QUESTION AND ANSWERS March – 2017
“amma”
II PUC
“Sree Guru”
MATHEMATICS (March – 2017)
Subject code: 35 (NS)
Time: 3 hours 15 minute
Max. Marks: 100
Instructions:
i. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
ii. Use the graph sheet for the question on linear programming in PART – E.
PART – A
Answer ALL the questions
10  1=10
1. Find the number of binary operation on the set {𝑎, 𝑏}.
Solution: A binary operation ∗ on the set {𝑎, 𝑏} is a function from
{𝑎, 𝑏} × {𝑎, 𝑏} → {𝑎, 𝑏} ⟹ {(𝑎, 𝑎), (𝑎, 𝑏), (𝑏, 𝑎), (𝑏, 𝑏)} → {𝑎, 𝑏}. That is 24 = 16.
−1
2. Find the principal value of cot −1 ( ).
Solution: Let
−1
cot −1 ( )
√3
√3
= 𝑦. Then cot 𝑦 =
−1
−1
√3
2𝜋
√3
3
Therefore principal value of cot −1 ( ) 𝑖𝑠
𝜋
𝜋
2𝜋
= − cot (3 ) = cot (𝜋 − 3 ) = cot ( 3 ) .
.
3. Define a Identity matrix.
Solution: A square matrix in which elements in the diagonal are all 1 and rest are
all zero is called an identity matrix. “OR” Let 𝐴 = [𝑎𝑖𝑗 ] 𝑛 × 𝑛 is an identity matrix, if
1 , 𝑖𝑓 𝑖 = 𝑗
𝑎𝑖𝑗 = {
.
0 , 𝑖𝑓 𝑖 ≠ 𝑗
cos 𝜃 − sin 𝜃
4. Show that |
| = 1.
sin 𝜃 cos 𝜃
cos 𝜃 − sin 𝜃
Solution: Let |
| = cos2 𝜃 + sin2 𝜃 = 1.
sin 𝜃 cos 𝜃
5. Find the derivative of 2sin 𝑥 .
𝒅𝒚
Solution: 𝒅𝒙 = 2sin 𝑥 × log 2 × cos 𝑥.
6. Evaluate ∫(1 − 𝑥)√𝑥 𝑑𝑥.
1
Solution: 𝐼 =
𝑥2
3
+1
1
+1
2
−
𝑥2
+1
3
+1
2
+𝐶
7. Find the values of x, y and z so that the vectors 𝑎⃗ = 𝑥𝑖̇̂ + 2𝑗̇̂ + 𝑧𝑘̂ 𝑎𝑛𝑑 𝑏⃗⃗ = 2𝑖̇̂ + 𝑦𝑗̇̂ + 𝑘̂
are equal.
Solution: Since 𝑎⃗ = 𝑥𝑖̇̂ + 2𝑗̇̂ + 𝑧𝑘̂ 𝑎𝑛𝑑 𝑏⃗⃗ = 2𝑖̇̂ + 𝑦𝑗̇̂ + 𝑘̂ are equal, their corresponding
components are equal ∴ 𝑥 = 2 , 𝑦 = 2, 𝑧 = 1.
8. Find the distance between the two planes 2𝑥 + 3𝑦 + 4𝑧 = 4 𝑎𝑛𝑑 4𝑥 + 6𝑦 + 8𝑧 = 12.
Solution: Given planes 2𝑥 + 3𝑦 + 4𝑧 = 4 𝑎𝑛𝑑 2𝑥 + 3𝑦 + 4𝑧 = 6 are parallel
𝑑 −𝑑
6−4
2
Distance, 𝑑 = |√𝑎2 2 2 1 2 | = |
|=
+𝑏 +𝑐
√4+9+16
√29
9. Define feasible solution in L.P.P.
Solution: A set of values of the variables satisfying all the constraints is known as
a feasible solution.
10. Given two independent events A and B such that 𝑃(𝐴) = 0.3 and 𝑃(𝐵) = 0.6. Find
𝑃(𝐴 𝑎𝑛𝑑 𝐵)
Solution: 𝑃(𝐴 𝑎𝑛𝑑 𝐵) = 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) × 𝑃(𝐵) = 0.3 × 0.6 = 0.18.
PART B
Answer any TEN questions:
10  2=20
1
11. Find 𝑔𝑜𝑓 and 𝑓𝑜𝑔, if 𝑓(𝑥) = 8 𝑥 3 𝑎𝑛𝑑 𝑔(𝑥) = 𝑥 3 .
1
Solution: Let (𝑔𝑜𝑓)(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(8 𝑥 3 ) = (8 𝑥 3 )3 = 2𝑥
1
1
3
and (𝑓𝑜𝑔)(𝑥) = 𝑓(𝑔(𝑥)) = 𝑓 (𝑥 3 ) = 8 (𝑥 3 ) = 8𝑥
12. Prove that tan−1 𝑥 + cot −1 𝑥 =
𝜋
2
, 𝑥∈𝑅
𝜋
Proof: Let tan−1 𝑥 = 𝑦. Then 𝑥 = tan 𝑦 = cot ( 2 − 𝑦)
𝜋
𝜋
There fore cot −1 𝑥 = 2 − 𝑦 = 2 − tan−1 𝑥
Hence tan−1 𝑥 + cot −1 𝑥 =
𝜋
2
.
13. Find the value of cos −1 (𝑐𝑜𝑠
Solution: Let cos−1 (𝑐𝑜𝑠
7𝜋
6
7𝜋
6
).
𝜋
𝜋
) = cos −1 [𝑐𝑜𝑠 (𝜋 + 6 )] = cos −1 (−𝑐𝑜𝑠 6 )
S U J M P U College, Harapanahalli – 583131.
Page 1
QUESTION AND ANSWERS March – 2017
“amma”
𝜋
= 𝜋 − cos −1 (𝑐𝑜𝑠 6 )
𝜋
6𝜋−𝜋
II PUC
[∵ cos−1 ( −𝑥) = 𝜋 − cos −1 (𝑥)]
5𝜋
=𝜋−6= 6 = 6 .
14. Find equation of line joining (3, 1) and (9, 3) using determinants.
Solution: Let 𝑃(𝑥, 𝑦) be any point on the line joining 𝐴(3, 1) and 𝐵(9, 3).
Therefore points A, P and B are collinear. Area of the triangle 𝐴𝑃𝐵 = 0
𝑥 𝑦 1
1
⟹ ∆= 2 |3 1 1| = 0 ⟹ 𝑥[1 − 3] − 𝑦[3 − 9] + 1[9 − 9] = 0.
9 3 1
−2𝑥 + 6𝑦 = 0 𝑜𝑟 3𝑦 = 𝑥. This is the required equation of the line AB.
dy
15. Find
, if y  cos(log x  e x ), x  0
dx
dy
d
1

Solution:
  sin(log x  e x ) (log x  e x )   sin(log x  e x )   e x  .
dx
dx
x

16. Prove that the function f given by 𝑓(𝑥) = |𝑥 − 1|, 𝑥 ∈ 𝑅 is not differentiable at 𝑥 = 1.
Solution:
〈∵ 𝑦 = −(𝑥 − 1) = 1 − 𝑥, ∀ 𝑥 < 0〉
Now 𝐿𝑓 ′ (1) = 0 − 1 = −1
〈∵ 𝑦 = 𝑥 − 1, ∀ 𝑥 ≥ 0〉
𝑅𝑓 ′ (1) = 1 − 0 = 1
Hence 𝐿𝑓 ′ (1) ≠ 𝑅𝑓 ′ (1). Thus 𝑓(𝑥) = |𝑥 − 1|, 𝑥 ∈ 𝑅 is not differentiable at 𝑥 = 1.
17. Find the maximum and minimum values of the function 𝑔(𝑥) = 𝑥 3 + 1.
Solution: Given 𝑔(𝑥) = 𝑥 3 + 1, then 𝑔′(𝑥) = 3𝑥 2 and 𝑔′′(𝑥) = 6𝑥.
For maximum or minimum 𝑔′(𝑥) = 0, we get 𝑥 = 0.
At 𝑥 = 0, 𝑔′′(𝑥) = 6 × 0 = 0
Hence at 𝑥 = 0, 𝑔(𝑥) is neither maxima not minima. It is a point of inflexion.
18.
 x log x dx
Find:
d
 log x 

 log x  x dx  log x x dx    dx  log x   x dx  dx
Solution:
x2
1 x2
x 2 log x 1  x 2 
x 2 log x x 2


dx 
  C 
 C
2
x 2
2
2 2 
2
4

𝑥
19. ∫ 𝑒 𝑥 (1+𝑥)2 𝑑𝑥.
𝑥+1−1
𝑥+1
1
Solution: = ∫ 𝑒 𝑥 (1+𝑥)2 𝑑𝑥 = ∫ 𝑒 𝑥 [(1+𝑥)2 − (1+𝑥)2 ] 𝑑𝑥
1
1
1
= ∫ 𝑒 𝑥 [(1+𝑥) − (1+𝑥)2 ] 𝑑𝑥 = 𝑒 𝑥 (1+𝑥) + 𝐶.
20. Find the order and degree of the differential equation, 𝑦 ′′′ + 2𝑦 ′′ + 𝑦′ = 0.
Solution: Order is 3 and Degree is 1.
21. Find the angle between two vectors 𝑎⃗ 𝑎𝑛𝑑 𝑏⃗⃗ with magnitudes √3 and 2 respectively
having 𝑎⃗ ∙ 𝑏⃗⃗ = √6.
Solution: Given |𝑎⃗| = √3, |𝑏⃗⃗| = 2 and 𝑎⃗ ∙ 𝑏⃗⃗ = √6
⃗⃗
𝑎⃗⃗∙𝑏
⃗⃗
𝑎⃗⃗∙𝑏
1
√6
𝜋
We have 𝑐𝑜𝑠𝜃 = |𝑎⃗⃗||𝑏⃗⃗| ⟹ 𝜃 = cos −1 (|𝑎⃗⃗||𝑏⃗⃗|) = cos −1 (2√3) = cos −1 ( ) = 4 .
√2
22. Find 𝜆 and 𝜇, if (2𝑖̇̂ + 6𝑗̇̂ + 27𝑘̂) × (𝑖̇̂ + 𝜆𝑗̇̂ + 𝜇𝑘̂) = ⃗⃗
0
𝑖̇̂ 𝑗̇̂ 𝑘̂
Solution: Consider (2𝑖̇̂ + 6𝑗̇̂ + 27𝑘̂) × (𝑖̇̂ + 𝜆𝑗̇̂ + 𝜇𝑘̂) = |2 6 27| = ⃗0⃗
1 𝜆 𝜇
⟹ 𝑖̇̂(6𝜇 − 27𝜆) − 𝑗̇̂(2𝜇 − 27) + 𝑘̂(2𝜆 − 6) = 0𝑖̇̂ + 0𝑗̇̂ + 0𝑘̂ ,
On comparing the corresponding components, we have
27
2𝜆 − 6 = 0 ⟹ 2𝜆 = 6 ⟹ 𝜆 = 3
and 2𝜇 − 27 = 0 ⟹ 𝜇 = 2 .
23. Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and z-axis
respectively.
𝑥
𝑦
𝑧
Solution: Let the equation of the plane be, 𝑎 + 𝑏 + 𝑐 = 1
Here 𝑎 = 2, 𝑏 = 3 𝑎𝑛𝑑 𝑐 = 4
𝑥
𝑦
𝑧
Required equation of a plane as, 2 + 3 + 4 = 1 𝑜𝑟 6𝑥 + 4𝑦 + 3𝑧 = 12.
24. An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X
represent the number of black balls. What are the possible values of X? Is X a
random variable?
S U J M P U College, Harapanahalli – 583131.
Page 2
QUESTION AND ANSWERS March – 2017
“amma”
II PUC
Solution: Two balls are selected can be represented as BB, BR, RB, RR, where B
represents a black ball and R represents a red ball and X represents the number
of black balls.
𝑋(𝐵𝐵) = 2, 𝑋(𝐵𝑅) = 1, 𝑋(𝑅𝐵) = 1 𝑎𝑛𝑑 𝑋(𝑅𝑅) = 0
Therefore, X can take the values 0, 1 or 2.
Thus, X is a random variable
PART C
Answer any TEN questions:
10  3=30
25. Show that the relation R defined in the set A of all triangles as
𝑅 = {(𝑇1 , 𝑇2 ) ∶ 𝑇1 𝑖𝑠 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝑇2 }, is equivalence relation.
Solution: Let A be the set of all triangles in a plane.
The relation is 𝑅 = {(𝑇1 , 𝑇2 ) ∶ 𝑇1 𝑖𝑠 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝑇2 }.
Reflexive: Every triangle is similar to itself, so R is reflexive.
Symmetric: If the triangle 𝑇1 is similar to the triangle 𝑇2 , then clearly 𝑇2 is similar
to 𝑇1 . Therefore R is symmetric.
Transitive: Let 𝑇1 is similar to the triangle 𝑇2 and 𝑇2 is similar to the triangle𝑇3 .
Then the triangle 𝑇1 is similar to the triangle 𝑇3 . Therefore R is transitive.
Hence R is an equivalence relation.
26. Solve, 2tan−1 (𝑐𝑜𝑠𝑥) = tan−1 (2𝑐𝑜𝑠𝑒𝑐 𝑥).
Solution: Given 2tan−1 (𝑐𝑜𝑠𝑥) = tan−1 (2𝑐𝑜𝑠𝑒𝑐 𝑥)
2𝑐𝑜𝑠𝑥
1
⟹ tan−1 (1−𝑐𝑜𝑠2 𝑥) = tan−1 (2 × 𝑠𝑖𝑛 𝑥)
⟹
2𝑐𝑜𝑠𝑥
𝑠𝑖𝑛2 𝑥
𝜋
∴𝑥=
2
𝜋
= 𝑠𝑖𝑛 𝑥 ⟹ sin 𝑥 = 𝑐𝑜𝑠𝑥 ⟹ tan 𝑥 = 1 = tan 4
𝜋
𝑜𝑟 𝑥 = 𝑛𝜋 + 4 , 𝑛 ∈ 𝐼.
1 0
3 0
27. If 𝐴 = [
] and 𝐵 = [
], then show that 𝐴𝐵 = 𝐵𝐴.
0 2
0 4
3+0 0+0
1 0 3 0
3 0
Solution: 𝐴𝐵 = [
][
]=[
]=[
]…….(1)
0+0 0+8
0 2 0 4
0 8
3+0 0+0
3 0 1 0
3 0
𝐵𝐴 = [
][
]=[
]=[
]…….(2)
0+0 0+8
0 4 0 2
0 8
From (1) and (2), 𝐴𝐵 = 𝐵𝐴.
dy
28. Find
, if x  a  cos   sin   , y  a sin    cos  .
dx
dx
 a ( sin  )   cos   sin  1  a[ sin    cos   sin  ]  a cos 
Solution:
d
dy
 a (cos  )   ( sin  )  cos  1  a  cos    sin   cos    a sin 
d
dy a sin 

 tan  .
Hence
dx a cos 
4
29. Verify Rolle’s theorem for the function 𝑓(𝑥) = 𝑥 2 + 2𝑥 – 8, 𝑥 ∈ [– 4, 2].
Solution: The function 𝑓(𝑥) = 𝑥 2 + 2𝑥 – 8 is continuous in [−4, 2] and
differentiable in (−4, 2). Also 𝑓(−4) = 𝑓(2) = 0 and hence the value of 𝑓(𝑥) at -4
and 2 coincide. Rolle’s theorem states that there is a point 𝑐 ∈ (−4, 2), where
𝑓 ′ (𝑥) = 2𝑥 + 2 ⟹ 𝑓 ′ (𝑐) = 0 ⟹ 2𝑐 + 2 = 0 ⟹ 𝑐 = −1
Since, we get 𝑐 = −1. Thus at 𝑐 = −1
we have 𝑓 ′ (−1) = 0 𝑎𝑛𝑑 𝑐 = −1 ∈ (−4, 2).
30. Find the equation of the tangent and normal to the curve 𝑦 = 𝑥 2 at (0, 0)
𝑑𝑦
Solution: Let 𝑦 = 𝑥 2 , then 𝑑𝑥 = 2𝑥.
𝑑𝑦
Therefore slope of the tangent at (0, 0) is 𝑚 = (𝑑𝑥 )
(0,0)
=2×0 =0
Therefore equation of the tangent is, 𝑦 − 0 = 0(𝑥 − 0) ⟹ 𝑦 = 0
1
1
Slope of the normal is, − 𝑚 = − 0 is not defined
1
Therefore equation of the normal is, 𝑦 − 0 = − 0 (𝑥 − 0) ⟹ 𝑥 = 0
𝜋
2
31. Prove that ∫02 sin3 𝑥 𝑑𝑥 = 3.
𝜋
Solution: Let ∫02 (1 − cos 2 𝑥) sin 𝑥 𝑑𝑥 put 𝑐𝑜𝑠 𝑥 = 𝑡 ⟹ sin 𝑥 𝑑𝑥 = −𝑑𝑡
𝜋
When 𝑥 = 0 then 𝑡 = 1 and 𝑥 = 2 then 𝑡 = 0
S U J M P U College, Harapanahalli – 583131.
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QUESTION AND ANSWERS March – 2017
“amma”
0
𝑡3
0
1
II PUC
2
∫1 (1 − 𝑡 2 )(−𝑑𝑡) = [ 3 − 𝑡] = 0 − [3 − 1] = 3 .
1
32. Evaluate
Solution:
dx
  x  1 x  2
1
 x  1 x  2

A
B
Real numbers A and B are to be determined

 x  1  x  2
1  A  x  2  B  x  1 Equating the coefficient of x and the constant terms,
We get, A  B  0 and 2 A  B  1 solving these equations, we get,
A  1 and B  1
I 
 1
1
  x  1 dx    x  2  dx
 log x  1  log x  2  C  log
x 1
C
x2
33. Find the area under the curve 𝑦 = 𝑥 4 and given lines 𝑥 = 1, 𝑥 = 5 and X-axis.
Solution: Let 𝑦 = 𝑥 4 . Here, x is even, so curve is symmetrical about Y-axis and
passes through the origin
Required area, 𝐴 = 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑦 = 𝑥 4 , X − axis and 𝑥 = 1, 𝑥 = 5
𝑥5
5
5
1
1
𝐴 = ∫1 𝑥 4 𝑑𝑥 = [ 5 ] = 5 [55 − 15 ] = 5 [55 − 1]
1
34. Form the differential equation representing family of curves 𝑦 = asin(𝑥 + 𝑏), where
a and b arbitrary constants.
Solution: Given 𝑦 = 𝑎 sin(𝑥 + 𝑏) ……(1)
𝑑𝑦
differentiating we get, 𝑑𝑥 = 𝑎 cos(𝑥 + 𝑏)
Again differentiating we get,
𝑑2 𝑦
𝑑2 𝑦
𝑑2 𝑦
𝑑𝑥 2
= −𝑎 sin(𝑥 + 𝑏).
= −𝑦 ⟹ 𝑑𝑥 2 + 𝑦 = 0 ( using equation 1)
35. If 𝑎⃗, 𝑏⃗⃗, 𝑐⃗ are unit vectors such that 𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗ = ⃗0⃗ , find the value of 𝑎⃗ ∙ 𝑏⃗⃗ + 𝑏⃗⃗ ∙ 𝑐⃗ + 𝑐⃗ ∙ 𝑎⃗.
Solution: Given 𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗ = ⃗0⃗
2
2
⃗⃗|
⟹ |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = |0
𝑑𝑥 2
2
⟹ |𝑎⃗|2 + |𝑏⃗⃗| + |𝑐⃗|2 + 2( 𝑎⃗ ∙ 𝑏⃗⃗ + 𝑏⃗⃗ ∙ 𝑐⃗ + 𝑐⃗ ∙ 𝑎⃗) = 0 〈∵ |𝑎⃗| = |𝑏⃗⃗| = |𝑐⃗| = 1 〉
⟹ 2( 𝑎⃗ ∙ 𝑏⃗⃗ + 𝑏⃗⃗ ∙ 𝑐⃗ + 𝑐⃗ ∙ 𝑎⃗) = −3
3
∴ 𝑎⃗ ∙ 𝑏⃗⃗ + 𝑏⃗⃗ ∙ 𝑐⃗ + 𝑐⃗ ∙ 𝑎⃗ = − 2 .
36. Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) as
its vertices.
⃗⃗⃗⃗⃗⃗ = 𝑗̇̂ + 2𝑘̂ and 𝐴𝐶
⃗⃗⃗⃗⃗⃗ = 𝑖̇̂ + 2𝑗̇̂.
Solution: We have 𝐴𝐵
̂𝑖̇ 𝑗̇̂ 𝑘̂
⃗⃗⃗⃗⃗⃗ × 𝐴𝐶
⃗⃗⃗⃗⃗⃗ = |0 1 3| = −4𝑖̇̂ + 2𝑗̇̂ − 𝑘̂
Now 𝐴𝐵
1 2 0
1
1
⃗⃗⃗⃗⃗⃗ × ⃗⃗⃗⃗⃗⃗
Thus, the area of the given triangle is |𝐴𝐵
𝐴𝐶 | = √(−4)2 + (2)2 + (−1)2 = √21
𝑥+3
𝑦−1
𝑧−5
2
𝑥+1
𝑦−2
𝑧−5
2
37. Show that the lines −3 = 1 = 5 and −1 = 2 = 5 are coplanar.
Solution: Here 𝑥1 = −3, 𝑦1 = 1, 𝑧1 = 5 and 𝑎1 = −3, 𝑏1 = 1, 𝑐1 = 5
𝑥2 = −1, 𝑦2 = 2, 𝑧2 = 5 and 𝑎2 = −1, 𝑏2 = 2, 𝑐2 = 5
𝑥2 − 𝑥1 𝑦2 − 𝑦1 𝑧2 − 𝑧1
𝑏1
𝑐1 | = 0
Condition for coplanar, if | 𝑎1
𝑎2
𝑏2
𝑐2
2 1 0
𝐿𝐻𝑆 = |−3 1 5| = 2[5 − 10] − 1[−15 + 5] + 0 = −10 + 10 = 0
−1 2 5
Therefore, the lines are coplanar.
38. Of the students in a college, it is known that 60% reside in hostel and 40% are
day scholars (not residing in hostel). Previous year results report that 30% of all
students who reside in hostel attain A grade and 20% of day scholars attain A
grade in their annual examination. At the end of the year, one student is chosen
at random from the college and he has an A grade, what is the probability that the
student is a hostler?
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QUESTION AND ANSWERS March – 2017
II PUC
Solution: Let E1: the event that the students is residing in hostel.
and E2: the event that the students is not residing in hostel.
Then E1 and E2 are mutually exclusive and exhaustive.
60
3
40
2
Moreover, 𝑃(𝐸1 ) = 60% = 100 = 5
and 𝑃(𝐸2 ) = 40% = 100 = 5
Let E: A student attains A grade,
30
3
20
2
then, 𝑃(𝐸|𝐸1 ) = 30% = 100 = 10 and 𝑃(𝐸|𝐸2 ) = 20% = 100 = 10
By using Baye’s theorem, we have
𝑃(𝐸1 |𝐸) =
𝑃(𝐸|𝐸1 )𝑃(𝐸1 )
𝑃(𝐸|𝐸1 )𝑃(𝐸1 )+𝑃(𝐸|𝐸2 )𝑃(𝐸2 )
3 3
×
10 5
3 3 2 2
× + ×
10 5 10 5
=
9
9
= 9+4 = 13 .
PART D
Answer any SIX questions:
𝟔  𝟓 = 𝟑𝟎
39. Let 𝑓: 𝑁 → 𝑌 be a function defined as 𝑓(𝑥) = 4𝑥 + 3, where,
𝑌 = {𝑦 ∈ 𝑁: 𝑦 = 4𝑥 + 3 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑥 ∈ 𝑁 }. Show that 𝑓 is invertible. Find the inverse.
Solution: Consider an arbitrary element 𝑦 of Y.
By definition of Y, Let us define 𝑦 = 4𝑥 + 3, for some 𝑥 in the domain N.
𝑦− 3
This shows that 4𝑥 = 𝑦 − 3 ⟹ 𝑥 = 4 .
Define 𝑔: 𝑌 → 𝑁 by 𝑔(𝑦) =
𝑦− 3
4
Now (𝑔𝑜𝑓)(𝑥) = 𝑔[𝑓(𝑥)] = 𝑔(4𝑥 + 3) =
(4𝑥+3) − 3
4
4
And (𝑓𝑜𝑔)(𝑦) = 𝑓[𝑔(𝑦)] = 𝑓 (
𝑦− 3
4
𝑦− 3
) = 4(
=
4𝑥
4
=𝑥
) + 3 = 𝑦 − 3 + 3 = 𝑦.
This shows that 𝑔𝑜𝑓 = 𝐼𝑁 𝑎𝑛𝑑 𝑓𝑜𝑔 = 𝐼𝑌 , which implies that 𝑓 is invertible and 𝑔 is the
inverse of 𝑓.
“OR”
Let 𝑥1 , 𝑥2 ∈ 𝑁. Let 𝑓(𝑥1 ) = 𝑓(𝑥2 ).
Then 4𝑥1 + 3 = 4𝑥2 + 3 ⟹ 4𝑥1 = 4𝑥2 ⟹ 𝑥1 = 𝑥2 ∴ 𝑓 is one-one.
Since the range of the function is same as the co domain of the function, the given
function is onto. Since the given function is both one-one and on-to the function is
invertible.
Let 𝑓 − 1 (𝑦) = 𝑥 ∈ 𝑁.
Then 𝑓(𝑥) = 𝑦 ⟹ 4𝑥 + 3 = 𝑦
𝑦− 3
⟹ 4𝑥 = 𝑦 − 3 ⟹ 𝑥 = 4 .
Thus the inverse of the function is 𝑓 − 1 ∶ 𝑌 → 𝑁, 𝑓 − 1 (𝑦) =
𝑦− 3
4
.
1 2 3
 3 1 2 
4 1 2




40. If A  5 0 2  , B   4 2 5  and C   0 3 2  , then compute
1 1 1 
 2 0 3 
1 2 3 
A  B and B  C . Also, verify that A   B  C    A  B  C .
1 2 3  3 1 2   4 1 1
Solution: Let A  B  5 0 2    4 2 5   9 2 7 
1 1 1   2 0 3   3 1 4 
 3 1 2   4 1 2   1 2 0 
B  C   4 2 5   0 3 2    4 1 3 
 2 0 3 1 2 3   1 2 0 
1 2 3  1 2 0  0 0 3
LHS  A   B  C   5 0 2    4 1 3  9 1 5 
1 1 1   1 2 0   2 1 1 
 4 1 1  4 1 2  0 0 3
RHS   A  B   C  9 2 7   0 3 2   9 1 5 
 3 1 4  1 2 3   2 1 1 
Hence, A   B  C    A  B  C is verified.
41. Solve the system of equation by matrix method:
3
2𝑥 + 𝑦 + 𝑧 = 1 , 𝑥 − 2𝑦 − 𝑧 = 2 , 3𝑦 − 5𝑧 = 9
Rewriting second and third equation as:
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QUESTION AND ANSWERS March – 2017
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II PUC
𝟐𝒙 − 𝟒𝒚 − 𝟐𝒛 = 𝟑 𝒂𝒏𝒅 𝟎 × 𝒙 + 𝟑𝒚 − 𝟓𝒛 = 𝟗
𝑥
2 1
1
1
Solution: Let 𝐴𝑋 = 𝐵 ⟹ 𝑋 = 𝐴−1 𝐵, where 𝐴 = [2 −4 −2] 𝑋 = [𝑦] and 𝐵 = [3]
𝑧
0 3 −5
9
2 1
1
Now |𝐴| = |2 −4 −2| = 2[20 + 6]— 1[−10 − 0] + 1[6 − 0] = 52 + 10 + 6 = 68 ≠ 0
0 3 −5
+(20 + 6)
−(−5 − 3) +(−2 + 4)
26
8
2
𝑎𝑑𝑗𝐴 = [−(−10 − 0) +(−10 − 0) −(−4 − 2)] = [10 −10
6 ]
6
−6 −10
+(6 − 0)
−(6 − 0)
+(−8 − 2)
1
𝑥
𝑥
26
8
2
1
26 + 24 + 18
68
1
1
1
1
𝑦
⟹ [𝑦] = 68 [10 −10
]
[
]
=
[
]
=
[
]
⟹
[
]
=
[
6
3
10 − 30 + 54
34
2]
68
68
−3
𝑧
𝑧
6
−6 −10 9
6 − 18 − 90
−102
2
1
3
∴ 𝑥 = 1, 𝑦 = 2 𝑎𝑛𝑑 𝑧 = − 2 .
−1 𝑥
42. If 𝑦 = 𝑒 acos
, −1 ≤ 𝑥 ≤ 1, show that (1 − 𝑥 2 )𝑦2 − 𝑥𝑦1 − 𝑎2 𝑦 = 0.
−1 𝑥
Solution: 𝑦 = 𝑒 acos
𝑦1 = 𝑒 acos
−1 𝑥
−1
× 𝑎 × √1−𝑥 2
−1
⟹ √1 − 𝑥 2 𝑦1 = −𝑎 𝑒 acos 𝑥
Again differentiating with respect to x, we get,
(√1 − 𝑥 2 ) × 𝑦2 + 𝑦1 ×
𝑥𝑦
(√1 − 𝑥 2 )𝑦2 − √1−𝑥1 2 =
⟹
(1−𝑥 2 )𝑦2 −𝑥𝑦1
√1−𝑥 2
1
2√1−𝑥2
[0 − 2𝑥] = −𝑎 [𝑒 acos
𝑎2 𝑒 acos
−1 𝑥
−1
× 𝑎 × √1−𝑥 2 ]
−1 𝑥
√1−𝑥2
𝑎2 𝑦
= √1−𝑥 2
𝑥 2 )𝑦2
∴ (1 −
− 𝑥𝑦1 − 𝑎2 𝑦 = 0.
43. A man of height 2 meters walks at a uniform speed of 5 km/hour, away from a
lamp post which is 6 meters high. Find the rate at which the length of his shadow
increases.
Solution: Let AB be the lamp-post, the lamp being at the position B and let MN be
the man at a particular time t and let 𝐴𝑀 = 𝑙 meters. Then MS is the shadow of the
man. Let 𝑀𝑆 = 𝑠 meters.
Given
𝑑𝑙
𝑑𝑡
= 5 𝑘𝑚/ℎ
Note that ∆𝑀𝑆𝑁~∆𝐴𝑆𝐵
Or
𝑀𝑆
𝐴𝑆
=
𝑀𝑁
𝐴𝐵
Therefore
𝑠
2
⟹ 𝑙+𝑠 = 6 ⟹ 𝑙 + 𝑠 = 3𝑠
𝑑𝑙
𝑑𝑡
𝑑𝑠
𝑑𝑠
∴ 𝑙 = 2𝑠
𝑑𝑠
= 2 𝑑𝑡 ⟹ 5 = 2 𝑑𝑡
5
∴ 𝑑𝑡 = 2 𝑘𝑚/ℎ.
Hence, the length of the shadow increases at the rate
44. Find the integral of
1
𝑥 2 − 𝑎2
5
2
km/hour.
with respect to x and evaluate ∫
𝑑𝑥
𝑥 2 − 16
.
1
Solution: Let 𝐼 = ∫ 𝑥 2 − 𝑎2 𝑑𝑥
Consider
1
1
𝑥 2 − 𝑎2
1
(𝑥+𝑎)−(𝑥−𝑎)
1
1
1
1
= (𝑥−𝑎)(𝑥+𝑎) = 2𝑎 [ (𝑥−𝑎)(𝑥+𝑎) ] = 2𝑎 [(𝑥−𝑎)] − 2𝑎 [(𝑥+𝑎)]
1
1
1
1
∴ 𝐼 = ∫ 2𝑎 [(𝑥−𝑎)] 𝑑𝑥 − ∫ 2𝑎 [(𝑥−𝑎)] 𝑑𝑥
1
1
𝑥−𝑎
= 2𝑎 [log|𝑥 − 𝑎| − log|𝑥 + 𝑎|] + 𝑐 = 2𝑎 log |𝑥+𝑎| + 𝑐
Consider∫
1
1
𝑥 2 − 16
𝑑𝑥 = ∫
1
𝑥 2 − 42
𝑑𝑥
𝑥−4
∴ 𝐼 = 2×4 log |𝑥+4| + 𝐶.
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QUESTION AND ANSWERS March – 2017
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II PUC
45. Find the area of the circle 𝑥 2 + 𝑦 2 = 𝑎2 by the method of integration and hence find
the area of the circle 𝑥 2 + 𝑦 2 = 6.
Solution: Required area, 𝐴 = 4(𝑎𝑟𝑒𝑎 𝐴𝑂𝐵)
𝑎
= 4 ∫0 𝑦 𝑑𝑥
𝑎
= 4 ∫0 √𝑎2 − 𝑥 2 𝑑𝑥
𝑥
= 4 [2 √𝑎2 − 𝑥 2 +
= 4 [0 +
𝑎2
2
(sin−1
𝑎2
2
𝑥
sin−1 𝑎]
1 − sin
𝑎
0
−1
0)] = 4 ×
𝑎2
2
𝜋
× 2 = 𝜋𝑎2
The area of the circle 𝑥 + 𝑦 = 6. Here 𝑎 = √6
46. Solve, y dx  ( x  y 2 )dy  0
2
2
2
∴ 𝐴 = 𝜋𝑎2 = 𝜋 × (√6) = 6𝜋
dx
x y2
 
dy
y y
dx 1
dx
 x  y . This is of the form,
PxQ
dy y
dy
Solution: y dx  ( x  y 2 )dy 
Where P 
1
and
y
Q y
1
I.F.  e
 p dy
 dy
log y
e y e
y
 x( I .F .)   Q ( I .F .) d y  C
y3
x  y   y  yd y  C  x y   y d y  C
 xy 
 C.
3
47. Derive the formula to find the shortest distance between two skew lines in both
vector and Cartesian form.
Proof: Line 𝐿1 passes through 𝐴(𝑎⃗1 ) and parallel to 𝑏⃗⃗1 .
Line 𝐿2 passes through 𝐵(𝑎⃗2 ) and parallel to 𝑏⃗⃗2.
Let SD be the shortest distance between lines 𝐿1 and 𝐿2 .
⟹ SD is  to lines lines 𝐿1 and 𝐿2 .
⟹ SD is  to both 𝑏⃗⃗1 and 𝑏⃗⃗2 .
Since, line 𝐿1 is parallel to 𝑏⃗⃗1 and line 𝐿2 is parallel to 𝑏⃗⃗2 .
But, 𝑏⃗⃗1 × 𝑏⃗⃗2 is  to both 𝑏⃗⃗1 and 𝑏⃗⃗2 .
⃗⃗⃗⃗⃗⃗ is parallel to 𝑏⃗⃗1 × 𝑏⃗⃗2.
∴ 𝑆𝐷
⃗⃗
⃗⃗
⃗⃗⃗⃗⃗⃗ = |𝑆𝐷
⃗⃗⃗⃗⃗⃗ | 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 = |𝑆𝐷
⃗⃗⃗⃗⃗⃗ | 𝑏1 × 𝑏2
⟹ 𝑆𝐷
2
⃗⃗1 × 𝑏
⃗⃗2 |
|𝑏
⃗⃗
⃗⃗
𝑏 ×𝑏
⃗⃗⃗⃗⃗⃗ | = Projection of ⃗⃗⃗⃗⃗⃗
|𝑆𝐷
𝐴𝐵 on ⃗⃗⃗⃗⃗⃗
𝑆𝐷 = (𝑎⃗2 − 𝑎⃗1 ) ∙ Unit vector of ⃗⃗⃗⃗⃗⃗
𝑆𝐷 = (𝑎⃗2 − 𝑎⃗1 ) ∙ |𝑏⃗⃗1 × 𝑏⃗⃗2 |
1
2
(𝑎⃗2 − 𝑎⃗1 ) ∙ (𝑏⃗⃗1 × 𝑏⃗⃗2 )
𝑆𝐷 = |
|
|𝑏⃗⃗1 × 𝑏⃗⃗2 |
Cartesian form: The shortest distance between the lines
𝑥−𝑥1
𝑎1
=
𝑦−𝑦1
𝑏1
=
𝑧−𝑧1
𝑐1
and
𝑥−𝑥2
𝑎2
=
𝑦−𝑦2
𝑏2
=
𝑧−𝑧2
𝑐2
is
𝑥2 −𝑥1 𝑦2 −𝑦1 𝑧2 −𝑧1
𝑏1
𝑐1 |
| 𝑎1
𝑎2
𝑏2
𝑐2
|
|.
2
2
√(𝑏1 𝑐2 −𝑏2 𝑐1 ) +(𝑐1 𝑎2 −𝑐2 𝑎1 ) +(𝑎1 𝑏2 −𝑎2 𝑏1 )2
48. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find
the probability of two successes.
Solution: It is a case of Bernoulli trials with 𝑛 = 4. Here, successes is getting a
Doublet.
When a pair of dice is thrown once there are 6 × 6 = 36 equally likely outcomes
⟹ Total number of cases = 36
And possible doublets are {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
⟹ Favourable number of cases = 6
∴ 𝑃 = 𝑃(𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠) = 𝑃(𝑔𝑒𝑡𝑡𝑖𝑛𝑔 𝑎 𝑑𝑜𝑢𝑏𝑙𝑒𝑡 𝑖𝑛 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡ℎ𝑟𝑜𝑤 𝑜𝑓 𝑎 𝑝𝑎𝑖𝑟 𝑜𝑓 𝑑𝑖𝑐𝑒)
Favourable number of cases
6
1
= Total number of cases = 36 = 6
1
5
𝑞 = 𝑃(𝑓𝑎𝑖𝑙𝑢𝑟𝑒) = 1 − 𝑝 = 1 − 6 = 6
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QUESTION AND ANSWERS March – 2017
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1
II PUC
5
Clearly, X has the binomial distribution with 𝑛 = 4, 𝑝 = 6 , 𝑞 = 6
Therefore, 𝑃(𝑋 = 𝑟) = 𝑛𝐶𝑟 𝑝𝑟 𝑞 𝑛−𝑟 , where 𝑟 = 0, 1, 2, … . 𝑛
1 𝑟
5 4−𝑟
6
6
𝑃(𝑋 = 𝑟) = 4𝐶𝑟 ( ) ( )
1 2 5 4−2
Now, 𝑃(2𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠) = 𝑃(𝑋 = 2) = 4𝐶2 (6) (6)
4×3
1
25
25
= 2×1 × 36 × 36 = 216.
PART E
10  1=10
Answer any ONE question:
𝑎
49. (a) Prove that ∫−𝑎 𝑓(𝑥)𝑑𝑥
𝑎
2 ∫ 𝑓(𝑥)𝑑𝑥,
={ 0
0,
𝑖𝑓 𝑓(𝑥) 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑖𝑓 𝑓(𝑥) 𝑖𝑠 𝑜𝑑𝑑𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝜋
and hence evaluate ∫ 2𝜋(𝑥 3 + 𝑥 cos 𝑥 + 𝑡𝑎𝑛5 𝑥 + 1) 𝑑𝑥.
–
2
Proof: Let
Putting
𝑎
∫–𝑎 𝑓(𝑥)𝑑𝑥
0
𝑎
= ∫−𝑎 𝑓(𝑥)𝑑𝑥 + ∫0 𝑓(𝑥)𝑑𝑥
𝑥 = −𝑡 in the first integral on the RHS ⟹ 𝑑𝑥 = −𝑑𝑡
Also, 𝑥 = −𝑎 ⟹ 𝑡 = 𝑎 ;
0
0
𝑎
𝑎
𝑥=0⟹𝑡=0
0
𝑎
∴ ∫−𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(−𝑡)(−𝑑𝑡) = − ∫𝑎 𝑓(−𝑡)𝑑𝑡 = ∫0 𝑓(−𝑥)𝑑𝑥
𝑎
∴ ∫−𝑎 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(−𝑥)𝑑𝑥 + ∫0 𝑓(𝑥)𝑑𝑥
Case(1): If 𝑓(𝑥) is even. ⟹ 𝑓(−𝑥) = 𝑓(𝑥)
𝑎
𝑎
𝑎
𝑎
𝑎
∴ ∫–𝑎 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑥)𝑑𝑥 + ∫0 𝑓(𝑥)𝑑𝑥 ⟹ ∫–𝑎 𝑓(𝑥)𝑑𝑥 = 2 ∫0 𝑓(𝑥)𝑑𝑥
Case(2): If 𝑓(𝑥) is odd. ⟹ 𝑓(−𝑥) = −𝑓(𝑥)
𝑎
𝑎
𝑎
𝑎
∴ ∫–𝑎 𝑓(𝑥)𝑑𝑥 = − ∫0 𝑓(𝑥)𝑑𝑥 + ∫0 𝑓(𝑥)𝑑𝑥 ⟹ ∫–𝑎 𝑓(𝑥)𝑑𝑥 = 0.
Hence the proof.
𝜋
2
𝜋
–
2
𝜋
2
𝜋
–
2
𝐼 = ∫ (𝑥 3 + 𝑥 cos 𝑥 + 𝑡𝑎𝑛5 𝑥 + 1) 𝑑𝑥. Here 𝑥 3 + 𝑥 cos 𝑥 + 𝑡𝑎𝑛5 𝑥 are odd functions.
𝜋
𝜋
𝐼 = ∫ (𝑥 3 ) 𝑑𝑥 = 0 ; 𝐼 = ∫ 2𝜋(𝑥 cos 𝑥) 𝑑𝑥 = 0 ; 𝐼 = ∫ 2𝜋(𝑡𝑎𝑛5 𝑥) 𝑑𝑥 = 0.
–
2
–
𝜋
2
𝜋
2
2
𝜋
Therefore 𝐼 = 0 + 0 + 0 + 2 ∫0 1 𝑑𝑥 = 2[𝑥]0 = 2 × 2 = 𝜋.
(b) Find all points of discontinuity of f, where f is defined by (𝑥) = {
𝑥10 − 1 , 𝑖𝑓 𝑥 ≤ 1
.
𝑥2 ,
𝑖𝑓 𝑥 > 1
Solution: When 𝑥 < 1, 𝑓(𝑥) = 𝑥10 − 1
When 𝑥 > 1, 𝑓(𝑥) = 𝑥 2 . Every polynomial function is continuous. So, the discontinuity
of given function may be occur at 𝑥 = 1.
At 𝒙 = 𝟏,
𝑓(1) = 110 − 1 = 0
𝑅𝐻𝐿 = lim+ 𝑓(𝑥) = 𝑥 2 = 1 ≠ 𝑓(1)
𝑛→1
Thus, f (x) is discontinuous at 𝑥 = 1.
So, 𝑥 = 1 is the only point of discontinuity of f.
50. (a) A co-operative society of farmers has 50 hectare of land to grow two crops X
and Y. The profit from crops X and Y per hectare are estimated as Rs 10,500 and Rs
9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and
Y at rates of 20 litres and 10 litres per hectare. Further no more than 800 litres of
herbicide should be used in order to protect fish and wild life using a pond which
collects drainage from this land. How much should be allocated to each crop so as to
maximize the total profit of the society?
Solution: Let 𝑥 hectare of land be allocated to crop X and 𝑦 hectare of land be allocated
to crop Y. obviously 𝑥 ≥ 0 , 𝑦 ≥ 0.
Profit per hectare on crop 𝑋 = 𝑅𝑠: 10,500.
Profit per hectare on crop 𝑌 = 𝑅𝑠: 9,000.
Therefore, total profit, = 10,500𝑥 + 9,000𝑦.
Now maximize: 𝑧 = 10,500𝑥 + 9,000𝑦
𝑥 + 𝑦 ≤ 50 ………..1 (constraint related to land)
S U J M P U College, Harapanahalli – 583131.
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“amma”
QUESTION AND ANSWERS March – 2017
II PUC
20𝑥 + 10𝑦 ≤ 800 “OR” 2𝑥 + 𝑦 ≤ 80…….2 (constraint related to use of herbicide)
For 𝑥 + 𝑦 = 50, Put 𝑦 = 0 ⟹ 𝑥 = 50, 𝑨 = (𝟓𝟎, 𝟎), Put 𝑥 = 0 ⟹ 𝑦 = 50, 𝑩 = (𝟎, 𝟓𝟎)
For 2𝑥 + 𝑦 = 80 , Put 𝑦 = 0 ⟹ 𝑥 = 40, 𝑪 = (𝟒𝟎, 𝟎) Put 𝑥 = 0 ⟹ 𝑦 = 80 𝑫 = (𝟎, 𝟖𝟎)
From 1 and 2, we get 𝑬 = (𝟑𝟎, 𝟐𝟎)
Corner points
0(0, 0)
𝐴(0, 50)
𝐷(40,0)
𝐸(30, 20)
𝑍 = 3𝑥 + 9𝑦
00
4,50,000
4,20,000
4,95,000
Therefore, Minimum value of 𝑍 = 4,95,000 at the point 𝐸(30, 20).
3𝑎
−𝑎 + 𝑏 −𝑎 + 𝑐
(b) Prove that |−𝑏 + 𝑎
3𝑏
−𝑏 + 𝑐 | = 3(𝑎 + 𝑏 + 𝑐)(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎).
−𝑐 + 𝑎 −𝑐 + 𝑏
3𝑐
Solution: Applying 𝐶1 ⟶ 𝐶1 + 𝐶2 + 𝐶3 , and taking (𝑎 + 𝑏 + 𝑐) as a common factor,
we get
1 −𝑎 + 𝑏 −𝑎 + 𝑐
= (𝑎 + 𝑏 + 𝑐) |1
3𝑏
−𝑏 + 𝑐 | Applying 𝑅2 ⟶ 𝑅2 − 𝑅1 ; 𝑅3 ⟶ 𝑅3 − 𝑅1 , we get
1 −𝑐 + 𝑏
3𝑐
1 −𝑎 + 𝑏 −𝑎 + 𝑐
= (𝑎 + 𝑏 + 𝑐) |0 2𝑏 + 𝑎 𝑎 − 𝑏 |
0 𝑎−𝑐
2𝑐 + 𝑎
= (𝑎 + 𝑏 + 𝑐){1[4𝑏𝑐 + 2𝑎𝑏 + 2𝑎𝑐 + 𝑎2 − 𝑎2 + 𝑎𝑏 + 𝑎𝑐 − 𝑏𝑐]}
= (𝑎 + 𝑏 + 𝑐)[3𝑏𝑐 + 3𝑎𝑏 + 3𝑎𝑐] = 3(𝑎 + 𝑏 + 𝑐)(𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎).
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PART – A - Q.N. 1 to 10 - Answer All the questions – Carries, 10 × 1 = 10
PART – B -Q.N. 11 to 24-Answer any Ten questions – Carries, 10 × 2 = 20
PART – C -Q.N. 25 to 38-Answer any Ten questions – Carries, 10 × 3 = 30P
PART – D -Q.N. 39 to 48-Answer any Six questions – Carries,
6 × 5 = 30
PART – E -Q.N. 49 and 50-Answer any One questions – Carries, 10 × 1 = 10.
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S U J M P U College, Harapanahalli – 583131.
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