1. Closure property – True 3.

𝐬𝐢𝐧(𝜶 + ⁡𝜷) = 𝒔𝒊𝒏𝜶⁡𝒄𝒐𝒔𝜷 + 𝒄𝒐𝒔𝜶⁡𝒔𝒊𝒏𝜷

2. Commutative property –True ⁡𝑎⁡
⃗⃗⃗⃗⃗ = cos 𝛼⁡⁡𝑖⁡
⃗⃗⃗ + 𝑠𝑖𝑛𝛼⁡𝑗⁡
⃗⃗⃗⃗
3. Associative Property – True ⃗⃗⃗⃗⃗ = cos 𝛽⁡⁡𝑖⁡
⁡𝑏⁡ ⃗⃗⃗⃗ − 𝑠𝑖𝑛𝛽⁡𝑗⁡
⃗⃗⃗⃗
4. Identify Property -True ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ………….. (1)
⁡𝑏⁡ ⁡ × ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ = sin(𝛼 + ⁡𝛽) ⁡𝑘⁡
1⁄ 1⁄
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⁡ …….. (2)
𝐸 =⁡( 2 2) ⁡ ∈ 𝑀 ⁡𝑏⁡ ⁡ × ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ = (𝑠𝑖𝑛𝛼⁡𝑐𝑜𝑠𝛽 + 𝑐𝑜𝑠𝛼⁡𝑠𝑖𝑛𝛽)⁡𝑘⁡
1⁄ 1⁄ From (1) and (2)
2 2
5. Inverse property – True , sin(𝛼 + ⁡𝛽) = 𝑠𝑖𝑛𝛼⁡𝑐𝑜𝑠𝛽 + 𝑐𝑜𝑠𝛼⁡𝑠𝑖𝑛𝛽
1 1 HENCE PROVED
Inverse element = (4𝑥
1
4𝑥
1 )⁡∈ 𝑀 4. 𝐬𝐢𝐧(𝜶 − ⁡𝜷) = 𝒔𝒊𝒏𝜶⁡𝒄𝒐𝒔𝜷 − 𝒄𝒐𝒔𝜶⁡𝒔𝒊𝒏𝜷
4𝑥 4𝑥 ⃗⃗⃗⃗⃗
⁡𝑎⁡ = cos 𝛼⁡⁡𝑖⁡
⃗⃗⃗ + 𝑠𝑖𝑛𝛼⁡𝑗⁡
⃗⃗⃗⃗
Questions Closure Commutative Associate Identity Inverse ⃗⃗⃗⃗⃗
⁡𝑏⁡ = cos 𝛽⁡⁡𝑖⁡
⃗⃗⃗⃗ + 𝑠𝑖𝑛𝛽⁡𝑗⁡
⃗⃗⃗⃗
Property Property Property Property Property ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑎⁡
⁡𝑏⁡ ⃗⃗⃗⃗⃗ ………….. (1)
⃗⃗⃗⃗⃗ ⁡ = sin(𝛼 − ⁡𝛽) ⁡𝑘⁡
23 N
⃗⃗⃗⃗⃗
⁡𝑏⁡ ⁡ × ⁡ ⁡𝑎⁡ ⃗⃗⃗⃗⃗⁡ ….. (2)
⃗⃗⃗⃗⃗ ⁡ = (𝑠𝑖𝑛𝛼⁡𝑐𝑜𝑠𝛽 − 𝑐𝑜𝑠𝛼⁡𝑠𝑖𝑛𝛽)⁡𝑘⁡
a * b = 𝑎𝑏  x x x x
24 Q From (1) and (2)
𝑎+𝑏   x x x sin(𝛼 − ⁡𝛽) = 𝑠𝑖𝑛𝛼⁡𝑐𝑜𝑠𝛽 − 𝑐𝑜𝑠𝛼⁡𝑠𝑖𝑛𝛽
𝑎 ∗ 𝑏 =⁡
2 Hence Proved.
25 Z 5.` Show that the attitudes of a triangle are concurrent by using vectors...
m * n = m + n - mn     x ⃗⃗⃗⃗⃗⃗
𝑂𝐴 = ⁡ ⁡𝑎⁡ ⃗⃗⃗⃗⃗, 𝑂𝐶
⃗⃗⃗⃗⃗⃗ = ⁡ ⁡𝑏⁡
⃗⃗⃗⃗⃗,⁡ 𝑂𝐵 ⃗⃗⃗⃗⃗⃗ = ⁡ ⁡𝑐⁡
⃗⃗⃗⃗
e=0
−𝑥 ⃗⃗⃗⃗⃗. (⁡𝑏⁡
⁡𝑎⁡ ⃗⃗⃗⃗⃗ − ⁡ ⁡𝑐⁡
⃗⃗⃗⃗⁡) = 0 ………….. (1)
26 Q – {1} 𝑥 −1 = ⁡
x * y = x + y - xy     1−𝑥 ⃗⃗⃗⃗⃗
⁡𝑏⁡.⁡⁡(⁡𝑐⁡⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗) = 0 ………….. (2)
e=0  (1) + (2)
⃗⃗⃗⃗. (⁡𝑎⁡
⁡𝑐⁡ ⃗⃗⃗⃗⃗ − ⁡𝑏⁡ ⃗⃗⃗⃗⃗) = 0
Chapter 6. APPLICATION OF VECTOR ALGEBRA ⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⊥ ⁡ ⁡𝐴𝐵⁡
⁡𝑂𝐶⁡ ⃗⃗⃗⃗⃗⃗⃗⃗⃗ Hence
5 MARKS The attitudes of a triangle are concurrent hence proved.
1. Prove by vector method 𝐜𝐨𝐬(𝜶 + ⁡𝜷) = 𝒄𝒐𝒔𝜶⁡𝒄𝒐𝒔𝜷 − 𝒔𝒊𝒏𝜶⁡𝒔𝒊𝒏𝜷⁡
6. ⃗⃗⃗⃗⃗
⁡𝒂⁡ = ⁡ ⁡𝒊⁡ ⃗⃗⃗⃗, ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − ⁡ ⁡𝒋⁡ ⁡𝒃⁡ = ⁡ ⁡𝒊⁡⃗⃗⃗⃗ − ⁡ ⁡𝒋⁡ ⃗⃗⃗⃗⃗, ⁡𝒄⁡
⃗⃗⃗⃗ − 𝟒⁡𝒌⁡ ⃗⃗⃗⃗ − ⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ = ⁡𝟑⁡𝒋⁡ ⃗⃗⃗⃗⃗ = ⁡𝟐⁡𝒊⁡
⁡𝒌⁡𝒂𝒏𝒅⁡⁡𝒅⁡ ⃗⃗⃗⃗ + 𝟓⁡⁡𝒋⁡ ⃗⃗⃗⃗⃗ verify that…
⃗⃗⃗⃗ + ⁡ ⁡𝒌⁡
⃗⃗⃗⃗⃗
⁡𝑎⁡ = cos 𝛼⁡⁡𝑖⁡
⃗⃗⃗⃗ + 𝑠𝑖𝑛𝛼⁡𝑗⁡
⃗⃗⃗⃗
⃗⃗⃗⃗⃗) ⁡ × ⁡ (⁡𝒄⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝒃⁡
(i) (⁡𝒂⁡ ⃗⃗⃗⃗⃗) = ⁡ [⁡𝒂⁡
⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝒅⁡ ⃗⃗⃗⃗⃗,⁡⁡⁡⁡𝒅⁡
⃗⃗⃗⃗⃗⁡, ⁡𝒃⁡ ⃗⃗⃗⃗⃗]⁡𝒄⁡
⃗⃗⃗⃗ − ⁡ [⁡𝒂⁡ ⃗⃗⃗⃗⃗,⁡⁡⁡⁡𝒄⁡
⃗⃗⃗⃗⃗⁡, ⁡𝒃⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗]⁡𝒅⁡
⃗⃗⃗⃗⃗
⁡𝑏⁡ = cos 𝛽⁡⁡𝑖⁡
⃗⃗⃗ − 𝑠𝑖𝑛𝛽⁡𝑗⁡
⃗⃗⃗⃗
Solution: LHS (⁡𝑎⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡ (⁡𝑐⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗⃗)
⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑑⁡
⃗⃗⃗⃗⃗
⁡𝑏⁡⁡. ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ = cos(𝛼 + ⁡𝛽) ………….. (1)
⃗⃗⃗⃗⃗⁡. ⁡𝑎⁡
⁡𝑏⁡ ⃗⃗⃗⃗⃗ ⁡ = 𝑐𝑜𝑠𝛼⁡𝑐𝑜𝑠𝛽 − 𝑠𝑖𝑛𝛼⁡𝑠𝑖𝑛𝛽⁡ …….. (2) ⃗⃗⃗
⁡𝑖⁡ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
⁡𝑗⁡ ⁡𝑘⁡
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = ⁡ | 1
⁡𝑎⁡ ⁡ × ⁡ ⁡𝑏⁡ −1 0 | ⁡ = ⁡⁡4⁡𝑖⁡ ⃗⃗⃗ + 4⁡𝑗⁡ ⃗⃗⃗⃗
From (1) and (2)
1 −1 −4
cos(𝛼 + ⁡𝛽) = 𝑐𝑜𝑠𝛼⁡𝑐𝑜𝑠𝛽 − 𝑠𝑖𝑛𝛼⁡𝑠𝑖𝑛𝛽⁡
HENCE PROVED ⃗⃗⃗ ⁡𝑗⁡
⁡𝑖⁡ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
⁡𝑘⁡
⃗⃗⃗⃗ ⁡ × ⁡ ⃗⃗⃗⃗⃗
⁡𝑐⁡ ⁡𝑑⁡ = ⁡ | 0 3 −1| ⁡ = ⁡⁡8⁡𝑖⁡ ⃗⃗⃗ − 2⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 6⁡𝑘⁡
2 5 1
2. 𝐜𝐨𝐬(𝜶 − ⁡𝜷) = 𝒄𝒐𝒔𝜶⁡𝒄𝒐𝒔𝜷 + 𝒔𝒊𝒏𝜶⁡𝒔𝒊𝒏𝜷
⃗⃗⁡𝑖⁡
⃗⃗ ⁡𝑗⁡ ⃗⃗⃗⃗ ⁡𝑘⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ = cos 𝛼⁡⁡𝑖⁡
⁡𝑎⁡ ⃗⃗⃗⃗ + 𝑠𝑖𝑛𝛼⁡𝑗⁡
⃗⃗⃗⃗ ⃗⃗⃗⃗⃗) ⁡ × ⁡ (⁡𝑐⁡
(⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗ ⁡ × ⁡ ⃗⃗⃗⃗⃗
⁡𝑑⁡) = ⁡ | 4 4 0 | ⁡ = ⁡ −24⁡𝑖⁡ ⃗⃗⃗ + 24⁡𝑗⁡ ⃗⃗⃗⃗⃗ ……. (1)
⃗⃗⃗⃗ − 40⁡𝑘⁡
⃗⃗⃗⃗⃗ = cos 𝛽⁡⁡𝑖⁡
⁡𝑏⁡ ⃗⃗⃗⃗ + 𝑠𝑖𝑛𝛽⁡𝑗⁡
⃗⃗⃗⃗ 8 −2 −6
⃗⃗⃗⃗⃗⁡. ⁡𝑎⁡
⁡𝑏⁡ ⃗⃗⃗⃗⃗ ⁡ = cos(𝛼 − ⁡𝛽) ………….. (1) ⃗⃗⃗⃗⃗,⁡⁡⁡⁡𝑑⁡
⃗⃗⃗⃗⃗]⁡𝑐⁡ ⃗⃗⃗⃗⃗,⁡⁡⁡⁡𝑐⁡ ⃗⃗⃗⃗⃗
RHS [⁡𝑎⁡ ⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡ ⃗⃗⃗⃗ − ⁡ [⁡𝑎⁡⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡ ⃗⃗⃗⃗]⁡𝑑⁡
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ⁡ = 𝑐𝑜𝑠𝛼⁡𝑐𝑜𝑠𝛽 + 𝑠𝑖𝑛𝛼⁡𝑠𝑖𝑛𝛽⁡ …….. (2)
⁡𝑏⁡⁡. ⁡𝑎⁡ 1 −1 0
From (1) and (2) ⃗⃗⃗⃗⃗⁡, ⃗⃗⃗⃗⃗
[⁡𝑎⁡ ⃗⃗⃗⃗⃗] ⁡ = ⁡ |1 −1 −4| ⁡ = 28
⁡𝑏⁡,⁡⁡⁡⁡𝑑⁡
cos(𝛼 − ⁡𝛽) = 𝑐𝑜𝑠𝛼⁡𝑐𝑜𝑠𝛽 + 𝑠𝑖𝑛𝛼⁡𝑠𝑖𝑛𝛽⁡ 2 5 1
1 −1 0
HENCE PROVED ⃗⃗⃗⃗⃗
[⁡𝑎⁡
⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡,⁡⁡⁡⁡𝑐⁡⃗⃗⃗⃗] ⁡ = ⁡ |1 −1 −4| ⁡ = 12
0 3 −1
[⁡𝑎⁡ ⃗⃗⃗⃗⃗,⁡⁡⁡⁡𝑑⁡
⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡ ⃗⃗⃗⃗⃗]⁡𝑐⁡
⃗⃗⃗⃗ − ⁡ [⁡𝑎⁡ ⃗⃗⃗⃗⃗,⁡⁡⁡⁡𝑐⁡
⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡ ⃗⃗⃗⃗⃗ =⁡ −24⁡𝑖⁡
⃗⃗⃗⃗]⁡𝑑⁡ ⃗⃗⃗⃗ + 24⁡𝑗⁡ ⃗⃗⃗⃗⃗ ……. (2)
⃗⃗⃗⃗ − 40⁡𝑘⁡
30 31
 From (1) & (2) 8. Find the non- parametricform form of vector equation and cartesian equation of the plane
(⁡𝑎⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡ (⁡𝑐⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗⃗) = ⁡ [⁡𝑎⁡
⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑑⁡ ⃗⃗⃗⃗⃗⁡, ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗]⁡𝑐⁡
⁡𝑏⁡,⁡⁡⁡⁡𝑑⁡ ⃗⃗⃗⃗ − ⁡ [⁡𝑎⁡ ⃗⃗⃗⃗⃗,⁡⁡⁡⁡𝑐⁡
⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗]⁡𝑑⁡ passing through the point (0, 1, -5) and parallel to the straight line
Hence proved ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ + 𝟐⁡𝒋⁡
⁡𝒓⁡ = ⁡ (⁡𝒊⁡ ⃗⃗⃗⃗⃗) + ⁡⁡𝒔(𝟐⁡𝒊⁡
⃗⃗⃗⃗ − 𝟒⁡𝒌⁡ ⃗⃗⃗⃗ + 𝟑⁡𝒋⁡ ⃗⃗⃗⃗⃗) and
⃗⃗⃗⃗ + 𝟔⁡𝒌⁡
For practice: ⃗⃗⃗⃗⃗ = ⁡ (⁡𝒊⁡
⁡𝒓⁡ ⃗⃗⃗⃗ − 𝟑⁡𝒋⁡ ⃗⃗⃗⃗⃗) + ⁡⁡𝒕(⁡𝒊⁡
⃗⃗⃗⃗ + 𝟓⁡𝒌⁡ ⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + ⁡𝒋⁡ ⁡𝒌⁡)
(⁡𝑎⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡ (⁡𝑐⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗ ⁡ × ⁡ ⃗⃗⃗⃗⃗
⁡𝑑⁡) = ⁡ [⁡𝑎⁡ ⁡𝑐⁡ ⃗⃗⃗⃗⃗]⁡𝑏⁡
⃗⃗⃗⃗⃗⁡, ⃗⃗⃗⃗,⁡⁡⁡⁡𝑑⁡ ⃗⃗⃗⃗⃗ − ⁡ [⁡𝑏⁡
⃗⃗⃗⃗⃗⁡, ⁡𝑐⁡ ⃗⃗⃗⃗⃗]⁡𝑎⁡
⃗⃗⃗⃗,⁡⁡⁡⁡𝑑⁡ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ ⁡ = ⁡𝑗⁡
Solution :⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑎⁡ ⃗⃗⃗⃗ − 5⁡𝑘⁡
7. ⃗⃗⃗⃗⃗ = 𝟐⁡𝒊⁡
If ⁡𝒂⁡ ⃗⃗⃗⃗ + 𝟑⁡𝒋⁡ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − ⁡ ⁡𝒌⁡⁡,⁡ ⁡𝒃⁡ = 𝟑⁡𝒊⁡ ⃗⃗⃗⃗ + 𝟓⁡𝒋⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 𝟐⁡𝒌⁡⁡, ⁡𝒄⁡ ⃗⃗⃗⃗ = −⁡𝒊⁡ ⃗⃗⃗⃗ − 𝟐⁡𝒋⁡ ⃗⃗⃗⃗⃗;
⃗⃗⃗⃗ + 𝟑⁡𝒌⁡ ⃗⃗⃗⃗⃗
⁡𝑏⁡ ⁡ = ⁡2⁡𝑖⁡⃗⃗⃗⃗ + 3⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 6⁡𝑘⁡
Then verify that ⃗⃗⃗⃗ ⁡ = ⁡ ⁡𝑖⁡
⁡𝑐⁡ ⃗⃗⃗ + ⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − ⁡𝑘⁡
(i) (⁡𝒂⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡𝒄⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝒃⁡ ⃗⃗⃗⃗ =⁡ (⁡𝒂⁡
⃗⃗⃗⃗⃗⁡. ⁡𝒄⁡ ⃗⃗⃗⃗⃗ − ⁡ (⁡𝒃⁡
⃗⃗⃗⃗)⁡𝒃⁡ ⃗⃗⃗⃗⃗⁡. ⁡𝒄⁡
⃗⃗⃗⃗)⁡𝒂⁡
⃗⃗⃗⃗⃗ (i) Parametric vector equation .
Solution : L H S (⁡𝑎⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡𝑐⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗ ⁡𝑟⁡ = ⁡ ⁡𝑎⁡
⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ + 𝑡⁡⁡𝑐⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗
⃗⃗⃗⃗
⁡𝑖⁡ ⁡𝑗⁡ ⁡𝑘⁡
⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ = ⁡ (⁡𝑗⁡
⁡𝑟⁡ ⃗⃗⃗⃗⃗) + 𝑠(2⁡𝑖⁡
⃗⃗⃗⃗ − 5⁡𝑘⁡ ⃗⃗⃗⃗ + 3⁡𝑗⁡⃗⃗⃗⃗ + 6⁡𝑘⁡ ⃗⃗⃗⃗⃗) + 𝑡(⁡𝑖⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗)
⁡𝑗⁡ − ⁡𝑘⁡
⁡𝑎⁡ ⃗⃗⃗⃗⃗ = ⁡ | 2
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑏⁡ 3 −1| = ⁡11⁡𝑖⁡ ⃗⃗⃗ − 7⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + ⁡𝑘⁡ 𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
3 5 2 (ii) Cartesian equation: | 𝑏1 𝑏2 𝑏3 | = 0
⃗⃗⃗⃗⃗ 𝑐1 𝑐2 𝑐3
⃗⃗⃗
⁡𝑖⁡ ⃗⃗⃗⃗ ⁡𝑘⁡
⁡𝑗⁡
(⁡𝑎⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡𝑐⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗ = ⁡ | 11 −7 1 | = ⁡ −19⁡𝑖⁡ ⃗⃗⃗⃗ − 34⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 29⁡𝑘⁡ ………. (1) 𝑥 − ⁡0 𝑦 − ⁡1 𝑧 + 5 (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (0, 1, −5)
−1 −2 3 | 2 3 6 |=0 (𝑏1 ⁡, 𝑏2 , 𝑏3 ) = (2,⁡⁡⁡⁡3,⁡⁡⁡6)
⃗⃗⃗⃗⃗ − ⁡ (⁡𝑏⁡⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡ 1 1 −1 (𝑐1 ⁡, 𝑐2 , 𝑐3 ) = (1,⁡⁡⁡1, −1)
RHS (⁡𝑎⁡ ⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⃗⃗⃗⃗)⁡𝑏⁡ ⃗⃗⃗⃗)⁡𝑎⁡⃗⃗⃗⃗⃗
(𝑥⁡ − 0)⁡(−3 − 6)⁡–⁡(𝑦 − 1)⁡(−2 − 6) ⁡ + ⁡ (𝑧⁡ + ⁡5)⁡(2⁡– ⁡3) ⁡ = ⁡0
⁡𝑎⁡
⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⃗⃗⃗⃗ = ⁡ −11 𝑥(−9) ⁡ + ⁡8(𝑦⁡ − 1)⁡– ⁡1(𝑧 + ⁡5) ⁡ = ⁡0
⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⁡𝑏⁡ ⃗⃗⃗⃗ = ⁡ −7 −9𝑥⁡ + ⁡8𝑦⁡– ⁡𝑧⁡– ⁡13⁡ = ⁡0
/ (⁡𝑎⁡
⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⃗⃗⃗⃗)⁡𝑏⁡⃗⃗⃗⃗⃗ − ⁡ (⁡𝑏⁡
⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⃗⃗⃗⃗)⁡𝑎⁡
⃗⃗⃗⃗⃗ = ⁡ −19⁡𝑖⁡ ⃗⃗⃗⃗ − 34⁡𝑗⁡ ⃗⃗⃗⃗⃗ ………. (2)
⃗⃗⃗⃗ − 29⁡𝑘⁡ / 9x – 8y + z + 13 = 0
(iii) Non – parametric vector equation..
From (1) & (2)
(⁡𝑟⁡
⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗). (⁡𝑏⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡
⃗⃗⃗⃗) = 0
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑏⁡
(⁡𝑎⁡ ⃗⃗⃗⃗⃗) ⁡ × ⃗⃗⃗⃗
⁡𝑐⁡ =⁡ (⁡𝑎⁡ ⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡ ⃗⃗⃗⃗⃗ − ⁡ (⁡𝑏⁡
⃗⃗⃗⃗)⁡𝑏⁡ ⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⃗⃗⃗⃗)⁡𝑎⁡
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⁡. (9⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗⃗ − 8⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + ⁡𝑘⁡) + 13 = 0
HENCE PROVED
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡ ⃗⃗⃗⃗⃗ − ⁡ (⁡𝑎⁡ 9. Find the non – parametric form of vector equation and Cartesian equation of the plane passing
ii. ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ × (⁡𝑏⁡ ⃗⃗⃗⃗) =⁡ (⁡𝑎⁡⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⃗⃗⃗⃗)⁡𝑏⁡ ⃗⃗⃗⃗⃗⁡. ⃗⃗⃗⃗⃗
⁡𝑏⁡)⁡𝑐⁡
⃗⃗⃗⃗
through the point (2,3,6) and parallel to the straight lines
LHS ⁡𝑎⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ ⁡ × (⁡𝑏⁡ ⁡ × ⁡ ⁡𝑐⁡ ⃗⃗⃗⃗) 𝒙−𝟏 𝒚+𝟏 𝒛−𝟑 𝒙+𝟑 𝒚−𝟑 𝒛+𝟏
=⁡ =⁡ & ⁡⁡⁡⁡ =⁡ =⁡ .
𝟐 𝟑 𝟏 𝟐 −𝟓 −𝟑
⁡𝑖⁡
⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
⁡𝑗⁡ ⁡𝑘⁡ Solution :
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡
⁡𝑏⁡ ⃗⃗⃗⃗ = ⁡ | 3 5 2 | = ⁡19⁡𝑖⁡
⃗⃗⃗⃗ − 11⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − ⁡𝑘⁡
⁡𝑎⁡ ⁡ = ⁡2⁡𝑖⁡
⃗⃗⃗⃗⃗ ⃗⃗⃗ + 3⁡𝑗⁡
⃗⃗⃗⃗ + 6⁡𝑘⁡⃗⃗⃗⃗⃗
−1 −2 3
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ⁡ = ⁡2⁡𝑖⁡
⁡𝑏⁡ ⃗⃗⃗⃗ + 3⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + ⁡𝑘⁡
⃗⃗⃗⃗
⁡𝑖⁡ ⃗⃗⃗⃗
⁡𝑗⁡ ⁡𝑘⁡
⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ × (⁡𝑏⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡
⃗⃗⃗⃗) = ⁡ | 2 3 −1| ⁡ = ⁡ −14⁡𝑖⁡ ⃗⃗⃗ − 17⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 79⁡𝑘⁡ ⃗⃗⃗⃗ ⁡ = ⁡2⁡𝑖⁡
⁡𝑐⁡ ⃗⃗⃗⃗ − 5⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 3⁡𝑘⁡
19 −11 −1 (i) Parametric vector equation : ⃗⃗⃗⃗ ⁡𝑟⁡ = ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗⃗ + 𝑡⁡⁡𝑐⁡ ⃗⃗⃗⃗
⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⁡𝑎⁡ ⃗⃗⃗⃗ = ⁡ −11 ⃗⃗⃗⃗⃗) + 𝑠(2⁡𝑖⁡ ⃗⃗⃗⃗⃗) + 𝑡(2⁡𝑖⁡
⃗⃗⃗⃗ = ⁡ (2⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗⃗ + 3⁡𝑗⁡
⃗⃗⃗⃗ + 6⁡𝑘⁡ ⃗⃗⃗ + 3⁡𝑗⁡ ⃗⃗⃗⃗ + ⁡𝑘⁡ ⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗ − 5⁡𝑗⁡ 3⁡𝑘⁡)
⃗⃗⃗⃗⃗⁡. ⃗⃗⃗⃗⃗
⁡𝑎⁡ ⁡𝑏⁡ = ⁡19 𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
RHS (⁡𝑎⁡
⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡ ⃗⃗⃗⃗⃗ − ⁡ (⁡𝑎⁡
⃗⃗⃗⃗)⁡𝑏⁡ ⃗⃗⃗⃗⃗)⁡𝑐⁡
⃗⃗⃗⃗⃗⁡. ⁡𝑏⁡ ⃗⃗⃗⃗ = ⁡ −14⁡𝑖⁡
⃗⃗⃗⃗ − 17⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 79⁡𝑘⁡ (ii) Cartesian Equation : | 𝑏1 𝑏2 𝑏3 | = 0
⃗⃗⃗⃗⃗ − ⁡ (⁡𝑎⁡ ⃗⃗⃗⃗⃗)⁡𝑐⁡ ⃗⃗⃗⃗⃗ 𝑐1 𝑐2 𝑐3
From (1) & (2) ⁡(⁡𝑎⁡
⃗⃗⃗⃗⃗⁡. ⁡𝑐⁡
⃗⃗⃗⃗)⁡𝑏⁡ ⃗⃗⃗⃗⃗⁡. ⁡𝑏⁡ ⃗⃗⃗⃗ = ⁡ −14⁡𝑖⁡
⃗⃗⃗⃗ − 17⁡𝑗⁡
⃗⃗⃗⃗ − 79⁡𝑘⁡
𝑥 − 2 𝑦 − ⁡3 𝑧 − 6 (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (2, 3, 6)
Hence proved. | 2 (𝑏1 ⁡, 𝑏2 , 𝑏3 ) = (2, 3, 1)
3 1 |=0
One Point & Two vectors 2 −5 −3 (𝑐1 ⁡, 𝑐2 , 𝑐3 ) = (2, −5, −3)
Parametric vector equation : ⃗⃗⃗⃗ = ⁡ ⁡𝑎⁡
⁡𝑟⁡ ⃗⃗⃗⃗⃗ + 𝑠⁡𝑐⁡
⃗⃗⃗⃗⃗ + 𝑡⁡𝑏⁡ ⃗⃗⃗⃗ (𝑥 − 2)⁡(−9 + 5)⁡–⁡(𝑦 − 3)⁡(−6 − 2) ⁡ + ⁡ (𝑧 − 6)⁡(−10 − 6) ⁡ = 0
(𝑥⁡ − 2)⁡(−4)⁡– (𝑦 − 3)⁡(−8) ⁡ + ⁡ (𝑧 − 6)(−16) ⁡ = ⁡0
Non – Parametric vector equation : (⁡𝑟⁡ ⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡
⃗⃗⃗⃗⃗)⁡. (⁡𝑏⁡ ⃗⃗⃗⃗) = 0 (𝑥 − 2)⁡(−1) ⁡⁡ + ⁡2(𝑦 − 3)⁡– ⁡4(𝑧⁡– ⁡6) ⁡ = ⁡0
𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1 −𝑥⁡ + ⁡2𝑦⁡– ⁡4𝑧⁡ + ⁡20⁡ = ⁡0
Cartesian equation : ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡| 𝑏1 𝑏2 𝑏3 | = 0 𝑥⁡– ⁡2𝑦⁡ + 4𝑧⁡ − 20⁡ = ⁡0
𝑐1 𝑐2 𝑐3 ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡
(iii) Non – parametric vector equation : (⁡𝑟⁡
⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡⃗⃗⃗⃗⃗). (⁡𝑏⁡ ⃗⃗⃗⃗) = 0
⃗⃗⃗⃗⁡. (⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗ − 2⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 4⁡𝑘⁡) − 20 = 0`
32 33
10. Find the vector ( parametric and non- parametric and Cartesian equation of the plane containing 12. Find the non – parametric from of vector equation and Cartesian equation of the plane
the line ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ − ⁡𝒋⁡
⁡𝒓⁡ = ⁡ (⁡𝒊⁡ ⃗⃗⃗⃗⃗) + ⁡⁡𝒕(𝟐⁡𝒊⁡
⃗⃗⃗⃗ + 𝟑⁡𝒌⁡ ⃗⃗⃗⃗ − ⁡𝒋⁡ ⃗⃗⃗⃗⃗) and perpendicular to the plane
⃗⃗⃗⃗ + 𝟒⁡𝒌⁡ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ − ⃗⃗⃗⃗
⁡𝒓⁡ = ⁡ (𝟔⁡𝒊⁡ ⃗⃗⃗⃗⃗) + 𝒔(−⁡𝒊⁡
⁡𝒋⁡ + ⁡𝒌⁡ ⃗⃗⃗⃗ + 𝟐⁡𝒋⁡ ⃗⃗⃗⃗⃗) + 𝒕(−𝟓⁡𝒊⁡
⃗⃗⃗⃗ + ⁡𝒌⁡ ⃗⃗⃗⃗ − 𝟒⁡𝒋⁡ ⃗⃗⃗⃗⃗⃗⃗⃗)
⃗⃗⃗⃗ − 𝟓⁡𝒌⁡
⃗⃗⃗⃗⃗. (⁡𝒊⁡
⁡𝒓⁡ ⃗⃗⃗⃗ + 𝟐⁡𝒋⁡ ⃗⃗⃗⃗⃗) = 𝟖.
⃗⃗⃗⃗ + ⁡𝒌⁡ Solution : ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ = ⁡6⁡𝑖⁡
⃗⃗⃗ − ⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + ⁡𝑘⁡ (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (6, −1, 1)
Solution : ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ = ⁡ ⁡𝑖⁡
⃗⃗⃗⃗ − ⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 3⁡𝑘⁡ (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (1, −1, 3) ⃗⃗⃗⃗⃗
⁡𝑏⁡ ⁡ = ⁡ −⁡𝑖⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 2⁡𝑗⁡ ⁡𝑘⁡ (𝑏1 ⁡, 𝑏2 , 𝑏3 ) = (−1, 2, 1)
⃗⃗⃗⃗⃗
⁡𝑏⁡ ⁡ = ⁡2⁡𝑖⁡ ⃗⃗⃗⃗ − ⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 4⁡𝑘⁡ (𝑏1 ⁡, 𝑏2 , 𝑏3 ) = (2, −1, 4) ⃗⃗⃗⃗
⁡𝑐⁡ ⁡ = −5⁡⁡𝑖⁡
⃗⃗⃗ − 4⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 5⁡𝑘⁡ (𝑐1 ⁡, 𝑐2 , 𝑐3 ) = (−5, −4, −5)
𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
⃗⃗⃗⃗
⁡𝑐⁡ ⁡ = ⁡ ⃗⃗⁡𝑖⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗ + 2⁡𝑗⁡ ⁡𝑘⁡ (𝑐1 ⁡, 𝑐2 , 𝑐3 ) = (1, 2, 1)
(i) Cartesian equation | 𝑏1 𝑏2 𝑏3 | = 0
(i) Parametric vector equation ⃗⃗⃗⃗ ⁡𝑟⁡ = ⁡𝑎⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡⃗⃗⃗⃗⃗ + 𝑡⁡⁡𝑐⁡⃗⃗⃗⃗ 𝑐1 𝑐2 𝑐3
⁡𝑟⁡
⃗⃗⃗⃗ = ⁡ (⁡𝑖⁡
⃗⃗⃗– ⁡𝑗⁡ ⃗⃗⃗⃗⃗) + 𝑠(2⁡𝑖⁡
⃗⃗⃗⃗ + 3⁡𝑘⁡ ⃗⃗⃗– ⃗⃗⃗⃗⁡𝑗⁡ + 4⁡𝑘⁡⃗⃗⃗⃗⃗) + 𝑡(⁡𝑖⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 2⁡𝑗⁡ ⁡𝑘⁡) 𝑥−6 𝑦+1 𝑧−1
𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1 | −1 2 1 |=0
(iii) Cartesian equation : | 𝑏1 𝑏2 𝑏3 | = 0 −5 −4 −5
𝑐1 𝑐2 𝑐3 (𝑥 − 6)(−10 + 4) − (𝑦 + 1)(5 + 5) + (𝑧 − 1)(4 + 10) = 0
𝑥−1 𝑦+1 𝑧−3 (𝑥 − 6)(−6) − (𝑦 + 1)(10) + (𝑧 − 1)(14) = 0
| 2 −1 4 |=0 −3𝑥 − 5𝑦 + 7𝑧 + 6⁡ = ⁡0
1 2 1 / 3𝑥 + 5𝑦 − 7𝑧 − 6⁡ = ⁡0
(𝑥⁡ − 1)⁡(−1 − 8)⁡–⁡(𝑦⁡ + ⁡1)⁡(2 − 4) ⁡ + ⁡ (𝑧 − 3)⁡(4⁡ + ⁡1) ⁡ = ⁡0
(ii) Non – parametric vector equation :
(𝑥⁡ − 1)⁡(−9)⁡–⁡(𝑦 + 1)⁡(−2) ⁡ + ⁡ (𝑧 − 3)⁡(5) ⁡ = ⁡0
(⁡𝑟⁡
⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡
⃗⃗⃗⃗⃗). (⁡𝑏⁡ ⃗⃗⃗⃗) = 0
−9𝑥⁡ + ⁡2𝑦⁡ + ⁡5𝑧⁡– ⁡4⁡ = ⁡0
⃗⃗⃗⃗⁡. (3⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗⃗ + 5⁡𝑗⁡ ⃗⃗⃗⃗⃗) − 6 = 0
⃗⃗⃗⃗ − 7⁡𝑘⁡
/ 9𝑥⁡– ⁡2𝑦 − 5𝑧 + 4⁡ = ⁡0
(iii) Non parametric vector equation (⁡𝑟⁡ ⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡
⃗⃗⃗⃗⃗). (⁡𝑏⁡ ⃗⃗⃗⃗) = 0
TWO POINTS AND ONE VECTORS
⁡𝑟⁡ (9⁡𝑖⁡
⃗⃗⃗⃗⁡. ⃗⃗⃗ − 2⁡𝑗⁡ ⃗⃗⃗⃗⃗) + 4 = 0
⃗⃗⃗⃗ − 5⁡𝑘⁡
1. Parametric vector equation : ⁡⁡⁡𝑟⁡ ⃗⃗⃗⃗ = (1 − 𝑠)⁡𝑎⁡ ⃗⃗⃗⃗⃗ + 𝑡⁡𝑐⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗
11. Find the vector (parametric and non – parametric) and Cartesian equations of the plane passing
2. Non – parametric vector : (⁡𝑟⁡⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗). [(⁡𝑏⁡ − ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗) ⁡ × ⁡𝑐⁡
⃗⃗⃗⃗] = 0
through the point (1,-2,4) and perpendicular to the plane ;
𝒙+𝟕 𝒚+𝟑 𝒛 𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
x + 2y – 3z = 11 and parallel to the line =⁡ = ⁡ 𝟏. 3. Cartesian equation: |𝑥2 − ⁡ 𝑥1 ⁡ 𝑦2 − ⁡ 𝑦1 𝑧2 − ⁡ 𝑧1 | = 0
𝟑 −𝟏
Solution : ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ = ⁡ ⁡𝑖⁡
⃗⃗⃗⃗ − 2⁡𝑗⁡⃗⃗⃗⃗ + 4⁡𝑘⁡ ⃗⃗⃗⃗⃗ (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (1, −2, 4) 𝑐1 𝑐2 𝑐3
⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 13. Find the parametric form of vector equation and Cartesian equation of the plane passing
⁡𝑏⁡ ⁡ = ⁡ ⁡𝑖⁡
⃗⃗⃗⃗ + 2⁡𝑗⁡⃗⃗⃗⃗ − 3⁡𝑘⁡ (𝑏1 ⁡, 𝑏2 , 𝑏3 ) = (1, 2, −3)
through the points ( 2,2,1) (9,3,6) and perpendicular to the plane 𝟐𝒙+6y+6z= 9.
⃗⃗⃗⃗
⁡𝑐⁡ ⁡ = 3⁡⁡𝑖⁡ ⃗⃗⃗ − ⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + ⁡𝑘⁡ (𝑐1 ⁡, 𝑐2 , 𝑐3 ) = (3, −1, 1)
Ans : ⁡𝑎⁡ ⃗⃗⃗⃗⃗ ⁡ = ⁡2⁡𝑖⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗ + 2⁡𝑗⁡ ⁡𝑘⁡ (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (2, 2, 1)
(i) Parametric vector equation : ⁡𝑟⁡ ⃗⃗⃗⃗ = ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗⃗ + 𝑡⁡⁡𝑐⁡⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
⁡𝑏⁡ ⁡ = ⁡9⁡𝑖⁡ ⃗⃗⃗⃗ + 3⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 6⁡𝑘⁡ (𝑥2 ⁡, 𝑦2 , 𝑧2 ) = (9, 3, 6)
⁡𝑟⁡
⃗⃗⃗⃗ = ⁡ (⁡𝑖⁡
⃗⃗⃗ − 2⁡𝑗⁡ ⃗⃗⃗⃗⃗) + 𝑠(⁡𝑖⁡
⃗⃗⃗⃗ + 4⁡𝑘⁡ ⃗⃗⃗ + 2⁡𝑗⁡ ⃗⃗⃗⃗ − 3⁡𝑘⁡ ⃗⃗⃗⃗⃗) + 𝑡(3⁡𝑖⁡
⃗⃗⃗⃗ − ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗)
⁡𝑗⁡ + ⁡𝑘⁡
⃗⃗⃗⃗
⁡𝑐⁡ ⁡ = 2⁡⁡𝑖⁡
⃗⃗⃗⃗ + 6⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 6⁡𝑘⁡ (𝑐1 ⁡, 𝑐2 , 𝑐3 ) = (2, 6, 6)
𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
(ii) Cartesian equation : | 𝑏1 𝑏2 𝑏3 | = 0 ⁡𝑟⁡ = (1 − 𝑠)⁡𝑎⁡
i. Parametric vector equation : ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ + 𝑡⁡𝑐⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗
𝑐1 𝑐2 𝑐3 ⁡𝑟⁡ = ⁡ (1 − 𝑠)(2⁡𝑖⁡
⃗⃗⃗⃗ ⃗⃗⃗ + 2⁡𝑗⁡ ⃗⃗⃗⃗⃗) + 𝑠(9⁡𝑖⁡
⃗⃗⃗⃗ + ⁡𝑘⁡ ⃗⃗⃗ + 3⁡𝑗⁡ ⃗⃗⃗⃗⃗) + 𝑡(2⁡𝑖⁡
⃗⃗⃗⃗ + 6⁡𝑘⁡ ⃗⃗⃗⃗ + 6⁡𝑗⁡ ⃗⃗⃗⃗⃗)
⃗⃗⃗⃗ + 6⁡𝑘⁡
𝑥−1 𝑦+2 𝑧−4 𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
| 1 2 −3 | = 0 ii. Cartesian equation : |𝑥2 − ⁡ 𝑥1 ⁡ 𝑦2 − ⁡ 𝑦1 𝑧2 − ⁡ 𝑧1 | = 0
3 −1 1 𝑐1 𝑐2 𝑐3
(𝑥 − 1)⁡(2 − 3)⁡–⁡(𝑦 + 2)⁡(1 + 9) ⁡ + ⁡ (𝑧 − 4)⁡(−1 − 6) ⁡ = ⁡0 𝑥−2 𝑦−2 𝑧−1
(𝑥⁡– ⁡1)⁡(−1)⁡–⁡(𝑦⁡ + 2)⁡(10) ⁡ + ⁡ (𝑧 − 4)⁡(−7) ⁡ = ⁡0 |9 − 2 3 − 2 6 − 1| = 0
−𝑥 − 10𝑦 − 7𝑧 + 9 = 0 2 6 6
/ 𝑥 + 10𝑦 + 7𝑧 − 9⁡ = ⁡0 𝑥−2 𝑦−2 𝑧−1
| 7 1 5 |=0
(iii) Non – Parametric vector equation : (⁡𝑟⁡ ⃗⃗⃗⃗ − ⁡ ⃗⃗⃗⃗⃗).
⁡𝑎⁡ (⁡𝑏⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑐⁡
⃗⃗⃗⃗) = 0 2 6 6
⃗⃗⃗⃗⁡. (⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗ + 10⁡𝑗⁡ ⃗⃗⃗⃗⃗) − 9 = 0
⃗⃗⃗⃗ + 7⁡𝑘⁡ (𝑥 − 2)⁡(6 − 30)⁡–⁡(𝑦 − 2)⁡(42 − 10) ⁡ + (𝑧 − 1)⁡(42 − 2) ⁡ = ⁡0
(𝑥 − 2)⁡(−24)⁡–⁡(𝑦 − 2)⁡(32) ⁡ + ⁡ (𝑧 − 1)(40) ⁡ = ⁡0
(𝑥 − 2)⁡(−3)⁡– (𝑦 − 2)⁡(4) ⁡ + ⁡ (𝑧 − 1)⁡(5) ⁡ = ⁡0
−3𝑥⁡– ⁡4𝑦⁡ + ⁡5𝑧⁡ + ⁡9⁡ = ⁡0
/ 3𝑥 + 4𝑦 − 5𝑧 − 9⁡ = ⁡0

34 35
 iii. Non – parametric vector equation : (⁡𝑟⁡ ⃗⃗⃗⃗ − ⁡ ⃗⃗⃗⃗⃗). ⃗⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡
⁡𝑎⁡ [(⁡𝑏⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡𝑐⁡
⃗⃗⃗⃗] = 0 (𝑥 − 2)(−20 + 8) − (𝑦 − 2)(−5 − 6) + (𝑧 − 1)(4 + 12) = 0
⃗⃗⃗⃗⃗ (𝑥 − 2)(−12) − (𝑦 − 2)(−11) + (𝑧 − 1)(16) = 0
⁡𝑟⁡
⃗⃗⃗⃗⁡. (3⁡𝑖⁡
⃗⃗⃗ + 4⁡𝑗⁡
⃗⃗⃗⃗ − 5⁡𝑘⁡) − 9 = 0
−12𝑥 + 11𝑦 + 16𝑧 − 14⁡ = ⁡0
/ 12𝑥 − 11𝑦 − 16𝑧 + 14⁡ = ⁡0
14. Find the vector parametric , vector non- parametric and carterian form of the equation of the
iii. Non – parametric vector equation : (⁡𝑟⁡
⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡ ⃗⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗). [(⁡𝑏⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡𝑐⁡
⃗⃗⃗⃗] = 0
plane passing through the points (-1, 2, 0) (2, 2, -1) and parallel to the straight line ;
𝒙−𝟏 𝟐𝒚+𝟏 𝒛+𝟏 ⃗⃗⃗⃗⁡. (12⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗ − 11⁡𝑗⁡ ⃗⃗⃗⃗⃗) = −14
⃗⃗⃗⃗ − 16⁡𝑘⁡
𝟏
=⁡ 𝟐
=⁡ −𝟏
.
Ans : ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ = ⁡ −⁡𝑖⁡
⃗⃗⃗⃗ + 2⁡𝑗⁡
⃗⃗⃗⃗ (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (−1, 2, 0)
⃗⃗⃗⃗⃗ 3 Points
⁡𝑏⁡ ⁡ = ⁡2⁡𝑖⁡ ⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 2⁡𝑗⁡ ⁡𝑘⁡ (𝑥2 ⁡, 𝑦2 , 𝑧2 ) = (2, 2, −1)
1. ⃗⃗⃗⃗ = (1 − 𝑠 − 𝑡)⁡𝑎⁡
Parametric vector equation: ⁡𝑟⁡ ⃗⃗⃗⃗⃗ + 𝑡⁡𝑐⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗
⃗⃗⃗⃗
⁡𝑐⁡ ⁡ = ⁡ ⃗⃗⁡𝑖⁡
⃗⃗ + ⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − ⁡𝑘⁡ (𝑐1 ⁡, 𝑐2 , 𝑐3 ) = (1, 1, −1)
i. Parametric vector equation: ⃗⃗⃗⃗ ⁡𝑟⁡ = (1 − 𝑠)⁡𝑎⁡ ⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗⃗ + 𝑡⁡𝑐⁡ ⃗⃗⃗⃗ 2. Non – parametric vector equation : [⁡𝑟⁡ ⃗⃗⃗⃗⃗, ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡ ⁡𝑏⁡ − ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗, ⁡𝑐⁡
⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗] = 0
⃗⃗⃗⃗ = ⁡ (1 − 𝑠)(−⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗⃗) + 𝑠(2⁡𝑖⁡
⃗⃗⃗⃗ + 2⁡𝑗⁡ ⃗⃗⃗⃗ + 2⁡𝑗⁡
⃗⃗⃗⃗ − ⁡𝑘⁡ ⃗⃗⃗⃗⃗) + 𝑡(⁡𝑖⁡ ⃗⃗⃗⃗ + ⁡𝑗⁡ ⃗⃗⃗⃗⃗)
⃗⃗⃗⃗ − ⁡𝑘⁡ 𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1 3. Cartesian equation: |𝑥2 − ⁡ 𝑥1 ⁡ 𝑦2 − ⁡ 𝑦1 𝑧2 − ⁡ 𝑧1 | = 0
ii. Cartesian equation: |𝑥2 − ⁡ 𝑥1 ⁡ 𝑦2 − ⁡ 𝑦1 𝑧2 − ⁡ 𝑧1 | = 0 𝑥3 − ⁡ 𝑥1 𝑦3 − ⁡ 𝑦1 𝑧3⁡ − ⁡ 𝑧1
𝑐1 𝑐2 𝑐3
𝑥+1 𝑦−2 𝑧−0 16. Find the vector (Parametric and non – parametric) and Cartesian from of the equations of the
|2 + 1 2 − 2 −1 − 0| = 0
plane (3, 6, -2), (-1, -2, 6) and (6, 4, -2)
1 1 −1
𝑥+1 𝑦−2 𝑧−0 ANS: ⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ = ⁡3⁡𝑖⁡
⃗⃗⃗ + 6⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 2⁡𝑘⁡ (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (3, 6, −2)
| 3 0 −1 | = 0 ⃗⃗⃗⃗⃗
⁡𝑏⁡ ⁡ = −⁡𝑖⁡⃗⃗⃗ − 2⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 6⁡𝑘⁡ (𝑥2 ⁡, 𝑦2 , 𝑧2 ) ⁡ = (−1, −2, 6)
1 1 −1 ⃗⃗⃗⃗⃗
⃗⃗⃗⃗
⁡𝑐⁡ ⁡ = 6⁡𝑖⁡
⃗⃗⃗ + 4⁡𝑗⁡
⃗⃗⃗⃗ − 2⁡𝑘⁡ (𝑥3 ⁡, 𝑦3 , 𝑧3 ) ⁡ = (6, 4, −2)
(𝑥 + 1)(0 + 1) − (𝑦 − 2)(−3 + 1) + 𝑧(3 − 0) = 0
1. Parametric vector equation: ⁡𝑟⁡ ⃗⃗⃗⃗ = (1 − 𝑠 − 𝑡)⁡𝑎⁡ ⃗⃗⃗⃗⃗ + 𝑡⁡𝑐⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗
(𝑥 + 1) − (𝑦 − 2)(−2) + 3𝑧 = 0
/ 𝑥 + 2𝑦 + 3𝑧 − 3⁡ = ⁡0 ⃗⃗⃗⃗ = ⁡ (1 − 𝑠 − 𝑡)3⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗⃗ + 6⁡𝑗⁡ ⃗⃗⃗⃗ − 2⁡𝑘⁡⃗⃗⃗⃗⃗ + 𝑠(−⁡𝑖⁡
⃗⃗⃗⃗⃗⃗⃗ − 2⁡𝑗⁡ ⃗⃗⃗⃗⃗) + 𝑡(6⁡𝑖⁡
⃗⃗⃗⃗ + 6⁡𝑘⁡ ⃗⃗⃗ + 4⁡𝑗⁡ ⃗⃗⃗⃗⃗)
⃗⃗⃗⃗ − 2⁡𝑘⁡
⃗⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡ 𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
iii. Non – parametric vector equation : (⁡𝑟⁡ ⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡⃗⃗⃗⃗⃗). [(⁡𝑏⁡ ⃗⃗⃗⃗⃗) ⁡ × ⁡𝑐⁡
⃗⃗⃗⃗] = 0
2. Cartesian equation : |𝑥2 − ⁡ 𝑥1 ⁡ 𝑦2 − ⁡ 𝑦1 𝑧2 − ⁡ 𝑧1 | = 0
⃗⃗⃗⃗⁡. (⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗⃗ + 2⁡𝑗⁡ ⃗⃗⃗⃗⃗) − 3 = 0
⃗⃗⃗⃗ + 3⁡𝑘⁡ 𝑥3 − ⁡ 𝑥1 𝑦3 − ⁡ 𝑦1 𝑧3⁡ − ⁡ 𝑧1
𝑥−3 𝑦−6 𝑧+2
15. Find parametric from of vector equation and Cartesian equation of the plane passing through |−1 − 3 −2 − 6 6 + 2 | = 0
the points (2, 2, 1) (1, -2, 3)and parallel to the straight line passing through the points (2, 1, -3) 6−3 4 − 6 −2 + 2
𝑥−3 𝑦−6 𝑧+2
and (-1, 5, -8).
| −4 −8 8 |=0
Ans : ⃗⃗⃗⃗⃗
⁡𝑎⁡ ⁡ = ⁡2⁡𝑖⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗ + 2⁡𝑗⁡ ⁡𝑘⁡ (𝑥1 ⁡, 𝑦1 , 𝑧1 ) = (2,2,1) 3 −2 0
⃗⃗⃗⃗⃗
⁡𝑏⁡ ⁡ = ⁡𝑖⁡
⃗⃗⃗⃗ − 2⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 3⁡𝑘⁡ (𝑥2 ⁡, 𝑦2 , 𝑧2 ) ⁡ = (1, −2, 3) (𝑥⁡ − 3)⁡(0 + 16)⁡–⁡(𝑦 − 6)⁡(0 − 24) ⁡ + ⁡ (⁡𝑧 + ⁡2)⁡(8⁡ + ⁡24) ⁡ = ⁡0
⁡𝑐⁡ ⁡ = (2, 1, −3) − (−1, 5, −8)
⃗⃗⃗⃗ (𝑥⁡ − 3)⁡(16)⁡–⁡(𝑦 − 6)⁡(−24) ⁡ + ⁡ (𝑧⁡ + 2)⁡(32) ⁡ = ⁡0
(𝑥⁡ − 3)⁡(2)⁡–⁡(𝑦⁡ − 6)⁡(−3) ⁡ + ⁡ (𝑧 + 2)⁡(4) ⁡ = ⁡0
= (2 + 1, 1 − 5, −3 + 8)
/ 2𝑥 + 3𝑦 + 4𝑧 − 16 = 0
= (3, −4, 5)⁡ 3. Non – parametric vector equation
/ ⁡𝑐⁡ ⃗⃗⃗⃗ = ⁡3⁡𝑖⁡
⃗⃗⃗ − 4⁡𝑗⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 5⁡𝑘⁡ (𝑐1 ⁡, 𝑐2 , 𝑐3 ) = (3, −4,5) ⃗⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡
[⁡𝑟⁡
⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗, ⁡𝑏⁡ ⃗⃗⃗⃗⃗, ⁡𝑐⁡
⃗⃗⃗⃗ − ⁡ ⁡𝑎⁡
⃗⃗⃗⃗⃗] = 0
i. Parametric vector equation: ⁡𝑟⁡ ⃗⃗⃗⃗ = (1 − 𝑠)⁡𝑎⁡ ⃗⃗⃗⃗⃗ + 𝑡⁡𝑐⁡
⃗⃗⃗⃗⃗ + 𝑠⁡𝑏⁡ ⃗⃗⃗⃗
⃗⃗⃗⃗⁡. (2⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗ + 3⁡𝑗⁡ ⃗⃗⃗⃗⃗) = 16
⃗⃗⃗⃗ + 4⁡𝑘⁡
⃗⃗⃗⃗ = ⁡ (1 − 𝑠)(2⁡𝑖⁡
⁡𝑟⁡ ⃗⃗⃗⃗ + 2⁡𝑗⁡ ⃗⃗⃗⃗⃗⁡) + 𝑠(⁡𝑖⁡
⃗⃗⃗⃗ + ⁡𝑘⁡ ⃗⃗⃗ − 2⁡𝑗⁡
⃗⃗⃗⃗ + 3⁡𝑘⁡⃗⃗⃗⃗⃗) + 𝑡(3⁡𝑖⁡
⃗⃗⃗⃗ − 4⁡𝑗⁡ ⃗⃗⃗⃗⃗)
⃗⃗⃗⃗ + 5⁡𝑘⁡ 17. Find the vector parametric and Cartesian equation of a straight line passing through the points
𝑥 − ⁡ 𝑥1 𝑦 − ⁡ 𝑦1 𝑧 − ⁡ 𝑧1
(-5, 7, -4) and (13, -5, 2). Find the point where the straight line crosses the xy- plane..
ii. Cartesian equation: |𝑥2 − ⁡ 𝑥1 ⁡ 𝑦2 − ⁡ 𝑦1 𝑧2 − ⁡ 𝑧1 | = 0
𝑐1 𝑐2 𝑐3 Ans : A (-5, 7, -4) B (13, -5, 2)
𝑥−2 𝑦−2 𝑧−1 AB (13 + 5, -5-7, 2+4) = AB (18, -12, 6)
|1 − 2 −2 − 2 3 − 1| = 0 = AB (3, -2, 1)
3 −4 5 1. Parametric vector equation:
𝑥−2 𝑦−2 𝑧−1 ⁡𝑟⁡ = ⁡ (−5⁡𝑖⁡
⃗⃗⃗⃗ ⃗⃗⃗⃗ + 7⁡𝑗⁡ ⃗⃗⃗⃗⃗) + 𝑡(3⁡𝑖⁡
⃗⃗⃗⃗ − 4⁡𝑘⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 2⁡𝑗⁡ ⁡𝑘⁡) and
| −1 −4 2 |=0
3 −4 5 ⁡𝑟⁡
⃗⃗⃗⃗ = ⁡ (13⁡𝑖⁡
⃗⃗⃗ − 5⁡𝑗⁡ ⃗⃗⃗⃗⃗) + 𝑠(3⁡𝑖⁡
⃗⃗⃗⃗ + 2⁡𝑘⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ − 2⁡𝑗⁡ ⁡𝑘⁡)
36 37
 2. Cartesian equation : 24. For any vectors ⃗⃗⃗⃗⃗, ⁡𝒂⁡ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⁡then proved that [⁡𝒂⁡
⁡𝒃⁡, ⁡𝒄⁡ ⃗⃗⃗⃗⃗ + ⁡𝒄⁡ ⃗⃗⃗⃗⃗ + ⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗, ⁡𝒂⁡ ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⁡𝒃⁡, ⁡𝒂⁡ ⁡𝒃⁡ + ⁡𝒄⁡ ⃗⃗⃗⃗⃗, ⃗⃗⃗⃗⃗
⃗⃗⃗⃗] = ⁡ [⁡𝒂⁡ ⃗⃗⃗⃗]
⁡𝒃⁡, ⁡𝒄⁡
𝑥+5 𝑦−7 𝑧+4 𝑥−13 𝑦+5 𝑧−2
3
=⁡ −2
=⁡ 1
and 3
=⁡ −2
=⁡ 1
Ans : L.H.S
𝑥+5 𝑦−7 𝑧+4 1 0 1
=⁡ =⁡ ⁡= 𝑡 [⁡𝑎⁡
⃗⃗⃗⃗⃗ + ⁡𝑐⁡
⃗⃗⃗⃗, ⃗⃗⃗⃗⃗,
⃗⃗⃗⃗⃗ + ⁡ ⁡𝑏⁡
⁡𝑎⁡ ⃗⃗⃗⃗⃗ + ⁡𝑏⁡
⁡𝑎⁡ ⃗⃗⃗⃗⃗ + ⁡𝑐⁡ ⃗⃗⃗⃗] = ⁡⁡ |1 1 0| [⁡𝑎⁡ ⃗⃗⃗⃗⃗, ⁡𝑐⁡
⃗⃗⃗⃗⃗, ⁡𝑏⁡ ⃗⃗⃗⃗⁡]
3 −2 1
(x, y, z) = (3t – 5, -2t + 7, t -4) 1 1 1
= [⁡𝑎⁡ ⃗⃗⃗⃗⃗, ⁡𝑐⁡
⃗⃗⃗⃗⃗, ⁡𝑏⁡ ⃗⃗⃗⃗⁡]
Crosses xy plane = 0
t–4=0⟹𝑡=4 [⁡𝑎⁡
⃗⃗⃗⃗⃗ + ⁡𝑐⁡
⃗⃗⃗⃗, ⁡𝑎⁡ ⃗⃗⃗⃗⃗, ⁡𝑎⁡
⃗⃗⃗⃗⃗ + ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗⃗ + ⁡𝑏⁡ ⃗⃗⃗⃗⃗ + ⁡𝑐⁡
⃗⃗⃗⃗] ⁡⁡ = ⁡ [⁡𝑎⁡ ⃗⃗⃗⃗⃗, ⁡𝑐⁡
⃗⃗⃗⃗⃗, ⁡𝑏⁡ ⃗⃗⃗⃗⁡] Hence proved.
Crosses the xy plane (x, y, z) = (7,-1,0) 25. Prove that the distance from the origin to the plane 𝟑𝒙⁡ + ⁡𝟔𝒚⁡ + ⁡𝟐𝒛⁡ + ⁡𝟕⁡ = ⁡𝟎 is 1 units..
2 Marks questions
Ans : origin (x, y, z) = (0, 0, 0)
18. Find the volume of the parallelopiped whose coterminus edges ( adjacent sides) are given by
3(0)+6(0)+2(0)+7 7
the vectors 𝟐⁡𝒊⁡ ⃗⃗⃗⃗ − 𝟑⁡𝒋⁡ ⃗⃗⃗⃗ + 𝟒⁡𝒌⁡ ⃗⃗⃗⃗⃗, ⁡𝒊⁡ ⃗⃗⃗⃗ + 𝟐⁡𝒋⁡ ⃗⃗⃗⃗ − ⁡𝒌⁡ ⃗⃗⃗⃗⃗ and 𝟑⁡𝒊⁡ ⃗⃗⃗⃗ − ⁡𝒋⁡ ⃗⃗⃗⃗⃗.
⃗⃗⃗⃗ + 𝟐⁡𝒌⁡ distance = | | = ⁡ |7| = 1 units.
√32 ⁡+⁡⁡62 ⁡+⁡⁡22
2 −3 4
Ans : ⃗⃗⃗⃗⃗, ⃗⃗⃗⃗⃗
[⁡𝑎⁡ ⁡𝑏⁡, ⁡𝑐⁡ ⃗⃗⃗⃗⁡] = ⁡ |1 2 −1| = ⁡ −7
3 −1 2 𝒙−𝟏 𝟐−𝒚 𝒛−𝟒 𝒙−𝟑 𝒚−𝟑 𝟓−𝟐
26. Show that the lines =⁡ =⁡ and =⁡ =⁡ are parallel.
Volume of parallelepiped = |−7| = ⁡⁡7 cu.units 𝟒 𝟔 𝟏𝟐 −𝟐 𝟑 𝟔

19. The volume of the parallelepiped whose determine edges are Ans : 4⁡𝑖⁡
⃗⃗⃗ − 6⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗ = −2(−2⁡𝑖⁡
⃗⃗⃗⃗⃗ + 12⁡𝑘⁡⁡ ⃗⃗⃗ + 3⁡𝑗⁡ ⃗⃗⃗⃗⃗⁡⁡)
⃗⃗⃗⃗ − 6⁡𝑘⁡
⃗⃗⃗⃗ + 𝝀⁡𝒋⁡
7⁡𝒊⁡ ⃗⃗⃗⃗ − 𝟑⁡𝒌⁡ ⃗⃗⃗⃗⃗, ⁡𝒊⁡ ⃗⃗⃗⃗ + 𝟐⁡𝒋⁡ ⃗⃗⃗⃗ − ⁡𝒌⁡ ⃗⃗⃗⃗⃗, −𝟑⁡𝒊⁡ ⃗⃗⃗⃗ + 𝟕⁡𝒋⁡ ⃗⃗⃗⃗ + 𝟓⁡𝒌⁡ ⃗⃗⃗⃗⃗ is 90 cubic units. Find the value of 𝝀. The given lines are parallel…
7 𝜆 −3
Ans : [⁡𝑎⁡ ⃗⃗⃗⃗⃗, ⁡𝑐⁡
⃗⃗⃗⃗⃗, ⁡𝑏⁡ ⃗⃗⃗⃗⁡] = ⁡ | 1 2 −1| = ⁡90 3 Marks
−3 7 5
27. Find the magnitude and the direction cosines of the torque about the point (2, 0, -1) of a force
𝜆 = ⁡ −5
⃗⃗⃗⃗ + ⁡𝒋⁡⁡
𝟐⁡𝒊⁡ ⃗⃗⃗⃗⃗⃗ whose line of action passes through the origin.
⃗⃗⃗⃗⃗ − ⁡𝒌⁡⁡
20. Show that the vectors ⃗⃗⃗⃗ ⁡𝒊⁡ + 𝟐⁡𝒋⁡ ⃗⃗⃗⃗ − 𝟑⁡𝒌⁡ ⃗⃗⃗⃗⃗, 𝟐⁡𝒊⁡ ⃗⃗⃗⃗ − ⃗⃗⃗⃗ ⁡𝒋⁡ + 𝟐⁡𝒌⁡ ⃗⃗⃗⃗⃗ and 𝟑⁡𝒊⁡ ⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + ⁡𝒋⁡ ⁡𝒌⁡ are coplanar.
1 2 −3 Ans: ⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 2⁡𝑖⁡⁡
⁡𝑂𝐴⁡ ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ − ⁡𝑘⁡⁡
Ans : ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗, ⁡𝑏⁡, ⁡𝑐⁡
[⁡𝑎⁡ ⃗⃗⃗⃗⁡] = |2 −1 2 | = ⁡0 ⃗⃗⃗⃗⃗⃗⃗⃗⃗ = −2⁡𝑖⁡⁡
⃗⃗⃗⃗ = ⁡𝐴𝑂⁡
⁡⁡⁡𝑟⁡ ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ + ⁡𝑘⁡⁡
3 1 −1 ⃗⃗⃗⃗⃗ ⁡ = ⁡2⁡𝑖⁡ ⃗⃗⃗⃗⃗⃗
⁡𝐹⁡ ⃗⃗⃗ + ⁡𝑗⁡⁡
⃗⃗⃗⃗⃗ − ⁡𝑘⁡⁡
⸫ Given vectors are coplanar.
21. ⃗⃗⃗⃗ − ⁡𝒋⁡
If 𝟐⁡𝒊⁡ ⃗⃗⃗⃗ + 𝟑⁡𝒌⁡ ⃗⃗⃗⃗⃗, 𝟑⁡𝒊⁡ ⃗⃗⃗⃗ + 𝟐⁡𝒋⁡ ⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⁡𝒌⁡, ⁡𝒊⁡ ⃗⃗⃗⃗ + 𝒎⁡𝒋⁡ ⃗⃗⃗⃗ + 𝟒⁡𝒌⁡ ⃗⃗⃗⃗⃗ are coplanar find the value of m. ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ ⁡𝑗⁡
⁡𝑖⁡ ⁡𝑘⁡
Torque ⁡𝑡⁡ ⃗⃗⃗⃗ × ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ = ⁡𝑟⁡ ⁡𝐹⁡⁡ = ⁡ |−2 0 1 | = −⁡⁡𝑖⁡⁡ ⃗⃗⃗⃗⃗⃗⁡
⃗⃗⃗⃗⃗ − 2⁡𝑘⁡⁡
2 −1 3
Ans : |3 2 1| = ⁡0 2 1 −1
1 𝑚 4 ⃗⃗⃗⃗| = ⁡ √12 + 0 + ⁡ 22 = ⁡ √5
Magnitude |⁡𝑡⁡
m = -3 −1 −2
Direction cosines; ⁡, 0,
√5 √5
22. For any vector ⁡𝒂⁡ ⃗⃗⃗⃗⃗ , prove that….
⃗⃗⃗⃗⃗ ⁡ × ⁡ (⁡𝒂⁡ ⃗⃗⃗⃗⃗) = 𝟐⁡𝒂⁡ 28. Forces of magnitudes 𝟓√𝟐 and 𝟏𝟎√𝟐 units acting in the directions
⃗⃗⃗⃗ ⁡ × ⁡ (⁡𝒂⁡
⁡𝒊⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝒊⁡
⃗⃗⃗⃗) + ⁡ ⁡𝒋⁡⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ (⁡𝒂⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝒋⁡ ⃗⃗⃗⃗) + ⁡ ⁡𝒌⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝒌⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 𝟒⁡𝒋⁡⁡
3⁡𝒊⁡ ⃗⃗⃗⃗⃗⃗ and 𝟏𝟎⁡𝒊⁡
⃗⃗⃗⃗⃗ + 𝟓⁡𝒌⁡⁡ ⃗⃗⃗⃗ + 𝟔⁡𝒋⁡⁡ ⃗⃗⃗⃗⃗⃗ respectively, act on a particle which is displaced from
⃗⃗⃗⃗⃗ − 𝟖⁡𝒌⁡⁡
Ans : ⃗⃗⃗⃗ ⁡ × ⁡ (⁡𝑎⁡
⁡𝑖⁡ ⃗⃗⃗) = ⁡ (⁡𝑖⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑖⁡ ⃗⃗⃗⃗⁡. ⁡𝑖⁡ ⃗⃗⃗⃗⃗⃗ − ⁡ (⁡𝑖⁡
⃗⃗⃗)⁡𝑎⁡⁡ ⃗⃗⃗. ⁡𝑎⁡
⃗⃗⃗⃗⃗)⁡𝑖⁡
⃗⃗⃗
the point with position vector 𝟒⁡𝒊⁡ ⃗⃗⃗⃗ − 𝟑⁡𝒋⁡⁡ ⃗⃗⃗⃗⃗⃗ to the point with position vector
⃗⃗⃗⃗⃗ − 𝟐⁡𝒌⁡⁡
L.H.S.
⃗⃗⃗⃗ + ⁡𝒋⁡⁡
𝟔⁡𝒊⁡ ⃗⃗⃗⃗⃗⃗ find the work done by the solution…
⃗⃗⃗⃗⃗ ⁡ − 𝟑⁡𝒌⁡⁡
⃗⃗ ⁡ × ⁡ (⁡𝑎⁡
⃗⁡𝑖⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑖⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ (⁡𝑎⁡
⃗⃗⃗⃗) + ⁡ ⁡𝑗⁡⁡ ⃗⃗⃗⃗) + ⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑗⁡ ⁡𝑘⁡ ⁡ × ⁡ (⁡𝑎⁡ ⃗⃗⃗⃗⃗) = 3⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑘⁡ ⃗⃗⃗⃗⃗ − ⁡ 𝑎⁡
⃗⃗⃗⃗
⃗⃗⃗⃗+4⁡𝑗⁡⁡
3⁡𝑖⁡ ⃗⃗⃗⃗⃗⃗⁡
⃗⃗⃗⃗⃗+5⁡𝑘⁡⁡ ⃗⃗⃗⃗+6⁡𝑗⁡⁡
10⁡𝑖⁡ ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗−8⁡𝑘⁡⁡
= 2⁡𝑎⁡
⃗⃗⃗⃗⃗ R.H.S ⃗⃗⃗⃗ = 5√2⁡(
Ans : Force ⁡𝐹⁡ ) + 10√2⁡( )
5√2 10√2
⃗⃗ ⁡ × ⁡ (⁡𝑎⁡
⃗⃗⁡𝑖⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑖⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ (⁡𝑎⁡
⃗⃗⃗) + ⁡ ⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑗⁡ ⃗⃗⃗⃗⃗ ⁡ × ⁡ (⁡𝑎⁡
⃗⃗⃗⃗) + ⁡ ⁡𝑘⁡ ⃗⃗⃗⃗⃗) = 2⁡𝑎⁡
⃗⃗⃗⃗⃗ ⁡ × ⁡ ⁡𝑘⁡ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ + 10⁡𝑖⁡ ⃗⃗⃗⃗⃗⃗
⁡⁡=⁡3⁡𝑖⁡
⃗⃗⃗ + 4⁡𝑗⁡⁡
⃗⃗⃗⃗⃗ + 5⁡𝑘⁡⁡ ⃗⃗⃗⃗ + 6⁡𝑗⁡⁡
⃗⃗⃗⃗⃗ − 8⁡𝑘⁡⁡
Hence proved.. ⃗⃗⃗⃗⃗ = ⁡13⁡𝑖⁡ ⃗⃗⃗⃗⃗⃗
⁡𝐹⁡ ⃗⃗⃗ + 10⁡𝑗⁡⁡
⃗⃗⃗⃗⃗ − 3⁡𝑘⁡⁡
23. Prove that [⁡𝒂⁡ ⃗⃗⃗⃗⃗, ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗–⁡⁡𝒃⁡ ⃗⃗⃗⃗, ⁡𝒄⁡
⁡𝒃⁡–⁡⁡𝒄⁡ ⃗⃗⃗⃗–⁡⁡𝒂⁡
⃗⃗⃗⃗⃗]⁡= 0 ⃗⃗⃗⃗⃗ = (6⁡𝑖⁡
Displacement ⁡𝑑⁡ ⃗⃗⃗ + ⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗) − ⁡ (4⁡𝑖⁡
⃗⃗⃗⃗⃗ ⁡ − 3⁡𝑘⁡⁡ ⃗⃗⃗ − 3⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗⁡)
⃗⃗⃗⃗⃗ − 2⁡𝑘⁡⁡
ANS: L.H.S
1 −1 0 = 2⁡𝑖⁡
⃗⃗⃗ + 4⁡𝑗⁡⁡⃗⃗⃗⃗⃗ ⁡ − ⃗⃗⃗⃗⃗⃗
⁡𝑘⁡⁡
[⁡𝑎⁡ ⃗⃗⃗⃗⃗,
⃗⃗⃗⃗⃗–⁡⁡𝑏⁡ ⃗⃗⃗⃗⃗–⁡⁡𝑐⁡
⁡𝑏⁡ ⃗⃗⃗⃗,
⃗⃗⃗⃗–⁡⁡𝑎⁡
⁡𝑐⁡ ⃗⃗⃗⃗⃗] = ⁡ | 0 1 −1|⁡[⁡𝑎⁡ ⃗⃗⃗⃗⃗, ⁡𝑐⁡
⃗⃗⃗⃗⃗, ⁡𝑏⁡ ⃗⃗⃗⃗⁡] = 0 Work done by the force = 𝑊 = ⁡ ⁡𝐹⁡ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⁡. ⁡𝑑⁡
−1 0 1 (13⁡𝑖⁡
⃗⃗⃗⃗ + 10⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗)⁡. (2⁡𝑖⁡
⃗⃗⃗⃗⃗ − 3⁡𝑘⁡⁡ ⃗⃗⃗⃗ + 4⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗) = 69 units.
⃗⃗⃗⃗⃗ ⁡ − ⁡𝑘⁡⁡
[⁡𝑎⁡ ⃗⃗⃗⃗⃗, ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗–⁡⁡𝑏⁡ ⁡𝑏⁡–⁡⁡𝑐⁡
⃗⃗⃗⃗, ⃗⃗⃗⃗–⁡⁡𝑎⁡
⁡𝑐⁡ ⃗⃗⃗⃗⃗] = 0
Hence proved…

38 39
29. Find the angle between the straight lines
𝒙+𝟑
𝟐
=⁡
𝒚−𝟏
𝟐
= ⁡ −𝒛 with coordinate axes. Chapter – 5
𝑥+3 𝑦−1 −𝑧
Ans : =
2
=⁡
2
=⁡
1
Two Dimensional Analytical Geometry
⃗⃗⃗⃗+2⁡𝑗⁡⁡
⃗⃗⃗⃗⃗⃗ ⁡⁡⁡⁡⁡⁡ = ⁡ 2⁡𝑖⁡ ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⁡−⁡𝑘⁡⁡ 1 1. Find the equation of the circle passing through the points (1, 1), (2, -1) and (3, 2) (5.10) March
⁡⁡𝑏⁡ |2⁡𝑖⁡
⃗⃗⃗⃗+2⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗|
⃗⃗⃗⃗⃗⁡−⁡𝑘⁡⁡
⁡ = ⁡ (2⁡𝑖⁡ ⃗⃗⃗⃗⃗ ⁡ − ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗ + 2⁡𝑗⁡⁡ ⁡𝑘⁡⁡)
3 20
𝛼 = ⁡ 𝑐𝑜𝑠 −1 ( ),⁡⁡ 𝛽 = ⁡ 𝑐𝑜𝑠 −1 (2⁄3),⁡ 𝛾⁡ = ⁡ 𝑐𝑜𝑠 −1 (−1⁄3)⁡⁡
2
𝑨(𝟏, 𝟏)⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑩(𝟐, −𝟏)⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑪(𝟑, 𝟐)⁡
3
Slope of AB = ⁡ −2⁡ = 𝑚1
30. If ⃗⃗⃗⃗⃗⁡,
⁡𝒂⁡ ⁡𝒃⁡ ⃗⃗⃗⃗⃗, ⁡𝒄⁡
⃗⃗⃗⃗⁡are coplanar. Then prove that ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⁡, ⁡𝒃⁡
⁡𝒂⁡ + ⁡ ⁡𝒃⁡ ⃗⃗⃗⃗⃗ + ⁡𝒄⁡⃗⃗⃗⃗⁡, ⁡𝒄⁡
⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⁡𝒂⁡ are coplanar..
1 1 0 Slope of AC = = ⁡ 1⁄2 = ⁡ 𝑚2
Ans : [⁡𝑎⁡ ⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡
⃗⃗⃗⃗⃗ + ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⁡,
⁡𝑐⁡ ⁡𝑐⁡
⃗⃗⃗⃗ + ⁡𝑎⁡
⃗⃗⃗⃗⃗] = ⁡ |0 1 1|⁡[⁡𝑎⁡ ⃗⃗⃗⃗⃗, ⁡𝑐⁡
⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡ ⃗⃗⃗⃗⁡] 𝑚1 𝑚2 = ⁡ −1
1 0 1 BC is diameter of circle (2> -1) (3> 2)
=0 𝑥1 ⁡⁡⁡⁡𝑦1 𝑥2 ⁡⁡⁡⁡𝑦2
= {⁡𝑎⁡ ⃗⃗⃗⃗⃗⁡, ⁡𝑏⁡
⃗⃗⃗⃗⃗ + ⁡ ⁡𝑏⁡ ⃗⃗⃗⃗⃗ + ⁡𝑐⁡
⃗⃗⃗⃗⁡, ⁡𝑐⁡
⃗⃗⃗⃗ + ⁡𝑎⁡⃗⃗⃗⃗⃗} are coplanar.. Equation of circle: (𝑥 − ⁡ 𝑥1 )⁡(𝑥 − ⁡ 𝑥2 ) + ⁡ (𝑦 − ⁡ 𝑦1 )⁡(𝑦 − ⁡ 𝑦2 ) = 0
𝑥 2 + ⁡ 𝑦 2 − 5𝑥 − 𝑦 + 4 = 0
31. Show that the points (2, 3, 4) (-1, 4, 5) and (8, 1, 2) are collinear
2 3 4 2. Find the equation of circle through the points (1, 0) (-1, 0) and (0, 1) (5.1-6) June 23, Sep 23
Ans : |−1 4 5| = 0 𝑨(⁡𝟎, 𝟏)⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑩(−𝟏, 𝟎)⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑪(𝟏, 𝟎)
8 1 2 Slope of AB = 1 = 𝑚1
The given points are collinear.. Slope of AC = −1 = ⁡ 𝑚2
𝒙−𝟒 𝒚 𝒛+𝟏 𝒙−𝟏 𝒚+𝟏 𝒛−𝟐
32. Find the angle between the straight lines =⁡ =⁡ and =⁡ =⁡ and state whether 𝑚1 𝑚2 = ⁡ −1
𝟐 𝟏 −𝟐 𝟒 −𝟒 𝟐
they are parallel (or) perpendicular: BC is diameter of circle (-1> 0) (1> 0)
Ans: Angle between the two st.lines 𝑥1 ⁡⁡⁡⁡𝑦1 𝑥2 ⁡⁡⁡⁡𝑦2
Equation of circle: (𝑥 − ⁡ 𝑥1 )⁡(𝑥 − ⁡ 𝑥2 ) + ⁡ (𝑦 − ⁡ 𝑦1 )⁡(𝑦 − ⁡ 𝑦2 ) = 0
𝑥2 + ⁡ 𝑦2 = 1
𝑎1 𝑏1 ⁡+⁡𝑎2 𝑏2 ⁡+⁡𝑎3 𝑏3
𝜃 = ⁡ 𝑐𝑜𝑠 −1 3. The maximum and minimum distance of the Earth from the sun respectively are 𝟏𝟓𝟐⁡ × ⁡ 𝟏𝟎𝟔
√𝑎1 2 +⁡⁡𝑎2 2 +⁡𝑎3 2 ⁡⁡⁡⁡⁡⁡√√𝑏1 2 +⁡⁡𝑏2 2 +⁡𝑏3 2 ⁡⁡ km and 𝟗𝟒. 𝟓⁡ × ⁡ 𝟏𝟎𝟔 km. The sun is at one focus of the elliptical orbit. Find the distance from
( ) the sun to the other focus (5.32) Mar 22 (5m), Mar – 23 (3m)
𝜃 = ⁡ 𝑐𝑜𝑠 −1 ⁡(0) = ⁡ 𝜋⁄2
The given lines are perpendicular 𝑎 + 𝑐 = ⁡152⁡ × ⁡ 106 km …………. (1)
33. Find the vector equation of a plane which is at a distance of 7 units from the origin having
𝑎 − 𝑐 = ⁡94.5⁡ × ⁡ 106 ⁡𝑘𝑚 ……………… (2)
3, -4, 5 as direction ratios of a normal to it ….
⁡
⃗⃗⃗⃗ (1) – (2) ⟹
Solution : 𝑟⁡⁡⁡. ⁡𝑛⁡
⃗⃗⃗⃗⃗ = 𝑃
⃗⃗⃗⃗⃗⃗)
⁡𝑟⁡⁡
⃗⃗⃗⃗⃗⃗⁡.
(3⁡𝑖⁡
⃗⃗⃗⃗−4⁡𝑗⁡⁡
⃗⃗⃗⃗⃗+5⁡𝑘⁡⁡
⁡ = ⁡⁡7 2𝑐 = 57.5⁡ × ⁡ 106 ⁡𝑘𝑚
√9+16+25

⁡𝑟⁡⁡
⃗⃗⃗⃗⃗⃗⁡.
(3⁡𝑖⁡
⃗⃗⃗⃗−4⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗)
⃗⃗⃗⃗⃗+5⁡𝑘⁡⁡
⁡= 7 Required distance = 575⁡ × ⁡ 105 ⁡𝑘𝑚
5√2
Cartesian equation is 3𝑥 − 4𝑦 + 5𝑧 = 35√2 4. Assume that water issuing from the end of horizontal pipe 7.5m above the ground describes a
34. Find the distance from the point (2, 5, -3) to the plane parabolic path. The vertex of the parabolic path is at the end of the pipe at a position 2.5m below
⃗⃗⃗⃗⃗⃗⁡. (𝟔⁡𝒊⁡
⁡𝒓⁡⁡ ⃗⃗⃗⃗ − 𝟑⁡𝒋⁡⁡ ⃗⃗⃗⃗⃗⃗) = 𝟓
⃗⃗⃗⃗⃗ + 𝟐⁡𝒌⁡⁡ the line of the pipe, the flow of water has curved outward 3m beyond the vertical line through
|⁡𝑢⁡⁡ |(2⁡𝑖⁡
⃗⃗⃗⃗+5⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗).(6⁡𝑖⁡
⃗⃗⃗⃗⃗−3⁡𝑘⁡⁡ ⃗⃗⃗⃗−3⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗)−5|
⃗⃗⃗⃗⃗+2⁡𝑘⁡⁡
the end of the pipe. How far beyond this vertical line will the water strike the ground? (5.5 – 8)
⃗⃗⃗⃗⃗⃗⁡.⁡𝑛⁡
⃗⃗⃗⃗⃗−𝑃|
Ans: 𝛿 =⁡ |⁡𝑛⁡
⃗⃗⃗⃗⃗|
=⁡ |(6⁡𝑖⁡
⃗⃗⃗⃗−3⁡𝑗⁡⁡ ⃗⃗⃗⃗⃗⃗)|
⃗⃗⃗⃗⃗+2⁡𝑘⁡⁡ Mar – 20 Mar – 24
𝛿 = 2 units. Equation of Parabola : 𝑥 2 = 4𝑎𝑦
Find the distance between the two planes. x + 2y – 2z + 1 = 0 and 2x + 4y – 4z + 5 = 0
(3, −2.5)𝑎⁡ = 9⁄10
35.
|𝑑 ⁡−⁡⁡𝑑2 |
Ans : 𝛿 = ⁡ √𝑎2 1 9
⁡+⁡⁡𝑏 2 ⁡+⁡⁡𝑐 2 𝑥 2 ⁡ = ⁡ −4⁡ × ⁡ 𝑦
10
2x + 4y – 4z + 5 = 0
5 (𝑥1 ⁡, −7.5)
x + 2y – 2z + 2 = 0
Required distance = 3√3 m
|1−⁡⁡5⁄2| 1
𝛿 = ⁡ √12 = ⁡ 2 units……
⁡+⁡⁡22 ⁡+⁡⁡22
40 51
5. On lighting a rocket cracker, it gets projected in a parabolic path and reaches a maximum height 𝒙𝟐 𝒚𝟐
9. Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation 𝟑𝟎𝟐
− ⁡ 𝟒𝟒𝟐 =
of 4m when it is 6m away from the point of projection. Finally it reaches the ground 12m away 𝟏 The tower is 150m tall and the distance from the top of the tower to the centre of the hyperbola
from the starting point. Find the angle of projection. (5.5 – 9) Sep – 21. is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter
Equation of parabola: 𝑥 2 = ⁡ −4𝑎𝑦 of the top and base of the tower (5.5 – 6)
𝑥2 𝑦2
−⁡ =1
(−6, −4)⁡⁡⁡⁡𝑎 = ⁡ 9⁄4 302 44 2

𝑥 2 = ⁡ −9𝑦 (𝑥1 ⁡, 50) 𝑥1 = 45.41

𝑑𝑦 −2𝑥 Top diameter = 90.82 m
𝑑𝑥
=⁡ 9
4 (𝑥2 ⁡, −100) 𝑥2 = 74.45
Angle = 𝜃 = ⁡ 𝑡𝑎𝑛−1 ( )
3
Bottom diameter = 148.9 m
6. Parabolic cable of a 60m portion of the road bed of a suspension bridge are positioned as shown
below. Vertical cables are to be space every 6m along the portion of the road bed. Calculate 10. A bridge has a parabolic arch that is 10m high at the centre and 30m wide at the bottom. Find
the height of the arch 6m from the centre on either sides? (5.5 – 1) June - 24
the lengths of the first two of these vertical cables from the vertex. (5.5 - 5)
Equation of parabola : 𝑥 2 = −⁡4𝑎𝑦
Equation of parabola:
225
𝑥 2 = ⁡4𝑎𝑦 (15, −10) 4𝑎⁡ = ⁡ 10

225
(30, 13)⁡⁡⁡4𝑎 = ⁡
900
𝑥 2 = −⁡ ⁡𝑦
13 10

8
900 (6, −𝑦1 ) 𝑦1 ⁡ = ⁡ 5 = ⁡1.6
𝑥2 = ⁡ 13
⁡𝑦⁡
Required height = 8.4 m
(6, 𝑦1 ) Req. Distance = 3.52⁡m
(𝒙−𝟏𝟏)𝟐 𝒚𝟐
11. If the equation of the ellipse is + ⁡ 𝟔𝟒 = 𝟏⁡⁡(𝒙 and y are measured in c.m) where to the
𝟒𝟖𝟒
(12, 𝑦2 ) Req. Distance = 5.08⁡m
nearest cm, should the patient’s kidney stone be placed so that the reflected sound hits the kidney
stone) (5.39)
7. A rod of length 1.2 m moves with its ends always touching the co-ordinate axes. The locus of a
point P on the rod, which is 0.3m from the end in contact with x – axis is an ellipse. Find the
eccentricity (5.5 - 7) Sep – 20

𝑥2 𝑦2
Equation of Ellipse : + ⁡ 0.32 = 1 𝑎2 = 484 𝑏 2 = 64 𝑐 2 = 𝑎2 − ⁡ 𝑏 2 = 420
0.92

𝑐 = ⁡ √420⁡ = 20.5⁡⁡𝑐𝑚⁡
𝑏2 2√2
𝑒 = ⁡ √1 − ⁡ 𝑎2 = ⁡
3 12. Show that the line 𝒙 − 𝒚 + 𝟒 = 𝟎 is a tangent to the ellipse 𝒙𝟐 + 𝟑𝒚𝟐 ⁡ = 𝟏𝟐. Also find the co-
ordinates of the point of contact. (5.4 -3)
8. A semi elliptical archway over a one-way road has a height of 3m and a width of 12m. The
𝑥 − 𝑦 + 4 = 0⁡ ⟹ ⁡⁡𝑚 = 1,⁡⁡⁡𝑐 = 4
truck has a width of 3m and a height of 2.7m. Will the truck clear the opening of the arch 𝑥2 𝑦2
+ ⁡ = 1⁡⁡⁡ ⟹ ⁡⁡⁡ 𝑎2 ⁡ = 12,⁡⁡⁡𝑏 2 ⁡ = ⁡⁡4
way? (5.31) 12 4
𝑎 𝑚 + ⁡ 𝑏 2 = 16 = ⁡ 𝑐 2
2 2

𝑥2 𝑦2 −𝑎2 𝑚 𝑏2
Equation of ellipse +⁡ =1 It is a tangent point of contact = ( ⁡, )⁡ = (−3, 1)
62 32 𝑐 𝑐

3
(2 ⁡, 𝑦1 ) 𝑦1 = 2.9 > 2.7

Truck will clear the archway.

52 53
13. Find the equations two tangents that can be drawn from (5, 2) to the ellipse 𝟐𝒙𝟐 + 𝟕𝒚𝟐 = 𝟏𝟒 18. Find the vertex, focus, directrix and length of the latus rectum of the parabola
(5.4 – 1) 𝒙𝟐 − 𝟒𝒙 − 𝟓𝒚 − 𝟏 = 𝟎 (5.19)
𝑎2 ⁡ = 7,⁡⁡⁡𝑏 2 ⁡ = ⁡⁡2 𝑥 2 − 4𝑥 − 5𝑦 − 1 = 0
𝑥 2 − 4𝑥 = 5𝑦 + 1
(5, 2)⁡⁡⁡⁡𝑦 = 𝑚𝑥⁡ ± √𝑎2 𝑚2 + ⁡ 𝑏 2
(𝑥 − 2)2 = 5(𝑦 + 1) 4𝑎 = 5 ⟹ 𝑎 = ⁡ 5⁄4
9𝑚2 − 10𝑚 + 1 = 0⁡⁡⁡⁡⁡⁡⁡⁡𝑚⁡ = 1, 1⁄9 Vertex:(2, −1) ⁡⁡⁡ = ⁡ (ℎ, 𝑘)
Equation of tangents: 𝑥 − 𝑦 − 3 = 0 Focus: ⁡(ℎ⁡ + ⁡0, 𝑘⁡ + ⁡𝑎) ⁡ = (2, −1 + ⁡ 5⁄4) = ⁡ (2, 1⁄4)
𝑥 − 9𝑦 + 13 = 0 Equation of Directrix: 𝑦 = 𝑘 − 𝑎 = ⁡ −1 − ⁡⁡ 5⁄4 ⁡ = ⁡ −⁡ 9⁄4
14. The parabolic communication antenna has a focus at 2m distance from the vertex of the antenna. Length of latus rectum = ⁡⁡4𝑎⁡ = ⁡5
Find the width of the antenna 3m from the vertex. (5.34) June 22
𝑎 = 2> Equation of parabola 𝑦 2 = 8𝑥 19. Identify the type of conic and find centre, foci, vertices and directrices of
𝟏𝟖𝒙𝟐 + 𝟏𝟖𝒙𝟐 + 𝟏𝟐𝒚𝟐 − 𝟏𝟒𝟒𝒙 + 𝟒𝟖𝒚 + 𝟏𝟐𝟎 = 𝟎 (5.2 -8(V)) Mar 23
(3, 𝑦1 ) 𝑦1 = 2√6
It is an ellipse.
Width = 4√6 m 18𝑥 2 − 144𝑥 + ⁡12𝑦 2 + 48𝑦 = ⁡ −120
15. Show that the equation of the parabola whose focus (−√𝟐, 𝟎) and directrix 𝒙 = ⁡ √𝟐 is 18(𝑥 2 − 8𝑥) + 12(𝑦 2 + 4𝑦) = ⁡ −120
𝟐
𝒚 = −𝟒√𝟐𝒙 Mar. 22, Sep – 21 (𝑥−4)2 (𝑦+2)2
12
+⁡ =1; 𝑐 2 = ⁡ 𝑎2 − ⁡ 𝑏 2
18
Focus: (−√2, 0) ; Directrix 𝑥 = ⁡ √2 ⟹ 𝑎 = ⁡ √2
𝑎2 = 18, 𝑏 2 = 12 𝑐2 = 6
2
Equation of parabola : 𝑦 = −4𝑎𝑥 𝑎 = 3√2 𝑐 = ⁡ √6
𝑦 2 = −4√2𝑥 𝑒 = ⁡ 𝑐⁄𝑎 = ⁡ 1⁄
√3
𝑥2 𝑦2
16. Find eccentricity, foci, vertices and centre of the ellipse +⁡ = 1. And draw approximate diagram Centre: (ℎ, 𝑘) ⁡⁡ = ⁡ (4, −2)
25 9
(created) June – 22, Sep – 21 Foci: (ℎ⁡ + ⁡0, 𝑘 ± 𝑐) = ⁡ (4, −2 ± √6)
𝑥2 𝑦2
+⁡ =1 2
𝑎 ⁡ = 25, 𝑏 = 9 2 Vertices: (ℎ + 0,⁡⁡⁡𝑘⁡ ± 𝑎) = (4, −2 ± 3√2)
25 9
Directrices: : 𝑦 = 𝑘 ± 𝑎⁄𝑒
𝑐 2 ⁡ = ⁡ 𝑎2 − ⁡ 𝑏 2 ⁡ = 16
𝑦 = −2 ± 3√6
𝑒 = ⁡ 𝑐⁄𝑎 = ⁡ 4⁄5
20. Find the foci, vertices and length of major and minor axis of the conic
Centre: (0, 0) 𝟒𝒙𝟐 + 𝟑𝟔𝒚𝟐 + 𝟒𝟎𝒙 − 𝟐𝟖𝟖𝒚 + 𝟓𝟑𝟐 = 𝟎 (5.22)
4𝑥 2 + 36𝑦 2 + 40𝑥 − 288𝑦 + 532 = 0
Foci: (±⁡𝑐, 0) ⁡ = ⁡ (±⁡4, 0)
4(𝑥 2 + 10𝑥) + 36(𝑦 2 − 8𝑦) = ⁡ −532
Vertices: (±⁡𝑎, 0) ⁡ = ⁡ (±⁡5, 0)
(𝑥+5)2 (𝑦−4)2
+⁡ =1; 𝑐 2 = ⁡ 𝑎2 − ⁡ 𝑏 2
𝟐 36 4
17. Find the vertex, focus, equation of directrix and length of the latus rectum of 𝒚 − 𝟒𝒚 − 𝟖𝒙 + 2 2
𝑎 = 36, 𝑏 =4 𝑐 2 = 32
𝟏𝟐 = 𝟎 (𝟓. 𝟐⁡ − 𝟒(𝑽)⁡⁡𝑴𝒂𝒓⁡– ⁡𝟐𝟒
𝑦 2 − 4𝑦 − 8𝑥 + 12 = 0 𝑎=6 𝑏=2 𝑐 = ⁡4√2
𝑦 2 − 4𝑦 = 8𝑥 − 12 Centre: (ℎ, 𝑘) ⁡⁡ = ⁡ (−5, 4)
(𝑦 − 2)2 = 8(𝑥 − 1) 4𝑎 = 8 ; 𝑎⁡ = ⁡2⁡ Foci: (ℎ⁡ ± 𝑐, 𝑘 + 0) = ⁡ (−5 ± 4√2⁡, 4)
Vertex:(ℎ, 𝑘) ⁡ = ⁡ (1, 2) Vertices: (ℎ ± 𝑎, 0) = (−5 + 6, 4)&⁡(−5 − 6, 4)
Focus: (ℎ⁡ + ⁡𝑎, 𝑘⁡ + ⁡0) ⁡⁡⁡ = ⁡ (3, 2) = (1, 4),⁡⁡⁡(−11, 4)
Equation of Directrix: 𝑥⁡ = ⁡ℎ⁡– 𝑎⁡ = ⁡ −1 Length of major axis = ⁡⁡2𝑎 = 12
Length of latus rectum = ⁡⁡4𝑎⁡ = ⁡8 Length of minor axis = 2𝑏 = 4

54 55
21. Find the centre, foci and eccentricity of the hyperbola 𝟏𝟏𝒙𝟐 − 𝟐𝟓𝒚𝟐 − 𝟒𝟒𝒙 + 𝟓𝟎𝒚 − 𝟐𝟓𝟔 = 𝟎 24. A circle of area 𝟗𝝅 sq. units has two of its diameters along the lines 𝒙 + 𝒚 = 𝟓 and 𝒙 − 𝒚 = 𝟏.
(𝟓. 𝟐𝟔)⁡⁡𝑺𝒆𝒑⁡– ⁡𝟐𝟎 Find the equation of the circle (5.1 – 7) Sep -20
11𝑥 2 − 25𝑦 2 − 44𝑥 + 50𝑦 − 256 = 0 𝜋⁡𝑟 2 = 9𝜋⁡⁡ ⟹ 𝑟 = 3
11(𝑥 2 − 4𝑥) − 25(𝑦 2 − 2𝑦) = 256 𝑥 + 𝑦 = 5⁡⁡,⁡⁡⁡𝑥 − 𝑦 = 1
(𝑥−2)2
−⁡
(𝑦−1)2
=1 ; 𝑐 2 = ⁡ 𝑎2 + ⁡ 𝑏 2 Centre: (ℎ, 𝑘) = ⁡ (𝑥, 𝑦) ⁡⁡ = ⁡ (3, 2)⁡
25 11
2 2
Equation of Circle: (𝑥 − ℎ)2 ⁡ + ⁡ (𝑦 − 𝑘)2 ⁡ = ⁡ 𝑟 2
𝑎 = 25, 𝑏 = 11 𝑐 2 = 36 (𝑥 − 3)2 ⁡ + ⁡ (𝑦 − 2)2 ⁡ = ⁡ 32
𝑎=6 𝑐 = ⁡6 𝑥 2 + ⁡ 𝑦 2 − 6𝑥 − 4𝑦 + 4 = 0
Centre: (ℎ, 𝑘) ⁡⁡ = ⁡ (2, 1) 25. Show that the equation of parabola whose focus (4, 0) and directrix x = 4 is 𝒚𝟐 = 𝟏𝟔𝒙
Foci: (ℎ⁡ ± 𝑐, 𝑘 + 0) = ⁡ (2 + 6, 1)⁡, (2 − 6, 1) (5.2-1(i)) March 21
= ⁡ (8, 1), (−4, 1) Focus = ⁡ (4, 0)
Vertices: (ℎ ± 𝑎, 𝑘 + 0) = (2 + 5, 4), (2 − 5, 1) Directrix: 𝑥⁡ = ⁡ −4 𝑎⁡ = ⁡4
Equation of Parabola : 𝑦 2 = 4𝑎𝑥
= (7, 1),⁡⁡⁡(−3, 1)
𝒚𝟐 = 𝟏𝟔𝒙
eccentricity = 𝑒 = ⁡ 𝑐⁄𝑎 = ⁡ 6⁄5 26. Find the equation of the parabola with vertex (-1, -2), axis parallel to y – axis and parsing
22. Identify the type of conic and find centre, foci, vertices and directrices of through (3, 6) (5.18) March 23
𝟗𝒙𝟐 − 𝒚𝟐 − 𝟑𝟔𝒙 − 𝟔𝒚 + 𝟏𝟖 = 𝟎 (5.2 – 8 (vi) Jun 24) Vertex = (-1, -2) = (h, k)
It is a hyperbola. Equation of Parabola (𝑥 − ℎ)2 = 4𝑎(𝑦 − 𝑘)
9𝑥 2 − 𝑦 2 − 36𝑥 − 6𝑦 = −18 (𝑥 + 1)2 = 4𝑎(𝑦 + 2)
9(𝑥 2 − 4𝑥) − (𝑦 2 + 6𝑦) = −18 (3, 6)⁡⁡⁡4𝑎⁡⁡ = ⁡2
(𝑥−2)2
−⁡
(𝑦+3)2
=1 (𝑥 + 1)2 = 2(𝑦 + 2)
1 9
𝒙𝟐 + 𝟐𝒙 − 𝟐𝒚 − 𝟑 = 𝟎
𝑎2 = 1, 𝑏2 = 9 𝑐 2 = ⁡ 𝑎2 + ⁡ 𝑏 2 ⁡ = 10
27. Find the general equation of the circle whose diameter is the line segment joining the
𝑎=1 𝑐 = ⁡ √10 points (-4, -2) and (1, 1) (5.4) March 22
Equation of Circle: (𝑥 − ⁡ 𝑥1 )⁡(𝑥 − ⁡ 𝑥2 ) + ⁡ (𝑦 − ⁡ 𝑦1 )⁡(𝑦 − ⁡ 𝑦2 ) = 0
𝑒 = ⁡ 𝑐⁄𝑎 = ⁡ √10 (𝑥⁡ + ⁡4)⁡(𝑥⁡– ⁡1) ⁡ + ⁡ (𝑦⁡ + ⁡2)⁡(𝑦⁡– ⁡1) ⁡ = ⁡0
Centre: (ℎ, 𝑘) ⁡⁡ = ⁡ (2, −3) 𝒙𝟐 + ⁡ 𝒚𝟐 + 𝟑𝒙 + 𝒚 − 𝟔 = 𝟎
28. Find centre and radius of the circle 𝒙𝟐 + ⁡ 𝒚𝟐 + 𝟔𝒙 − 𝟒𝒚 + 𝟒 = 𝟎 June 22
foci: (ℎ⁡ ± 𝑐, 𝑘 + 0) = ⁡ (2 ± √10⁡, −3) Centre = (−𝑔, −𝑓) ⁡⁡ = ⁡⁡ (−3, 2)
Radius = √𝑔2 + ⁡ 𝑓 2 − 𝑐
Vertices: (ℎ ± 𝑎⁡⁡, 𝑘 = 0) = (2 + 1, −3)⁡(2 − 1, −3) = ⁡ √9 + 4 − 4 ⁡ = 3⁡𝑚
= (3, −3),⁡⁡⁡(1, −3) 29. If the equation 𝟑𝒙𝟐 + (𝟑 − 𝒑)𝒙𝒚 + 𝒒𝒚𝟐 − 𝟐𝒑𝒙 = 𝟖𝒑𝒒 represents a circle, find p and q. Also
determine the centre and radius of the circle (5.1 – 12)
Direction: 𝑥 = ℎ ± 𝑎⁄𝑒 3𝑥 2 + (3 − 𝑝)𝑥𝑦 + 𝑞𝑦 2 − 2𝑝𝑥 = 8𝑝𝑞
𝑥 = 2 ± 1⁄ co – eff of 𝑥 2 = co. eff. of 𝑦 2 ⟹ 𝑞 = 3
√10 co.eff of 𝑥𝑦 = ⁡0 ⇒ 𝑝 = 3
3𝑥 2 + 3𝑦 2 − 6𝑥 − 72 = 0
3 Marks:
𝑥 2 + 𝑦 2 − 2𝑥 − 24 = 0
23. A concrete bridge is designed as a parabolic arch. The road over bridge is 40m long and the centre: (−𝑔, −𝑓) ⁡⁡⁡⁡ = ⁡⁡ (1, 0)
maximum height of the arch is 15m. Write the equation of the parabolic arch.
radius = √𝑔2 + ⁡ 𝑓 2 − 𝑐⁡ ⁡ = ⁡⁡5
(5.33) Mar – 20
30. Find the equation of ellipse whose foci (±𝟑, 𝟎) and eccentricity = ⁡ 𝟏⁄𝟐 (5.2-2(i) Jun 2024
Equation of parabola: 𝒙𝟐 = ⁡ −𝟒𝒂𝒚
400 80 Foci = (±3, 0) 𝑒 = ⁡ 1⁄2
(20, −15)⁡⁡⁡⁡⁡⁡⁡⁡4𝑎 = ⁡ ⁡=⁡
15 3 Centre: = (0, 0),⁡⁡⁡⁡𝑐 = 3
2 80
𝑥 =⁡− 3
⁡𝑦⁡ 𝟑𝒙𝟐 ⁡ = ⁡ −𝟖𝟎𝒚 𝑐⁡ = ⁡𝑎𝑒 ⟹ 𝑎⁡ = ⁡6
𝑐 2 = ⁡ 𝑎2 − ⁡ 𝑏 2 ⁡ ⟹ 36 − ⁡⁡ 𝑏 2 = 9
2
𝑏 = 27
𝑥2 𝑦2
Equation of ellipse :
36
+ ⁡ 27 = 1

56 57
31. Prove that the point of intersection of the tangents at ′𝒕𝟏 ′ and ′𝒕𝟐 ′ on the parabola 𝒚𝟐 = 𝟒𝒂𝒙 Chapter 11. Probability Distribution
is (𝒂𝒕𝟏 ⁡𝒕𝟐 ⁡, 𝒂(𝒕𝟏 + ⁡ 𝒕𝟐 )) (5.4 -7) June 23 2 Marks:
Equation of tangent : 𝑦𝑡 = 𝑥 + 𝑎𝑡 2 1. Suppose two coine are tossed once. If X denotes the number of tails. (i) Write down the sample
𝑦𝑡1 = 𝑥 + 𝑎𝑡12 ………………. (1) space (ii) Find the inverse image of 1 (iii) the values of a random variable and number of
𝑦𝑡2 = 𝑥 + 𝑎𝑡22 ………………. (2) elements in its inverse images
Ans:
(1) - (2) 𝑦 = ⁡𝑎(𝑡1 + ⁡ 𝑡2 )
(𝑖)⁡𝑆 = (𝐻𝐻, 𝐻𝑇,⁡⁡⁡𝑇𝐻,⁡⁡⁡𝑇𝑇} 𝑛(𝑆) = 4
Sub in (1) , 𝑥⁡ = ⁡𝑎𝑡1 ⁡𝑡2
(𝑖𝑖)⁡𝑋 −1 ⁡{1} = {𝑇𝐻,⁡⁡⁡𝐻𝑇}
/ Printing of intersection is (𝑎𝑡1 ⁡𝑡2 ⁡, 𝑎(𝑡1 + ⁡ 𝑡2 ))
(iii)
2 Marks:
Values of the random variables 0 1 2 Total
32. If 𝒚⁡ = ⁡𝟒𝒙⁡ + ⁡𝒄 is a tangent to the circle 𝒙𝟐 + ⁡ 𝒚𝟐 = 𝟗, find c (5.12) March 23
Number of elements in inverse image 1 2 1 4
𝑦⁡ = ⁡4𝑥⁡ + ⁡𝑐⁡ = ⁡𝑚𝑥⁡ + ⁡𝑐⁡⁡ ⇒ ⁡⁡⁡𝑚⁡ = 4⁡⁡⁡⁡⁡⁡
𝑐⁡ = 𝑐⁡⁡
𝑥 2 + ⁡ 𝑦 2 = 9⁡ = ⁡ 𝑎2 2. Suppose a pair of unbiased dice is rolled once. If X denotes the total score of two dice, write
𝑐 2 = ⁡ 𝑎2 ⁡(1 + ⁡ 𝑚2 ) = 9⁡(1 + 16)⁡ down (i) the sample space (ii) the values taken by the random variable X (iii) the inverse image
𝑐 = ⁡ ±3√17 of 10, and (iv) the number of elements in inverse image of X.
33. Find the equation of the hyperbola with vertices (𝟎,⁡⁡⁡ ± 𝟒) and foci (𝟎,⁡⁡⁡ ± 𝟔) Ans:
(𝟓. 𝟐𝟒)⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑱𝒖𝒏. 𝟐𝟑 (𝑖)⁡ 𝑛(𝑆) = 36
Vertices = (0,⁡⁡⁡ ± 4) 𝑎⁡ = ⁡4 𝑎2 = 16 (ii) Then the random variable X takes on the values: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
foci = (0,⁡⁡⁡ ± 6) 𝑐⁡ = ⁡6 𝑐 2 = 36 (iii) The inverse images of 10 is {(4, 6), (5, 5), (6, 4)}
Centre (0, 0) 𝑐 2 = ⁡ 𝑎2 + ⁡ 𝑏 2 ⁡⁡ ⟹ ⁡ 𝑏 2 = 20 (iv)
𝑦2 𝑥2 Values of the random variable 2 3 4 5 6 7 8 9 10 11 12 Total
Equation of hyperbola : − ⁡ 20 = 1 Number of elements in inverse image 1 2 3 4 5 6 5 4 3 2 1 36
16
34. Find the general equation of a circle with centre (−𝟑, −𝟒) and radius 3 units.
(𝟓. 𝟏) March 24
Centre = (ℎ, 𝑘) = ⁡ (−3, −4)⁡⁡
3. Suppose X is the number of tails occurred when three fair coins are tossed once simultaneously.
radius = ⁡𝑟 = ⁡3
Find the volume of the random variable X and number of points into inverse images.
Equation of circle : (𝑥 − ℎ)2 + ⁡ (𝑦 − 𝑘)2 = ⁡ 𝑟 2
Ans:
(𝑥 + 3)2 + ⁡ (𝑦 + 4)2 = ⁡ 32
𝑆 = {𝐻𝐻𝐻, 𝐻𝐻𝑇, 𝐻𝑇𝐻, 𝑇𝐻𝐻, 𝐻𝑇𝑇, 𝑇𝐻𝑇, 𝑇𝑇𝐻, 𝑇𝑇𝑇}
𝑥 2 + ⁡ 𝑦 2 + 6𝑥 + 8𝑦 + 16 = 0
𝑛(𝑆) ⁡ = ⁡8
35. Obtain the equation of the circle for which (3, 4) and (2, -7) are the ends of a diameter. Values of the random variable 0 1 2 3 Total
(5.1-5) Jun 24 Number of elements in inverse image 1 3 3 1 8
Equation of circle: (𝑥 − ⁡ 𝑥1 )⁡(𝑥 − ⁡ 𝑥2 ) + ⁡ (𝑦 − ⁡ 𝑦1 )⁡(𝑦 − ⁡ 𝑦2 ) = 0

(𝑥 − ⁡3)⁡(𝑥⁡– ⁡2) ⁡ + ⁡ (𝑦⁡ − 4)⁡(𝑦 + 7) ⁡ = ⁡0 𝑲⁡𝒙𝒆−𝟐𝒙 𝒇𝒐𝒓⁡𝒙 > 𝟎

4. The Probability density function of X is given by 𝒇(𝒙) = ⁡ { . Find the value
𝟎 𝒇𝒐𝒓⁡𝒙⁡ ≤ 𝟎
𝒙𝟐 + ⁡ 𝒚𝟐 − 𝟓𝒙 + 𝟑𝒚 − 𝟐𝟐 = 𝟎
of K?
36. Examine the position of the point (2, 3) with respect to the circle
Ans:
𝑥 2 + ⁡ 𝑦 2 − 6𝑥 − 8𝑦 + 12 = 0

sub (2, 3)
= 4 + 9 − 12 − 24 + 12 = ⁡ −11 < 0
 f ( x)dx  1


(2, 3) lies inside of the circle. 

𝐾  xe2 x dx  1
0

𝐾⁡ = ⁡4

58 59
 3 Marks 9. If X is the random variable with distribution function 𝑭(𝒙) given by,
5. An urn contains 5 mangoes and 4 apples. here fruits are taken at random. If the number of 𝟎 , −∞ < 𝒙 < 𝟎
𝟏
apples taken is a random variable, then find the values of the random variable and number of 𝑭(𝒙) = ⁡ { (𝒙𝟐 + 𝒙) ,
𝟐
𝟎≤𝒙<𝟏
points in its inverse images. 𝟎 , 𝟏 < 𝒙 < ⁡∞
Ans: then find (i) the probability density function 𝒇(𝒙)⁡
Values of the random variable 0 1 2 3 Total (ii) 𝑷(𝟎. 𝟑⁡ ≤ 𝑿⁡ ≤ 𝟎. 𝟔)
Number of elements in 5𝐶3 5𝐶2 ⁡ × ⁡ 4𝐶1 5𝐶1 ⁡ × ⁡ 4𝐶2 4𝐶3 84 Ans:
inverse image = 10 = 40 = 30 =4 0 , 𝑥<0
1
(𝑖)⁡⁡⁡⁡𝑓(𝑥) = ⁡⁡ 𝐹 ′ (𝑥) = ⁡ { (2𝑥 + 1) , 0 ≤ 𝑥 < 1
2
6. Two fair coins are tossed simultaneously (equivalent to a fair coin is tossed twice). Find the 0 , 1<𝑥
probability mass function for number of heads occurred. (𝑖𝑖)⁡⁡⁡𝑃(0.3⁡ ≤ 𝑋⁡ ≤ 0.6) = 𝐹(0.6) − 𝐹(0.3) ⁡ = ⁡⁡0.285
Ans: 10. Find the mean and variance of a random variable X, whose probability density function is
𝑆 = {𝐻𝐻, 𝐻𝑇,⁡⁡⁡𝑇𝐻, 𝑇𝑇} 𝑛(𝑆) ⁡ = 4
𝝀𝒆−𝝀𝒙 , 𝒙≥𝟎
𝒇(𝒙) = ⁡⁡ {
Values of the random variable X 0 1 2 Total 𝟎 , 𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
Number of elements in inverse image 1 2 1 4 Ans:

Mean: 𝐸(𝑋) =⁡  xf ( x)dx
Probability mass function: 
1
𝑥 0 1 2 =⁡
𝜆
𝑓(𝑥) 1 1 1 

4 2 4 Variance: 𝐸(𝑋 2 ) =⁡  x 2 f ( x)dx


2
7. A pair of fair dice is tossed once. Find the probability of mass function to get the number of =⁡ 2
𝜆
forms. 𝑉𝑎𝑟⁡(𝑋) = ⁡𝐸(𝑋 2 ) − ⁡ [𝐸(𝑋)]2
2 1 2
Ans: = ⁡ 𝜆 2 − ⁡ (𝜆 )
1
Values of Random variable X 0 1 2 Total = ⁡ 𝜆2
Number of elements in inverse images 25 10 1 36 1 1
Mean:
𝜆
> Variance = 𝜆2
The probability mass function is presented as
𝑥 0 1 2 5 Marks
𝑓(𝑥) 25 10 1 11. A six sided die is marked ‘1’ on one face, ‘2’ on two of its faces, and ‘3’ on remaining three faces.
36 36 36 The die is rolled twice. If X denotes the total score in two throws.
(i) Find the probability mass function
8. If X is the random variable with distribution function F(x) given by (ii) Find the cumulative distribution function
0 , 𝑥<0 (iii) Find 𝑷(𝟑⁡ ≤ 𝑿 < 𝟔)
𝐹(𝑥) = ⁡ {𝑥 , 0 ≤ 𝑥 < 1 (iv) Find 𝑷(𝑿⁡ ≥ 𝟒)
1 , 1≤𝑥 Ans:
then find (i) the probability density function 𝑓(𝑥) (ii) 𝑃(0.2⁡ ≤ 𝑋⁡ ≤ 0.7) 1 2 2 3 3 3
Ans: I II
0 , 𝑥<0
(i) 𝑓(𝑥) = ⁡⁡ 𝐹 ′ (𝑥) = ⁡ {1 , 0 ≤ 𝑥 < 1
1 2 3 3 4 4 4
0 , 1≤𝑥
1 , 0≤𝑥<1 2 3 4 4 5 5 5
𝑓(𝑥) = ⁡ { 2 3 4 4 5 5 5
0 , 𝑂𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
3 4 5 5 6 6 6
(𝑖𝑖)⁡⁡⁡𝑃(0.2⁡ ≤ 𝑋⁡ ≤ 0.7) ⁡⁡ = 𝐹(0.7) − 𝐹(0.2)⁡
3 4 5 5 6 6 6
= 0.7 − 0.2
3 4 5 5 6 6 6
= 0.5

60 61
 (i) Probability mass function is: (ii) Cumulative distribution function:
𝑥 2 3 4 5 6 Total 𝑥 2 4 6 8 10
𝑓(𝑥) 1 4 10 12 9 1 𝐹(𝑥) 1 5 15 27 36
36 36 36 36 36 36 36 36 36 36
4 10 12 26 13
(ii) Cumulative distribution function (𝑖𝑖𝑖) 𝑃(4⁡ ≤ 𝑋 < 10) = ⁡ 36 ⁡ + ⁡⁡ ⁡ + ⁡⁡ ⁡=⁡ ⁡=⁡
36 36 36 18
10 12 9 31
𝑥 2 3 4 5 6 (𝑖𝑣) 𝑃(𝑋⁡ ≥ 6) = ⁡ ⁡ + ⁡⁡ ⁡ + ⁡⁡ ⁡=⁡
36 36 36 36
𝐹(𝑥) 1 5 15 27 36
36 36 36 36 36 14. The cumulative distribution function of a discrete random variable is given by
4 10 12 26 𝟎 , −∞ < 𝒙 < −𝟏
(𝑖𝑖𝑖) 𝑃(3⁡ ≤ 𝑋 < 6) = ⁡ 36 ⁡ + ⁡⁡ ⁡ + ⁡⁡ ⁡=⁡ 𝟎. 𝟏𝟓 , −𝟏 < 𝒙 < 𝟎
36 36 36
(𝑖𝑣) 𝑃(𝑋⁡ ≥ 4) = ⁡
10
⁡ + ⁡⁡
12
⁡ + ⁡⁡
9
⁡=⁡
31 𝟎. 𝟑𝟓 , 𝟎⁡ ≤ 𝒙 < 𝟏
𝑭(𝒙) = ⁡
36 36 36 36 𝟎. 𝟔𝟎 ⁡⁡, ⁡⁡𝟏 ≤ 𝒙 < ⁡𝟐
12. A random variable X has the following probability mass function 𝟎. 𝟖𝟓⁡⁡ , 𝟐≤𝒙<𝟑
𝒙 1 2 3 4 5 6 { 𝟏 , 𝟑⁡ ≤ 𝒙 < ⁡∞
𝒇(𝒙) K 2K 6K 5K 6K 10K Find (i) the probability mass function (ii) 𝑷(𝑿 < 𝟏) and 𝑷(𝑿⁡ ≥ 𝟐)
Ans: (i) Probability mass function is
Find (𝒊) 𝑷(𝟐 < 𝑿 < 𝟔) (𝒊𝒊) 𝑷(𝟐 ≤ 𝑿 < 𝟓) 𝑥 -1 0 1 2 3
(𝒊𝒊𝒊) 𝑷(𝑿⁡ ≤ 𝟒) (𝒊𝒗) 𝑷(𝟑 < 𝑿) 𝑓(𝑥) 0.15 0.20 0.25 0.25 0.15
Ans: Σ𝑓(𝑥) = 1
𝐾⁡ + ⁡2𝐾⁡ + ⁡6𝐾⁡ + ⁡5𝐾⁡ + ⁡6𝐾⁡ + ⁡10𝐾⁡⁡ = ⁡1 (𝑖𝑖) 𝑃(𝑋 < 1) ⁡ = 0.14⁡ + 0.20⁡ = 0.35
1 (𝑖𝑖) 𝑃(𝑋 ≥ 2) ⁡ = 0.25 + 0.15⁡ = 0.40
30𝐾 = 1⁡ ⟹ 𝐾 = ⁡ 30
17
(𝑖) 𝑃(2 < 𝑋 < 6) ⁡ = ⁡6𝐾 + 5𝐾 + 6𝐾 = 17𝐾 = ⁡ 30 15. A random variable X has the following probability mass function
13 𝒙
(𝑖𝑖) 𝑃(2 ≤ 𝑋 < 5) = ⁡2𝐾 + 6𝐾 + 5𝐾 = 13𝐾⁡ = ⁡ 1 2 3 4 5
30
14 𝒇(𝒙) 𝑲𝟐 𝟐𝑲𝟐 𝟑𝑲𝟐 𝟐𝑲 𝟑𝑲
(𝑖𝑖𝑖) 𝑃(𝑋⁡ ≤ 4) = ⁡𝐾 + 2𝐾 + 6𝐾 + 5𝐾 = 14𝐾⁡ = ⁡ 30⁡
21
(𝑖𝑣) 𝑃(3 < 𝑋) = ⁡5𝐾 + 6𝐾 + 10𝐾 = 21𝐾⁡ = ⁡ 30 Find (i) the value of K (𝒊𝒊) 𝑷(𝟐⁡ ≤ 𝑿 < 𝟓) (𝒊𝒊𝒊)⁡𝑷(𝟑 < 𝒙)
Ans: (𝑖) Σ𝑓(𝑥) = 1
13. A six sided die is marked ‘1’ on one face, ‘3’ on two of its faces, and ‘5’ on remaining three
faces. The die is thrown twice. If X denotes the total score in two throws, find 6𝐾 2 + 5𝐾 = 1 ⟹ 6𝐾 2 + 5𝐾 − 1 = 0
1
(i) the probability mass function (ii) the cumulative distribution function 𝐾 = ⁡ −1 (𝑁𝑜𝑡⁡𝑃𝑜𝑠𝑠𝑖𝑏𝑒) 𝐾 =⁡6
(𝒊𝒊𝒊) 𝑷(𝟒 ≤ 𝑿 < 𝟏𝟎) (𝒊𝒗) 𝑷(𝑿⁡ ≥ 𝟔) (𝑖𝑖)⁡𝑃(2⁡ ≤ 𝑋 < 5) ⁡ = ⁡2𝐾 + 3𝐾 + 2𝐾 = 5𝐾 2 + 2𝐾⁡⁡⁡
2 2

Ans: 5 2
= 36 + ⁡ ⁡
1 3 3 5 5 5 6
5 12 17
I II = ⁡ 36 + ⁡ 36 ⁡ = ⁡ 36
5
(𝑖𝑖𝑖) ⁡𝑃(3 < 𝑥) = ⁡⁡2𝐾 + 3𝐾 = 5𝐾 = ⁡ 6
1 2 4 4 6 6 6
3 4 6 6 8 8 8
16. The cumulative distribution function of a discrete random variable is given by
3 4 6 6 8 8 8 𝟎 , −∞ < 𝒙 < 𝟎
5 6 8 8 10 10 10 𝟏
, 𝟎≤𝒙<𝟏
5 6 8 8 10 10 10 𝟐
𝟑
5 6 8 8 10 10 10 , 𝟏⁡ ≤ 𝒙 < 𝟐
𝑭(𝒙) = ⁡ 𝟓𝟒
(i) The probability mass function is: 𝟓
, ⁡⁡𝟐 ≤ 𝒙 < ⁡𝟑
𝟗
𝑥 2 4 6 8 10 Total ⁡⁡ ,𝟑≤𝒙<𝟒
𝟏𝟎
𝑓(𝑥) 1 4 10 12 9 1 { 𝟏 , 𝟒⁡ ≤ 𝒙 < ⁡∞
36 36 36 36 36 Find (i) the probability mass function (𝒊𝒊)⁡⁡⁡𝑷(𝑿 < 𝟑) and (iii) 𝑷(𝑿 ≥ 𝟐)
62 63
 Ans: 19. The mean and variance of a binomial variate 𝒙 are respectively 2 and 1.5.
(𝑖) Find (𝒊)⁡𝑷(𝒙 = 𝟎) (𝒊𝒊)⁡𝑷(𝒙 = 𝟏) (𝒊𝒊𝒊)⁡𝑷(𝒙 ≥ 𝟏)
𝑥 0 1 2 3 4 Ans: Mean = 𝑛𝑝⁡ = ⁡2 Variance = ⁡𝑛𝑝𝑞⁡ = ⁡1.5
𝑛𝑝𝑞 1.5 3 1
𝑓(𝑥) 1 1 1 1 1 ⁡=⁡ ⁡=⁡4 = 𝑞 𝑝 =⁡4 𝑛⁡ = ⁡8
𝑛𝑝 2
2 10 5 10 10 𝑃(𝑋 = 𝑥) ⁡ = ⁡𝑛𝑐𝑥 ⁡𝑝 𝑥 𝑞 𝑛−𝑥
1 1 1 8 4
(𝑖𝑖) 𝑃(𝑋 < 3) = ⁡ + ⁡ + ⁡ = ⁡ = ⁡ 1 𝑥 3 8−𝑥
2 10 5 10 5 = 8𝑐𝑥 ⁡⁡( ) ⁡( ) 𝑥 = 0, 1, 2, … … … . 8
4 4
1 1 1 4 2
(𝑖𝑖𝑖) 𝑃(𝑋 ≥ 2) = ⁡ 5 + ⁡ 10 + ⁡ 10 = ⁡ 10 = ⁡ 5 1 0 3 8−0 3 8
(𝑖)⁡𝑃(𝑥 = 0) ⁡ = ⁡⁡8𝑐0 ⁡⁡⁡( ) ⁡⁡( ) ⁡ = ⁡⁡⁡ ( ) ⁡
4 4 4
17. Suppose the amount of milk sold daily at a milk 600 is distribution with a minimum of 200 litres
3 7
and a maximum of 600 litres with probability density function (𝑖𝑖)⁡𝑃(𝑥 = 1) = ⁡⁡2 ( )
4
𝑲 , 𝟐𝟎𝟎⁡ ≤ 𝒙 < 𝟔𝟎𝟎 3 8
𝒇(𝒙) = ⁡ { (𝑖𝑖𝑖)⁡𝑃(𝑥 ≥ 1) = ⁡⁡1 − 𝑃(𝑥 < 1) ⁡⁡ = ⁡⁡1 − ( ) ⁡
𝟎 , 𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆 4

Find (i) the value of K (ii) the distribution function 20. If 𝑿⁡~⁡𝑩(𝒏, 𝒑) such that 𝟒𝑷(𝑿⁡ = 𝟒) ⁡⁡ = ⁡⁡𝑷(𝑿⁡ = 𝟐) and n = 6. Find the distribution, mean
(iii) the probability that daily sales will fall between 300 litres on 500 litres. standard deviation of X
Ans: Ans:
1 2
 𝑛=6 𝑝 =⁡ 𝑞 =⁡
3 3
(𝑖) 

f ( x)dx  1
𝑝(𝑥 = 𝑥) = ⁡⁡𝑓(𝑥) = ⁡𝑛𝑐𝑥 ⁡⁡⁡𝑝 𝑥 ⁡⁡𝑞 𝑛−𝑥 𝑥 = 0, 1, 2, … … … . 𝑛
600 1 𝑥 2 6−𝑥
𝐾⁡ ∫200 𝑑𝑥 =1 𝑓(𝑥) = ⁡⁡6𝑐𝑥 ⁡⁡⁡(3) ⁡⁡( ) ⁡ 𝑥 = 0, 1, 2, … … … . 6
3
1 1
𝐾⁡ = Mean = ⁡𝑛𝑝 = 6⁡ × ⁡ 3 ⁡ = ⁡⁡2
400
0 , 𝑥 < 200 1 2
𝑥 1 Standard deviation = ⁡ √𝑛𝑝𝑞 = ⁡ √6⁡ × ⁡ ⁡ × ⁡
(𝑖𝑖) 𝐹(𝑥) = ⁡ {400 − ⁡ 2 , 200⁡ ≤ 𝑥⁡ ≤ 600 3 3
2
1 , 𝑥 > 600 =⁡
√3
500
200 1
 f ( x)dx
(𝑖𝑖𝑖) = 400 = ⁡ 2 21. If the probability that a fluorescent light has a useful life of at least 600 hours in 0.9, find the
300 probabilities that among 12 such lights
18. For the random variable X with the given probability mass function as below, find the mean and i. Exactly 10 will have a useful life of at least 600 hours;
variance ii. at least 11 will have useful life of at least 600 hours;
𝟐(𝒙 − 𝟏) , 𝟏 < 𝒙 < 𝟐 iii. at least 2 will not have a useful life of at least 600 hours.
𝒇(𝒙) = ⁡ {
𝟎 , 𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆 Ans:
9 9 1
Ans: 𝑝 = 0.9 = ⁡ 10 𝑞 = 1 − 𝑝 = 1 − ⁡ 10 ⁡ = ⁡ 10⁡ 𝑛 = 12

Mean: 𝐸(𝑥) =⁡  xf ( x)dx 𝑃(𝑋 = 𝑥) = 𝑛𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 ,⁡⁡⁡⁡⁡𝑥 = 0, 1, 2, … … … . , 𝑛
 9 10 1 12−10
(𝑖)⁡⁡𝑃(𝑋 = 10) = 12𝐶10 ( ) (10)
2 10
= ⁡2  ( x 2  x)dx = 12𝐶10 (0.9)10 (0.1)12−10
1 (𝑖𝑖)⁡⁡⁡𝑃(𝑋⁡ ≥ 11) = 𝑃(𝑋 = 11) + 𝑃(𝑋 = 12)
5 5
= ⁡2 (6) ⁡ = ⁡ 3 9 11 1 12−11 9 12 1 12−12
= 12𝐶11 (10) (10) − ⁡12𝐶12 (10) (10)

𝐸(𝑥 2 ) =⁡  x 2 f ( x)dx 9 11 1 1 9 12
= 12 (10) (10) +⁡⁡(10)

2 9 11 1 9
= ⁡2  x 2 ( x  1)dx = ⁡ (10) ⁡[12⁡ × ⁡ 10 + ⁡ 10⁡]
1 = ⁡ (0.9)11 [1.2 + 0.9]
17
=⁡ 6 = 2.1(0.9)11
Variance: 𝑉𝑎𝑟(𝑥) = 𝐸(𝑥 2 ) ⁡ − ⁡ [𝐸(𝑥)]2 (𝑖𝑖𝑖)⁡𝑃⁡(𝑋 ≥ 2) = ⁡⁡1 − ⁡𝑃(𝑋⁡ ≥ 11)
17 25
= ⁡ 6 ⁡ − ⁡⁡ 9 = 1 − ⁡2.1(0.9)11
1
= ⁡ 18
64 65
22. A multiple choice examination has ten questions, each question has four distance with exactly (iii) Factorize: (𝑦 + 1)(𝑦 + 7) = 0
one correct answer. Suppose a student answers by guessing and if X denotes the number of 𝑦⁡ = ⁡ −1,⁡⁡⁡𝑦⁡ = ⁡ −7
correct answers, find (i) binomial distribution (ii) probability that the student will get seven (iv) If 𝑦⁡ = −1⁡then If 𝑦⁡ = ⁡ −7 then
correct answers (iii) the probability of getting at least one correct answer. 6𝑥 2 − 13𝑥 = ⁡ −1 6𝑥 2 − 13𝑥 = −7
Ans: 6𝑥 2 − 13𝑥 + ⁡1 = 0 6𝑥 2 − 13𝑥 + 7⁡ = 0
1 1 3
𝑛 = 10; ⁡⁡⁡⁡⁡𝑝 = ⁡ ⁡ ⟹ 𝑞 = 1 − 𝑝 = 1 − ⁡ = ⁡ 𝑎 = ⁡6,⁡⁡⁡⁡𝑏⁡ = −13,⁡⁡⁡𝑐⁡ = ⁡1 Factorize
4 4 4
−𝑏⁡±⁡√𝑏2 −4𝑎𝑐
(𝑖)⁡⁡⁡𝑃(𝑋 = 𝑥) ⁡ = ⁡𝑛𝐶𝑥 𝑝 𝑞 𝑥 𝑛−𝑥
,⁡⁡⁡⁡⁡𝑥 = 0, 1, 2, … … … . , 𝑛⁡ 𝑥 =⁡ (𝑥 − 1)⁡(𝑥 − 7⁄6) = 0
2𝑎
𝑃(𝑋 = 𝑥) = ⁡10𝐶𝑥 𝑝 𝑥 𝑞10−𝑥 ,⁡⁡⁡⁡⁡𝑛 = 0, 1, 2, … … … . , 10 13⁡±⁡√(−13)2 −4(6)(1)
1 7 3 10−7
=⁡ 2⁡×6
𝑥 = 1,⁡⁡⁡⁡
(𝑖𝑖)⁡⁡⁡𝑃(𝑋 = 7) ⁡ = ⁡⁡10𝐶7 ( ) ( )
4 4
𝑥 = ⁡ 7⁄6
13⁡±⁡√145
1 7 3 3
𝑥 =⁡
12
=⁡ 10𝐶3 ( ) ( )
Thus 1, 7⁄6 ⁡,
4 4 13+⁡√145 13−⁡√145
Result: , are the solutions of the given equation.
10.9.8 1 7 3 3 12 12
=⁡ 1.2.3
. (4) (4) TRY THIS SUMS:
33 1. Solve: (𝑥 − 5)(𝑥 − 7)(𝑥 + 6)(𝑥 + 4) = 504
= 120 (410 )
2. Solve: (𝑥 − 4)(𝑥 − 7)(𝑥 − 2)(𝑥 + 1) = 16
(𝑖𝑖𝑖)⁡⁡𝑃(𝑋⁡ ≥ 1) = 1 − 𝑃(𝑋 < 1)
3. Solve: (2𝑥⁡ − 1)⁡(𝑥⁡ + ⁡3)(𝑥 − 2)(2𝑥 + 3) + 20 = 0
= 1 − 𝑃(𝑋 = 0)
1 0 3 10−0
= 1 − ⁡10𝐶0 ( ) ( )
4 4
10. Ordinary Differential Equation
3 10
= 1 − ⁡ (4) 2 Marks:
23. On the average, 20% of the products manufactured by ABC Company are found to be defective. 1. The order and Degree of the Differential equation 𝒚′ + ⁡ (𝒚′′ )𝟐 = 𝒙⁡(𝒙 + 𝒚′′)𝟐
If we select 6 of these products at random and X denotes the number of defective products find Ans:
the probability that (i) two products are defective (ii) at most one product is defective Order -2, Degree – 2
(iii) at least two products are defective. 𝒅𝟐 𝒚
𝟐
𝒅𝒚 𝟐 𝒅𝟐 𝒚
Ans: 2. The order and Degree of the Differential equation (𝒅𝒙𝟐 ) + ⁡ (𝒅𝒙) = 𝒙⁡𝒔𝒊𝒏 (𝒅𝒙𝟐 )
20 1 1 4
𝑝 = 20%⁡ = ⁡⁡ 100 = ⁡ 5 𝑞 = 1 − 𝑝 = 1 −⁡5 =⁡5 𝑛=6 Ans:
𝑥 𝑛−𝑥 Order – 2, Degree is not Defined.
𝑃(𝑋 = 𝑥) ⁡ = ⁡𝑛𝐶𝑥 𝑝 𝑞 ,⁡⁡⁡⁡⁡𝑥 = 0, 1, 2, … … … . , 𝑛
1 2 4 6−2 𝒅𝒚 𝟏−⁡𝒚𝟐
(𝑖) 𝑃(𝑋 = 2) ⁡ = ⁡⁡6𝐶2 (5) (5) 3. Show that the solution of = ⁡ √𝟏−⁡𝒙𝟐 is 𝒔𝒊𝒏−𝟏 𝒚 = ⁡ 𝒔𝒊𝒏−𝟏 𝒙 + 𝒄 (OR)
𝒅𝒙

6.5 1 2 4 4 𝒅𝒚 𝟏−⁡𝒚𝟐
= ⁡ 1.2 . (5) (5) Solve: = ⁡ √𝟏−⁡𝒙𝟐
𝒅𝒙
44
= 15 ( 6 ) Ans:
5
𝑑𝑦 √1−⁡𝑦 2
(𝑖𝑖)⁡(𝑋⁡ ≤ 1) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) =⁡
𝑑𝑥 √1−⁡𝑥 2
1 0 4 6−0 1 1 4 6−1 𝑑𝑦 𝑑𝑥
= 6𝐶0 (5) (5) + ⁡6𝐶1 (5) (5) = ⁡ √1−⁡𝑥 2
√1−⁡𝑦2
4 6 1 1 4 5 𝑑𝑦 𝑑𝑥
= (5) + 6 (5) (5) ∫ √1−⁡𝑦 2 = ⁡ ∫ √1−⁡𝑥2
45
= ⁡ 56 ⁡(4 + 6) 𝑠𝑖𝑛−1 𝑦 = ⁡ 𝑠𝑖𝑛−1 𝑥 + 𝑐
45 4. 𝒚 = 𝒂𝒆𝒙 + 𝒃𝒆−𝒙 is a solution of the differential equation 𝒚′′ − 𝒚 = 𝟎
= ⁡ 56 ⁡⁡ × 10 Ans:
4 5 𝑦 = ⁡𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥
= 2 (5)
𝑦 ′ = ⁡𝑎𝑒 𝑥 − 𝑏𝑒 −𝑥 …………………… (1)
(𝑖𝑖𝑖)⁡𝑃⁡(𝑋 ≥ 2) = ⁡⁡1 − ⁡𝑃(𝑋⁡ ≤ 1)
𝑦 ′′ = ⁡𝑎𝑒 𝑥 − 𝑏𝑒 −𝑥 ⁡(−1)
4 5
= 1 − ⁡2 (5) 𝑦 ′′ = ⁡𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥
𝑦 ′′ = 𝑦
𝑦 ′′ − 𝑦 = 0
66 80
5. Form the differential equation of the curve 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 where a, b and c are arbitrary 11. Show that the differential equation corresponding to 𝒚 = 𝑨⁡𝒔𝒊𝒏𝒙,⁡ where A is an arbitrary
constant. constant is 𝒚 = 𝒚′ 𝐭𝐚𝐧 𝒙
Ans: Ans:
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑦 = 𝐴⁡𝑠𝑖𝑛𝑥
𝑦 ′ = 2𝑎𝑥 + 𝑏 ………………….. (1) 𝑦′ = 𝐴⁡𝑐𝑜𝑠𝑥
𝑥 2 − ⁡ 𝑦 ′′ − 2𝑥𝑦 ′ + 2𝑦 = ⁡0 𝑦′
6. The Population P of a city increase at a rate proportional to the product of population and to the cos 𝑥
=𝐴
difference between 5,00,000 and the population physical statements in the form of differential 𝑦 = 𝑦 ′ tan 𝑥
equation.
Ans:
𝑑𝑃 3 Marks:
𝑑𝑡
⁡𝛼⁡𝑃(500000 − ⁡𝑃)
𝑑𝑃
12. Show that each of the following expression is a solution of the corresponding given Differential
= 𝐾𝑃⁡(500000 − 𝑃) equation 𝒚 = 𝒂𝒆𝒙 + 𝒃𝒆−𝒙 ; 𝒚′′ − 𝒚 = 𝟎
𝑑𝑡
where 𝐾 is constant Ans:
7. Find the differential equation of the family of parabola 𝒚𝟐 = 𝟒𝒂𝒙 where a is an arbitrary 𝑦 = ⁡𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥
constant. 𝑦 ′ = ⁡𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥 (−1) …………………… (1)
Ans: 𝑦 ′ = ⁡𝑎𝑒 𝑥 − 𝑏𝑒 −𝑥
𝑦 2 = 4𝑎𝑥 ………………… (1)
𝑑𝑦 𝑦 ′′ = ⁡𝑎𝑒 𝑥 − 𝑏𝑒 −𝑥 ⁡(−1)
2𝑦⁡ = 4𝑎 …………………. (2)
𝑑𝑥
𝑑𝑦 𝑦2
𝑦 ′′ = ⁡𝑎𝑒 𝑥 + 𝑏𝑒 −𝑥
2𝑦⁡ 𝑑𝑥 = ⁡
𝑥 𝑦 ′′ = 𝑦
𝑑𝑦 𝑦2 𝑦
=⁡ =⁡ 𝑦 ′′ − 𝑦 = 0
𝑑𝑥 2𝑥𝑦 2𝑥
𝑑𝑦 𝑦
/ = ⁡ 2𝑥 13. Solve : (𝒆𝒚 ⁡ + 𝟏) 𝐜𝐨𝐬 𝒙⁡𝒅𝒙 + ⁡ 𝒆𝒚 ⁡𝒔𝒊𝒏𝒙⁡𝒅𝒚 = 𝟎
𝑑𝑥
Ans:
8. Show that the solution of the differential equation 𝒚𝒙𝟑 𝒅𝒙 + ⁡ 𝒆−𝒙 ⁡𝒅𝒚 = 𝟎 is
(𝑒 𝑦 ⁡ + 1) cos 𝑥⁡𝑑𝑥 = ⁡ −𝑒 𝑦 ⁡𝑠𝑖𝑛𝑥⁡𝑑𝑦
(𝒙𝟑 − 𝟑𝒙𝟐 ⁡ + 𝟔𝒙 − 𝟔)⁡𝒆𝒙 + 𝐥𝐨𝐠 𝒚 = 𝒄
cos 𝑥 𝑒𝑦
Ans: 𝑑𝑥 = ⁡ − 𝑒 𝑦 ⁡+1 𝑑𝑦
𝑑𝑦 sin 𝑥
3 𝑥
𝑥 𝑒 𝑑𝑥 + ⁡ 𝑦
=0 cos 𝑥 𝑒𝑦
∫ sin 𝑥 ⁡𝑑𝑥 ⁡ = ⁡ −⁡∫ 𝑒 𝑦 ⁡+1 𝑑𝑦⁡
(𝑥 3 2
− 3𝑥 ⁡ + 6𝑥 − 6)⁡𝑒 + log 𝑦 = 𝑐 𝑥
log sinx = ⁡ − log(𝑒 𝑦 ⁡ + 1) + log 𝑐
𝒅𝒚
9. Solve: + 𝒚 = ⁡ 𝒆−𝒙 (𝑒 𝑦 ⁡ + 1) sin 𝑥 = 𝑐
𝒅𝒙

Ans: 14. Form the differential equations by eliminating arbitrary constants given in bracket
𝐼. 𝐹⁡ = ⁡ 𝑒 ∫ 𝑝𝑑𝑥 = ⁡ 𝑒 𝑥 𝒚 = ⁡𝒆𝟑𝒙 ⁡(𝑪⁡𝒄𝒐𝒔𝟐𝒙⁡ + 𝑫 𝐬𝐢𝐧 𝟐𝒙), {C, D}
𝑥 Ans:
Solution 𝑦𝑒 ⁡ = ⁡⁡𝑥 + 𝑐
10. Show that the differential equation for the function 𝒚 = ⁡ 𝒆−𝒙 + 𝑴𝒙 + 𝒏 where M and n are 𝑦 = ⁡𝑒 3𝑥 ⁡(𝐶⁡𝑐𝑜𝑠2𝑥⁡ + 𝐷 sin 2𝑥)
𝒅𝟐 𝒚 𝑦⁡𝑒 −3𝑥 = 𝐶⁡𝑐𝑜𝑠2𝑥⁡ + 𝐷𝑠𝑖𝑛⁡2𝑥
arbitrary constants 𝒆𝒙 (𝒅𝒙𝟐 ) − 𝟏 = 𝟎
Ans: Difference W. r to x
𝑦 = ⁡ 𝑒 −𝑥 + 𝑀𝑥 + 𝑛 𝑦⁡𝑒 −3𝑥 (−3)
+ ⁡ ⁡𝑒 −3𝑥 𝑦 ′ = −2⁡𝐶⁡𝑠𝑖𝑛2𝑥 + 2𝐷⁡𝑐𝑜𝑠2𝑥
𝑑𝑦
= ⁡ −𝑒 −𝑥 + 𝑀 ⁡𝑒 −3𝑥
⁡[−3𝑦 + 𝑦′] = ⁡ −2𝐶⁡𝑠𝑖𝑛2𝑥⁡ + 2𝐷⁡𝑐𝑜𝑠2𝑥
𝑑𝑥
𝑑2 𝑦 𝑑2 𝑦
𝑑𝑥 2
=⁡ 𝑒 −𝑥 ⁡⁡ ⟹⁡ 𝑑𝑥 2 =⁡ 1⁄𝑒 𝑥 Difference r to x
𝑑2 𝑦
𝑒 𝑥 ⁡(𝑑𝑥 2 ) − 1 = 0 ⁡𝑒 −3𝑥 ⁡[𝑦 ′′ − 3𝑦 ′ ] + [𝑦 ′ − 3𝑦]𝑒 −3𝑥 ⁡(−3) = ⁡ −4⁡𝐶 cos 2𝑥 − 4𝐷⁡𝑠𝑖𝑛2𝑥
𝑦 ′′ − 6𝑦 ′ + 13𝑦 = 0

81 82
15. Assume that a spherical rain drop evaporates at a rate proportional to its surface area. Form a 𝒅𝒚 𝒚
19. Solve the Differential equation: + ⁡ 𝒙 = 𝒔𝒊𝒏𝒙
𝒅𝒙
differential equation involving the rate of change of the radius of the rain drop.
Ans:
Ans: 𝑑𝑦 𝑦
+ ⁡ = ⁡𝑠𝑖𝑛𝑥
𝑉 = ⁡ 4⁄3 ⁡𝜋𝑟 3 - Volume of sphere 𝑑𝑥 𝑥

𝐴 = 4𝜋𝑟 2 - surface Area ∫ 𝑃𝑑𝑥⁡ = ⁡⁡ ∫ 1⁄𝑥 𝑑𝑥 = log 𝑥

𝑑𝑉
𝑑𝑡
= −𝑘𝐴 𝐼. 𝐹 = ⁡ 𝑒 ∫ 𝑃𝑑𝑥 = ⁡ 𝑒 log 𝑥 = 𝑥
𝑑(4⁄3⁡𝜋𝑟 3 )
∫ 𝑑𝑡
⁡ = ⁡ −𝑘⁡(4⁡𝜋𝑟 2 ) Solution: 𝑦𝑒 ∫ 𝑃𝑑𝑥 = ⁡ ∫ 𝑄𝑒 ∫ 𝑃𝑑𝑥 𝑑𝑥 + 𝑐
4⁄ ⁡𝜋⁡3𝑟 2 ⁡ 𝑑𝑣 ⁡ = ⁡ −𝑘(4⁡𝜋𝑟 2 ) 𝑥𝑦 + 𝑥 cos 𝑥⁡ = sin 𝑥 + 𝑐
3 𝑑𝑡
𝑑𝑉
𝑑𝑡
= −𝑘 20. If F is the constant force Generated by the motor of an automobile of mass M its velocity V is
𝒅𝑽
16. Find the differential equation corresponding to the family of curves represented by the equation given by 𝑴 = 𝑭 − 𝑲𝑽 where K is a constant, Express V interms of t given that v = 0
𝒅𝒕
𝒚 = 𝑨𝒆𝟖𝒙 + 𝑩𝒆−𝟖𝒙 where A and B are arbitrary constants when t = 0
Ans: Ans:
𝑦 = ⁡𝐴𝑒 8𝑥 + 𝐵𝑒 −8𝑥 𝑑𝑉
𝑀 𝑑𝑡 = 𝐹 − 𝐾𝑉
′ 8𝑥 −8𝑥
𝑦 = ⁡8𝐴𝑒 − 8𝐵𝑒 𝑑𝑉 𝑑𝑡
𝐹−𝐾𝑉
⁡ = ⁡⁡ 𝑀
𝑦 ′′ = ⁡8𝐴(8)𝑒 8𝑥 − 8𝐵⁡(−8)𝑒 −8𝑥
𝑑𝑉 𝑑𝑡
𝑦′′ = ⁡64𝐴𝑒 8𝑥 + 64𝐵𝑒 −8𝑥 ∫ 𝐹−𝐾𝑉 ⁡ = ⁡ ∫ 𝑀

= ⁡ 1⁄𝑀 ⁡𝑡 + log 𝑐
log(𝐹−𝐾𝑉)
𝑦′′ = ⁡64(𝑦)
−𝐾
𝑦 ′′ − 64𝑦 = 0 𝐾𝑡
log(𝐹 − 𝐾𝑉) = ⁡ − 𝑀 ⁡ + log 𝑐

5 Marks: 𝑡 = 0,⁡⁡⁡⁡⁡𝑉 = 0
𝒅𝒚 −𝑀
17. Show that the solution of the Differential equation (𝟏 + ⁡ 𝒙𝟐 ) 𝒅𝒙 ⁡ = 𝟏 +⁡𝒚 𝟐
is 𝑐 =⁡ log 𝐹
𝐾
−𝟏 −𝟏 −𝟏 −𝟏
𝒕𝒂𝒏 𝒚 = ⁡ 𝒕𝒂𝒏 𝒙 + 𝒄 (or) 𝒕𝒂𝒏 𝒙 = ⁡ 𝒕𝒂𝒏 𝒚 + 𝒄 𝑘𝑡

Ans: 𝐹 = (𝐹 − 𝐾𝑉)𝑒 𝑚
𝑑𝑦 21. The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the
(1 + ⁡ 𝑥 2 ) ⁡ = 1 + ⁡ 𝑦2
𝑑𝑥 number present. Given that the number triples in 5 hours, find how man bacteria will be present
𝑑𝑦 𝑑𝑥 after 10 hours?
= ⁡ 1+⁡𝑥 2
1+⁡𝑦 2
Ans:
𝑑𝑦 𝑑𝑥
∫ 1+⁡𝑦 2 ⁡ = ⁡ ∫ 1+⁡𝑥 2 𝐴 = 𝐶𝑒 𝐾𝑡
𝑡 𝐴
𝑡𝑎𝑛−1 𝑦 = ⁡ 𝑡𝑎𝑛−1 𝑥 + 𝑐 𝑡 = 0,⁡⁡⁡⁡𝐴 = ⁡ 𝐴0 0 𝐴0
𝒅𝒚 𝐴0 = 𝐶𝑒 𝐾(0) 5 3𝐴0
18. Solve: 𝒅𝒙
= ⁡ 𝒆𝒙+𝒚 + ⁡ 𝒙𝟑 𝒆𝒚 10 ?
𝐶 = ⁡ 𝐴0 ⁡ ⟹ 𝐴⁡ = ⁡ 𝐴0 𝑒 𝐾𝑡
Ans:
𝑑𝑦
= ⁡ 𝑒 𝑦 [𝑒 𝑥 ⁡ + ⁡⁡ 𝑥 3 ] 𝑡 = 5,⁡⁡⁡⁡⁡𝐴 = 3𝐴0
𝑑𝑥
𝑑𝑦 3𝐴0 ⁡ = ⁡⁡ 𝐴0
⁡𝑒 𝑦
= ⁡ [𝑒 𝑥 ⁡ + ⁡⁡ 𝑥 3 ]𝑑𝑥
𝑒 5𝑘 ⁡ = 3
∫ 𝑒 −𝑦 𝑑𝑦⁡ = ⁡ ∫(𝑒 𝑥 ⁡ + ⁡⁡ 𝑥 3 )𝑑𝑥
𝑡 = 10,⁡⁡⁡⁡𝐴 =⁡⁡⁡?⁡⁡⟹ ⁡⁡𝐴 = ⁡⁡ 𝐴0 (𝑒 5𝐾 )
𝑥4
−𝑒 −𝑦 ⁡ = ⁡ 𝑒 𝑥 ⁡ + ⁡⁡ +𝑐 𝐴 = 9𝐴0
4
𝑥4
𝑒 𝑥 + ⁡ 𝑒 −𝑦 + ⁡ ⁡ = 𝑐⁡
4

83 84
22. Find the population of a city at any time t given that the rate of increase of population is 25. A radioactive isotope has an initial mass 200mg which two years later is 150mg. Find the
proportional to the population at that instant and that in 𝒂 period of 40 years the population expression for the isotope remaining at any time. What is its half – life? (half – life means the
increased from 3,00,000 to 4,00,000 time taken for the radioactivity of a specified isotope to fall to half its original value)
Ans: Ans:
𝐴 = 𝐶𝑒 𝐾𝑡 𝐴 = 𝐶𝑒 𝐾𝑡
𝑡 𝐴
𝑡 = 0,⁡⁡⁡⁡𝐴 = ⁡3,00,000 𝑡 = 0,⁡⁡⁡⁡𝐴 = ⁡200
𝑡 𝐴 𝐾(0)
0 200
200 = 𝐶𝑒 ⁡ ⟹ ⁡𝐶 = ⁡200
3,00,000 = 𝐶𝑒 𝐾(0) 2 150
0 3,00,000 𝐴 = 200⁡𝑒 𝐾𝑡
? 100
𝐶 = ⁡3,00,000 𝑡 = 2,⁡⁡⁡⁡⁡𝐴 = 150
40 4,00,000
𝐴 = 3,00,000⁡𝑒 𝐾𝑡 150 = 200𝑒 2𝐾
𝐾 = ⁡ −1⁄2 log(4⁄3)
𝑡 = 40,⁡⁡⁡⁡⁡𝐴 = 4,00,000
4
𝐴 = 100⁡, 𝑡 =⁡?
𝑒 40𝑘 ⁡ = 𝑡 4⁄ )
3 𝐴 = 200⁡𝑒 − ⁄2 ⁡log( 3 ⁡
1 2 log(1⁄2)
4
(𝑒 𝐾 )40 = ⁡ ⁡ ⟹ ⁡ 𝑒 𝐾 = ⁡ ( )
4 40 𝑡 =⁡ 4
3 3 𝑙𝑜𝑔( )
3

𝑘 = ⁡ 1⁄40 𝑙𝑜𝑔(4⁄3)
26. Water at temperature 𝟏𝟎𝟎𝒐 𝑪 cools in 10 minutes to 𝟖𝟎𝒐 𝑪 in a room temperature of 𝟐𝟓𝒐 𝑪.
𝑡
4 40 Find (i) The temperature of water after 20 minutes
⟹𝐴= 3,00,000⁡ ( )
3 (ii) The time when the temperature is 𝟒𝟎𝒐 𝑪
23. Suppose a person deposits Rs.10,000 in a bank account at the rate of 5% per annum compounded Ans:
continuously. How much money will be in his bank account 18 months later? 𝑇 − 𝑆 = 𝐶𝑒 𝐾𝑡
Ans: 𝑇 − 25 = 𝐶𝑒 𝐾𝑡 𝑡 𝑇
𝐴 = 𝐶𝑒 𝐾𝑡 𝑡 = 0,⁡⁡⁡⁡𝑇 = ⁡ 100𝑜 𝐶 0 1000 𝐶
5
𝐾 = 5% = ⁡ 100 = 0.05 100 − 25 = 𝐶⁡𝑒 𝐾(0) 10 800 𝐶
𝑡 𝐴
0.05𝑡 𝐶⁡ = ⁡75 20 ?
𝐴 = 𝐶𝑒 0 10000
𝑡 = 0,⁡⁡⁡⁡𝐴 = ⁡10000 1.5 ? 𝑇 − 25 = 75⁡𝐶𝑒 𝐾𝑡 ? 40
10000 = 10000⁡𝑒 0.05𝑡 𝑡 = 10,⁡⁡⁡⁡⁡𝑇 = ⁡⁡ 80𝑜 𝐶
𝑡 = 1.5,⁡⁡⁡⁡⁡𝐴 =?
55⁡ = ⁡⁡75⁡𝑒 −10𝐾
𝐴⁡ = ⁡⁡10000⁡𝑒 0.05(1.5)
𝐾 = ⁡ −1⁄10 ⁡𝑙𝑜𝑔 (11)
15
𝐴 = 10000⁡𝑒 0.075
24. Assume that the rate at which radioactive nuclei decay is proportional to the number of such
𝑡 = 20,⁡⁡⁡⁡𝑇⁡ =⁡?
nucleic that are present in each sample. In a certain sample 10% of the original number of
radioactive nuclei have undergone disintegration in a period of 100 years. What percentage of 𝑇 − 25 = ⁡75⁡𝑒 20𝐾
the original radioactive nuclei will remain after 1000 years? 𝑇 − 25 = ⁡75⁡(𝑒 −10𝐾 )2
Ans:
11 2
𝐴 = 𝐶𝑒 𝐾𝑡 𝑇 − 25 = ⁡75 (15)
𝑡 𝐴
𝑡 = 0,⁡⁡⁡⁡𝐴 = ⁡100 0 100 𝑇 = 65.330 𝐶
100 = 𝐶𝑒 𝐾(0) 100 9/100 𝑡 = 40,⁡⁡⁡40 − 25⁡ = ⁡⁡75𝑒 −𝐾𝑡
𝐶 = ⁡100 1000 ?
𝑒 −𝐾𝑡 ⁡ = ⁡⁡ 15⁄75 = ⁡ 1⁄5
𝐴 = 100⁡𝑒 𝐾𝑡
𝐾𝑡 = log 5
9
𝑡 = 100,⁡⁡⁡⁡⁡𝑒 100𝐾 = ⁡ 10 1.6095
𝑡 = ⁡ 0.03101 = 53.46 minus
910
𝑡 = 1000, 𝐴 = ⁡ 108 ⁡%

85 86
27. A pot of boiling water at 𝟏𝟎𝟎𝒐 𝑪 is removed from a stove at time t = 0 and left the cool in the 29. A tank initially contains 50 liters of pure water starting at time t = 0 a brine containing with 2
kitchen after 5 minutes the water temperature has decreased to 𝟖𝟎𝒐 𝑪 and another 5 minutes grams of dissolved salt per liters flows into the tank at the rate of 3 liters per minute. The mixture
later it has dropped to 𝟔𝟓𝒐 𝑪. Determine the temperature of the kitchen? is kept uniform by stirring and the well – stirred mixture simultaneously flows out of the tank at
Ans: the same rate. Find the amount of salt present in the tank at any time t > 0
𝑑𝑇 Ans:
𝑑𝑇
⁡ = 𝐾(𝑇 − 𝑆)
𝑑𝑥
𝑇 = 𝑆 + 𝐶𝑒 𝐾𝑡 𝑑𝑡
= 𝐼𝑁 - 𝑂𝑈𝑇
𝑑𝑥 3
𝑡 = 0,⁡⁡⁡𝑇 = ⁡⁡ 1000 𝑡 𝑇 = 6 −⁡ 𝑥
𝑑𝑡 50
𝑇 − 𝑆 = ⁡𝐶𝑒 𝐾𝑡 0 1000 𝐶 −3𝑡
𝑥 = 100 + 𝐶𝑒 100
100 − 𝑆 = ⁡𝐶𝑒 𝐾(0) 5 800 𝐶
𝑡⁡ = 0,⁡⁡⁡⁡⁡⁡⁡𝐶⁡ = ⁡ −100
100 − 𝑆 = 𝐶 10 650 𝐶
Amount of salt at time 𝑡
𝑇 − 𝑆 = (100 − 𝑆)𝑒 𝐾𝑡 −3𝑡
𝑡 = 5,⁡⁡⁡𝑇 = ⁡ 800 ⁡ 𝑥 = 100 − 100 𝑒 50
80 − 𝑆 = (100 − 𝑆)⁡𝑒 5𝐾
80−𝑆 Type: 1 Increase (or) Decrease in the Amount quantity t is the Amount of quantity A
𝑒 5𝐾 = ⁡ 𝑑𝐴 𝑑𝐴
100−𝑆
0 ⁡ ∝ 𝐴⁡ ⟹ ⁡ 𝑑𝑡 ⁡ = 𝐾𝐴⁡ ⟹ 𝐴 = 𝐶𝑒 𝐾𝑡
Temperature of the Kitchen S = 20 𝑑𝑡

28. In Murder investigation, a corpse was found by detective at exactly 8 P.M. Being alert, the Increase if K > 0
detective also measured the body temperature and found it to be 𝟕𝟎𝟎 ⁡𝑭 . Two hours later the Decrease if K < 0
detective measured the body temperature again and found it to be 𝟔𝟎𝟎 ⁡𝑭 . If the room Type: 2 Newton’s law of cooling / warming
𝑑𝑇
temperature is 𝟓𝟎𝟎 ⁡𝑭 assuming that the body temperature of the person before death was 98.6F ⁡𝛼⁡𝑇 − 𝑆 S – Room Temperature
𝑑𝑡
at what time did the murder occur? 𝑑𝑇
⟹ ⁡ 𝑑𝑡 = 𝐾(𝑇 − 𝑆)
Ans:
𝐥𝐨𝐠⁡(𝟐.𝟒𝟑)
⟹ 𝑇 = 𝑆 + ⁡𝐶𝑒 𝐾𝑡
[ 𝒍𝒐𝒈𝟐
= ⁡𝟏. 𝟐𝟖] Type: 3 Mixture Problems
𝑑𝑇 𝑑𝑥
𝑑𝑡
= 𝑘(𝑇 − 𝑆) Letting x to denote the amount of S present at time t and the derivative
𝑑𝑡
to denote the rate of
𝑡 𝑇
𝑇 − 𝑆 = ⁡⁡𝐶𝑒 𝐾𝑡 change of x w.r. to. x
0 700 𝐶 8⁡𝑃𝑀
𝑇 − 50 =⁡= ⁡𝐶𝑒 𝐾𝑡 If IN Denotes the rate at which S enters the mixture and OUT denote the rate at which it leaves,
2 600 𝐶 10⁡𝑃𝑀 𝑑𝑥
𝑡 = 0,⁡⁡⁡⁡⁡⁡⁡𝑇 = 70 then we have the equation = IN – OUT
? 98.6 𝑑𝑡
70 − 50⁡ =⁡⁡= ⁡𝐶𝑒 𝐾(0) S.No. PROBLEMS ANSWERS
𝐶 = 20
1 Number of Bacteria 9 times the original number of bacteria
𝑇 − 50⁡ = 20⁡𝑒 𝐾𝑡
𝑡 = 2,⁡⁡⁡⁡⁡𝑇 = 60 𝐴⁡– ⁡9𝐴0
60 − 50⁡ = ⁡⁡20𝑒 2𝐾 2 The Population of a city 4
𝑡⁄
40
𝐴⁡ = 3,00,000 (3)
𝑒 2𝐾 ⁡ = ⁡ 1⁄2
3 Electromotice force for an Electric circuit 𝐸 𝑅𝑡
𝐾 = ⁡ 1⁄2 𝑙𝑜𝑔(1⁄2) 𝐼 =⁡ + 𝐶𝑒 − 𝐿
𝑅
𝑇 = 98.6,⁡⁡⁡⁡⁡⁡⁡⁡𝑡 =⁡? 4 Engine of a Motor boat moving at 10m/s is shut off 𝑉 = 10𝑒 −2
1
𝑙𝑜𝑔( )𝑡⁄2
98.6 − 50⁡ = ⁡⁡20⁡𝑒 2 5 At the rate of 5% per annum compounded continuously 𝐴 = 10000⁡𝑒 0.075
𝑡⁄ log(1⁄ ) = log (48.6) 6 The rate at which ratioactive nuclei Decay 910
2 2 108
% of ratioactive nuclei will remain
20
48.6
log( ) after 1000 years
𝑡=2 20
1
𝑙𝑜𝑔( ) log 3
2 7 The growth of a population
𝑡 = 50⁡ ( )
𝑡 = 2.56⁡⁡⁡⁡(𝑜𝑟) ⁡⁡ − 2.30⁡ℎ𝑜𝑢𝑟𝑠 log 2
Time of death is 5.30 P.M. (8.00 – 2.30 PM)
87 88
 8 A radioactive isotope 1 4 5. Use the linear approximation to find approximate value of √𝟐𝟔
𝟑
𝐾 = ⁡ 𝑙𝑜𝑔 ( )
2 3 Solution:
3
9 Room Temperature of 25𝑜 𝐶 𝑡 = 53.46⁡𝑚𝑖𝑛𝑢𝑡𝑒𝑠 𝑓(𝑥) = ⁡ √𝑥 Given: 𝑥 = 26, 𝑥0 = 27
1
10 At 10.00 AM a woman took a cup of hot instant coo coffee 𝑇 = ⁡151.4⁡𝐹 𝑓 ′ (𝑥) = ⁡ 3 2 𝐿(𝑥) = ⁡𝑓(𝑥0 ) + 𝑓 ′ (𝑥0 )(𝑥 − 𝑥0 )
3( √𝑥)
3 1
√26 = 3 + 27 (−1) = 2.963
𝑜 𝑜
11 A pot of boiling water at 100 𝐶 Room Temperature S = 20 𝐶 6. Assuming 𝐥𝐨𝐠 𝟏𝟎 𝒆 = 𝟎. 𝟒𝟑𝟒𝟑 , find an approximate value of 𝐥𝐨𝐠 𝟏𝟎 𝟏𝟎𝟎𝟑
12 In a Murder investigation corpse The person was Murdered = 5.30 PM Solution:
(𝑥) = log10 𝑥⁡ Given: 𝑥 = 1003, 𝑥0 = 1000⁡
13 A tank initially contains 50 litres of pure water 𝐴 = 100 − 100⁡𝑒 −3𝑡/50 1
𝑓 ′ (𝑥) = ⁡ 𝑥 log10 𝑒 𝐿(𝑥) = ⁡𝑓(𝑥0 ) + 𝑓 ′ (𝑥0 )(𝑥 − 𝑥0 )
14 A tank contains 1000 litres of water in which 100 gram of 𝐴⁡ = 5000 − 4900⁡𝑒 −0.01𝑡 3
log10 1003 = 3 + × 0.4343 = 3.0013
salt 1000
7. The time T, taken for a complete oscillation of a simple pendulum with length 𝒍, is given by the
𝒍
equation 𝑻 = 𝟐𝝅√𝒈 , where g is a constant. Find the approximate percentage error in the
Chapter 8. Differentials and Partial derivative
calculated value of T corresponding to an error of 2 percent in the value of 𝒍.
2 and 3 Marks:
Solution:
1. Use linear approximation to find an approximate value of √𝟗. 𝟐⁡ without using a calculator. 2𝜋 1
Solutions: log 𝑇 = ⁡ log ( 𝑔) + ⁡ 2 log 𝑙
√
𝑓(𝑥) = ⁡ √𝑥 Given: x = 9.2, 𝑥0 = 9 𝑑𝑇 1 𝑑𝑙
× 100 = 2 ⁡( 𝑙 × 100)
1 𝑇
𝑓 ′ (𝑥) = ⁡ 2√𝑥 𝐿⁡(𝑥) = 𝑓(𝑥0 ) + 𝑓 ′ (𝑥0 )(𝑥 − 𝑥0 ) 1
Percentage error in T = × 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒⁡𝑒𝑟𝑟𝑜𝑟⁡𝑖𝑛⁡𝑙
0.2 2
√9.2 = 3 + 6
= 3.033 1
= 2 × 2% = 1%
𝟏
𝟑 8. Show that the percentage error in the 𝒏𝒕𝒉 root of a number in approximately times the
2. Let 𝒇(𝒙) = ⁡ √𝒙. Find the linear approximation at x = 27. Use the linear approximation to 𝒏
𝟑
approximate √𝟐𝟕. 𝟐 percentage error in the number.
Solution: Solution:
1⁄
3
𝑓(𝑥) = ⁡ √𝑥 Given: 𝑥 = 27.2, 𝑥0 = 27 𝑦=𝑥 𝑛

1 1
𝑓 ′ (𝑥) = ⁡ 2 𝐿(𝑥) = ⁡𝑓(𝑥0 ) + 𝑓 ′ (𝑥0 )(𝑥 − 𝑥0 ) log 𝑦 = 𝑛 ⁡log 𝑥
3
3( √𝑥) 𝑑𝑦 1 𝑑𝑥
3 0.2 × 100 = 𝑛 ⁡( 𝑥 × 100)
√27.2 = 3 + 27 = 3.0074 𝑦
1
Percentage error in y = 𝑛
× Percentage error in x
𝟐⁄
3. Use the linear approximation to find approximate value of (𝟏𝟐𝟑) 𝟑 9. Let 𝒈(𝒙) = 𝐱 𝟐 + 𝐬𝐢𝐧 𝒙. Calculate the differential dg.
Ans: Solution:
2⁄
𝑓(𝑥) = ⁡ 𝑥 3 Given: 𝑥 = 123⁡, 𝑥0 = 125⁡ 𝑔(𝑥) = x 2 + sin 𝑥
′ (𝑥) 2
𝑓 =⁡ 3 𝐿(𝑥) = ⁡𝑓(𝑥0 ) + 𝑓 ′ (𝑥0 )(𝑥 − 𝑥0 ) 𝑑𝑔 = ⁡ 𝑔′ (𝑥)𝑑𝑥
3 √𝑥
2⁄ 2
(123) 3 = 25 + 15 (−2) = 24.73 𝑑𝑔 = ⁡ (2𝑥 + 𝑐𝑜𝑠𝑥)𝑑𝑥
10. Find df for⁡𝒇(𝒙) = 𝐱 𝟐 + 𝟑𝒙 and evaluate it for (i) 𝒙 = 𝟐, 𝒂𝒏𝒅⁡⁡𝒅𝒙 = 𝟎. 𝟏
𝟒
4. Use the linear approximation to find approximate value of √𝟏𝟓 (ii) 𝒙 = 𝟑⁡𝒂𝒏𝒅⁡⁡𝒅𝒙 = 𝟎. 𝟎𝟐
Solution: Solution:
4
𝑓(𝑥) = ⁡ √𝑥 Given: 𝑥 = 15, ⁡⁡⁡𝑥0 = 16⁡ 𝑓(𝑥) = x 2 + 3𝑥 (i) 𝑥 = 2, 𝑑𝑥 = 0.1
1
𝑓 ′ (𝑥) = ⁡ 4 3 𝐿(𝑥) = ⁡𝑓(𝑥0 ) + 𝑓 ′ (𝑥0 )(𝑥 − 𝑥0 ) 𝑑𝑓 = 𝑓 ′ (𝑥)𝑑𝑥 𝑑𝑓 = 7 × 0.1 = 0.7
4( √𝑥)
= (2𝑥 + 3)𝑑𝑥 (ii) 𝑥 = 3, 𝑑𝑥 = 0.02
4 1
√15 = 2 + 32 (−1) = 1.9688 𝑑𝑓 = 9 × 0.02 = 0.18

89 90
11. If the radius of a sphere, with radius 10cm, has to decrease by 0.1cm, approximately how much 𝒙𝟐 +𝒚𝟐 𝝏𝒗 𝝏𝒗
16. If 𝒗(𝒙, 𝒚) = 𝐥𝐨𝐠⁡( 𝒙+𝒚
) , prove that 𝒙 𝝏𝒙 + 𝒚 𝝏𝒚 = 𝟏
will its volume decrease?
Solution: Solution:
4 3 𝑥 2 +𝑦 2
𝑣 = ⁡ 𝜋𝑟 𝑓(𝑥, 𝑦) = = 𝑒𝑣
3 𝑥+𝑦
Given: 𝑑𝑟 = −0.1, 𝑟 = 10
f is homogeneous with degree n = 1
𝑑𝑣 = 4𝜋𝑟 2 𝑑𝑟 = 4𝜋(100)(−0.1) = −40𝜋𝑐𝑚2
𝝏𝑼 𝝏𝑼 𝝏𝑼 By Euler’s theorem,
12. If ∪ (𝒙, 𝒚, 𝒛) = ⁡ 𝐥𝐨𝐠(𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 ), find 𝝏𝒙
+ 𝝏𝒚 + 𝝏𝒛
𝜕𝑓 𝜕𝑓
Solution: 𝑥 +𝑦 = 𝑛𝑓
𝜕𝑥 𝜕𝑦
𝜕𝑈 𝜕𝑈 𝜕𝑈 3𝑥 2 3𝑦 2 3𝑧 2
+ + = + + 𝑥
𝜕(𝑒 𝑣 )
+𝑦
𝜕(𝑒 𝑣 )
= 1. 𝑒 𝑣
𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑥 3 +𝑦 3 +𝑧 3 𝑥 3 +𝑦 3 +𝑧 3 𝑥 3 +𝑦 3 +𝑧 3
𝜕𝑥 𝜕𝑦
3(𝑥 2 +𝑦 2 +𝑧 2 )
= 𝜕𝑣 𝜕𝑣
𝑥 3 +𝑦 3 +𝑧 3 𝑥𝑒 𝑣 + 𝑦𝑒 𝑣 = 𝑒𝑣
𝜕𝑥 𝜕𝑦
𝜕𝑣 𝜕𝑣
5 Marks: 𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 = 1
𝒙
13. If 𝒇(𝒙, 𝒚) = 𝒕𝒂𝒏−𝟏 ( ) , find 𝒇𝒙 , 𝒇𝒚 and show that 𝒇𝒙𝒚 = ⁡ 𝒇𝒚𝒙
𝒚 17. Prove that 𝒇(𝒙, 𝒚) = 𝒙𝟑 − 𝟐𝒙𝟐 𝒚 + 𝟑𝒙𝒚𝟐 + 𝒚𝟑 is homogeneous what is the degree? verify Euler’s
Solution:
𝑦 −𝑥 theorem for f
𝑓𝑥 = ⁡ 𝑓𝑦 =
𝑥 2 +𝑦 2 𝑥 2 +𝑦 2
Solution:
𝑥 2 −𝑦2 𝑥 2 −𝑦 2
𝑓𝑥𝑦 = (𝑥 2 -(1) 𝑓𝑦𝑥 = (𝑥 2 2)2 -(2);
+𝑦 2 )2 +𝑦 𝑓(𝜆𝑥, 𝜆𝑦) = 𝜆3 𝑓(𝑥, 𝑦)
/ 𝑓𝑥𝑦 = ⁡ 𝑓𝑦𝑥 f is homogeneous with degree n = 3
𝒙+𝒚 𝝏𝒖 𝝏𝒖 𝟏
14. If 𝒖 = ⁡ 𝐬𝐢𝐧−𝟏 [ ] , show that 𝒙 +𝒚 = 𝐭𝐚𝐧 𝒖 𝜕𝑓 𝜕𝑓
√𝒙+√𝒚 𝝏𝒙 𝝏𝒚 𝟐 By Euler’s theorem, 𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 = 3𝑓
Solution:
𝜕𝑓
𝑓(𝑥, 𝑦) =
𝑥+𝑦
= sin 𝑢 𝜕𝑥
= 3𝑥 2 − 4𝑥𝑦 + 3𝑦 2
√𝑥+√𝑦
1 𝜕𝑓
f is homogeneous with degree 𝑛 = 2 𝜕𝑦
= −2𝑥 2 + 6𝑥𝑦 + 3𝑦 2

By Euler’s theorem, 𝜕𝑓
𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 = 3𝑓
𝜕𝑓

𝜕𝑓 𝜕𝑓
𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 = 𝑛𝑓
Hence verified.
𝜕(sin 𝑢) 𝜕(sin 𝑢) 1 𝟓𝒙𝟑 𝒚𝟒 +𝟕𝒚𝟐 𝒙𝒛𝟒 −𝟕𝟓𝒚𝟑 𝒛𝟒 𝝏𝒘 𝝏𝒘 𝝏𝒘
𝑥 𝜕𝑥
+𝑦 𝜕𝑦
= 2 sin 𝑢 18. If 𝒘(𝒙, 𝒚, 𝒛) = 𝐥𝐨𝐠⁡( ), find 𝒙 𝝏𝒙 + 𝒚 𝝏𝒚 + 𝒛 𝝏𝒛
𝒙𝟐 +𝒚𝟐
𝜕𝑢 𝜕𝑢 1
𝑥𝑐𝑜𝑠⁡𝑢 𝜕𝑥 + 𝑦𝑐𝑜𝑠⁡𝑢 𝜕𝑦 = 2 sin 𝑢 Solution:
5𝑥 3 𝑦 4 +7𝑦2 𝑥𝑧 4 −75𝑦 3 𝑧 4
𝜕𝑢 𝜕𝑢
𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 = 2 tan 𝑢
1 𝑓= = ⁡ 𝑒𝑤
𝑥 2 +𝑦 2

𝒙𝟐 +𝒚𝟐 𝝏𝒖 𝝏𝒖 𝟑 f is homogenous with degree n=5

15. If 𝒖(𝒙, 𝒚) = , prove that 𝒙 𝝏𝒙 + 𝒚 𝝏𝒚 = 𝟐 𝒖
√𝒙+𝒚 By Euler’s theorem,
Solution: 𝜕𝑓 𝜕𝑓 𝜕𝑓
𝑥 2 +𝑦 2 𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 + 𝑧 𝜕𝑧 = 𝑛𝑓
𝑢(𝑥, 𝑦) =
√𝑥+𝑦
3 𝜕(𝑒 𝑤 ) 𝜕(𝑒 𝑤 ) 𝜕(𝑒 𝑤 )
u is homogeneous with degree 𝑛 = 2 𝑥 +𝑦 +𝑧 = 5𝑒 𝑤
𝜕𝑥 𝜕𝑦 𝜕𝑧
By Euler’s theorem, 𝜕𝑤 𝜕𝑤 𝜕𝑤
𝜕𝑓 𝜕𝑓
𝑥𝑒 𝑤 ⁡ 𝜕𝑥 + 𝑦𝑒 𝑤 + 𝑧𝑒 𝑤 = 5𝑒 𝑤
𝜕𝑦 𝜕𝑧
𝑥 𝜕𝑥 + 𝑦 = 𝑛𝑓
𝜕𝑦
𝜕𝑤 𝜕𝑤 𝜕𝑤
𝜕𝑢 𝜕𝑢 3 𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 + 𝑧 𝜕𝑧 = 5
𝑥 𝜕𝑥 + 𝑦 𝜕𝑦 = 2 𝑢

91 92
19.
𝒚
Prove that 𝒈(𝒙, 𝒚) = 𝒙 𝐥𝐨𝐠 (𝒙) is homogeneous, what is the degree? Verify Euler’s theorem Chapter – 7
for 𝒈 Applications of Differential calculus
Solution: 5 Marks:
𝑔(𝜆𝑥⁡,⁡⁡⁡𝜆𝑦) ⁡ = ⁡ 𝜆1 𝑔(𝑥, 𝑦) 1. Example: 7.5
A particle is fired straight up from the ground to reach a height of s feet in t seconds,
𝑔 is homogeneous with degree n = 1
where s(t) = 𝟏𝟐𝟖𝒕 − 𝟏𝟔𝒕𝟐
By Euler’s theorem, (i) Compute the maximum height of the particle reached
𝜕𝑔 𝜕𝑔 (ii) What is the velocity when the particle hits the ground
𝑥 𝜕𝑥 + 𝑦 ⁡ = ⁡⁡𝑔
𝜕𝑦
Solution:
𝜕𝑔 𝑦 i. At maximum height 𝑣(𝑡) = 0
= log (𝑥 ) − 1
𝜕𝑥
𝑑𝑠
𝜕𝑔 𝑥 𝑣(𝑡) = 𝑑𝑡 = 128 − 32𝑡
=⁡
𝜕𝑦 𝑦
𝜕𝑔 𝜕𝑔
⁡⁡⁡⁡⁡⁡𝑣(𝑡) = 0
𝑥 +𝑦 ⁡ = ⁡⁡𝑔
𝜕𝑥 𝜕𝑦 128 − 32𝑡 = 0 ⟹⁡⁡⁡t = 4
Hence verified. The height t = 4 is s(t) = 128(4) – 16(4)2 = 256 ft.
ii. When the particle hits the ground then S = 0
3 Marks: S=0
20. Assume that the cross section of the artery of human is circular. A drug is given to a patient to 128𝑡 − 16𝑡 2 = 0
dilate his arteries. If the radius of an artery is increased from 2mm to 2.1mm, how much is cross
t = 0, t = 8 seconds
– sectional area increased approximately?
Solution: The particle hits the ground at t = 8 seconds
𝑟 = 2, 𝑑𝑟 = 2.1 − 2 = 0.1 The velocity when it hit the ground is 𝑣(8) = 128 − 32(8)
𝐴 = 𝜋𝑟 2
= ⁡ −128⁡𝑓𝑡/𝑠
𝑑𝐴 = 2𝜋𝑟⁡𝑑𝑟
= 2𝜋(2)(0.1) 2. Exercise 7.1 -2
= 0.4𝜋⁡⁡⁡𝑚𝑚2 A camera is accidentally knocked off an edge of a cliff 400ft high. The camera falls a distance of
21. An egg of a particular bird is very nearly spherical. If the radius to the inside of the shell is S = 𝟏𝟔𝒕𝟐 in t seconds
5mm and radius to the outside of the shell is 5.3 m.m, find the volume of the shell approximately i. How long does the camera fall before it hits the ground?
Solution: ii. What is the average velocity high which the camera falls during the rest 2 seconds?
𝑟 = 5, 𝑑𝑟 = 5.3 − 5 = 0.3 iii. What is the instantaneous velocity of the camera when it hits the ground?
4 Solution:
𝑉 = 3 𝜋𝑟 3
i. To hit the ground the camera has to travel 400ft
𝑑𝑉 = 4𝜋𝑟 2 𝑑𝑟 = 4𝜋(5)2 (0.3) = 30𝜋 𝑚𝑚3 S = 400
22. A circular plate expands uniformly under the influence of heat. If it’s radius increases from 16t2⁡= ⁡400
10.5cm to 10.75cm, then find an approximate change in the area and the approximate percentage 400
t2⁡= 16
= 25
change in the area.
Solution: 𝑡 = 5⁡𝑠𝑒𝑐𝑜𝑛𝑑
𝑟 = 10.5, 𝑑𝑟 = 10.75 − 10.5 = 0.25 ii. The average velocity in last 2 seconds
s(5)−s(3)
𝐴 = 𝜋𝑟 2 =
5−3
𝑑𝐴 = 2𝜋𝑟⁡𝑑𝑟 = 128⁡𝑓𝑡/𝑠𝑒𝑐
= 2𝜋(10.5)(0.25) = 5.25𝜋 𝑐𝑚2 iii. The instantaneous velocity when it hits the ground
𝑑𝐴 𝑑𝑠
Approximate percentage change = ⁡ ⁡ × 100% V = 𝑑𝑡 = 32𝑡
𝐴
= 4.76% when t = 5, ; 𝑣 = 32⁡ × ⁡5
𝑣⁡ = 160𝑓𝑡/𝑠𝑒𝑐

93 94
3. Exercise: 7.1 -3 5. Exercise: 7.1-10
A particle moves along a line according to the law 𝒔(𝒕) = 𝟐𝒕𝟑 − 𝟗𝒕𝟐 + 𝟏𝟐𝒕 − 𝟒 where 𝒕 ≥ ⁡𝟎 A police jeep approaching an orthogonal intersection from the northern direction is
i. At what times the particle changes direction? chasing a speeding car that has turned and moving straight east. When the jeep is 0.6km north
ii. Find the total distance travelled by the particle in the first 4 seconds of the intersection and the car is 0.8km to the east. The police determine with a radar that the
iii. Find the particle acceleration each time the velocity is zero distance between them and the car is increasing at 20 km/hr. It the jeep is moving at 60 km/hr
Solution: at the instant of measurement, what is the speed fo the car?
𝑣(𝑡) = 6𝑡 2 − 18𝑡 + 12 Solution:
𝑑𝑦 𝑑𝑧
𝑣(𝑡) = 6(𝑡 − 1)(𝑡 − 2) Given = −60⁡𝑘𝑚/ℎ𝑟,⁡ = 20⁡𝑘𝑚/ℎ𝑟
𝑑𝑡 𝑑𝑡
𝑎(𝑡) = 12𝑡 − 18 𝑧2 = 𝑥2 + 𝑦2
i. 𝑣(𝑡) = 0 =⁡ (0.8)2 + (0.6)2
6(𝑡 − 1)(𝑡 − 2) ⁡ = ⁡0 𝑧2 = 1
𝑡 = 1, 2 𝑧=1
The particle change its direction at 𝑡 = 1 sec and 𝑡 = 2 sec
𝑥2 + 𝑦2 = ⁡ 𝑧2
ii. The total distance travelled by the particle 𝑑𝑥 𝑑𝑦 𝑑𝑧
2𝑥 𝑑𝑡 + 2𝑦 𝑑𝑡 ⁡ = ⁡⁡2𝑧⁡ 𝑑𝑡
= ⁡ |𝑠(0) − 𝑠(1)| + |𝑠(1) − 𝑠(2)| + |𝑠(2) − 𝑠(4)|
𝑑𝑥
= |−4 − 1| + |1 − 0| + |0 − 28| (0.8) + (0.6)(−60) = (1)(20)
𝑑𝑡
= 5 + 1 + 28 𝑑𝑥
= 70⁡𝑘𝑚/ℎ𝑟
= 34 meters 𝑑𝑡
6. Exercise: 7.1-9. A ladder 17 metre long is learning against the wall. The base of the ladder
iii. The acceleration when v = 0 is 𝑎(1) = −6⁡ 𝑚⁄𝑠 2
is pulled away from the wall at a rate of 5m/s when the base of the ladder is 8 metres from the
𝑎(2) = 6⁡ 𝑚⁄𝑠 2
well
Do yourself:
i. How fast is the top fo the ladder moving down the well?
Example 7.6
ii. At what rate the area of the triangle formed by the ladder, wall and the floor is changing?
A particle moves along a horizontal line such that its position at any time 𝑡 > 0 is given by
Solution:
𝑠(𝑡) = ⁡ 𝑡 3 − 6𝑡 2 + 9𝑡 + 1 where 𝑠 is measured in metres and⁡𝑡 is seconds?
(𝑖) 𝑥 2 + 𝑦 2 = 172
i. At what time the particle is at rest? 𝑑𝑥 𝑑𝑦
ii. At what time the particle changes its direction? 2𝑥 𝑑𝑡 + 2𝑦 𝑑𝑡 = 0 𝑥2 + 𝑦2 = 𝑧2
𝑑𝑦 −𝑥 𝑑𝑥
iii. Find the total distance travelled by the particle in the first 2 seconds/
𝑑𝑡
= 𝑦 𝑑𝑡
82 + 𝑦 2 = 172
4. Example: 7.9 𝑑𝑦 −5𝑥
Salt is poured from a conveyer belt at a rate of 30 cubic metre per minute forming a conical pile = 𝑦 = 15
𝑑𝑡 𝑦

= −8⁄3 ⁡𝑚/𝑠
with a circular base whose height and diameter of base are always equal. How fast is the height 𝑑𝑦 −5(8)
=
𝑑𝑡 15
of the pile increasing when the pile is 10 metre high? 1
Solution: 𝑖𝑖). 𝐴 = 2 ⁡𝑥𝑦
𝑑𝐴 1 𝑑𝑦 𝑑𝑥
2𝑟 = ℎ = 2 (𝑥 𝑑𝑡 + 𝑦 𝑑𝑡 )⁡
𝑑𝑡
1 −8
𝑟 = ⁡ ℎ⁄2 = ⁡ 2 [8⁡( 3 ) + 15(5)]
1
1
V=3 𝜋𝑟 ℎ 2 = ⁡ 6 [−64 + 225]
161
1 =
𝑉 = ⁡ 12 ⁡𝜋ℎ3 6
= 26.83⁡ 𝑚2 ⁄𝑠𝑒𝑐
dv 1 dh
= 4 ⁡πh2 ⁡ dt Do yourself:
dt
Example: 7.7 -If we blow air into a balloon of spherical shape at a rate of 1000 cm3 per second, at what rate
𝑑ℎ 𝑑𝑣 1
= 4 𝑑𝑡 .⁡⁡ 𝜋ℎ2 the radius of the balloon changes when the radius is 7cm. Also compute the rate at which the
𝑑𝑡
𝑑ℎ 1 surface area changes.
𝑑𝑡
= 4⁡ × 30⁡ × ⁡ 100𝜋
Example: 7.8: The price of a product is related to the number of units available (supply) by the equation px +
𝑑ℎ 6
= 5𝜋 ⁡𝑚/𝑚𝑖𝑛 3p – 16x = 234, where p is the price of the product per unit in rupees (Rs). and x is the number
𝑑𝑡
of units. Find the rate at which the price is changing with respect to time when 90 units are
available and the supply is increasing at the rate of 15 units / week.
95 96
Example: 7.10: A road running north to south crosses a road going east to west at the point P. Car A 9. Example 7.18:
is driving north along the first road, and car B is driving east along the second road. At Prove that ellipse 𝒙𝟐 + 𝟒𝒚𝟐 = 𝟖 and hyperbola 𝒙𝟐 − 𝟐𝒚𝟐 = 𝟒 intersect orthogonally
a particular time car A is 10 kilometres to the north of P and traveling of 80km/hr while Solution: Let point of intersection be (a, b)
car B is 15 kilometres to the east of P and braseling at 100km/hr. How fast is the 𝑎2 + 4𝑏 2 = 8
distance between the two cars changing?
𝑎2 − 2𝑏 2 = 4
Exercise: 7.1 – 7: A beacon makes one revolution every 10 seconds. It is located on a ship which is
anchored 5km from a straight shore line. How fast is the beam moving along the shore (-) (+) (-)
line when it makes an angle of 45o with the shore? 𝟔𝒃𝟐 ⁡⁡ = 𝟒
Exercise: 7.1-8: A conical water tank with vertex down of 12 metres height has a radius of 5 metres at 𝒃𝟐 = 𝟒/𝟔
the top. If water flows into the tank at a rate 10 cubic m/min, how fast is the depth of 𝒃𝟐 = 𝟐/𝟑
the water increases when the water is 8 metres deep? 𝒂𝟐 = 𝟏𝟔/𝟑
7. Example: 7.15 𝒙𝟐 + 𝟒𝒚𝟐 = 𝟖
Find the angle between the curves 𝒚 = 𝒙𝟐 and 𝒙 = 𝒚𝟐 of their point of intersection (0, 0) 𝑑𝑦
2𝑥 + 8𝑦⁡ ⁡= 0
and (1, 1) 𝑑𝑥
𝑑𝑦
Solution: 𝑚1 = ⁡⁡ 𝑑𝑥 = ⁡ −𝑎/4𝑏
(𝑎,𝑏)
The tangent at (0, 0) are x axis and y axis. Angle between x axis 2 2
and y axis is 90o. 𝑥 − 2𝑦 = 4
𝑑𝑦
𝜃⁡ = 900 2𝑥 − 4𝑦 =0
2 𝑑𝑥
𝑦=𝑥 𝑦2 = 𝑥
𝑑𝑦
𝑑𝑦
= 2𝑥
𝑑𝑦
= 2𝑦
1 𝑚2 = ⁡⁡ 𝑑𝑥 = ⁡𝑎/2𝑏
𝑑𝑥 𝑑𝑥 (𝑎,𝑏)
𝑑𝑦 𝑑𝑦 1
𝑚1 = 𝑑𝑥⁡(1,1)⁡ = 2 𝑚2 = 𝑑𝑥⁡(1,1)⁡ = 2 𝑚1 ⁡𝑚2 = ⁡ (− 4𝑏) (2𝑏)
𝑎 𝑎

1
2−2 𝑎2
𝑡𝑎𝑛⁡𝜃 = ⁡ | | = ⁡ − 8𝑏2
1
1 + 2(2) 16
−
𝑡𝑎𝑛⁡𝜃 = 3/4 =⁡ 3
2
8( )
𝜃 = 𝑡𝑎𝑛−1 (3/4)⁡ 3

8. Example: 7.17 𝑚1 ⁡𝑚2 = ⁡ −1

If the curves 𝒂𝒙𝟐 + 𝒃𝒚𝟐 = 𝟏 and 𝒄𝒙𝟐 + 𝒅𝒚𝟐 = 𝟏 intersect each other orthogonally then
𝟏 𝟏 𝟏 𝟏
The curves cut orthogonally
show that 𝒂⁡- 𝒃
= 𝒄−𝒅 10. Exercise: 7.2 – 9
Solution: Find the angle between the curves 𝒙𝒚 = 𝟐 and 𝒙𝟐 + 𝟒𝒚 = 𝟎
Let the two curves intersect at (𝑥1 , 𝑦1 ) Solution:
(𝑎 − 𝑐)𝑥12 + (𝑏 − 𝑑)𝑦12 =⁡0 …………………….. (1) The point of intersection of two curve is (-2, -1)
𝑎𝑥 2 + 𝑏𝑦 2 = 1 𝑥 2 + 𝑑𝑦 2 = 1 𝑥𝑦 = 2
𝑑𝑦 𝑎𝑥 𝑑𝑦 𝑐𝑥
= − 𝑏𝑦 = ⁡ −⁡ 𝑑𝑦 𝑥𝑦 ′ + 𝑦 = 0
𝑑𝑥 𝑑𝑥 −𝑦
𝑑𝑦 𝑎𝑥𝑖 𝑑𝑦 𝑐𝑥 𝑦′ = ⁡ 𝑥
𝑚1 = ⁡ 𝑑𝑥 ⁡(𝑥1 , 𝑦1 ) = ⁡ − 𝑏𝑦 𝑚2 = ⁡ 𝑑𝑥 ⁡(𝑥1 , 𝑦1 ) = ⁡ −⁡ 𝑑𝑦1
𝑖 1 𝑚1 = 𝑦 ′ (−2, −1) = ⁡ −1/2⁡
Two curve cut orthogonally 𝑥 2 + 4𝑦 = 0
𝑚1 𝑚2 = −1 2𝑥 + 4𝑦 ′ = 0
−𝑎𝑥 −𝑐𝑥
( 𝑏𝑦 1 ) ⁡⁡ × ( 𝑑𝑦 1 ) = −1 𝑦′ = ⁡ −
2𝑥
1 1 4
𝑎𝑐𝑥12 + 𝑏𝑑𝑦12 = 0⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡ …………………………… (2)
𝑚2 = 𝑦 ′ (−2, −1) = 1
(1) 𝑎−𝑐 𝑏−𝑑
⇒⁡ = −1
(2) 𝑎𝑐 𝑏𝑑 𝑡𝑎𝑛𝜃 = | 2
−1 −3/2
| ⁡ = ⁡⁡ | 1/2 |
−1
1 1 1 1 1+( )(1)
⇒⁡𝑐 − 𝑎 = 𝑑 − 𝑏 2

1
⇒⁡𝑎 − 𝑏 = 𝑐 − 𝑑
1 1 1 𝜽 = 𝒕𝒂𝒏−𝟏 (𝟑)

97 98
11. Exercise 7.2-10 13. Exercise : 7.5 -10
Show that the two curves 𝒙𝟐 − 𝒚𝟐 = 𝒓𝟐 and 𝒙𝒚 = 𝒄𝟐 where c, r are constant, cut 𝒍𝒊𝒎
orthogonally. Evaluate : 𝒙 → 𝝅⁡(𝒔𝒊𝒏𝒙)𝒕𝒂𝒏𝒙
𝟐
Solution:
Let the point of intersection be (𝑥1 , 𝑦1 ) Solution:
Let y = (𝑠𝑖𝑛𝑥)𝑡𝑎𝑛𝑥
𝑥2 − 𝑦2 = 𝑟2
2𝑥 − 2𝑦𝑦 ′ = ⁡0 𝑙𝑜𝑔⁡𝑦 = 𝑡𝑎𝑛⁡𝑥⁡𝑙𝑜𝑔⁡𝑠𝑖𝑛⁡𝑥
𝑙𝑜𝑔𝑠𝑖𝑛𝑥
𝑦′ = ⁡ 𝑦
𝑥 log 𝑦 =⁡ 𝑐𝑜𝑡𝑥

′ 𝑥1 𝑙𝑖𝑚 𝑙𝑖𝑚 𝑙𝑜𝑔𝑠𝑖𝑛𝑥 0

𝑚1 = 𝑦 =⁡
𝑥 → 2 log 𝑦 = 𝑥 → 2 ⁡ 𝑐𝑜𝑡𝑥 ⁡⁡(⁡0 ⁡𝑓𝑜𝑟𝑚)
𝜋 𝜋
(𝑥1 ,𝑦1 ) 𝑦1
2
𝑥𝑦 = 𝑐 𝑙𝑖𝑚 1
.𝑐𝑜𝑠𝑥
= 𝑥 → 𝜋 ⁡⁡ 𝑠𝑖𝑛𝑥 2
𝑥𝑦 ′ + 𝑦 = 0⁡ 2
−𝑐𝑜𝑠𝑒𝑐 𝑥

𝑦
𝑦′ = ⁡ − 𝑙𝑖𝑚
𝑥 = ⁡ 𝑥 → 𝜋⁡– 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥
𝑦1 2
𝑚2 = 𝑦 ′ (𝑥 =⁡−
1 ,𝑦1 ) 𝑥1 𝑙𝑖𝑚
𝑥 → log 𝑦 = 0
𝜋
𝑥 𝑦
𝑚1 ⁡𝑚2 = ⁡ ( 1 )⁡(− 1 ) 2
𝑦 1 𝑥1
𝑙𝑖𝑚
𝑚1 ⁡𝑚2 = ⁡ −1 𝑙𝑜𝑔𝑒 ⁡𝑥 → 𝜋⁡⁡⁡𝑦 = 0
2
Two curves cut orthogonally.
𝑙𝑖𝑚 0
𝑥 → 2 ⁡𝑦 = 𝑒
𝜋

12. Exercise: 7.4.1(V)

𝑙𝑖𝑚 𝑡𝑎𝑛𝑥
𝑥 → 2 ⁡(𝑠𝑖𝑛𝑥) =1
𝜋
Write the Maclaurin series expansion of 𝒕𝒂𝒏−𝟏 𝒙: −𝟏 ≤ 𝒙 ≤ 𝟏

Solution: Do yourself:
𝑙𝑖𝑚 1
Example: 7.43: Using the 𝐿′ ℎ𝑜̂⁡𝑝𝑖𝑡𝑎𝑙 rule, prove that (1 + 𝑥) ⁄𝑥 = 𝑒
𝑓(𝑥) = 𝑡𝑎𝑛−1 𝑥⁡⁡⁡⁡⁡⁡⁡ 𝑓(0) = 0 𝑥 → 0+
1
𝑙𝑖𝑚
1 Example: 7.44: Evaluate : (1 + 2𝑥)2𝑙𝑜𝑔𝑥
𝑓 ′ (𝑥) = 1+𝑥 2 = (1 + 𝑥 2 )−1 = 1 − 𝑥 2 + 𝑥 4 − 𝑥 6 + ⋯⁡⁡ 𝑓 ′ (0) = 1 𝑥→∞
𝑙𝑖𝑚 1
Example 7.45: Evaluate: ⁡⁡𝑥 1−𝑥
𝑓 ′′ (𝑥) = −2𝑥 +4𝑥 3 − 6𝑥 5 +…. 𝑓 ′′ (0) = 0 𝑥→1
𝑙𝑖𝑚
𝑓 ′′′ (𝑥) = −2 + 12𝑥 2 − 30𝑥 4 + ⋯ 𝑓 ′′′ (0) = −2 Exercise: 7.5-8 Evaluate: ⁡𝑥 𝑥
𝑥 → 0+
𝑙𝑖𝑚 1
𝑓 4 (𝑥) = 24𝑥 − 120𝑥 3 + ⁡ … 𝑓 4 (0) = 0 Exercise: 7.5-9: Evaluate: (1 + 𝑥)𝑥
𝑥→∞
1
𝑓 5 (𝑥) = ⁡24⁡ − ⁡360𝑥 2 + ⋯ .. 𝑓 5 (0) = 24 𝑙𝑖𝑚
Exercise: 7.5-11 Evaluate: (𝑐𝑜𝑠𝑥)𝑥2
𝑥 → 0+
𝑥 𝑥 𝑥 2
𝑥 𝑥 3 4 5 Exercise: 7.5 – 12
𝑡𝑎𝑛−1 𝑥 = 0 − ⁡⁡ ⁡(1) + ⁡ (0) + ⁡ (−2) + ⁡ ⁡(0) + ⁡ (24) + ⋯ ….
1! 2! 3! 4! 5! If an initial amount 𝐴𝑜 of money is invested at an interest rate r compounded n times a year,
𝑥3 𝑥5 𝑟 𝑛𝑡
𝑡𝑎𝑛−1 𝑥 = 𝑥 − ⁡ + −⁡… … … … the value of the investment after t years in 𝐴 = ⁡ 𝐴𝑜 ⁡(1 + ⁡ 𝑛) . If the interest is compounded
3 5

continuously, (that is as x  ∞), show that the amount after t years is 𝐴 = ⁡ 𝐴𝑜 ⁡𝑒 𝑟𝑡

99 100
14. Example: 7.60 16. Exercise: 7.8-(8)
Find the local extrema of the function 𝒇(𝒙) = 𝟒𝒙𝟔 − 𝟔𝒙𝟒 Prove that among all the rectangles of the given perimeter, the square has the maximum
Solution: area.
𝑓 ′ (𝑥) = 24𝑥 5 − 24𝑥 3 Solution:
= 24𝑥 3 (𝑥 2 − 1) Let x and y be the dimensions of rectangle.
𝑓 ′ (𝑥) = 24𝑥 3 (𝑥 + 1)(𝑥 − 1) 𝑃 = 2𝑥 + 2𝑦
𝑓 ′ (𝑥) = 0 𝑃−2𝑥
𝑥 = −1, 0, 1 𝑦 =⁡ 2
𝑓 ′′ (𝑥) = 120𝑥 4 − 72𝑥 2 Interval Sign of 𝑓′(𝑥) Monotonicity
𝐴 = 𝑥𝑦
𝑓 ′′ (𝑥) = 24𝑥 2 (5𝑥 2 − 3) 𝑥⁡ ∈⁡(−∞, −1) - Strictly decreasing
𝑃−2𝑥
𝑓 ′′ (−1) = 48 > 0, 𝑥⁡ ∈⁡(−1, 0) + Strictly increasing 𝐴(𝑥) = 𝑥 ( )
2
𝑓 ′′ (1) = 48 > 0 , 𝑥⁡ ∈⁡(0, 1) - Strictly decreasing
𝐴(𝑥) = 𝑥 − 𝑥
𝑝 2
2
⁡𝑓 ′′ (0) = 0 𝑥⁡ ∈⁡(1, ∞) + Strictly increasing
𝑝
𝐴′ (𝑥) = 2 − 2𝑥
𝑓(𝑥) has local maximum of 𝑥 = ⁡ −1 and 𝑥 = 1
𝐴′′ (𝑥) = −2
Local minimum value is −2
𝑓(𝑥) has local maximum at 𝑥 = 0 𝐴′ (𝑥) = 0
Local maximum value is 0 𝑝
𝑥 = ⁄4
Do yourself:
𝑝
Exercise: 7.7-3 When 𝑥 = ⁄4 𝐴′′ (𝑥) < 0
Find the function 𝑓(𝑥) = 4𝑥 3 + 3𝑥 2 − 6𝑥 + 1 find the intervals of monotonicity, local 𝑝
When 𝑥 = ⁄4 The area is maximum
extrema intervals of conctrity and point of inflection.
𝑝 𝑝
15. Exercise: 7.8 -5: When 𝑥 = ⁄4, 𝑦 = ⁄4
A rectangular page is to contain 24cm2 of print. The margins of the top and bottom of
Thus it is a square
the page are 1.5cm and the margins at other sides of the page is 1cm. What should be the
dimensions of the page so that the area of the paper used is minimum? 17. Exercise 7.8-9.
Solution: Find the dimensions of the largest rectangle that can be inscribed in a semicircle of
Let x and y be the dimensions of the printed portion radius⁡𝒓⁡𝒄𝒎
The poster dimensions are (𝑥 + 2)(𝑦 + 3) Solution:
𝑃𝑄 = 2𝑟 cos 𝜃
𝐴 = (𝑥 + 2)(𝑦 + 3)
24 𝑄𝑅 = 𝑟 sin 𝜃
𝐴(𝑥) = (𝑥 + 2) ( + 3)
𝑥 Area of the rectangle 𝐴 = 2𝑟𝑐𝑜𝑠𝜃. 𝑟𝑠𝑖𝑛𝜃
48
𝐴(𝑥) = 24 + + 3𝑥 + 6 (𝜃) = 𝑟 2 𝑠𝑖𝑛2𝜃
𝑥
−48
𝐴′ (𝑥)
= +3 𝐴′ (𝜃) = 2𝑟 2 𝑐𝑜𝑠2𝜃
𝑥2
′′ (𝑥) 96 𝐴′′ (𝜃) = −4𝑟 2 𝑠𝑖𝑛2𝜃
𝐴 = 𝑥3
𝜋
𝐴′ (𝜃) = 0 => 𝑐𝑜𝑠2𝜃 = 0 => 𝜃 = 4
𝐴′ (𝑥) = 0
𝜋
𝑥 = ±4 When 𝜃 = , 𝐴′′ (𝜃) < 0
4
𝜋
When 𝑥 = 4 𝐴′′ (𝑥) > 0 When 𝜃 = 4
, The area is maximum
When 𝑥=4 A is minimum The dimensions are 2𝑟 cos 4 ⁡, 𝑟 sin
𝜋 𝜋
4
When 𝑥 = 4 𝑦=6 𝑟
The dimensions are √2𝑟, 𝑐𝑚
The dimension of the poster are 6cm, 9cm. √2

101 102
18. Example: 7.65 Do yourself:
Prove that among all the rectangles of the given area square has the least perimeter. Example: 7.13
Solution:
Find the equation of the tangent and normal at any point to the Lissejons curve given by
Area of the rectangle is 𝑥𝑦⁡ = ⁡𝑘
𝑥 = 2⁡𝑐𝑜𝑠⁡3𝑡 and 𝑦 = 3⁡𝑠𝑖𝑛⁡2𝑡 , 𝑡⁡ ∈ 𝑅
Perimeter of the rectangle 𝑝(𝑥) = 2(𝑥 + 𝑦)
20. Exercise 7.2-2:
𝑝(𝑥) = 2(𝑥 + 𝑘⁄𝑥 )
Find the point on the curve 𝒚 = 𝒙𝟐 − 𝟓𝒙 + 𝟒 at which the tangent is parallel to the line
𝑝 ′ (𝑥)
= 2 (1 − 𝑘⁄𝑥 2 ) 𝟑𝒙 + 𝒚 = 𝟕
𝑝′ (𝑥) = 0 Solution:
𝑥 = ±√𝑘 The slope of the tangent is 𝑚⁡ = ⁡ −3

𝑝 ′′ (𝑥)
= 4𝑘⁄𝑥 3 𝑦 = 𝑥 2 − 5𝑥 + 4
𝑑𝑦
= 2𝑥 − 5
𝑝′′ (√𝑘) > 0 𝑑𝑥

2𝑥 − 5 = −3
P(x) has its minimum value at 𝑥⁡ = ⁡ √𝑘
𝑥=1
𝑥 = √𝑘, y= √𝑘
𝑥 = 1⁡ ⟹ 𝑦 = 0
The minimum perimeter rectangle of a given area is a square
The point is (1, 0)
Do yourself:
21. Exercise: 7.3 – 8
Exercise: 7.8 – 12
Does there exist a differentiable function 𝒇(𝒙) such that 𝒇(𝟎) = −𝟏; 𝒇(𝟐) = 𝟒 and
A hollow cone with base radius a cm and height b cm is placed on a table show that the volume
⁡𝒇′ (𝒙) ≤ 𝟐 for all x. Justify your answer
of the largest cylinder that can be hidden underneath is 4/9 times volume of the cone.
Solution:
𝑓(2)−𝑓(0)
3 Marks: 𝑓 ′ (𝑥) =
2−0
19. Example 7.11 4+1
𝑓 ′ (𝑥) = = 2.5
Find the equation of the tangent and normal to the curve 𝒚 = 𝒙𝟐 + 𝟑𝒙 − 𝟐 at the point 2

Does not exist since it is given that 𝑓 ′ (𝑥) ≤ 2

(1, 2).
Solution: 22. Example: 7.30
𝑑𝑦 Expand 𝐥𝐨𝐠(𝟏 + 𝒙) as a Maclaurin’s series upto 4 non-zero terms for −𝟏 < 𝒙 ≤ 𝟏
= 2𝑥 + 3
𝑑𝑥
Solution:
𝑑𝑦
𝑚 = 𝑑𝑥 =5 𝑓(𝑥) = log(1 + 𝑥) ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑓(0) = 0
(1,2)

1
The equation of the tangent is 𝑓 ′ (𝑥) = 1+𝑥 ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑓 ′ (0) = 1
(𝑦 − 2) = 5(𝑥 − 1) 1
𝑓 ′′ (𝑥) = − (1+𝑥)2 ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑓 ′′ (0) = −1
5𝑥 − 𝑦 − 3 = 0
2
The equation of the normal is 𝑓 ′′′ (𝑥) = (1+𝑥)3 ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑓 ′′′ (0) = 2
1 −6
𝑦⁡– ⁡2⁡ = - ⁡(𝑥 − 1) 𝑓 4 (𝑥) = ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑓 4 (0) = −6⁡
5 (1+𝑥)4

𝑥 + 5𝑦 − 11 = 0 Substituting the values and on simplification

𝑥2 𝑥3 𝑥4
log(1 + 𝑥) = 𝑥 − ⁡ + − + ⋯ … . −1 < 𝑥 ≤ 1
2 3 4

103 104
23. Example: 7.37 2 Marks:
𝟏−𝒄𝒐𝒔𝒎𝜽
If 𝐥𝐢𝐦 ( ) = 𝟏⁡ then prove that 𝒎 = ±𝒏 27. Exercise: 7.1 -4:
𝜽→𝟎 𝟏−𝒄𝒐𝒔𝒏𝜽
If the volume of a cube of side length 𝑥 is 𝒗 = 𝒙𝟑. Find the rate of change of the volume with
Solution:
1−𝑐𝑜𝑠𝑚𝜃 ⁡𝑚⁡𝑠𝑖𝑛𝑚𝜃 respect to x when x = 5 units.
lim ( 1−𝑐𝑜𝑠𝑛𝜃 ) = lim
𝜃→0 𝜃→0 𝑛⁡𝑠𝑖𝑛𝑛𝜃
Solution:
𝑠𝑖𝑛𝑚𝜃
⁡𝑚
= ⁡ lim ⁡ × 𝜃
𝑠𝑖𝑛𝑛𝜃
𝑑𝑣
= 3𝑥 2
𝜃→0 𝑛 𝑑𝑥
𝜃

𝑚2 = 3(25)
=
𝑛2
𝑑𝑣
1−𝑐𝑜𝑠𝑚𝜃 = 75
𝑑𝑥
lim ( )=1
𝜃→0 1−𝑐𝑜𝑠𝑛𝜃
28. Example: 7.16
𝑚2
=1 Find the angle of intersection of the curve y = sinx with the positive x axis.
𝑛2
2 2
𝑚 =𝑛 Solution:
𝑚 = ±𝑛 𝑦 = 𝑠𝑖𝑛𝑥 𝑦=0
𝑑𝑦
24. Example: 7.40 = 𝑐𝑜𝑠𝑥 𝑥 = 𝑛𝜋
𝑑𝑥
Evaluate: 𝐥𝐢𝐦+𝒙⁡𝒍𝒐𝒈𝒙
𝒙→𝟎 𝑚1 = 𝑠𝑙𝑜𝑝𝑒⁡𝑎𝑡⁡𝑥 = 𝑛𝜋⁡𝑎𝑟𝑒⁡𝑐𝑜𝑠𝑛𝜋 = (−1)𝑛 ⁡ 𝑚2 = 𝑠𝑙𝑜𝑝𝑒⁡𝑜𝑓⁡𝑥⁡𝑎𝑥𝑖𝑠
Solution:
(−1)𝑛 −0
log 𝑥
lim ⁡𝑥⁡𝑙𝑜𝑔𝑥 = lim+ ( 1/𝑥 ) = (∞ 𝑓𝑜𝑟𝑚)
∞ tan 𝜃 = |1+(−1)𝑛(0)| = 1 ∀⁡𝑛 𝑚2 = 0
𝑥⁡→⁡0+ 𝑥⁡→⁡0
𝜋
1 𝜃=
4
= lim+ ( 𝑥
1 )
𝑥⁡→⁡0 − 2
𝑥 29. Exercise: 7.3 –1 (i)
= lim (−𝑥) 𝟏
𝑥→0 Explain why Rolle’s theorem is not applicable to the function 𝒇(𝒙) = |𝒙| : 𝒙 ∈ ⁡ [−𝟏, 𝟏]
=0 Solution:
25. Exercise: 7.6-1(i)
Find the absolute extrema of 𝒇(𝒙) = 𝒙𝟐 − 𝟏𝟐𝒙 + 𝟏𝟎;⁡⁡⁡⁡[𝟏, 𝟐] 𝑓(𝑥) is not continuous at x= 0 in [-1, 1]
Solution: Do yourself:
𝒇(𝒙) = 𝒙𝟐 − 𝟏𝟐𝒙 + 𝟏𝟎 Exercise: 7.1 – 1(ii)
𝑓 ′ (𝑥) = 2𝑥 − 12
Explain why Rolle’s theorem is not applicable to the function 𝑓(𝑥) = 𝑡𝑎𝑛𝑥, 𝑥⁡ ∈ [0, 𝜋]
𝑓 ′ (𝑥) = 0
𝑥 = 6⁡ ∉ ⁡ [1,2] 30. Exercise: 7.3 – 3(i)
𝑓(1) = −1 Explain why Lagrange’s mean value theorem is not applicable to the function
𝑓(2) = −10 𝒙+𝟏
𝒇(𝒙) = ⁡ ; ⁡𝒙 ∈ ⁡ [−𝟏, 𝟐]
Absolute maximum is -1 𝒙
Absolute minimum is -10 Solution:
26. Example: 7.52
𝑓(𝑥) is not continuous at 𝑥 = 0⁡⁡ ∈ [−1,2]
Prove that the function 𝒇(𝒙) = 𝒙 − 𝒔𝒊𝒏𝒙 is increasing on the real line. Also discuss for
the existence of local extrema. Do yourself:
Solution: Exercise: 7.3 -3(ii)
𝑓 ′ (𝑥) = 1 − 𝑐𝑜𝑠𝑥 ≥ 0 Explain why Lagrange’s mean value theorem is not applicable to the function
𝑓 ′ (𝑥) = 0
𝑓(𝑥) = |3𝑥 + 1|; 𝑥 ∈ ⁡ [−1,3]
𝑥 = 2𝑛𝜋,⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑛𝜖𝑧
The function is increasing an the real line no sign change in 𝑓 ′ (𝑥) first derivative best there is
no local extrema
105 106
31. Example 7.21 35. Exercise: 7.5 – 2
𝒙𝟐 +𝟔 𝒍𝒊𝒎 𝟐𝒙𝟐 −𝟑
Compute the value of c satisfied by Rolle’s theorem for the function ⁡⁡𝒇(𝒙) = 𝐥𝐨𝐠 ( ) Evaluate: ⁡
𝟓𝒙 𝒙⁡ → ⁡∞ 𝒙𝟐 −𝟓𝒙+𝟑
in the interval [2, 3] Solution:
3
Solution: 𝑙𝑖𝑚 2𝑥 2 −3 𝑙𝑖𝑚 2− 2
⁡ 𝑥 2 −5𝑥+3 = ⁡ ⁡ 5𝑥 3
𝑓(𝑥) is continuous in [2, 3] 𝑥⁡ → ⁡∞ 𝑥⁡ → ⁡∞ 1−⁡𝑥+⁡𝑥2
= ⁡⁡2
𝑓(𝑥) is differentiable in (2, 3) Do yourself:
𝑓(2) ⁡ = ⁡𝑓(3) ⁡ = ⁡0 𝑙𝑖𝑚 1−𝑐𝑜𝑠𝑥
Exercise 7.5 -1 Evaluate: ⁡( 2 )
𝑥⁡ → 0 𝑥

𝑓 ′ (𝑥) =
𝑥 2 −6 𝑙𝑖𝑚 𝑥
𝑥(𝑥 2 +6)
Exercise: 7.5 -3 Evaluate: ⁡
𝑥⁡ → ∞ log 𝑥
𝑙𝑖𝑚 sec 𝑥
𝑓 ′ (𝐶) = 0
𝑥⁡ → ⁡ tan 𝑥
Exercise: 7.5 -4 Evaluate: 𝜋−
𝐶 2 −6 2
=0 𝑙𝑖𝑚 1 1
𝐶(𝐶 2 +6) Exercise: 7.5 – 6 Evaluate: ⁡( ⁡ − ⁡⁡ 𝑥)
𝐶 = ±√6 𝑥⁡ → 0 sin 𝑥
Exercise: 7.35
𝐶 = −√6 ⁡ ∉ (2,3) 𝑙𝑖𝑚 sin 𝑚𝑥
Evaluate the limit: ⁡
𝑥⁡ → 0 𝑥
𝐶 = √6 satisfies the Rolle’s theorem Exercise: 7.39
𝑙𝑖𝑚 1 1
32. Exercise: 7.4 -1(1) Evaluate: ⁡( − ⁡ )
𝑥⁡ → 0+ 𝑥 𝑒 𝑥 ⁡−1
Write the Maclaurin series expansion of the function 𝒆𝒙 36. Example: 7.46
Solution: Prove that the function 𝒇(𝒙) ⁡ = 𝒙𝟐 + 𝟐 is strictly increasing in the interval (2,7) and
strictly decreasing in the interval (-2, 0)
𝑓(𝑥) = ⁡ 𝑒 𝑥 𝑓(0) = ⁡1 Solution:
𝑓′(𝑥) = ⁡ 𝑒 𝑥 𝑓′(0) = ⁡1 𝑓 ′ (𝑥) = 2𝑥 > 0⁡⁡⁡;⁡⁡⁡⁡⁡⁡∀𝑥⁡ ∈ (2, 7)
𝑥
𝑓 ′ (𝑥) = 2𝑥 < 0⁡⁡;⁡⁡⁡⁡⁡∀𝑥⁡ ∈ (−2, 0)
𝑓′′(𝑥) = ⁡ 𝑒 𝑓′′(0) = ⁡1 and hence the proof is completed.
𝑥 ′ (0) 𝑥2 37. Example: 7.47:
𝑓(𝑥) = 𝑓(0) + ⁡ 1! 𝑓 + ⁡ 2⁡! 𝑓 ′′ (0) + ⋯ … ….
Prove that the function 𝒇(𝒙) ⁡ = 𝒙𝟐 − 𝟐𝒙 − 𝟑 is strictly increasing in (𝟐, ∞)
𝑥 𝑥2
𝑒 𝑥 = 1 + ⁡ 1! + ⁡ 2! ⁡ + ⋯ … … .. Solution:
𝑓(𝑥) ⁡ = 𝑥 2 − 2𝑥 − 3
Do yourself: 𝑓 ′ (𝑥) = 2𝑥 − 2 > 0⁡⁡⁡;⁡⁡⁡⁡⁡⁡∀𝑥⁡ ∈ (2, ∞)
Exercise: 7.4 – 1(ii)
𝑓(𝑥) is strictly increasing in (𝟐, ∞)
Write the Maclaurin series expansion of the function sinx
38. Example: 7.50
Exercise : 7.4 1(iii)
Find the intervals of monotonicity and hence find the local extrema for the function
Write the Maclaurin series expansion of the function cosx
𝒇(𝒙) ⁡ = 𝒙𝟐 − 𝟒𝒙 + 𝟒
33. Example: 7.33
Solution:
𝒍𝒊𝒎 𝒙𝟐 −𝟑𝒙+𝟐
Evaluate : ( ) 𝑓(𝑥) =⁡= 𝑥 2 − 4𝑥 + 4
𝒙⁡ → 𝟏 𝒙𝟐 −𝟒𝒙+𝟑
Solution: 𝑓(𝑥) = ⁡ (𝑥 − 2)2
𝑙𝑖𝑚 𝑥 2 −3𝑥+2 𝑙𝑖𝑚 2𝑥−3 𝑓 ′ (𝑥) = 2(𝑥 − 2) = 0
( ) ⁡ = ⁡⁡ ( )
𝑥⁡ → 1 𝑥 2 −4𝑥+3 𝑥⁡ → 1 2𝑥−4 𝑥 = 2⁡⁡
1
= ⁡ ⁄2
34. Example 7.34 Interval Sign of 𝑓′(𝑥) Monotonicity
𝒍𝒊𝒎 𝒙𝒏 ⁡−⁡⁡𝒂𝒏 𝑥⁡ ∈ (−∞, 2) − Strictly decreasing
Compute the limit : ⁡( )
𝒙⁡ → 𝒂 𝒙−𝒂 𝑥⁡ ∈ ⁡ (2, ∞) + Strictly increasing
Solution:
𝑙𝑖𝑚 𝑥 𝑛 ⁡−⁡⁡𝑎𝑛 𝑙𝑖𝑚 𝑛𝑥 𝑛−1 ⁡
⁡( 𝑥−𝑎 ) ⁡ = ⁡⁡ ⁡( 1 ) Local minimum at 𝑥⁡ = ⁡2
𝑥⁡ → 𝑎 𝑥⁡ → 𝑎
= 𝑛𝑎𝑛−1 Local minimum value is⁡𝑓(2) ⁡ = ⁡0

107 108
 CHAPTER - 9 
2
Applications of Integral Calculus 2. Evaluate:  cos7 xdx
Important points: 0

b Solution:
 The curve y = f(x), Area lies above the x - axis A =  ydx n = 7,
6 4 2
𝐼7 = ⁡ 7 . . . 1 = ⁡
16
a 5 3 35
b 
 The curve y = f(x) Area lies below the x - axis A =   ydx
 
2

a 3. Evaluate::  sin 2 x  cos4 x dx

d 0

 The curve x = g(y), Area lies right of y - axis A=  xdy Solution:

c
 
d
2 2
 The curve x = g(y), Area lies Left of y – axis A =   xdy I =  sin xdx ⁡+⁡  cos4 xdx
2

c
0 0
 Common area of the region bounded by the curves 𝑦𝑢 ⁡ = ⁡𝑓(𝑥)⁡, 𝑦𝐿 ⁡ = ⁡𝑔(𝑥)about x-axis A=
b
= 𝐼2 ⁡ + ⁡⁡ 𝐼4
 y
a
u  y L dx 1 𝜋 3 1 𝜋
= ⁡ ⁡. 2 ⁡ + ⁡ 4 ⁡. 2 ⁡. 2 ⁡ = ⁡ 4 + ⁡ 16 ⁡ = ⁡⁡ 16
2
𝜋 3𝜋 7𝜋


 Common area of the region bounded by the curves 𝑥𝑢 = ⁡𝑓(𝑦), ⁡𝑥𝐿 ⁡ = ⁡𝑔(𝑦)about y-axis A =
 
2
d 4. Evaluate  sin 2 x  cos4 x dx
 x
c
u  xL dy 0

Solution:
 If 𝑓(−𝑥) = −𝑓(𝑥)then f is an odd function 

 If 𝑓(−𝑥) =⁡𝑓(𝑥)then f is an even function

 
2


I =  1  cos2 x cos4 xdx
𝑛!
 x e dx =
n  ax 0
 Gamma Integral : 𝑎𝑛+1  
0
2 2

 cos xdx −  cos6 xdx

a a 4
=
  f ( x)dx   f (a  x)dx
0 0
0 0

= 𝐼4 − ⁡⁡ 𝐼6
b b
  f ( x)dx   f (a  b  x)dx
a a
3 1 𝜋 5 3 1 𝜋
= ⁡ 4 ⁡. 2 ⁡. 2 − ⁡ 6 ⁡⁡.⁡⁡ 4 ⁡. 2 ⁡. 2 ⁡ = ⁡⁡ 16 (1 − ⁡ 6) ⁡ = ⁡⁡ 32
3𝜋 5 𝜋

1 
𝑚!⁡⁡×𝑛!
 x (1  x) n dx =(𝑚+𝑛+1)! cos4 x 7
m 2

0
5. Evaluate::  dx
0 sin5 x 3

2 and 3 marks question Solution:

 Reduction formulae: 

 
2
 
2 2
𝑛−1 𝑛−3 2
… … ..⁡⁡ 3 . 1⁡;⁡ 𝑛⁡𝑖𝑠⁡𝑜𝑑𝑑 I =  3 cos4 x  7 sin5 x dx
𝐼𝑛 =⁡  sin n xdx   cosn xdx = {𝑛−1
𝑛 𝑛−2⁡
𝑛−3 1 𝜋 0
0 0
𝑛
… … . . 2 . 2 ⁡; 𝑛⁡𝑖𝑠⁡𝑒𝑣𝑒𝑛⁡
𝑛−2⁡  
 2 2

 cos xdx − 7  sin xdx

4 5
2 =3
1. Evaluate:  sin10 xdx 0 0
0
= 3𝐼4 − ⁡⁡7𝐼5
Solution:
9 7 5 3 1 𝜋 63⁡𝜋 3 1 𝜋 4 2 9𝜋 56
n = 10, 𝐼10 = ⁡ 10 . . . . ⁡. = ⁡ = ⁡3⁡ (4 ⁡. ⁡. ) − ⁡7⁡ ( ⁡. ⁡. 1) ⁡⁡⁡ = ⁡⁡ − ⁡
8 6 4 2 2 512 2 2 5 3 16 15

120 121
  

x e
3 x 2
4
10. Evaluate dx = 32,  > 0 Find 
6. Evaluate:  sin 6 (2 x)dx 0
0
x 0 𝜋⁄4 
x 0 ∞
𝜋⁄2 I =  x 2 e x ( xdx) = 32
2
t 0 Solution::
Solution: 0 t 0 ∞
𝑑𝑡 𝑑𝑡
𝐿𝑒𝑡⁡⁡⁡𝑡 = 2𝑥, = 2, = 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑑𝑥 2 𝑡 = ⁡ 𝑥 2 , = 2x, 2 = x dx
  𝑑𝑥
2
dt 1 2 1 ∞ 𝑑𝑡 1 ∞
I =  sin t. = ⁡  sin 6 t.dt = ⁡⁡ 𝐼6
6
I =∫0 𝑡. 𝑒 −𝛼𝑡 . 2 = 32 ∫0 𝑡1 𝑒 −𝛼𝑡 ⁡𝑑𝑡 = 32
2
2 2 2
0 0
∞
1 5 3 1 𝜋 5𝜋 ∫0 𝑡1 𝑒 −𝛼𝑡 ⁡𝑑𝑡⁡ = ⁡64 (By Gamma Integral)
⁡𝐼 = ⁡ (6 ⁡⁡.⁡⁡ 4 ⁡. 2 ⁡. 2 ) = ⁡⁡⁡⁡ 64
2
1⁡! 1 1
 = 64  𝛼 2 = ⁡ 64  = 8 (∵ ⁡𝛼 > 0)
6 𝛼1+1
7. Evaluate:  sin (3 x)dx
5
1
𝑚⁡!⁡⁡×⁡⁡𝑛⁡!
 x (1  x) n dx  (𝑚+𝑛+1)!(Where m and n – are positive integers)
m
0
x 0 𝜋⁄6
0
t 0 𝜋⁄2
Solution: 1

𝐿𝑒𝑡⁡⁡⁡𝑡 = 3𝑥,
𝑑𝑡
= 3,
𝑑𝑡
= 𝑑𝑥 11. Evaluate:  x 3 (1  x) 4 dx
𝑑𝑥 3 0
 
2 2 Solution: m= 3 , n=4 m+n+1=8
dt 1 1
I =  sin 5 t. = ⁡ 3  sin5 t.dt = ⁡⁡ 3 𝐼5 I=
3⁡!⁡⁡×⁡⁡4⁡!
=
6⁡⁡×⁡4⁡!
⁡
1
= 8.7.5⁡ = 280
1
0
3 0 8⁡! (8.7.6.5)×4⁡!
1

x (1  x) 3 dx
2
12. Evaluate:
1 4 2 8
⁡𝐼 = ⁡ 3 (5 ⁡⁡.⁡⁡ 3 ⁡. 1⁡) = ⁡⁡⁡⁡ 45 0

2 Solution: m = 2, n = 3, m + n + 1 = 6
7 x x 0 2𝜋
8. Evaluate:  sin dx I=
2⁡!⁡⁡×⁡⁡3⁡! 2⁡⁡×6
= 720 =720= 60
12 1

0
4 t 0 𝜋⁄2 6⁡!
 Properties of Integral:
Solution: a a b b

𝐿𝑒𝑡⁡⁡⁡𝑡 = 4,
𝑥 𝑑𝑡
𝑑𝑥
1
= 4 , 4𝑑𝑡 = 𝑑𝑥 i.  f ( x)dx =  f (a  x)dx
0 0
ii.  f ( x)dx =  f (a  b  x)dx
a a
  a
f ( x)

2 2
13. Evaluate: dx
I =  sin 7 t (4dt ) = ⁡4  sin 7 t.dt = ⁡⁡4⁡𝐼7 0
f ( x)  f (a  x)
0 0
a
f ( x)
Solution I = dx …………… (1)
0
f ( x)  f (a  x)
6 4 2 64
⁡𝐼 = ⁡⁡4⁡⁡ [7 ⁡⁡.⁡⁡ 5 ⁡. 3 ⁡.1⁡] = ⁡⁡⁡⁡ 35 a a
By property,  f ( x)dx =  f (a  x)dx

𝑛!
x e
0 0
n  ax
 Gamma Integration: dx = , a>0
𝑎𝑛+1
f (a  x)
a
0
I = dx …………………… (2)
 0
f (a  x)  f ( x)
x e
5 3 x
9. Evaluate: dx x a – x
0
f ( x)  f (a  x)
a

Solution: n = 5, a = 3 (1) + (2)  2I =  dx

0
f ( x)  f (a  x)
5⁡! 5⁡!
I =35+1 = a
36
2I =  dx =[𝑥]0 𝑎 ⁡ = 𝑎
0

𝑎
I =2
122 123
 3
x 0 ;𝑓 𝑖𝑠⁡𝑎𝑛⁡𝑜𝑑𝑑⁡𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛⁡
14. Evaluate:  dx a


a
5 x  x  f ( x)dx ={
2
a
2  f ( x)dx ;𝑓 is⁡an⁡even⁡function⁡
3
x 0
Solution I = dx ………………. (1)
2 5 x  x
b b 
By property  f ( x)dx =  f (a  b  x)dx 4

 sin
2
a a 16. Evaluate xdx
5 x
3 

 …………………… (2) 𝑥⁡⁡⁡5⁡– ⁡𝑥 4

I= dx
2 x  5 x Solution: 𝑓(𝑥) = ⁡ 𝑠𝑖𝑛2 𝑥 = ⁡ (sin 𝑥)2
x  5 x
3
(1) + (2)  2I =  dx 𝑓(−𝑥) = ⁡ [sin(−𝑥)]2=(− sin 𝑥)2 = ⁡ 𝑠𝑖𝑛2 𝑥⁡ = 𝑓⁡(𝑥)
2 x  5 x
f(-x) = f(x) / fis an even function
3
3
2I =  dx =(𝑥)2 ⁡ = 3 − 2 = 1  
4 4
(1−cos 2𝑥)
𝐼 =2  sin 2 xdx = 2 
2
1 2
⁡⁡𝑑𝑥
I= 0 0
2
3 𝜋⁄
sin 2𝑥 4 𝜋 1 𝜋⁡−⁡⁡2
8
1 I = [𝑥 − ⁡ ] ⁡ = ⁡⁡ ( ⁡ − ⁡ ) − 0 = ⁡
15. Evaluate: 
 1 tan x
dx 2 0 4 2 4

8
3 
8
1 2
Solution I= 
 sin x
17. Evaluate:  x cos xdx
8
1  
cos x 2

3
8
Solution: 𝑓(𝑥) = 𝑥⁡𝑐𝑜𝑠𝑥 𝑓(−𝑥) = (−𝑥)⁡𝑐𝑜𝑠⁡(−𝑥) = ⁡ −𝑥𝑐𝑜𝑠⁡𝑥
cos x
I= 
 cos x  sin x
dx ………………… (1) /𝑓(−𝑥) ⁡ = ⁡ −𝑓(𝑥)
8 𝑓(𝑥) is an odd function
b b

By property,  f ( x)dx =  f (a  b  x)dx 2
a
3
a
I=  x cos xdx
8 
sin x 2
I = 
 sin x  cos x
dx …………………. (2)
8
4
sin x  cos x 𝒅𝒙
3
18. Evaluate: 
3
𝒙𝟐 ⁡−𝟒
8
cos x  sin x
(1) ⁡ + ⁡ (2)⁡⁡⁡2𝐼⁡ =⁡  dx 4 𝑑𝑥 𝑑𝑥 1 𝑥−𝑎
 cos x  sin x Solution: I = ∫3 ∵ ∫ 𝑥 2 ⁡−𝑎2 = ⁡ 2𝑎 log (𝑥+𝑎))
𝑥 2 ⁡−22
8
1 𝑥−2 4
3
8
I = [2(2) log (𝑥+2))] a=2
3
=⁡  dx
 1 2 1
8
=4 [log 6 − log 5]
⁄
= (𝑥)3𝜋
𝜋⁄8
8
1
1

3𝜋 𝜋 2𝜋
= 4 [log ( 31 )]
2I = 8 − ⁡ = ⁡ 5
8 8
2𝜋 𝜋 1 5
2I = 8
, I=8 =4 log (3)

124 125
  4
3
𝐬𝐞𝐜 𝒙 𝐭𝐚𝐧 𝒙 22. Evaluate:  |𝒙 + 𝟑|𝒅𝒙
19. Evaluate: 
0
𝟏+𝒔𝒆𝒄𝟐 𝒙⁡
⁡𝒅𝒙 4

𝜋⁄3 −(𝑥 + 3) ; 𝑥 < ⁡ −3

x 0 Solution: |𝑥 + 3|={
𝑥+3 ; 𝑥⁡ ≥ ⁡ −3
𝑑𝑡 t 1 2
Solution: 𝑡 = sec 𝑥 ,𝑑𝑥 = sec 𝑥⁡𝑡𝑎𝑛𝑥,⁡⁡⁡dt = secx tan dx 3 4
I=  −(𝑥 + 3)𝑑𝑥 +  (𝑥 + 3)𝑑𝑥
2 4 3
𝑑𝑡 𝑑𝑥
I=  1+⁡𝑡 2
∵∫
1+⁡𝑥 2
= ⁡ 𝑡𝑎𝑛−1 𝑥
𝑥2
−3
𝑥2
4
1
=-[ + 3𝑥] ⁡⁡ + [ + 3𝑥]
2 −4 2 −3
2
=[𝑡𝑎𝑛−1 (𝑡)]1 =𝑡𝑎𝑛−1 (2)⁡– 𝑡𝑎𝑛−1 (1) 9 9
𝜋 =-[( − 9) − (8 − 12)] + ⁡ [(8 + 12) − ⁡ (2 − 9)]
2
= 𝑡𝑎𝑛−1 (2)⁡–⁡ 4
9 9
= − [− 2 + 4] + ⁡ [20 + ⁡ 2]
9 9
9
𝟏
=2 − ⁡4 + 20 + ⁡ 2 ⁡ = ⁡⁡16 + 9 = 25
20. Evaluate: 
0
𝒙+⁡√𝒙
⁡𝒅𝒙
 If n is odd and m is any positive integer (even or odd)
𝑑𝑡 
Solution: √𝑥 = 𝑡, x = ⁡ 𝑡 2 , = 2𝑡, 𝑑𝑥⁡ = 2𝑡𝑑𝑡
𝑑𝑥 x 0 9 2
𝑛−1 𝑛−3 𝑛⁡−⁡⁡5 2 1
3 3
  𝑠𝑖𝑛𝑚 𝑥⁡𝑐𝑜𝑠 𝑛 𝑥⁡𝑑𝑥 = ⁡ 𝑚+𝑛 ⁡⁡.⁡⁡ 𝑚+𝑛−2 ⁡⁡.⁡⁡ 𝑚+⁡𝑛⁡−4 ⁡… … . 𝑚+3 ⁡⁡.⁡⁡ 𝑚+1
𝐼 =⁡ 
1 𝑑𝑡 3 3

(2𝑡⁡𝑑𝑡)= =2⁡[log(1 + 𝑡)]0 t 0 0
𝑡 2 ⁡+𝑡 1⁡⁡+𝑡
0 0  
2 2
I = 2[log 4 – log 1] = 2 log 4 – 0 = log 42 = log 16  
0
𝑠𝑖𝑛𝑚 𝑥⁡𝑐𝑜𝑠 𝑛 𝑥⁡𝑑𝑥 =⁡  𝑠𝑖𝑛𝑛 𝑥⁡𝑐𝑜𝑠 𝑚 𝑥⁡𝑑𝑥
0

𝑚−1 𝑚−3 𝑚⁡−⁡⁡5 2 1

I= ⁡⁡.⁡⁡ 𝑛+𝑚−2 ⁡⁡.⁡⁡ 𝑛+𝑚−2 ⁡… … . 𝑛+3 ⁡⁡.⁡⁡ 𝑛+1
𝑛+𝑚
2
𝒙  If n or m any one odd
21. Evaluate: 
1
(𝒙+𝟏)(𝒙+𝟐)
⁡𝒅𝒙

2 
𝑥
Solution: 𝐼⁡ =⁡  ⁡𝑑𝑥 2

1
(𝑥+1)(𝑥+2) 23. Evaluate: 0
𝒔𝒊𝒏𝟑 𝒙⁡𝒄𝒐𝒔𝟓 𝒙⁡𝒅𝒙
𝑥 𝐴 𝐵 −1 2
𝐵𝑦⁡𝑝𝑎𝑟𝑡𝑖𝑎𝑙⁡𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠, (𝑥+1)(𝑥+2)
⁡= 𝑥+1 + ⁡ 𝑥+2 = ⁡ 𝑥+1 + ⁡ 𝑥+2 Solution: Here m = 3, n = 5 , m + n = 8
4 2 1 1
2
−1 2 I =8 ⁡⁡.⁡⁡ 6 ⁡⁡.⁡⁡ 4 ⁡⁡ = ⁡⁡ 24
𝐼⁡ =⁡⁡  (𝑥+1 + ⁡ 𝑥+2) 𝑑𝑥
1

=[− log(𝑥 + 1) + 2 log(𝑥 + 2)]12  

2 2
=[log(𝑥 + 2)2 − log⁡(𝑥 + 1)]12 24. Evaluate:  𝒔𝒊𝒏𝟓 𝒙⁡𝒄𝒐𝒔𝟒 𝒙⁡𝒅𝒙 =  𝒔𝒊𝒏𝟒 𝒙⁡𝒄𝒐𝒔𝟓 𝒙⁡𝒅𝒙
2 0 0
(𝑥+2)2
=[log ]
(𝑥+1) 1 Solution: 𝑚⁡ = ⁡5, 𝑛⁡ = ⁡4⁡, 𝑚⁡ + ⁡𝑛⁡ = ⁡9
16 9 4 2 1 8
=log ( 3 ) − log (2) I =9 ⁡⁡.⁡⁡ 7 ⁡⁡.⁡⁡ 5 ⁡⁡ = ⁡⁡ 315
16/3 32
=log ( 9/2 ) =𝑙𝑜𝑔 (27)

126 127
 Five marks questions 4
Eqn. of the curve 𝐴 = 𝐴1 + 𝐴2
Equation of the Diagram Formula Area 𝑏
𝑦 = |cos 𝑥|
Lines /curves 𝐴 = ∫ 𝑦 𝑑𝑥
2
𝑎
1 Eqn.of parabola
𝜋
2 2 𝜋
𝑦 = 4𝑎𝑥
𝑏 8 2 𝐴 = ∫ cos 𝑥 𝑑𝑥 + ∫ − cos 𝑥 𝑑𝑥
eqn.of Latus rectum; 𝑎
𝐴 = ∫ 𝑦 𝑑𝑥 3 0 𝜋
2
𝑥 = 𝑎 𝑎
5
Eqn. of the curve
2+x-x2+y=0 𝐴 = 𝐴1 + 𝐴2 + 𝐴3
2
−1 2 3
Eqn. of the Ellipse: y = x2 –x -2
𝑏 𝐴 = ∫ 𝑦 𝑑𝑥 + ∫ −𝑦 𝑑𝑥 + ∫ 𝑦 𝑑𝑥
𝑥2 𝑦2 15
+ = 1 𝐴 = ∫ 𝑦 𝑑𝑥
−3 −1 2
𝑎2 𝑏2 𝜋𝑎𝑏
𝑎

6 𝐴 = 𝐴1 + 𝐴2
3 𝑏 𝑐
Eqn. of the curves
Eqn. of the curve 𝐴 = ∫ 𝑦 𝑑𝑥 + ∫ 𝑦 𝑑𝑥
𝐴 = 𝐴1 + 𝐴2 y = tan x
𝑦 = sin 𝑥 𝑎 𝑏 𝐿𝑜𝑔 2
𝑏 𝑐 4 y = cot x 𝜋 𝜋
4 2
𝐴 = ∫ 𝑦 𝑑𝑥 + ∫ −𝑦 𝑑𝑥
𝐴 = ∫ tan 𝑥 𝑑𝑥 + ∫ 𝑐𝑜𝑡𝑥 𝑑𝑥
𝑎 𝑏
0 𝜋
4

139 140
7 10 Eqn. of the curves
Eqn. of the curves y = sinx
𝑑 4
𝑥 2 + 𝑦 2 = 16 (4𝜋 + √3) x = cos x
𝐴 = ∫(𝑥𝑢 − 𝑥𝐿 )𝑑𝑦 3
𝑦 2 = 16 𝑥 𝑐 𝑥=
𝜋
and 𝑥 =
5𝜋 𝑏
4 4 2√2
𝐴 = ∫(𝑦𝑢 − 𝑦𝐿 )𝑑𝑥
about x-axis 𝑎

8
Eqn. of the curves
11 Eqn. of the curve
𝑦2 = 4 𝑥 𝑏 16 𝑥2 = 𝑦
2
𝑥 = 4𝑦 𝐴 = ∫(𝑦𝑢 − 𝑦𝐿 )𝑑𝑥 3 Eqn. of the lines 𝑏
1
𝑎
𝑦 = |𝑥| 𝐴 = ∫(𝑦𝑢 − 𝑦𝐿 )𝑑𝑥
3
𝑎

9
Eqn. of the curves
𝑦 = 𝑠𝑖𝑛 𝑥 𝑏
𝑦 = 𝑐𝑜𝑠 𝑥 𝐴 = ∫(𝑦𝑢 − 𝑦𝐿 )𝑑𝑥 2√2
𝑥 = 0, 𝑎𝑛𝑑 𝑥 =  𝑎

About x-axis

141 142
12 Eqn. of the line 15 Eqn. of the circle
ℎ
𝑦 = 𝑥 2 − 2𝑥 +5 𝑥2 + 𝑦2 = 4 𝑎2 𝑐𝑜𝑠 −1 ( )
𝑏 𝑏
𝑎
Eqn. of the curve Eqn. of the line
𝐴 = ∫(𝑦𝑢 − 𝑦𝐿 )𝑑𝑥 36 𝐴 = 2 ∫ 𝑦𝑑𝑥 − ℎ√𝑎2 − ℎ2
y = x2 -2x x=h
𝑎 𝑎

13 Eqn. of the line 16 Eqn. of the curves

𝑦 2 = 4𝑥 , 𝐴 = 𝐴1 + 𝐴2 + 𝐴3 16
x +y =3 𝐴1 =
𝑏 𝑏 𝑏
3
Eqn. of the curve 𝐴 = 𝐴1 + 𝐴2 𝑥 2 = 4𝑦
16
2 3
𝐴 = ∫ 𝑦 𝑑𝑥 + ∫(𝑦𝑢 − 𝑦𝐿 )𝑑𝑥 + ∫ 𝑥𝑑𝑦 𝐴2 =
𝑦 2 = 4𝑥 7
x = 0, x = 4 and 3
𝑎 𝑎 𝑎
𝐴 = ∫ 𝑥𝑑𝑦 + ∫ 𝑥𝑑𝑦 16
6 y = 0, y = 4 𝐴3 =
0 2 3

14 Eqn. of the line

17 Eqn. of the curve
y = x-2
𝑑
𝑦 = (𝑥 − 2)2 + 1
Eqn. of the curve
9 𝑏
𝐴 = ∫(𝑥𝑢 − 𝑥𝐿 )𝑑𝑦 Slope of
4
𝑦2 = 𝑥 2 𝐴 = ∫(𝑦𝑢 − 𝑦𝐿 )𝑑𝑥
𝑐 PQ = 2 3
𝑎

143 144
18 Tangent and normal
to the circle
x2 + y2 = 4
𝑑
drawn at
2√3
𝐴 = ∫(𝑥𝑢 − 𝑥𝐿 )𝑑𝑦
(1, √3) 𝑐

19 Eqn. of the lines

5x – 2y = 15
x + y = -4
𝑑
35
𝐴 = ∫(𝑥𝑢 − 𝑥𝐿 )𝑑𝑦
2
𝑐

20 The triangle ABC

formed by the
vertices 𝐴 = 𝐴1 + 𝐴2 − 𝐴3
𝑏 𝑐 𝑐
(-1, 1) (3, 2) 15
𝐴 = ∫ 𝑦 𝑑𝑥 + ∫ 𝑦𝑑𝑥 − ∫ 𝑦𝑑𝑥
and (0, 5) 2
𝑎 𝑏 𝑎

145