparing the coefficients of ๐ฅ2 , ๐ฅ and constant terms, 1 = A + B 1 = 2B + C 1 = A + 2C 1 โ B = A 1 โ 2B = C 1 = 1 โ B + 2 1 โ 2B โฆ ๐๐ โฆ ๐๐๐ [From (ii) and (iii)]
1 โ2
5= A 1 โ 2
2
5 = C 1 = 1 โ B + 2 โ 4B
[From ๐๐ฃ] [From ๐๐ฃ] 1 โ 1 โ 2 = โ5B โ 2 = โ5B
โด A =3
5 C =
1
5 B =
2
5 โฆ (๐๐ฃ)
Putting the value of A, B and C in (i), we get
๐ฅ2+๐ฅ+1
๐ฅ+2 ๐ฅ2+1 =
3
5
๐ฅ+2+
2
5๐ฅ+
1
5
๐ฅ2+1
โด ๐ฅ2+๐ฅ+1
๐ฅ+2 ๐ฅ2+1 ๐๐ฅ =
3
5
๐๐ฅ
๐ฅ+2+
1
5
2๐ฅ
๐ฅ2+1๐๐ฅ +
1
5
๐๐ฅ
๐ฅ2+1
=3
5
๐๐ฅ
๐ฅ+2+
1
5
๐๐
๐+
1
5
๐๐ฅ
๐ฅ2+1 โฆ
Let ๐ = ๐ฅ2 + 1โด ๐๐ = 2๐ฅ ๐๐ฅ
=3
5log ๐ฅ + 2 +
1
5log ๐ +
1
5tanโ1 ๐ฅ + ๐
=๐
๐๐ฅ๐จ๐ ๐ + ๐ +
๐
๐๐ฅ๐จ๐ ๐๐ + ๐ +
๐
๐๐ญ๐๐งโ๐ ๐ + ๐
โต ๐ ๐
๐ ๐ ๐ ๐ = ๐ฅ๐จ๐ ๐ ๐ + ๐
Q. 15. Find : ๐ ๐
๐๐จ๐ฌ๐ ๐ ๐ ๐ฌ๐ข๐ง ๐๐
๐ /๐
๐
Sol.
๐๐ฅ
cos 3 ๐ฅ 2 sin 2๐ฅ
๐/4
0
= ๐๐ฅ
cos 3 ๐ฅ 2.2 tan ๐ฅ
1+tan 2 ๐ฅ
๐/4
0 โต ๐ฌ๐ข๐ง๐๐ =
๐ ๐ญ๐๐ง ๐
๐+๐ญ๐๐ง๐ ๐
= ๐๐ฅ
cos 3 ๐ฅ .2 tan ๐ฅ
sec 2 ๐ฅ
๐/4
0 โต ๐ + ๐ญ๐๐ง๐ ๐ = ๐ฌ๐๐๐ ๐
= ๐๐ฅ
2cos 3 ๐ฅ tan ๐ฅ .cos ๐ฅ
๐/4
0
=1
2
sec 2 ๐ฅ . sec 2 ๐ฅ .
tan ๐ฅ
๐/4
0๐๐ฅ =
1
2
1+tan 2 ๐ฅ sec 2 ๐ฅ ๐๐ฅ
tan ๐ฅ
๐/4
0
=1
2
1+๐2
๐๐๐
1
0
๐๐๐ญ ๐ = ๐ญ๐๐ง๐
๐ ๐ = ๐ฌ๐๐๐ ๐ ๐ ๐ ๐๐ก๐๐ง ๐ = ๐ /๐, ๐ = ๐๐๐ก๐๐ง ๐ = ๐, ๐ = ๐
=1
2
1
๐+
๐2
๐ ๐๐
1
0
=1
2 ๐โ
1
2 + ๐3
2 ๐๐1
0=
1
2
2
1๐1/2 +
2
5๐5/2
0
1
=1
2 2 1 1/2 +
2
5 1 5/2 โ
1
2 2 0 +
2
5 0
=1
2 2 +
2
5 =
1
2
10+2
5
=1
2
12
5 =
๐
๐
Q. 16. Find : ๐ฅ๐จ๐ ๐
๐+๐ ๐๐ ๐
Sol.
log ๐ฅ
๐ฅ+1 2๐๐ฅ
= log ๐ฅ. ๐ฅ + 1 โ2
I II ๐๐ฅ
Integrating by parts taking log ๐ฅ as Ist function, we have
= log ๐ฅ. ๐ฅ + 1 โ2 ๐๐ฅ โ ๐
๐๐ฅ log ๐ฅ ๐ฅ + 1 โ2 ๐๐ฅ ๐๐ฅ
= log ๐ฅ. ๐ฅ+1 โ1
โ1โ
1
๐ฅ. ๐ฅ+1 โ1
โ1๐๐ฅ
=โ log ๐ฅ
๐ฅ+1+
1
๐ฅ ๐ฅ+1 ๐๐ฅ
=โ log ๐ฅ
๐ฅ+1+
๐๐ฅ
๐ฅ2+๐ฅ
=โ log ๐ฅ
๐ฅ+1+
๐๐ฅ
๐ฅ2+๐ฅ+ 1
2
2+
1
2
2
=โ log ๐ฅ
๐ฅ+1+
1
2.1
2
log ๐ฅ +
1
2 โ
1
2
๐ฅ + 1
2 +
1
2
+ C
โต ๐ ๐
๐๐+๐๐ =๐
๐๐๐ฅ๐จ๐
๐โ๐
๐+๐ + ๐
=โ๐ฅ๐จ๐ ๐
๐+๐+ ๐ฅ๐จ๐
๐
๐+๐ + ๐
Q. 17. If ๐ = ๐ + ๐ + ๐ , ๐ = ๐๐ + ๐ and ๐ = ๐๐ โ ๐๐ โ ๐๐ , then find a unit vector
perpendicular to both of the vectors ๐ โ ๐ and ๐ โ ๐ .
Sol.
๐ โ ๐ = ๐ + 2๐ + 1๐ โ 2๐ + ๐
= โ๐ + ๐ + ๐
๐ โ ๐ = 3๐ โ 4๐ โ 5๐ โ 2๐ + ๐
= โ๐ โ 5๐ โ 5๐
Let D = ๐ โ ๐ ร ๐ โ ๐
= ๐ ๐ ๐
โ1 1 1 1 โ5 โ5
= ๐ โ5 + 5 โ ๐ 5 โ 1 + ๐ 5 โ 1
= โ4๐ + 4๐
D = 16 + 16 = 2 ร 16 = 4 2
Required vector = ยฑD
D
= ยฑ โ4๐ +4๐
4 2
= ยฑ โ4๐
4 2+
4๐
4 2
= ยฑ โ๐
๐๐ +
๐
๐๐
Q. 18. Find the equation of a line passing though the point ๐, ๐, โ๐ and perpendicular to
two lines ๐ = ๐๐ โ ๐๐๐ + ๐๐๐ + ๐ ๐๐ โ ๐๐๐ + ๐๐ and ๐ = ๐๐๐ + ๐๐๐ +
๐๐ + ๐ ๐๐ + ๐๐ โ ๐๐
Sol. The required line is perpendicular to the lines which are parallel to the vectors
๐1 = 3๐ โ 16๐ + 7๐ and ๐2
= 3๐ + 8๐ โ 5๐ respectively
So, it is parallel to the vector ๐ = ๐1 ร ๐2
๐ = ๐ ๐ ๐
3 โ16 73 5 โ5
๐ = ๐ 80 โ 56 โ ๐ โ15 โ 21 + ๐ 24 + 48
= 24๐ + 36๐ + 72๐ Thus, the required line passes thought the (1,2, โ4) and is parallel to the vector
๐ = 24๐ + 36๐ + 72๐ and ๐ = ๐ + 2๐ โ 4๐
Vector equation of this line is ๐ = ๐ + ๐๐
๐ = ๐ + 2๐ โ 4๐ + ๐ 24๐ + 36๐ + 72๐
or ๐ = ๐ + ๐๐ โ ๐๐ + ๐ ๐๐ + ๐๐ + ๐๐ โฆ ๐ฐ๐ก๐๐ซ๐ ๐ = ๐๐ ๐
Or
Find the equation of a line passing though the points โ ๐, ๐, ๐ , ๐, ๐, โ๐ and
parallel to the line ๐โ๐
๐=
๐๐+๐
๐=
๐+๐
โ๐.
Sol. Let ๐, ๐, ๐ be direction of ratios of the normal to the plane.
Equations of plane through point โ 1,2,0 is
๐ ๐ฅ + 1 + ๐ ๐ฆ โ 2 + ๐ ๐ง โ 0 = 0 โฆ (๐) Point 2,2, โ1 lies on (๐) ๐ 2 + 1 + ๐ 2 โ 2 + ๐ โ1 โ 0 = 0 3๐ + 0. ๐ โ ๐ = 0 โฆ (๐๐)
Given line : ๐ฅ+1
1=
2 ๐ฆ + 1
2
2=
๐ง+1
โ1
๐ฅโ1
1=
๐ฆ + 1
2
1=
๐ง+1
โ1
Direction ratios of given line are 1,1, โ1 Required plane is parallel to the given line ๐ + ๐ + ๐ = 0 โฆ (๐๐๐) 3๐ + 0. ๐ โ ๐ = 0 ๐ + ๐ โ ๐ = 0
๐
0+1=
โ๐
โ3+1=
๐
3โ0
๐
1=
๐
2=
๐
3= ๐(let)
๐ = ๐, ๐ = 2๐, ๐ = 3๐ Putting the value of ๐, ๐, ๐ in (๐), we have 1 ๐ฅ + 1 + 2 ๐ฆ โ 2 + 3๐ง = 0 ๐ฅ + 1 + 2๐ฆ โ 4 + 3๐ง = 0 โด ๐ + ๐๐ + ๐๐ โ ๐ = ๐ is the equation of the required plane.
Q. 19. 'Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence find the mean of the distribution. Sol. The Number of spades is a random variable Let it be denoted by X. X can take values 0,1,2 or 3
Let ๐ = P(a spade card) =13
52=
1
4
โด ๐ = 1 โ1
4=
3
4, Using P r = nCr
. ๐๐โ๐ . ๐๐
P X = 0 = 3C0. ๐3โ0. ๐0 = ๐3 =
3
4
3=
27
64
P X = 1 = 3C1. ๐3โ1๐1 = 3๐2๐ = 3
3
4
2
1
4 =
27
64
P X = 2 = 3C2. ๐3โ2. ๐2 = 32๐๐2 = 3
3
4
1
4
2=
9
64
P X = 3 = 3C3. ๐3โ3. ๐3 = ๐3 =
1
4
3=
1
64
๐๐ 0 1 2 3 ๐๐ 27
64
27
64
9
64
1
64
๐๐ ๐๐ ๐๐๐๐
0 27
64 0
1 27
64
27
64
2 9
64
18
64
3 1
64
3
64
P๐X๐ =
48
64
Mean= P๐X๐ =๐๐๐
๐๐๐=
๐
๐
Or For 6 trials of an experiment, let X be a binomial variants which satisfies the relation ๐๐ ๐ = ๐ = ๐ ๐ = ๐ Find the probability of success. Sol. Let ๐ be the probability of success. ๐ = 6 P ๐ = nCr
. ๐๐โ๐๐๐
9P X = 4 = P X = 2
9 6C4. ๐6โ4๐4 = 6C4
. ๐6โ2๐2
9. 6C4. ๐2๐4 = 6C4
. ๐4๐2 โต 6C4= 6C2
9๐2 = ๐2 Taking square-root, we get ๐ = 3๐ As 0 โค ๐ โค 1 0 โค ๐ โค 1 Now ๐ + ๐ = 1 ๐ + 3๐ = 1 4๐ = 1
๐ =1
4
โด ๐๐ก๐ ๐ฉ๐ซ๐จ๐๐๐๐ข๐ฅ๐ข๐ญ๐ฒ ๐จ๐ ๐ฌ๐ฎ๐๐๐๐ฌ๐ฌ, ๐ =๐
๐
Section โ C Question numbers 20 to 26 carry 6 marks each.
Q. 20. Consider ๐: ๐+ โ [โ๐, โ] given by ๐(๐) = ๐๐๐ + ๐๐ โ ๐. Prove that ๐ is invertible
with ๐โ๐(๐) = ๐๐+๐๐โ๐
๐ .
Sol. ๐(๐ฅ) = 5๐ฅ2 + 6๐ฅ โ 9 (given) Let ๐ ๐ฅ = ๐ฆ โ ๐ฆ = 5๐ฅ2 + 6๐ฅ โ 9 5๐ฆ = 25๐ฅ2 + 30๐ฅ โ 45 (Multiplying both sides by 5) 5๐ฆ = 25๐ฅ2 + 30๐ฅ + 3 2 โ 45 โ 3 2 5๐ฆ = 5๐ฅ + 3 2 โ 54 โฆ (๐) 5๐ฆ + 54 = 5๐ฅ + 3 2
5๐ฆ + 54 = 5๐ฅ + 3 (Square root on both sides)
5๐ฆ + 54 โ 3 = 5๐ฅ
๐ฅ = 5๐ฆ+54โ3
5
g ๐ฆ = 5๐ฆ+54โ3
5 โฆ (๐๐)
โ g๐๐ ๐ฅ = g ๐ ๐ฅ
= g 5๐ฅ2 + 6๐ฅ โ 9
= 5 5๐ฅ2+6๐ฅโ9 +54โ3
5
= 25๐ฅ2+30๐ฅโ45+54โ3
5
= 25๐ฅ2+30๐ฅ+9โ3
5
g๐๐ ๐ฅ = 5๐ฅ+3 2โ3
5=
5๐ฅ+3โ3
5= ๐ฅ โฆ (๐๐๐)
and ๐๐g ๐ฆ = ๐ g ๐ฆ
= ๐ 5๐ฆ+54โ3
5 [From (๐๐)
5
5๐ฆ+54โ3
5 +3
2
โ54
5
๐ ๐ซ๐จ๐ฆ ๐
๐๐ = ๐๐ + ๐ ๐ โ ๐๐
๐ = ๐๐+๐ ๐โ๐๐
๐
๐ ๐ = ๐๐+๐ ๐โ๐๐
๐
= 5๐ฆ+54
2โ54
5=
5๐ฆ+54โ54
5=
5๐ฆ
5= ๐ฆ
๐๐g ๐ฆ = ๐ฆ โฆ (๐๐ฃ) From (๐๐๐) and (๐๐ฃ), ๐ is invertible and ๐โ1 = g
๐โ1 ๐ฆ = 54+5๐ฆโ3
5 [From (๐๐)
Or A binary operation * is defined on the set ๐ = ๐ โ {โ๐} by ๐ โ ๐ = ๐ + ๐ + ๐๐, โ ๐, ๐ โ ๐. Check whether ยป is commutative and associative. Find the identity element and also find the inverse of each element of X. Sol. (i) ๐ฅ โ ๐ฆ = ๐ฅ + ๐ฆ + ๐ฅ๐ฆ, , โ ๐ฅ, ๐ฆ โ X
๐ฆ โ ๐ฅ = ๐ฆ + ๐ฅ + ๐ฆ๐ฅ = ๐ฅ + ๐ฆ + ๐ฅ๐ฆ = ๐ฅ โ ๐ฆ Hence โ*โis associative
(ii) ๐ฅ โ ๐ฆ โ ๐ง = ๐ฅ โ ๐ฆ + ๐ง + ๐ฆ๐ง
= ๐ฅ + ๐ฆ + ๐ง + ๐ฆ๐ง + ๐ฅ ๐ฆ + ๐ง + ๐ฆ๐ง
= ๐ฅ + ๐ฆ + ๐ง + ๐ฆ๐ง + ๐ฅ๐ฆ + ๐ง๐ฅ + ๐ฅ๐ฆ๐ง โฆ (๐)
๐ฅ โ ๐ฆ โ ๐ง = ๐ฅ + ๐ฆ + ๐ฅ๐ฆ โ ๐ง
= ๐ฅ + ๐ฆ + ๐ฅ๐ฆ + ๐ง + ๐ฅ + ๐ฆ + ๐ฅ๐ฆ ๐ง
= ๐ฅ + ๐ฆ + ๐ฅ๐ฆ + ๐ง + ๐ฅ๐ง + ๐ฆ๐ง + ๐ฅ๐ฆ๐ง โฆ (๐๐)
From (๐) and ๐๐ ๐ฅ โ ๐ฆ โ ๐ง = ๐ฅ โ ๐ฆ โ ๐ง
Hence โ*โis associative
Let โฒ๐โฒ be the identity element of X
Then ๐ฅ โ ๐ = ๐ฅ + ๐ + ๐ฅ๐ = ๐ฅ โ ๐ฅ โ X
๐ฅ + ๐ + ๐ฅ๐ = ๐ฅ
โ ๐ 1 + ๐ฅ = 0
โ ๐ = 0 or 1 + ๐ฅ = 0
๐ = 0 or ๐ฅ = โ1, Which is not possible because ๐ฅ โ R โ โ1 (given)
โด ๐ = 0
Clearly X โ 0 = ๐ฅ + 0 + 0 = ๐ฅ = 0 + ๐ฅ, โ ๐ฅ โ X
โด โ0โ is the identity element of X โฆ (๐)
Let ๐ฆ โ ๐ฅ be the inverse of ๐ฅ โ X
Then ๐ฅ โ ๐ฆ = ๐ฆ โ ๐ฅ = ๐
๐ฅ + ๐ฆ + ๐ฅ๐ฆ = 0
๐ฅ + ๐ฆ 1 + ๐ฅ = 0
๐ฆ 1 + ๐ฅ = โ๐ฅ
โ ๐ฆ =โ๐ฅ
1+๐ฅ or ๐ฅโ1 =
โ๐ฅ
1+๐ฅ
Q. 21. Find the value of ๐ for when the curves ๐๐ = ๐๐(๐ โ ๐) and ๐๐ = ๐(๐ + ๐) cut each other at right angles. Sol. ๐ฅ2 = 9๐ 9 โ ๐ฆ . . (๐) ๐ฅ2 = ๐ ๐ฆ + 1 โฆ (๐๐) From (๐) and ๐๐ , 9๐ 9 โ ๐ฆ = ๐ ๐ฆ + 1 โ 81 โ 9๐ฆ = ๐ฆ + 1 โ 81 โ 1 = ๐ฆ + 9๐ฆ โ 80 โ 10๐ฆ โ ๐ฆ = 8 Putting the value of ๐ฆ in (๐), we have ๐ฅ2 = 9๐ 9 โ 8
๐ฅ2 = 9๐ โ ๐ฅ = ยฑ3 ๐
โด Point of intersection A ยฑ3 p, 8
๐ฅ2 = 9๐ 9 โ ๐ฆ ๐ฅ2 = ๐ ๐ฆ + 1 Differentiating w.r.t. ๐ฅ, we have Differentiating w.r.t. ๐ฅ, we have
2๐ฅ = 9๐ โ๐๐ฆ
๐๐ฅ 2๐ฅ = ๐
๐๐ฆ
๐๐ฅ
๐1 =๐๐ฆ
๐๐ฅ=
โ2๐ฅ
9๐ ๐2 =
๐๐ฆ
๐๐ฅ=
2๐ฅ
๐
If the curve cut at right-angles, then ๐1 ร ๐2 = โ1
=โ2๐ฅ
9๐ร
2๐ฅ
๐= โ1
โ โ9๐2 = โ4๐ฅ2
โ 9๐2 = 4 9๐ Put ๐ฅ = ยฑ p
โ 9๐2 โ 36๐ = 0 โ 9๐ ๐ โ 4 = 0 โ 9๐ = 0 or ๐ โ 4 = 0 โ ๐ = ๐ or ๐ = ๐
Q. 22. Using integration, prove that the curves ๐๐ = ๐๐ and ๐๐ = ๐๐ divide the area of the square bounded by ๐ = ๐, ๐ = ๐, ๐ = ๐ and ๐ = ๐ into three equal parts. Sol.
๐ฆ2 = 4๐ฅ โ ๐ฆ = 2 ๐ฅ โฆ (๐)
๐ฅ2 = 4๐ฆ โ ๐ฆ =๐ฅ2
4 โฆ (๐๐)
For point of intersection ๐ฅ2 = 4๐ฆ
๐ฆ2
4
2
= 4๐ฆ from ๐ ๐ฆ2 = 4๐ฅ or ๐ฆ2
4= ๐ฅ
๐ฆ4 = 64๐ฆ ๐ฆ4 โ 64๐ฆ = 0 โ ๐ฆ ๐ฆ3 โ 64 = 0 โด ๐ฆ = 0 or ๐ฆ3 โ 64 โ ๐ฆ = 4
Shaded area I = ๐ฅ ๐๐ฆ4
0
= ๐ฆ2
4๐๐ฆ
4
0
=1
4.
1
3 ๐ฆ3 0
4
=1
12 43 โ 03
=64
12=
๐๐
๐ sq. units โฆ (๐)
Shaded area II = 2 ๐ฅ ๐๐ฅ โ ๐ฅ2
4 ๐๐ฅ
4
0
4
0
= 2.2
3 ๐ฅ3/2
0
4โ
1
4.
1
3 ๐ฅ3 0
4
=4
3 43/2 โ 0
0
4โ
1
12 43 โ 0
=4
3ร 2 ร
3
2โ
1
12 64
=32
3โ
16
3=
๐๐
๐ sq. units โฆ (๐๐)
Shaded area III = ๐ฅ2
4๐๐ฅ
4
0
=1
4.
1
3 ๐ฅ3 0
4
=1
12 43 โ 0
=64
12=
๐๐
๐ sq. units โฆ (๐๐๐)
From (๐),(๐๐) and ๐๐๐ ,
Area bounded by curves ๐๐ = ๐๐ and ๐๐ = ๐๐ divides the area of square in three equal parts.
Q. 23. Show that the differential equation ๐ ๐
๐ ๐=
๐๐
๐๐โ๐๐ is homogeneous and also solve it.
Sol.
๐๐ฆ
๐๐ฅ=
๐ฆ2
๐ฅ๐ฆโ๐ฅ2 โฆ (๐)
Dividing numerator and denominator by ๐ฅ2,
๐๐ฆ
๐๐ฅ=
๐ฆ
๐ฅ
2
๐ฆ
๐ฅโ1
= g ๐ฆ
๐ฅ โฆ (๐๐)
R.H.S. of differential equation (๐๐) is of the from g ๐ฆ
๐ฅ and so it is a homogeneous.
Therefore, Equation (๐) is a homogeneous differential equation
๐๐๐ญ ๐ = ๐๐ โ
๐
๐= ๐
๐ ๐
๐ ๐= ๐. ๐ + ๐.
๐ ๐
๐ ๐
๐ฃ + ๐ฅ๐๐ฆ
๐๐ฅ=
๐ฃ2
๐ฃโ1
๐ฅ๐๐ฆ
๐๐ฅ=
๐ฃ2
๐ฃโ1โ ๐ฃ
๐ฅ๐๐ฆ
๐๐ฅ=
๐ฃ2โ๐ฃ2+๐ฃ
๐ฃโ1
๐ฃโ1
๐ฃ๐๐ฃ =
๐๐ฆ
๐ฅ โฆ (๐๐๐)
Integrating both sides of equation (๐๐๐), we get
๐ฃ1
๐ฃโ
1
๐ฃ ๐๐ฃ =
๐๐ฅ
๐ฅ
๐ฃ โ log ๐ฃ = log ๐ฅ + C
๐ฆ
๐ฅโ log
๐ฆ
๐ฅ โ log ๐ฅ = C
๐ฆ
๐ฅโ log
๐ฆ
๐ฅ. ๐ฅ = C โต ๐ฅ๐จ๐ ๐ + ๐ฅ๐จ๐ ๐ = ๐ฅ๐จ๐ ๐๐
๐ฆ
๐ฅโ log ๐ฆ = C
๐ฆโlog ๐ฆ
๐ฅ= C
๐ โ ๐ ๐ฅ๐จ๐ ๐ = ๐๐ which is the general solution of the given differential solution. Or
Find the particular solution of the differential equation (๐ญ๐๐งโ๐ ๐ โ ๐)๐ ๐ = (๐ +
๐๐) ๐ ๐, given that ๐ = ๐ when ๐ = ๐ถ. Sol. tanโ1 ๐ฆ โ ๐ฅ ๐๐ฆ = 1 + ๐ฆ2 ๐๐ฅ
tan โ1 ๐ฆโ๐ฅ
1+๐ฆ2 =๐๐ฅ
๐๐ฆ
๐๐ฅ
๐๐ฆ=
tan โ1 ๐ฆโ๐ฅ
1+๐ฆ2 โ๐ฅ
1+๐ฆ2
๐๐ฅ
๐๐ฆ+
๐ฅ
1+๐ฆ2 =tan โ1 ๐ฆ
1+๐ฆ2
Comparing with ๐๐ฆ
๐๐ฅ+ P๐ฅ = Q
โด P =1
1+๐ฆ2 , Q =tan โ1 ๐ฆ
1+๐ฆ2
I. F. = ๐ P ๐๐ฆ = ๐
tan โ1 ๐ฆ
1+๐ฆ2 = ๐ tan โ1 ๐ฆ Hence the solution is ๐ฅ I. F. = Q. I. F. ๐๐ฆ
๐ฅ. ๐ P ๐๐ฆ = tan โ1 ๐ฆ
1+๐ฆ2 . ๐ tan โ1 ๐ฆ๐๐ฆ
= ๐ก. ๐๐ก๐๐กI II
๐๐๐ญ ๐ = ๐ญ๐๐งโ๐ ๐
๐ ๐ =๐
๐+๐๐ ๐ ๐
๐ฅ. ๐ tan โ1 ๐ฆ = ๐ก. ๐๐ก โ 1. ๐๐ก๐๐ก = ๐ก . ๐๐ก โ ๐๐ก + C = ๐๐ก ๐ก โ 1 + C
๐ฅ. ๐ tan โ1 ๐ฆ = ๐ tan โ1 ๐ฆ tanโ1 ๐ฆ โ 1 + C
๐ฅ = tanโ1 ๐ฆ โ 1 +C
๐ tan โ1 ๐ฆ โฆ (๐) โต ๐ฅ = 1, ๐ฆ = 0
1 = tanโ1 0 โ 1 +C
๐ tan โ1 ๐ฆ
2 = 0 +C
1 โต ๐0 = 1
C = 2 Putting the value of C in (๐), we have
๐ฅ = tanโ1 ๐ฆ โ 1 +2
๐ tan โ1 ๐ฆ
๐ = ๐ญ๐๐งโ๐ ๐ โ ๐ + ๐. ๐๐ญ๐๐งโ๐ ๐ which is the particular solution of given differential equation.
Q. 24. Find the distance of the point ๐(๐, ๐, ๐) from the point, where the line joining the points ๐(๐, โ๐, โ๐) and ๐(๐, โ๐, ๐) intersects the plane ๐๐ + ๐ + ๐ = ๐. Sol. Direction ratios of line AB are ๐ฅ2 โ ๐ฅ1, ๐ฆ2 โ ๐ฆ1, ๐ง2 โ ๐ง1 2 โ 3, โ3 + 4, 1 + 5 โ1, 1 6 Equation of line AB is
๐ฅโ๐ฅ1
๐=
๐ฆโ๐ฆ1
๐=
๐งโ๐ง1
๐
๐ฅโ3
โ1=
๐ฆ+4
1=
๐ง+5
6= ๐ let (Using Pt. A)
Let point C โ๐ + 3, ๐ โ 4, 6๐ โ 5 โฆ (๐) be the point of intersection between line AB and given plane So point C lies on the given plane 2๐ฅ + ๐ฆ + ๐ง = 7 2 โ๐ + 3 + ๐ โ 4 + 6๐ โ 5 โฆ (๐) โ2๐ + 6 + ๐ โ 4 + 6๐ โ 5 = 7
5๐ โ 3 = 7 โ 5๐ + 10 โ ๐ =10
5= 2
Putting the value of ๐ in (๐), Coordinates of point C = โ2 + 3,2 โ 4,12 โ 5 or 1, โ2,7 Point P is 3,4,4, Required distance = CP
= 3 โ 1 2 + 4 + 2 2 + 4 โ 7 2
= 4 + 36 + 9 = 49 = ๐ units
Q. 25. A company manufacturers three kinds of calculators : A, B and C in its two factories ๐ and ๐๐. The company has got an order for manufacturing at least 6400 calculators of kind A, 4000 of kind B and 4800 of kind C. The daily output of factory ๐ is of 50 calculators of kind A, 50 calculators of kind B, and 30 calculators of kind C. The daily output of factory ๐๐ is of 40 calculators of kind A, 20 of kind B and 40 of kind C. The cost per day to run factory ๐ is Rs. 12,000 and of factory ๐๐ is Rs. 5,000. How many days do the two factories have to be in operation to produce the order with the minimum cost? Formulate this problem as an LPP and solve it graphically. Sol.
๐๐ ๐๐ ๐๐๐ฆ๐
๐ถ๐๐๐๐ข๐ก๐๐ก๐๐๐ ๐ถ๐๐ ๐ก A B C
Fecories I Fecories II
๐ฅ ๐ฆ
50 40
50 20
30 40
Rs. 12,000 Rs. 15,000
6400 6400 4800
At least At least At least
Zmin = 12000๐ฅ + 15000๐ฆ Subjects to the constraints : 50๐ฅ + 40๐ฆ โฅ 6400 50๐ฅ + 20๐ฆ โฅ 4000
30๐ฅ + 40๐ฆ โฅ 4800 ๐ฅ, ๐ฆ โฅ 0 50๐ฅ + 40๐ฆ โฅ 6400 50๐ฅ + 20๐ฆ โฅ 4000 30๐ฅ + 40๐ฆ โฅ 4800 Let 50๐ฅ + 40๐ฆ โฅ 6400 Let 50๐ฅ + 20๐ฆ โฅ 4000 Let 30๐ฅ + 40๐ฆ โฅ 4800 or 5๐ฅ + 4๐ฆ = 640 or 5๐ฅ + 2๐ฆ = 400 or 3๐ฅ + 4๐ฆ = 480 โฆ (๐) โฆ (๐๐) โฆ (๐๐๐)
๐ฅ 0 128 ๐ฅ 0 80 ๐ฅ 0 160
๐ฆ 160 0 ๐ฆ 200 0 ๐ฆ 120 0
Om solving (๐) and ๐๐ 5๐ฅ + 4๐ฆ = 640 โฆ (๐) _ 5๐ฅ ยฑ 2๐ฆ = 400 โฆ (๐๐) 2๐ฆ = 240 โ ๐ฆ = 120
Putting the value of ๐ฆ in (๐๐), 5๐ฅ + 240 = 400; 5๐ฅ = 160 โ ๐ฅ = 32 โด Point B 32,120 Om solving (๐) and ๐๐ 5๐ฅ + 4๐ฆ = 640 โฆ (๐) _ 3๐ฅ ยฑ 4๐ฆ = 480 โฆ (๐๐๐) 2๐ฅ = 160 โ ๐ฅ = 80 Putting the value of ๐ฅ in (๐), 5 80 + 4๐ฆ = 640 4๐ฆ = 640 โ 400
๐ฆ =240
4= 60
โด Point C 80,60
Corner Points
Z = 12,000๐ฅ + 15,000๐ฆ
A 0,200 15000 200 = 30,00,000 B 32,120 12000 32 + 15000 120 = 21,84,000 C 80,60 12000 80 + 15000 60 = 10,60,000 โ Smallest D 160,0 12000 160 = 19,20,000
From, the table, we find 18,60,000 is the smallest value of Z the corner point C 80,60 Here is the feasible region is unbounded. Therefore, 18,60,000 may or may not be the minimum value of Z to decide this issue, graph the inequality 12000๐ฅ + 15000๐ฆ < 1860000 Let 12000๐ฅ + 15000๐ฆ = 1860000 or 12๐ฅ + 15๐ฆ = 1860 or 4๐ฅ + 5๐ฆ = 620 It has not common points So, Rs. 18,60,000 will be the C 80,60 Factory ๐ = ๐๐ Days Factory ๐๐ = ๐๐ Days Minimum cost = ๐๐ฌ. ๐๐, ๐๐, ๐๐๐
Q. 26. In a factory which manufacturers bolts, machine A, B and C manufacture respectively 30%, 50% and 20% of the bolts. Of their outputs 3, 4 and 1 percent respectively are defective bolts. A bolt is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B. Sol. Let E1 : the Bolt is manufactured by machine A Let E2 : the Bolt is manufactured by machine B Let E3 : the Bolt is manufactured by machine C
โด P E1 =30
100, P E2 =
50
100, P E3 =
20
100
=3
10 =
5
10 =
2
10
Let the event A be the bolt is defective
P A/E1 =3
100, P A/E2 =
4
100 , P A/E3 =
1
100
P (bolt is not manufactured by machine B) = P (bolt is manufactured by machine A or C) = P(E2
/A)