RELATIONS AND FUNCTIONS EACH OF THE FOLLOWING QUESTIONS CARRIES 1...

1

RELATIONS AND FUNCTIONS

EACH OF THE FOLLOWING QUESTIONS CARRIES 1 MARK:

Q 1) Define bijective function.

A function 𝑓: 𝐴 → 𝐵 is said to be a bijective function if it is both one one and onto function.

Q 2) Define binary operation on a set.

An operation ∗ defined on a set 𝐴 is said to be a binary operation if ∀ 𝑎, 𝑏 ∈ 𝐴 , 𝑎 ∗ 𝑏 ∈ 𝐴 .

Q 3) Define equivalence relation.

A relation 𝑅 defined on a set 𝐴 is said to be an equivalence relation if it is reflexive, symmetric

and transitive.

Q 4) Operation ∗ is defined by 𝑎 ∗ 𝑏 = 𝑎 . Is ∗ a binary operation on 𝑍+

Solution: ∀ 𝑎, 𝑏 ∈ 𝑍+ , 𝑎 ∗ 𝑏 = 𝑎 ∈ 𝑍+

∴ ∗ is a binary operation on 𝑍+ .

Q 5) Verify whether the operation ∗ defined on 𝑍 by 𝑎 ∗ 𝑏 = 𝑎𝑏 + 1 is binary or not.

Solution: ∀ 𝑎, 𝑏 ∈ 𝑍, 𝑎 ∗ 𝑏 = 𝑎𝑏 + 1 ∈ 𝑍

∴ ∗ is a binary operation on 𝑍 .

Q 6) Let ∗ be the binary operation on 𝑁 given by 𝑎 ∗ 𝑏 = 𝐿𝐶𝑀 of 𝑎 and 𝑏 , find 20 ∗ 16 .

Solution: 20 ∗ 16 = 𝐿𝐶𝑀 of 20 and 16 = 80 .

EACH OF THE FOLLOWING QUESTIONS CARRIES 2 MARKS:

Q 1) A binary operation ^ on the set {1, 2, 3, 4, 5} is defined by 𝑎^𝑏 = min{𝑎, 𝑏}.

Write the operation table for the operation ^.

Solution:

^ 1 2 3 4 5

1 1 1 1 1 1

2 1 2 2 2 2

3 1 2 3 3 3

4 1 2 3 4 4

5 1 2 3 4 5

Q 2) Verify whether the operation ∗ defined on 𝑄 by 𝑎 ∗ 𝑏 =𝑎𝑏

4 is associative or not.

Solution: 𝑎 ∗ 𝑏 ∗ 𝑐 = 𝑎𝑏

4 ∗ 𝑐 =

𝑎𝑏𝑐

16

𝑎 ∗ (𝑏 ∗ 𝑐) = 𝑎 ∗ 𝑏𝑐

4 =

𝑎𝑏𝑐

16

𝑎 ∗ 𝑏 ∗ 𝑐 = 𝑎 ∗ (𝑏 ∗ 𝑐) ∴ ∗ is associative on 𝑄 .

2

Q 3) Let ∗ be the binary operation on 𝑄 given by 𝑎 ∗ 𝑏 =𝑎𝑏

4 . Find the identity element.

Solution: 𝑎 ∗ 𝑒 = 𝑎 = 𝑒 ∗ 𝑎

𝑎 ∗ 𝑒 = 𝑎 ∴𝑎𝑒

4= 𝑎 ⟹ 𝑒 = 4


Q 1) Show that the relation 𝑅 in the set of integers given by 𝑅 = 𝑎, 𝑏 : 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎 − 𝑏 is an

equivalence relation.

Solution: 𝒂 − 𝒂 = 𝟎

5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 0

⟹ 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎 − 𝑎

∴ ∀ 𝑎 ∈ 𝑍, (𝑎, 𝑎) ∈ 𝑅

∴ 𝑅 is a reflexive relation.

Let 𝑎, 𝑏 ∈ 𝑅 ⟹ 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎 − 𝑏

⟹ 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 − 𝑎 − 𝑏

⟹ 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑏 − 𝑎

⟹ 𝑏, 𝑎 ∈ 𝑅

∴ 𝑅 is a symmetric relation.

Let 𝑎, 𝑏 ∈ 𝑅 𝑎𝑛𝑑 𝑏, 𝑐 ∈ 𝑅

⟹ 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎 − 𝑏 𝑎𝑛𝑑 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑏 − 𝑐

⟹ 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 (𝑎 − 𝑏 + 𝑏 − 𝑐)

⟹ 5 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎 − 𝑐

⟹ 𝑎, 𝑐 ∈ 𝑅

∴ 𝑅 is a transitive relation.

Since 𝑅 is reflexive, symmetric and transitive, it is an equivalence relation.

Q 2) Determine whether the relation 𝑅 in the set 𝐴 = 1, 2, 3, 4, 5, 6 as

𝑅 = { 𝑥, 𝑦 : 𝑦 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 𝑥} is an equivalence relation.

Solution: 𝑥 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 𝑥

∴ ∀ 𝑥 ∈ 𝐴, (𝑥, 𝑥) ∈ 𝑅

∴ 𝑅 is a reflexive relation.

Now 2, 4 ∈ 𝑅

⟹ 4 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 2

But 2 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4

⟹ (4, 2) ∉ 𝑅

∴ 𝑅 is not a symmetric relation.

Let 𝑥, 𝑦 ∈ 𝑅 𝑎𝑛𝑑 𝑦, 𝑧 ∈ 𝑅

⟹ 𝑦 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 𝑥 𝑎𝑛𝑑 𝑧 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 𝑦

⟹ 𝑧 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 𝑥

⟹ (𝑧, 𝑥) ∈ 𝑅

∴ 𝑅 is a transitive relation.

Since 𝑅 is reflexive and transitive but not symmetric it is not an equivalence relation.

Q 3) If ∗ is a binary operation defined on 𝐴 = 𝑁 × 𝑁 by 𝑎, 𝑏 ∗ 𝑐, 𝑑 = 𝑎 + 𝑐, 𝑏 + 𝑑 , prove

that ∗ is both commutative and associative. Find the identity, if it exists.

Solution: 𝑎, 𝑏 ∗ 𝑐, 𝑑 = 𝑎 + 𝑐, 𝑏 + 𝑑

3

𝑐, 𝑑 ∗ 𝑎, 𝑏 = 𝑐 + 𝑎, 𝑑 + 𝑏

∀ 𝑎, 𝑏 , 𝑐, 𝑑 ∈ 𝐴 , 𝑎, 𝑏 ∗ 𝑐, 𝑑 = 𝑐, 𝑑 ∗ 𝑎, 𝑏

∴ ∗ is commutative.

𝑎, 𝑏 ∗ 𝑐, 𝑑 ∗ 𝑕, 𝑘 = 𝑎 + 𝑐, 𝑏 + 𝑑 ∗ 𝑕, 𝑘

= 𝑎 + 𝑐 + 𝑕, 𝑏 + 𝑑 + 𝑘

𝑎, 𝑏 ∗ 𝑐, 𝑑 ∗ 𝑕, 𝑘 = 𝑎 ∗ 𝑏 ∗ 𝑐 + 𝑕, 𝑑 + 𝑘

= 𝑎 + 𝑐 + 𝑕, 𝑏 + 𝑑 + 𝑘

𝑎, 𝑏 ∗ 𝑐, 𝑑 ∗ 𝑕, 𝑘 = 𝑎, 𝑏 ∗ 𝑐, 𝑑 ∗ 𝑕, 𝑘

∴ ∗ is associative.

If possible let 𝑒, 𝑒′ be the identity element.

∴ 𝑎, 𝑏 ∗ 𝑒, 𝑒′ = 𝑎, 𝑏 = 𝑒, 𝑒′ ∗ 𝑎, 𝑏

⟹ 𝑎 + 𝑒, 𝑏 + 𝑒′ = 𝑎

⟹ 𝑎 + 𝑒 = 𝑎

⟹ 𝑒 = 0 ∉ 𝑁

The identity element does not exist.


Q 1) Verify whether the function 𝑓: 𝑁 → 𝑌 defined by 𝑓 𝑥 = 4𝑥 + 3 where

𝑌 = 𝑦: 𝑦 = 4𝑥 + 3, 𝑥 ∈ 𝑁 is invertible or not. Write the inverse of 𝑓 𝑥 if it exists.

Solution: Let 𝑓 𝑥1 = 𝑓(𝑥2) where 𝑥1, 𝑥2 ∈ 𝑁

4𝑥1 + 3 = 4𝑥2 + 3

⟹ 4𝑥1 = 4𝑥2

⟹ 𝑥1 = 𝑥2

∴ 𝑓 𝑥 is a one one function.

Let 𝑦 = 𝑓 𝑥

⟹ y = 4x + 3

⟹ 𝑥 =𝑦−3

4∈ 𝑁

For every 𝑦 ∈ 𝑌 there exists an 𝑥 =𝑦−3

4∈ 𝑁 such that 𝑓 𝑥 = 𝑦.

∴ 𝑓 𝑥 is an onto function.

∵ 𝑓 𝑥 is both one one and onto, it is invertible.

The inverse of 𝑓 𝑥 exists and is given by 𝑓−1 𝑦 =𝑦−3

4 .

Q 2) If 𝑅+ is the set of all non negative real numbers, prove that the function 𝑓: 𝑅+ → 4, ∞

defined by 𝑓 𝑥 = 𝑥2 + 4 is invertible. Also write the inverse of 𝑓 𝑥 .


⟹ 𝑥12 + 4 = 𝑥2

2 + 4

⟹ 𝑥12 = 𝑥2

2

⟹ 𝑥1 = 𝑥2



⟹ y = 𝑥2 + 4

𝑥 = 𝑦 − 4 ∈ 𝑅+

4

For every 𝑦 ∈ 4,∞ there exists an 𝑥 = 𝑦 − 4 ∈ 𝑅+ such that 𝑓 𝑥 = 𝑦.



The inverse of 𝑓 𝑥 exists and is given by 𝑓−1 𝑦 = 𝑦 − 4 .

Q 3) Let 𝑓: 𝑁 → 𝑅 be defined by 𝑓 𝑥 = 4𝑥2 + 12𝑥 + 15. Show that 𝑓: 𝑁 → 𝑆, where 𝑆 is the

range of the function is invertible. Also find the inverse of 𝑓.


⟹ 4𝑥12 + 12𝑥1 + 15 = 4𝑥2

2 + 12𝑥2 + 15

⟹ 2𝑥1 + 3 2 + 6 = 2𝑥2 + 3 2 + 6

⟹ 2𝑥1 + 3 2 = 2𝑥2 + 3 2

⟹ 2𝑥1 + 3 = 2𝑥2 + 3

⟹ 2𝑥1 = 2𝑥2

⟹ 𝑥1 = 𝑥2



⟹ 𝑦 = 4𝑥2 + 12𝑥 + 15

⟹ 2𝑥 + 3 2 + 6 = 𝑦

⟹ 2𝑥 + 3 2 = 𝑦 − 6

⟹ 2𝑥 + 3 = 𝑦 − 6

⟹ 𝑥 = 𝑦−6 − 3

2∈ 𝑁

For every 𝑦 ∈ 𝑆 there exists an 𝑥 = 𝑦−6 − 3

2∈ 𝑁 such that 𝑓 𝑥 = 𝑦.



The inverse of 𝑓 𝑥 exists and is given by 𝑓−1 𝑦 = 𝑦−6 − 3

2 .

Q 4) If 𝑅+ is the set of all non negative real numbers prove that the function

𝑓: 𝑅+ → [−5, ∞) defined by 𝑓 𝑥 = 9𝑥2 + 6𝑥 − 5 is invertible. Also find 𝑓−1(𝑥).

Solution: Let 𝑓 𝑥1 = 𝑓(𝑥2) where 𝑥1, 𝑥2 ∈ 𝑅+

⟹ 9𝑥12 + 6𝑥1 − 5 = 9𝑥2

2 + 6𝑥2 − 5

⟹ 3𝑥1 + 1 2 − 6 = 3𝑥2 + 1 2 − 6

⟹ 3𝑥1 + 1 2 = 3𝑥2 + 1 2

⟹ 3𝑥1 + 1 = 3𝑥2 + 1

⟹ 3𝑥1 = 3𝑥2

⟹ 𝑥1 = 𝑥2



⟹ 𝑦 = 9𝑥2 + 6𝑥 − 5

⟹ 3𝑥 + 1 2 − 6 = 𝑦

⟹ 3𝑥 + 1 2 = 𝑦 + 6

⟹ 3𝑥 + 1 = 𝑦 + 6

⟹ 𝑥 = 𝑦+6 − 1

3∈ 𝑅+

5

For every 𝑦 ∈ [−5, ∞) there exists an 𝑥 = 𝑦+6 − 1

3∈ 𝑅+ such that 𝑓 𝑥 = 𝑦.



The inverse of 𝑓 𝑥 exists and is given by 𝑓−1 𝑦 = 𝑦+6 − 1

3 .

Q 5) Let 𝑓: 𝐴 → 𝐴 be defined by 𝑓 𝑥 =4𝑥+3

6𝑥−4 where 𝐴 = 𝑅 −

2

3 , show that 𝑓 is invertible and

𝑓−1 = 𝑓.

Solution: Let 𝑓 𝑥1 = 𝑓(𝑥2) where 𝑥1, 𝑥2 ∈ 𝐴

⟹4𝑥1+3

6𝑥1−4=

4𝑥2+3

6𝑥2−4

⟹ 4𝑥1 + 3 6𝑥2 − 4 = 4𝑥2 + 3 6𝑥1 − 4

⟹ 24𝑥1𝑥2 − 16𝑥1 + 18𝑥2 − 12 = 24𝑥1𝑥2 + 18𝑥1 − 16𝑥2 − 12

⟹ 34𝑥1 = 34𝑥2

⟹ 𝑥1 = 𝑥2



⟹ 𝑦 =4𝑥+3

6𝑥−4

⟹ 6𝑥𝑦 − 4𝑦 = 4𝑥 + 3

⟹ 6𝑦 − 4 𝑥 = 4𝑦 + 3

⟹ 𝑥 =4𝑦+3

6𝑦−4∈ 𝐴

For every 𝑦 ∈ 𝐴 there exists an 𝑥 =4𝑦+3

6𝑦−4∈ 𝐴 such that 𝑓 𝑥 = 𝑦.



The inverse of 𝑓 𝑥 exists and is given by 𝑓−1 𝑦 =4𝑦+3

6𝑦−4 .

𝑓−1 𝑥 =4𝑥+3

6𝑥−4= 𝑓(𝑥)

6

CONTINUITY AND DERIVATIVES


Q 1) If 𝑦 = 𝑒3 log 𝑥 then show that 𝑑𝑦

𝑑𝑥= 3𝑥2.

Solution: 𝑦 = 𝑒log 𝑥3= 𝑥3

𝑑𝑦

𝑑𝑥= 3𝑥2.

Q 2) Differentiate log cos 𝑒𝑥 with respect to 𝑥 .

Solution: 𝑦 = log cos 𝑒𝑥 𝑑𝑦

𝑑𝑥=

−𝑒𝑥 sin 𝑒𝑥

cos 𝑒𝑥

Q 3) Differentiate sin 𝑥 with respect to 𝑥 .

Solution: 𝑦 = sin 𝑥 𝑑𝑦

𝑑𝑥=

cos 𝑥

2 𝑥

Q 4) Find the derivative of cos 𝑥2 with respect to 𝑥 .

Solution: 𝑦 = cos 𝑥2 𝑑𝑦

𝑑𝑥= −2𝑥 sin 𝑥2

Q 5) If 𝑦 = tan 2𝑥 + 3 𝑓𝑖𝑛𝑑 𝑑𝑦

𝑑𝑥 .

Solution: 𝑑𝑦

𝑑𝑥= 2𝑠𝑒𝑐2(2𝑥 + 3)

Q 6) If 𝑥 − 𝑦 = 𝜋 𝑓𝑖𝑛𝑑 𝑑𝑦

𝑑𝑥 .

Solution: 1 −𝑑𝑦

𝑑𝑥= 0 ⟹

𝑑𝑦

𝑑𝑥= 1

Q 7) If 𝑦 = 𝑒cos 𝑥 then find 𝑑𝑦

𝑑𝑥

Solution: 𝑑𝑦

𝑑𝑥= − sin 𝑥 𝑒cos 𝑥

Q 8) Write the points of discontinuity for the function 𝑓 𝑥 = 𝑥 , − 3 < 𝑥 < 3 .

Solution: Points of discontinuity of 𝑓 𝑥 are −2, −1, 0, 1, 2.

Q 9) The greatest integer function is not differentiable at integral points. Give reason.

The greatest integer function is not continuous at integral points. Therefore it is not

differentiable at integral points.

Q 10) Check the continuity of the function given by 𝑓 𝑥 = 2𝑥 + 3 at 𝑥 = 1 .

Solution: 𝑓 1 = 5

lim𝑥→1 𝑓(𝑥) = lim𝑥→1 2𝑥 + 3 = 5

lim𝑥→1 𝑓(𝑥) = 𝑓 1 = 5

∴ 𝑓 𝑥 is continuous at 𝑥 = 1.

7


Q 1) Find 𝑑𝑦

𝑑𝑥 if 2𝑥 + 3𝑦 = sin 𝑦

Solution: Differentiating w r t 𝑥

2 + 3𝑑𝑦

𝑑𝑥= cos 𝑦

𝑑𝑦

𝑑𝑥

𝑑𝑦

𝑑𝑥=

2

cos 𝑦−3

Q 2) If 𝑥 = 𝑎𝑡2 , 𝑦 = 2𝑎𝑡 show that 𝑑𝑦

𝑑𝑥=

1

𝑡

Solution: 𝑑𝑥

𝑑𝑡= 2𝑎𝑡 ,

𝑑𝑦

𝑑𝑡= 2𝑎

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡

𝑑𝑥

𝑑𝑡

=2𝑎

2𝑎𝑡=

1

𝑡

Q 3) Find 𝑑𝑦

𝑑𝑥 if 𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑦 = 𝑘 where 𝑘 is a constant.


2 sin 𝑥 cos 𝑥 − 2 cos 𝑦 sin 𝑦𝑑𝑦

𝑑𝑥= 0

⟹ 𝑑𝑦

𝑑𝑥=

sin 𝑥 cos 𝑥

sin 𝑦 cos 𝑦

Q 4) Find the derivative of 𝑥 + 𝑦 = 9 at 4, 9 .


1

2 𝑥+

1

2 𝑦

𝑑𝑦

𝑑𝑥= 0

𝑑𝑦

𝑑𝑥= − 𝑦

𝑥 ⟹

𝑑𝑦

𝑑𝑥

4,9 = −

3

2

Q 5) Find 𝑑𝑦

𝑑𝑥 if 𝑥 = 4𝑡 and 𝑦 =

4

𝑡

Solution: 𝑑𝑥

𝑑𝑡= 4 ,

𝑑𝑦

𝑑𝑡= −

4

𝑡2

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡

𝑑𝑥

𝑑𝑡

= −

4

𝑡2

4= −

1

𝑡2

Q 6) Find the derivative of 3𝑥2 − 7𝑥 + 3 5

2 with respect to 𝑥.

Solution: 𝑦 = 3𝑥2 − 7𝑥 + 3 5

2

Differentiating w r t 𝑥

𝑑𝑦

𝑑𝑥=

5

2 3𝑥2 − 7𝑥 + 3

3

2(6𝑥 − 7)

Q 7) If 𝑥 + 𝑦 = 𝑎 prove that 𝑑𝑦

𝑑𝑥= −

𝑦

𝑥


1

2 𝑥+

1

2 𝑦

𝑑𝑦

𝑑𝑥= 0

8

𝑑𝑦

𝑑𝑥= − 𝑦

𝑥 = −

𝑦

𝑥

Q 8) If 𝑦 = sin log 𝑥 prove that 𝑑𝑦

𝑑𝑥=

1−𝑦2

𝑥


𝑑𝑦

𝑑𝑥=

cos log 𝑥

𝑥

= 1−𝑠𝑖𝑛 2 log 𝑥

𝑥 =

1−𝑦2

𝑥

Q 9) Find the slope of the tangent to the curve 𝑦 = 𝑥3 − 3𝑥 + 2 at a point whose 𝑥 coordinate

is 3.

Solution: 𝑑𝑦

𝑑𝑥= 3𝑥2 − 3

The slope of the tangent is 𝑑𝑦

𝑑𝑥

𝑥=3= 3 3 2 − 3 = 24

Q 10) Find the slope of the tangent to the curve 𝑦 = 3𝑥4 − 4 at 𝑥 = 4.

Solution: 𝑑𝑦

𝑑𝑥= 12𝑥3

The slope of the tangent is 𝑑𝑦

𝑑𝑥

𝑥=4= 12 4 3 = 768

Q 11) If 𝑦 = 𝑐𝑜𝑠−1 sin 𝑥 then prove that 𝑑𝑦

𝑑𝑥= −1 .

Solution: 𝑦 = 𝑐𝑜𝑠−1 cos 𝜋

2− 𝑥 =

𝜋

2− 𝑥

𝑑𝑦

𝑑𝑥= −1

Q 12) If 𝑥2 + 𝑥𝑦 + 𝑦2 = 100 find 𝑑𝑦

𝑑𝑥 .


2𝑥 + 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 + 2𝑦

𝑑𝑦

𝑑𝑥= 0

⟹ (𝑥 +2y)𝑑𝑦

𝑑𝑥= −2𝑥 − 𝑦

⟹𝑑𝑦

𝑑𝑥=

−2𝑥−𝑦

(𝑥 +2y)

Q 13) Find the derivative of cos log 𝑥 + 𝑒𝑥 with respect to 𝑥.

Solution: y = cos log 𝑥 + 𝑒𝑥


𝑑𝑦

𝑑𝑥= − sin log 𝑥 + 𝑒𝑥

1

𝑥+ 𝑒𝑥

9


Q 1) Prove that if a function is differentiable at a point ′𝑎′ then it is also continuous at that point.

Proof: Let a function 𝑓 𝑥 be differentiable at 𝑥 = 𝑎 .

∴ 𝑓 ′ 𝑎 = lim𝑥→𝑎 𝑓 𝑥 −𝑓(𝑎)

𝑥−𝑎 exists

lim𝑥→𝑎 𝑓 𝑥 − 𝑓 𝑎 = lim𝑥→𝑎 𝑓 𝑥 −𝑓 𝑎 (𝑥−𝑎)

𝑥−𝑎

= lim𝑥→𝑎 𝑓 𝑥 −𝑓(𝑎)

𝑥−𝑎 lim𝑥→𝑎(𝑥 − 𝑎)

= 𝑓 ′ 𝑎 × 0

= 0

⟹ lim𝑥→𝑎 𝑓 𝑥 − 𝑓(𝑎) = 0

⟹ lim𝑥→𝑎 𝑓 𝑥 = 𝑓(𝑎)

∴ 𝑓 𝑥 is continuous at 𝑥 = 𝑎.

Q 2) Verify mean value theorem for the function 𝑓 𝑥 = 𝑥2 − 4𝑥 − 3 in the interval [1, 4].

Solution: 𝑓 𝑥 is a polynomial function ∴ it is continuous in [1, 4] and differentiable in (1, 4).

𝑓 1 = −6, 𝑓 4 = −3, 𝑓 ′ 𝑐 =𝑓 𝑏 −𝑓(𝑎)

𝑏−𝑎

2𝑐 − 4 =−3+6

3 and 𝑐 = 2.5 ∈ 1, 4

Q 3) Verify Rolle’s Theorem for the function 𝑓 𝑥 = 𝑥2 + 2𝑥 − 8, 𝑥 ∈ −4, 2 .

Solution: 𝑓 𝑥 is a polynomial function ∴ it is continuous in −4, 2 & differentiable in(−4, 2)

and 𝑓 −4 = 𝑓 2 = 0

𝑓 ′ 𝑐 = 0 ⟹ 2𝑐 + 2 = 0

𝑐 = −1 ∈ −4, 2

Q 4) Verify Mean value theorem for the function 𝑓 𝑥 = 𝑥3 − 5𝑥2 − 3𝑥 in the interval 𝑎, 𝑏

where 𝑎 = 1 and 𝑏 = 3. Find all 𝑐 ∈ 1, 3 for which 𝑓 ′ 𝑐 = 0.

Solution: 𝑓 𝑥 is a polynomial function ∴ it is continuous in 1, 3 and differentiable in (1, 3).

𝑓 1 = −7, 𝑓 3 = −27, 𝑓 ′ 𝑐 =𝑓 𝑏 −𝑓(𝑎)

𝑏−𝑎

3𝑐2 − 10𝑐 − 3 = −10 ⟹ 3𝑐2 − 10𝑐 + 7 = 0

Solving and getting 𝑐 = 7

3∈ (1, 3) (∵ 𝑐 ≠ 1)

𝑓 ′ 𝑐 = 0 ⟹ 3𝑐2 − 10𝑐 − 3 = 0

Solve and find the roots.

∴ there exists no 𝑐 ∈ 1, 3 such that 𝑓 ′ 𝑐 = 0.

10


Q 1) Define the continuity of a function at a point. Find all the points of discontinuity of the

function defined by 𝑓 𝑥 = 𝑥 − 𝑥 + 1 .

Solution: A function 𝑓 𝑥 is said to be continuous at a point 𝑥 = 𝑎 if it satisfies the following

conditions: (i) 𝑓 𝑎 exists

(ii) lim𝑥→𝑎 𝑓(𝑥) exists

and (iii) lim𝑥→𝑎 𝑓(𝑥) = 𝑓(𝑎)

Let 𝑔 𝑥 = 𝑥 and 𝑕 𝑥 = 𝑥 + 1 . As modulus functions are continuous, 𝑔 𝑥 and 𝑕 𝑥 are

continuous.

The difference of two continuous functions is continuous. ∴ 𝑓 𝑥 is continuous.

There is no point of discontinuity.

Q 2) Find the value of 𝑘 if 𝑓 𝑥 = 𝑘𝑥 + 1, 𝑥 ≤ 53𝑥 − 5, 𝑥 > 5

is continuous at 𝑥 = 5.

Solution: Since 𝑓 𝑥 is continuous at 𝑥 = 5, lim𝑥→5 𝑓(𝑥) = 𝑓(5)

l h l = r h l

⟹ lim𝑥→5− 𝑓(𝑥) = lim𝑥→5+ 𝑓(𝑥)

⟹ lim𝑥→5−(𝑘𝑥 + 1) = lim𝑥→5+(3𝑥 − 5)

⟹ 5𝑘 + 1 = 10 ⟹ 𝑘 =9

5

Q 3) Find all points of discontinuity of 𝑓 𝑥 where 𝑓 𝑥 is defined by 𝑓 𝑥 = 𝑥3 − 3, 𝑥 ≥ 2

𝑥2 + 1, 𝑥 < 2 .

Solution: 𝑓 2 = 5

l h l = lim𝑥→2− 𝑓(𝑥) = lim𝑥→2−(𝑥2 + 1) = 5

r h l = lim𝑥→2+ 𝑓(𝑥) = lim𝑥→2+(𝑥3 − 3) = 5

l h l = r h l

lim𝑥→2 𝑓(𝑥) = 𝑓 2 = 5

∴ 𝑓 𝑥 is continuous at 𝑥 = 2.

There is no point of discontinuity.

Q 4) Find the relationship between 𝑎 and 𝑏 so that the function defined by

𝑓 𝑥 = 𝑎𝑥 + 1, 𝑥 ≤ 3𝑏𝑥 + 3, 𝑥 > 3

is continuous at 𝑥 = 3.

Solution: Since 𝑓 𝑥 is continuous at 𝑥 = 3, lim𝑥→3 𝑓(𝑥) exists.

l h l = r h l

⟹ lim𝑥→3− 𝑓 𝑥 = lim𝑥→3+ 𝑓(𝑥)

⟹ 3𝑎 + 1 = 3𝑏 + 3

⟹ 3𝑎 − 3𝑏 = 2 ⟹ 𝑎 − 𝑏 =2

3

Q 5) Find the value of 𝑘 if 𝑓 𝑥 = 𝑘𝑥2 , 𝑥 ≤ 2

3, 𝑥 > 2 is continuous at 𝑥 = 2.

Solution: Since 𝑓 𝑥 is continuous at 𝑥 = 2, lim𝑥→2 𝑓(𝑥) = 𝑓(2)

l h l = r h l

⟹ lim𝑥→2− 𝑓(𝑥) = lim𝑥→2+ 𝑓(𝑥)

⟹ lim𝑥→2−(𝑘𝑥2) = lim𝑥→2+(3)

⟹ 4𝑘 = 3 ⟹ 𝑘 =3

4

11

Q 6) Find the value of 𝑘 if 𝑓 𝑥 = 𝑘𝑥 + 1 𝑖𝑓 𝑥 ≤ 𝜋𝑐𝑜𝑠𝑥 𝑖𝑓 𝑥 > 𝜋

is continuous at 𝑥 = 𝜋.

Solution: Since 𝑓 𝑥 is continuous at 𝑥 = 𝜋, lim𝑥→𝜋 𝑓(𝑥) = 𝑓(𝜋)

l h l = r h l

⟹ lim𝑥→𝜋− 𝑓(𝑥) = lim𝑥→𝜋+ 𝑓(𝑥)

⟹ lim𝑥→𝜋−(𝑘𝑥 + 1) = lim𝑥→𝜋+(𝑐𝑜𝑠 𝜋)

⟹ 𝑘𝜋 + 1 = −1 ⟹ 𝑘 =−2

𝜋


Q 1) If 𝑦 = 3 cos 𝑙𝑜𝑔𝑥 + 4 sin 𝑙𝑜𝑔𝑥 then show that 𝑥2 𝑦2 + 𝑥 𝑦1 + 𝑦 = 0.

Solution: Differentiating w r t 𝑥 :

𝑦1 = −3sin log 𝑥

𝑥+

4 cos log 𝑥

𝑥

⟹ 𝑥𝑦1 = −3sin log 𝑥 + 4 cos log 𝑥


𝑥𝑦2 + 𝑦1 =−3cos log 𝑥

𝑥−

4 sin log 𝑥

𝑥

⟹ 𝑥(𝑥𝑦2 + 𝑦1) = −(3 cos 𝑙𝑜𝑔𝑥 + 4 sin 𝑙𝑜𝑔𝑥 )

⟹ 𝑥2 𝑦2 + 𝑥 𝑦1 + 𝑦 = 0

Q 2) If 𝑦 = 𝑐𝑜𝑠−1𝑥 2 then prove that 1 − 𝑥2 𝑦2 − 𝑥𝑦1 − 2 = 0.

Solution: Differentiating w r t 𝑥:

𝑦1 = − 2 𝑐𝑜𝑠−1𝑥

1−𝑥2

⟹ ( 1 − 𝑥2 ) 𝑦1 = −2 𝑐𝑜𝑠−1𝑥


⟹ ( 1 − 𝑥2 ) 𝑦2 + 𝑦1(−2𝑥)

2 1−𝑥2=

2

1−𝑥2

Taking LCM and simplifying

⟹ 1 − 𝑥2 𝑦2 − 𝑥𝑦1 − 2 = 0

Q 3) If 𝑦 = 𝑡𝑎𝑛−1𝑥 2 then show that 𝑥2 + 1 2 𝑦2 + 2𝑥 𝑥2 + 1 𝑦1 = 2.


𝑦1 = 2 𝑡𝑎𝑛 −1𝑥

(1+𝑥2)

⟹ (1 + 𝑥2 ) 𝑦1 = 2 𝑡𝑎𝑛−1𝑥


⟹ (1 + 𝑥2 ) 𝑦2 + 𝑦12𝑥

(1+𝑥2)=

2

(1+𝑥2)


𝑥2 + 1 2 𝑦2 + 2𝑥 𝑥2 + 1 𝑦1 = 2

Q 4) If 𝑦 = 3𝑒2𝑥 + 2𝑒3𝑥 then prove that 𝑑2𝑦

𝑑𝑥2 − 5𝑑𝑦

𝑑𝑥+ 6𝑦 = 0.


𝑑𝑦

𝑑𝑥= 6𝑒2𝑥 + 6𝑒3𝑥

Differentiating w r t 𝑥:

𝑑2𝑦

𝑑𝑥2 = 12𝑒2𝑥 + 18𝑒3𝑥

12

𝑑2𝑦

𝑑𝑥2− 5

𝑑𝑦

𝑑𝑥+ 6𝑦 = 12𝑒2𝑥 + 18𝑒3𝑥 − 5 6𝑒2𝑥 + 6𝑒3𝑥 + 6(3𝑒2𝑥 + 2𝑒3𝑥)

= 12𝑒2𝑥 + 18𝑒3𝑥 − 30𝑒2𝑥 − 30𝑒3𝑥 + 18𝑒2𝑥 + 12𝑒3𝑥

= 0

Q 5) If 𝑦 = 𝑠𝑖𝑛−1𝑥 then prove that 1 − 𝑥2 𝑦2 − 𝑥𝑦1 = 0.


𝑦1 =1

1−𝑥2

( 1 − 𝑥2 ) 𝑦1 = 1


( 1 − 𝑥2 ) 𝑦2 + 𝑦1(−2𝑥)

2 1−𝑥2= 0


1 − 𝑥2 𝑦2 − 𝑥𝑦1 = 0

Q 6) If 𝑦 = 𝐴𝑒𝑚𝑥 + 𝐵𝑒𝑛𝑥 then prove that 𝑦2 − 𝑚 + 𝑛 𝑦1 + 𝑚𝑛𝑦 = 0.


𝑦1 = 𝐴𝑚𝑒𝑚𝑥 + 𝐵𝑛𝑒𝑛𝑥


𝑦2 = 𝐴𝑚2𝑒𝑚𝑥 + 𝐵𝑛2𝑒𝑛𝑥

𝑦2 − 𝑚 + 𝑛 𝑦1 + 𝑚𝑛𝑦

= 𝐴𝑚2𝑒𝑚𝑥 + 𝐵𝑛2𝑒𝑛𝑥 − 𝑚 + 𝑛 𝐴𝑚𝑒𝑚𝑥 + 𝐵𝑛𝑒𝑛𝑥 + 𝑚𝑛(𝐴𝑒𝑚𝑥 + 𝐵𝑒𝑛𝑥 )

= 𝐴𝑚2𝑒𝑚𝑥 + 𝐵𝑛2𝑒𝑛𝑥 − 𝐴𝑚2𝑒𝑚𝑥 − 𝐴𝑚𝑛𝑒𝑚𝑥 − 𝐵𝑚𝑛𝑒𝑛𝑥 − 𝐵𝑛2𝑒𝑛𝑥 + 𝐴𝑚𝑛𝑒𝑚𝑥 + 𝐵𝑚𝑛𝑒𝑛𝑥

= 0

Q 7) If 𝑒𝑦 𝑥 + 1 = 1, prove that 𝑑2𝑦

𝑑𝑥2 = 𝑑𝑦

𝑑𝑥

2

.


𝑒𝑦 + 𝑥 + 1 𝑒𝑦 𝑑𝑦

𝑑𝑥= 0

⟹ 𝑒𝑦 + 𝑑𝑦

𝑑𝑥= 0

⟹𝑑𝑦

𝑑𝑥= −𝑒𝑦


𝑑2𝑦

𝑑𝑥2= −𝑒𝑦 𝑑𝑦

𝑑𝑥

⟹ 𝑑2𝑦

𝑑𝑥2=

𝑑𝑦

𝑑𝑥

𝑑𝑦

𝑑𝑥

⟹𝑑2𝑦

𝑑𝑥2 = 𝑑𝑦

𝑑𝑥

2

13

INTEGRALS


Q 1) Find the integral of 1

𝑥2−𝑎2

Solution: 1

𝑥2−𝑎2 =

1

𝑥−𝑎 𝑥+𝑎 =

1

2𝑎

𝑥+𝑎 −(𝑥−𝑎)

𝑥−𝑎 (𝑥+𝑎) =

1

2𝑎

1

𝑥−𝑎 −

1

𝑥+𝑎

∴ 𝑑𝑥

𝑥2−𝑎2 = 1

2𝑎

1

𝑥−𝑎 𝑑𝑥 −

1

2𝑎

1

𝑥+𝑎 𝑑𝑥 =

1

2𝑎 log 𝑥 − 𝑎 − log | (𝑥 + 𝑎) + c

∴ 𝑑𝑥

𝑥2−𝑎2 = 1

2𝑎 log

𝑥−𝑎

𝑥+𝑎 + C


𝑎2−𝑥2

Solution: 1

𝑎2−𝑥2 = 1

𝑎−𝑥 𝑎+𝑥 =

1

2𝑎

𝑎+𝑥 +(𝑎−𝑥)

𝑎−𝑥 (𝑎+𝑥) =

1

2𝑎

1

𝑎−𝑥+

1

𝑎+𝑥

∴ 1

𝑎2−𝑥2 = 1

2𝑎

1

𝑎−𝑥 𝑑𝑥 +

1

2𝑎

1

𝑎+𝑥 𝑑𝑥 =

1

2𝑎 − log 𝑎 − 𝑥 + log | (𝑎 + 𝑥) + C

∴ 𝑑𝑥

𝑎2−𝑥2 = 1

2𝑎 log

𝑎+𝑥

𝑎−𝑥 + C

Q 3) Find the integral of 𝑑𝑥

𝑥2−𝑎2 .

Solution: Put 𝑥 = 𝑎 sec 𝜃 then 𝑑𝑥 = 𝑎 𝑠𝑒𝑐 𝜃 tan 𝜃 𝑑𝜃

∴ 𝑑𝑥

𝑥2−𝑎2 =

𝑎 sec 𝜃 tan 𝜃 𝑑𝜃

𝑎2𝑠𝑒𝑐 2𝜃−𝑎2 = sec 𝜃 𝑑𝜃 = log | sec 𝜃 + tan 𝜃 | + C

= log 𝑥

𝑎+

𝑥2

𝑎2 − 1 + C = log 𝑥 + 𝑥2 − 𝑎2 + C



𝑎2+𝑥2 and hence evaluate 𝑑𝑥

𝑥2− 6𝑥+13

Solution: Put 𝑥 = 𝑎 tan 𝜃 then 𝑑𝑥 = 𝑎 𝑠𝑒𝑐 2 𝜃 𝑑𝜃

∴ 1

𝑎2+𝑥2 dx = 𝑎 𝑠𝑒𝑐 2𝜃𝑑𝜃

𝑎2𝑡𝑎𝑛 2𝜃+𝑎2 = 1

𝑎 𝑑 𝜃 =

1

𝑎 tan−1 𝑥

𝑎+ C

To evaluate 𝑑𝑥

𝑥2− 6𝑥+13

𝑥2 − 6𝑥 + 13 = 𝑥2 − 6𝑥 +32 − 32 + 13 = 𝑥 − 3 2 + 4

14

∴ 𝑑𝑥

𝑥2− 6𝑥+13 =

1

(𝑥−3)2+ 22 dx =

1

4 tan−1 𝑥−3

2+ C


𝑎2−𝑥2 and hence evaluate

1

7−6𝑥−𝑥2 dx

Solution: Put 𝑥 = 𝑎 sin 𝜃 𝑑𝑥 = 𝑎 cos 𝜃 𝑑𝜃

∴ 𝑑𝑥

𝑎2−𝑥2 =

𝑎 cos 𝜃 𝑑𝜃

𝑎2−𝑎2𝑠𝑖𝑛 2𝜃 = 𝑑𝜃 = 𝜃 + C = sin−1 𝑥

𝑎+ C

To evaluate 1

7−6𝑥−𝑥2 dx

7 − 6𝑥 − 𝑥2 = − ( 𝑥2 + 6𝑥 − 7) = − 𝑥2 + 6𝑥 + 32 − 32 − 7 = −[(𝑥 + 3)2 − 16]

= 42 − (𝑥 + 3)2

∴ 1

7−6𝑥−𝑥2 dx =

1

42− (𝑥+3)2 dx = sin−1 (𝑥+3)

4 + C


𝑥2+𝑎2 and hence evaluate

1

2+2𝑥+𝑥2 dx

Solution: Put 𝑥 = 𝑎 tan 𝜃 then 𝑑𝑥 = 𝑎 𝑠𝑒𝑐 2 𝜃 𝑑𝜃

∴ 𝑑𝑥

𝑥2+𝑎2 =

𝑎𝑠𝑒𝑐 2𝜃 𝑑𝜃

𝑎2𝑡𝑎𝑛 2𝜃+𝑎2 = sec 𝜃 𝑑𝜃 = log | sec 𝜃 + tan 𝜃 | + C

= log 𝑥

𝑎+

𝑥2

𝑎2 + 1 + C = log 𝑥 + 𝑥2 + 𝑎2 + C

To evaluate 1

2+2𝑥+𝑥2 dx

2 + 2𝑥 + 𝑥2 = 𝑥2 + 2𝑥 + 2 = 𝑥2 + 2𝑥 + 12 − 12 + 2 = (𝑥 + 1)2 + 1

∴ 1

2+2𝑥+𝑥2 dx =

1

(𝑥+1)2+ 1

= log (𝑥 + 1) + (𝑥 + 1)2 + 1 + C

Q 7) Find the integral of 𝑥2 + 𝑎2 and hence evaluate 4𝑥2 + 32 dx

Solution: Let 𝐼 = 𝑥2 + 𝑎2 𝑑𝑥 = 𝑥2 + 𝑎2 .1 𝑑𝑥

= 𝑥2 + 𝑎2 . 𝑥 − 𝑥 1

2 𝑥2+𝑎2 2𝑥 𝑑𝑥

= 𝑥 𝑥2 + 𝑎2 − 𝑥2

𝑥2+𝑎2 𝑑𝑥

= 𝑥 𝑥2 + 𝑎2 − 𝑥2+𝑎2−𝑎2


= 𝑥 𝑥2 + 𝑎2 − 𝑥2+𝑎2

𝑥2+𝑎2 𝑑𝑥 +

𝑎2


= 𝑥 𝑥2 + 𝑎2 − 𝑥2 + 𝑎2 𝑑𝑥 + 𝑎2 log 𝑥 + 𝑥2 + 𝑎2 + C

15

= 𝑥 𝑥2 + 𝑎2 − 𝐼 + 𝑎2 log 𝑥 + 𝑥2 + 𝑎2 + C

2𝐼 = 𝑥 𝑥2 + 𝑎2 + 𝑎2 log 𝑥 + 𝑥2 + 𝑎2 + C

∴ 𝐼 = 𝑥

2 𝑥2 + 𝑎2 +

𝑎2

2log 𝑥 + 𝑥2 + 𝑎2 + C

To evaluate 4𝑥2 + 32 dx

4𝑥2 + 32 𝑑𝑥 = 2 𝑥2 + 3

2

2

𝑑𝑥 = 𝑥 𝑥2 + 3

2

2

+ 9

4log 𝑥 + 𝑥2 +

3

2

2

+ C

Q 8) Find the integral of 𝑥2 − 𝑎2 and hence evaluate 𝑥2 + 4𝑥 − 5 dx

Solution: Let 𝐼 = 𝑥2 − 𝑎2 𝑑𝑥 = 𝑥2 − 𝑎2 .1 𝑑𝑥

= 𝑥2 − 𝑎2 . 𝑥 − 𝑥 1

2 𝑥2−𝑎2 2𝑥 𝑑𝑥

= 𝑥 𝑥2 − 𝑎2 − 𝑥2

𝑥2−𝑎2 𝑑𝑥

= 𝑥 𝑥2 − 𝑎2 − 𝑥2−𝑎2+𝑎2

𝑥2−𝑎2 𝑑𝑥 = 𝑥 𝑥2 − 𝑎2 −

𝑥2−𝑎2

𝑥2−𝑎2 𝑑𝑥 −

𝑎2

𝑥2−𝑎2 𝑑𝑥

= 𝑥 𝑥2 − 𝑎2 − 𝑥2 − 𝑎2 𝑑𝑥 − 𝑎2 log 𝑥 + 𝑥2 − 𝑎2 + C

= 𝑥 𝑥2 − 𝑎2 − 𝐼 − 𝑎2 log 𝑥 + 𝑥2 − 𝑎2 + C

2𝐼 = 𝑥 𝑥2 − 𝑎2 − 𝑎2 log 𝑥 + 𝑥2 − 𝑎2 + C

∴ 𝐼 = 𝑥

2 𝑥2 − 𝑎2 −

𝑎2

2log 𝑥 + 𝑥2 − 𝑎2 + C

To evaluate 𝑥2 + 4𝑥 − 5 𝑑𝑥

By using complete the square method: 𝑥2 + 4𝑥 − 5 = 𝑥2 + 4𝑥 + 22 − 22 − 5

= (𝑥 + 2)2 −4 − 5

= (𝑥 + 2)2 − 32

∴ 𝑥2 + 4𝑥 − 5 𝑑𝑥

= (𝑥 + 2)2 − 32 𝑑𝑥

= (𝑥+2)

2 (𝑥 + 2)2 − 32 −

32

2log ( 𝑥 + 2) + (𝑥 + 2)2 − 32 + C

Q 9) Find the integral of 𝑎2 − 𝑥2 and hence evaluate 1 − 4𝑥 − 𝑥2 𝑑𝑥

Solution: Let 𝐼 = 𝑎2 − 𝑥2 𝑑𝑥 = 𝑎2 − 𝑥2 .1 𝑑𝑥

= 𝑎2 − 𝑥2 . 𝑥 − 𝑥 1

2 𝑎2−𝑥2 . (−2𝑥) 𝑑𝑥

= 𝑥 𝑎2 − 𝑥2 − −𝑥2

𝑎2−𝑥2 𝑑𝑥

= 𝑥 𝑎2 − 𝑥2 − 𝑎2−𝑥2−𝑎2

𝑎2−𝑥2 𝑑𝑥 = 𝑥 𝑎2 − 𝑥2 −

𝑎2−𝑥2

𝑎2−𝑥2 𝑑𝑥 +

𝑎2

𝑎2−𝑥2 𝑑𝑥

16

= 𝑥 𝑎2 − 𝑥2 − 𝑎2 − 𝑥2 𝑑𝑥 + 𝑎2 log 𝑥 + 𝑎2 − 𝑥2 + C

= 𝑥 𝑎2 − 𝑥2 − 𝐼 + 𝑎2 log 𝑥 + 𝑎2 − 𝑥2 + C

2𝐼 = 𝑥 𝑎2 − 𝑥2 + 𝑎2 log 𝑥 + 𝑎2 − 𝑥2 + C

∴ 𝐼 = 𝑥

2 𝑎2 − 𝑥2 +

𝑎2

2log 𝑥 + 𝑎2 − 𝑥2 + C

To evaluate 1 − 4𝑥 − 𝑥2 𝑑𝑥

By using complete the square method: 1 − 4𝑥 − 𝑥2 = − (𝑥2 + 4𝑥 − 1)

= − (𝑥2 + 4𝑥 + 22 − 22 − 1)

= −[(𝑥 + 2)2 − ( 5)2

]

= ( 5)2 − (𝑥 + 2)2

∴ 𝑥2 + 4𝑥 − 5 𝑑𝑥

= 52

− (𝑥 + 2)2 𝑑𝑥

= (𝑥+2)

2 5

2− (𝑥 + 2)2 +

52

2log (𝑥 + 2) + 5

2− (𝑥 + 2)2 + C

LINEAR PROGRAMMING


Q 1) Define linear objective function in Linear Programming Problem.

Linear function 𝑧 = 𝑎𝑥 + 𝑏𝑦 where 𝑎 and 𝑏 are constants which has to be maximized or

minimized is called a linear objective function.

Q 2) Define the term constraints in LPP.

The linear inequalities or restrictions on the variables of a Linear Programming Problem are

called constraints.

Q 3) Define the term corner point in the LPP.

A corner point of a feasible region is a point in the region which is the intersection of two

boundary lines.

Q 4) Define optimal solution in Linear Programming Problem.

Any point in the feasible region that gives the optimal value (maximum or minimum) of the

objective function is called an optimal solution.

Q 5) Define feasible region in Linear Programming Problem.

The common region determined by all the constraints including non-negative constraints

𝑥, 𝑦 ≥ 0 of a linear programming problem is called the feasible region.

17

THE FOLLOWING QUESTION CARRIES 6 MARKS:

Q 1) A company owned by shree group produces two products P and Q. Each P requires 4 hours of

grinding and 2 hours of polishing and each Q requires 2 hours of grinding and 5 hours of

polishing. The total available hours for grinding are Rs 20 and for polishing is Rs. 24. Profit

per unit of P is Rs.6 and that of Q is Rs.8. Formulate L.P.P and solve graphically.

Solution:

Let ‘x’ be the units of P manufactured and ‘y’ be the units of Q manufactured. Consider

table formed

Products Grinding hours Polishing hours Profir per unit

P 4 2 6

Q 2 5 8

Availability of time 20 24

Formulation of L.P.P

Maximize 𝑧 = 6𝑥 + 8𝑦

Subject to constraints 4𝑥 + 2𝑦 ≤ 20

2𝑥 + 5𝑦 ≤ 24

𝑥, 𝑦 ≥ 0

Consider the equation 4𝑥 + 2𝑦 = 20

X 0 5

Y 10 0

∴ The points are (0, 10) and (5, 0)

Consider the equation 2𝑥 + 5𝑦 = 24

X 0 12

Y 4.8 0

∴ The points are (0, 4.8) and (12, 0)

Using the obtained points we can draw the graph of lines separately

X axis 1cm=5 unit

Y axis 1 cm =5unit

15

10 feasible region

5A B

18

O C 5 10 15 20 25

At O (0, 0) Z = 6(0) + 8(0) = 0

At A (0, 4.8) Z = 6(0) + 8(4.8) = 38.4

At B (3.3, 3.5) Z = 6(3.3) + 8(3.5) =47.8

At C (5, 0) Z = 6(5) + 8(0) = 30

Thus the optimal solution of the graphical problem is obtained as

When x = 3.3 and y = 3.5 the maximum value of Z= 47.8

MATRICES


1. Define a diagonal matrix.

Ans. A square matrix is called a diagonal matrix if all the elements, except those in the diagonal are

zero.

2. Construct a 2×3 matrix whose elements are given by 𝑎𝑖𝑗 = 𝑖 − 𝑗 .

Ans. 0 1 21 0 1

3. What is the number of the possible square matrices of order 3 with each entry zero or one?

Ans. 29 = 512.

4. Define scalar matrix.

Ans. a diagonal matrix whose diagonal elements are equal is said to be a scalar matrix.


1. Express 1 23 4

as the sum of symmetric and skew symmetric matrices.

Ans: 1

2 𝐴 + 𝐴′ =

1

2 2 55 8

1

2 𝐴 − 𝐴′ =

1

2 0 −11 0

𝐴 =1

2 2 55 8

+ 1

2 0 −11 0

2. Using elementary transformations, find the inverse of the matrix. A= 1 −22 1

Ans: 𝐴−1= 1/5 2/5

−2/5 1/5

3. If A and B are square matrices of the same order then show that 𝐴𝐵 −1 = 𝐵−1𝐴−1.

Ans: Let (𝐴𝐵) 𝐴𝐵 −1 = 𝐼

𝐴−1 𝐴𝐵 𝐴𝐵 −1 = 𝐴−1𝐼

(𝐴−1𝐴)𝐵 𝐴𝐵 −1 = 𝐴−1

𝐼𝐵 𝐴𝐵 −1 = 𝐴−1

𝐵−1𝐵 𝐴𝐵 −1 = 𝐵−1𝐴−1

𝐴𝐵 −1 = 𝐵−1𝐴−1

19

4. Find the value of x, y and z in the following matrices.

𝑥 + 𝑦 25 + 𝑧 𝑥𝑦

= 6 25 8

Ans: 𝑥 + 𝑦 = 6 , 𝑥𝑦 = 8 , 5 + 𝑧 = 5

𝑥 = 4,2 𝑎𝑛𝑑 𝑦 = 2,4.

5. Find the value of x and y in 𝑥 + 2𝑦 2

4 𝑥 + 𝑦 −

3 24 1

= 𝑂 where O is a null matrix.

Ans: 𝑥 + 2𝑦 − 3 = 0, 𝑥 + 𝑦 − 1 = 0 , 𝑥 = −1 𝑎𝑛𝑑 𝑦 = 2.


1. If 𝐴 = 1 22 1

, 𝐵 = 2 01 3

𝑎𝑛𝑑 𝐶 = 1 12 3

. calculate AB, AC and A (B+C). Verify that

AB+AC=A (B+C).

Ans: AB= 4 65 3

, AC= 5 74 5

, B+C= 3 13 6

, A (B+C) = 9 139 8

l.h.s: AB+AC = 9 139 8

r.h.s: A (B+C) = 9 139 8

l.hs=r.h.s

2. If 𝐴 = −245

𝑎𝑛𝑑 𝐵 = 1 3 −6 𝑣𝑒𝑟𝑖𝑓𝑦 𝑡𝑕𝑎𝑡 (𝐴𝐵)′ = 𝐵′𝐴′ .

Ans: 𝐴𝐵 ′ = −2 4 5−6 12 1512 −24 −30

𝐵′𝐴′ = −2 4 5−6 12 1512 −24 −30

3. If 𝐴 = 2 0 10 −3 00 0 4

verify A3-3A

2-10A+24I=O, where O is a zero matrix of order 3X3.

Ans: A2=

4 0 60 9 00 0 16

, A3=

8 0 280 −27 00 0 64

, 3A2=

12 0 180 27 00 0 48

, 10A= 20 0 100 −30 00 0 40

,

24I= 24 0 00 24 00 0 24

, A3-3A

2-10A+24I=

0 0 00 0 00 0 0

.

DETERMINANTS


1. Find the value of x for which 3 𝑥𝑥 1

= 3 24 1

Ans: 𝑥 = ± 8

2. If A= 1 24 2

, find 2𝐴

20

Ans: 2𝐴 =-24.

3. If A is a square matrix with 𝐴 =6, find the value of 𝐴𝐴′

Ans: 6×6=36


1. Find the area of the triangle whose vertices are (3, 8), (-4, 2) and (5, 1) using determinants.

Ans: 1

2

3 8 1−4 2 15 1 1

=61

2 Sq units

2. If each element of a row is expressed as sum of two elements then verify for a third order

determinant that the determinant can be expressed as sum of two determinants.

Ans. As given in the text book.

3. Find the equation of the line passing through (1,2) and (3,6) using determinants.

Ans: 1

2 1 2 13 6 1𝑥 𝑦 1

=0 ⟹ 4x-2y=0

4. If the area of the triangle with vertices (2,-6), (5,4) and (k,4) is 35 sq units find the value of k

using determinant method.

Ans: 1

2 2 −6 15 4 1𝑘 4 1

= ±35, k= -2 and k=12.


Solve by matrix method: 𝑥 − 𝑦 + 2𝑧 = 3, 2𝑥 + 𝑧 = 1, 3𝑥 + 2𝑦 + 𝑧 = 4

Solution:

The system of equation can be written in matrix form AX=B where

A= 1 −1 22 0 13 2 1

, X= 𝑥𝑦𝑧 , B=

314

𝐴 = 1(0 - 2) + 1(2 - 3) + 2(4 - 0)

= 1(-2) + 1(-1) + 2(4)

= -2 - 1 + 8

= 5 ≠ 0 ∴ 𝐴−1 exists

adjA=

+

0 12 1

− 2 13 1

+ 2 03 2

− −1 22 1

+ 1 23 1

− 1 −13 2

+ −1 20 1

− 1 22 1

+ 1 −12 0

𝑇

21

adjA= −2 1 45 −5 −5

−1 3 2

𝑇

adjA= −2 5 −11 −5 34 −5 2

∴ 𝐴−1 = 1

𝐴 adjA

= 1

5 −2 5 −11 −5 34 −5 2

Solution is X = 𝐴−1B

= 1

5

−2 5 −11 −5 34 −5 2

314

= 1

5 −6 + 5 − 43 − 5 + 1212 − 5 + 8

X= 1

5 −51015

𝑥𝑦𝑧 =

−123

∴ 𝑥 = −1, 𝑦 = 2, 𝑧 = 3.

22

VECTORS


1. If the vectors 2 î +3ĵ-6𝑘 and 4î-mĵ-12𝑘 are parallel find m.

Ans: 𝑖 𝑗 𝑘

2 3 −64 −𝑚 −12

= 0 , 𝑚 = −6.

2. Write the vector joining the points A (2, 3, 0) and B (-1,-2,-4)

Ans: 𝐴𝐵 = 𝑂𝐵 − 𝑂𝐴 = −3𝐼 − 5𝐽 − 4𝑘 .

3. Define collinear vectors.

Ans: Two or more vectors are said to be collinear vectors if they lie on the same support line or

parallel support lines.

4. Find the direction ratios of the vectors, joining the points P(2,3,0) and Q(-1,-2,-3) direction from

P to Q.

Ans: 𝑃𝑄 = 𝑂𝑄 − 𝑂𝑃 = -3î-5ĵ-3𝑘

Direction ratios are -3,-5 and -3.

5. Find the unit vector in the direction of the vector 𝑎 = 𝑖 + 𝑗 + 2𝑘 .

Ans: 𝑎 =𝑎

𝑎 =

𝑖

6+

𝑗

6+

2𝑘

6.

6. Find the angle between the two vectors 𝑎 and 𝑏 such that 𝑎 = 1, 𝑏 =1 and 𝑎 .𝑏 =1.

Ans: 𝜃=cos−1 𝑎 .𝑏

𝑎 𝑏 =0°

7. If 𝑎 is a non zero vector of magnitude 𝑎 and 𝜆𝑎 is a unit vector, find the value of 𝜆.

Ans: Unit vector is defined as a vector whose magnitude is 1. ∴ 𝜆= ±1

𝑎

8. Find the vector components of the vector with initial point (2,1) and terminal point (-5,7).

Ans. Vector components are -7𝑖 and 6𝑗


1. If the position vectors of the points A and B respectively are 𝑖 +2𝑗 -3𝑘 and 𝑗 -𝑘 find the direction

cosines of 𝐴𝐵 .

Ans: direction cosines −1

6 ,

−1

6,

2

6.

2. Find a vector of magnitude 8units in the direction of the vector 𝑎 =5𝑖 -𝑗 +2𝑘 .

Ans: Required vector =8.𝑎

=40𝑖

30 -

8𝑗

30 +

16𝑘

30.

3. If 𝑎 is a unit vector such that (𝑥 - 𝑎 ).(𝑥 +𝑎 )=8 find 𝑥 .

Ans: 𝑥 2 − 𝑎 2 = 8, 𝑥 = 3.

4. Show that the vector 𝑖 + 𝑗 + 𝑘 is equally inclined to the positive direction of the axes.

23

Ans; cos 𝛼 = ±1

3 cos 𝛽 = ±

1

3 cos 𝛾 = ±

1

3.

∴ 𝛼 = 𝛽 = 𝛾

5. Find k if 𝑖 +3𝑗 +𝑘 , 2𝑖 -𝑗 -𝑘 and k𝑖 + 7𝑗 + 3𝑘 are coplanar.

Ans: 1 3 12 −1 −1𝑘 7 3

= 0 , 𝑘 = 0.

6. Find the area of the parallelogram whose adjacent sides are the vectors 3𝑖 +𝑗 +4𝑘 and 𝑖 -𝑗 +𝑘 .

Ans: 𝑎 × 𝑏 = 𝑖 𝑗 𝑘

3 1 41 −1 1

= 5𝑖 + 𝑗 − 4𝑘 , 𝑎 × 𝑏 = 42 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠.

7. If 𝑎 = 5𝑖 -𝑗 -3𝑘 and 𝑏 = 𝑖 + 3𝑗 − 5𝑘 then show that the vectors 𝑎 + 𝑏 and 𝑎 − 𝑏 are

perpendicular.

Ans: (𝑎 + 𝑏 ). 𝑎 − 𝑏 = 24 − 8 − 16 = 0.

8. Find the projection of the vector 𝑖 + 3𝑗 + 7𝑘 𝑖𝑛 𝑡𝑕𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 7𝑖 − 𝑗 + 8𝑘 .

Ans: 𝑎 . 𝑏 =60

114

9. Show that the points A(2,-1,1), B(1,-3,-5) and C(3,-4,-4) are vertices of a right angled triangle.

Ans: 𝐴𝐵 = −𝑖 − 2𝑗 − 6𝑘

𝐵𝐶 = 2𝑖 − 𝑗 + 𝑘

𝐶𝐴 = −𝑖 + 3𝑗 + 5𝑘 𝐴𝐵 = 41 , 𝐵𝐶 = 6, 𝐶𝐴 = 35.

𝐴𝐵 2= 𝐵𝐶

2+ 𝐶𝐴

2.

10. Find the angle 𝜃 between the vectors 𝑎 = 𝑖 +𝑗 -𝑘 and 𝑏 = 𝑖 -𝑗 +𝑘 .

Ans: cos 𝜃 =𝑎 .𝑏

𝑎 𝑏 = −

1

3 = cos−1 −

1

3 .

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RELATIONS AND FUNCTIONS EACH OF THE FOLLOWING QUESTIONS CARRIES 1...

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Transcript of RELATIONS AND FUNCTIONS EACH OF THE FOLLOWING QUESTIONS CARRIES 1...