Mathematical Logic Class 12 Maths 1 Exercise 1.3 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.3 Questions And Answers Maharashtra Board

Question 1.
If A = {3, 5, 7, 9, 11, 12}, determine the truth value of each of the following.
(i) Ǝ x ∈ A such that x – 8 = 1
Solution:
Clearly x = 9 ∈ A satisfies x – 8 = 1. So the given statement is true, hence its truth value is T.

(ii) Ɐ x ∈ A, x2 + x is an even number
Solution:
For each x ∈ A, x2 + x is an even number. So the given statement is true, hence its truth value is T.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) Ǝ x ∈ A such that x2 < 0
Solution:
There is no x ∈ A which satisfies x2 < 0. So the given statement is false, hence its truth value is F.

(iv) Ɐ x ∈ A, x is an even number
Solution:
x = 3 ∈ A, x = 5 ∈ A, x = 7 ∈ A, x = 9 ∈ A, x = 11 ∈ A do not satisfy x is an even number. So the given statement is false, hence its truth value is F.

(v) Ǝ x ∈ A such that 3x + 8 > 40
Solution:
Clearly x = 11 ∈ A and x = 12 ∈ A satisfies 3x + 8 > 40. So the given statement is true, hence its truth value is T.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) Ɐ x ∈ A, 2x + 9 > 14
Solution:
For each x ∈ A, 2x + 9 > 14. So the given statement is true, hence its truth value is T.

Question 2.
Write the duals of each of the following.
(i) p ∨ (q ∧ r)
Solution:
The duals of the given statement patterns are :
p ∧ (q ∨ r)

(ii) p ∧ (q ∧ r)
Solution:
p ∨ (q ∨ r)

(iii) (p ∨ q) ∧ (r ∨ s)
Solution:
(p ∧ q) ∨ (r ∧ s)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) p ∧ ~q
Solution:
p ∨ ~q

(v) (~p ∨ q) ∧ (~r ∧ s)
Solution:
(~p ∧ q) ∨ (~r ∨ s)

(vi) ~p ∧ (~q ∧ (p ∨ q) ∧ ~r)
Solution:
~p ∨ (~q ∨ (p ∧ q) ∨ ~r)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vii) [~(p ∨ q)] ∧ [p ∨ ~(q ∧ ~s)]
Solution:
[ ~(p ∧ q)] ∨ [p ∧ ~(q ∨ ~s)]

(viii) c ∨ {p ∧ (q ∨ r)}
Solution:
t ∧ {p ∧ (q Ar)}

(ix) ~p ∨ (q ∧ r) ∧ t
Solution:
~p ∧ (q ∨ r) ∨ c

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(x) (p ∨ q) ∨ c
Solution:
(p ∧ q) ∧ t

Question 3.
Write the negations of the following.
(i) x + 8 > 11 or y – 3 = 6
Solution:
Let p : x + 8 > 11, q : y — 3 = 6.
Then the symbolic form of the given statement is p ∨ q.
Since ~(p ∨ q) ≡ ~p ∧ ~q, the negation of given statement is :
‘x + 8 > 11 and y – 3 ≠ 6’ OR
‘x + 8 ≮ 11 and y – 3 ≠ 6’

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) 11 < 15 and 25 > 20
Solution:
Let p: 11 < 15, q : 25 > 20.
Then the symbolic form of the given statement is p ∧ q.
Since ~(p ∧ q) ≡ ~p ∨ ~q, the negation of given statement is :
’11 ≮ 15 or 25 > 20.’ OR
’11 ≯ 15 or 25 ≮ 20.’

(iii) Qudrilateral is a square if and only if it is a rhombus.
Solution:
Let p : Quadrilateral is a square.
q : It is a rhombus.
Then the symbolic form of the given statement is p ↔ q.
Since ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p), the negation of given statement is :
‘ Quadrilateral is a square but it is not a rhombus or quadrilateral is a rhombus but it is not a square.’

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) It is cold and raining.
Solution:
Let p : It is cold.
q : It is raining.
Then the symbolic form of the given statement is p ∧ q.
Since ~(p ∧ q) ≡ ~p ∨ ~q, the negation of the given statement is :
‘It is not cold or not raining.’

(v) If it is raining then we will go and play football.
Solution:
Let p : It is raining.
q : We will go.
r : We play football.
Then the symbolic form of the given statement is p → (q ∧ r).
Since ~[p → (q ∧ r)] ≡ p ∧ ~(q ∧ r) ≡ p ∧ (q ∨ ~r), the negation of the given statement is :
‘It is raining and we will not go or not play football.’

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) \(\sqrt {2}\) is a rational number.
Solution:
Let p : \(\sqrt {2}\) is a rational number.
The negation of the given statement is
‘ ~p : \(\sqrt {2}\) is not a rational number.’

(vii) All natural numbers are whole numers.
Solution:
The negation of the given statement is :
‘Some natural numbers are not whole numbers.’

(viii) Ɐ n ∈ N, n2 + n + 2 is divisible by 4.
Solution:
The negation of the given statement is :
‘Ǝ n ∈ N, such that n2 + n + 2 is not divisible by 4.’

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ix) Ǝ x ∈ N such that x – 17 < 20
Solution:
The negation of the given statement is :
‘Ɐ x ∈ N, x – 17 ≯ 20.’

Question 4.
Write converse, inverse and contrapositive of the following statements.
(i) If x < y then x2 < y2 (x, y ∈ R)
Solution:
Let p : x < y, q : x2 < y2.
Then the symbolic form of the given statement is p → q.
Converse : q → p is the converse of p → q.
i.e. If x2 < y2, then x < y.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If x ≯ y, then x2 ≯ y2. OR
If x ≮ y, then x2 ≮ y2.
Contrapositive : ~q → p is the contrapositive of
p → q i.e. If x2 ≯ y2, then x ≯ y. OR
If x2 ≮ y2, then x ≮ y.

(ii) A family becomes literate if the woman in it is literate.
Solution:
Let p : The woman in the family is literate.
q : A family become literate.
Then the symbolic form of the given statement is p → q
Converse : q → p is the converse of p → q.
i.e. If a family become literate, then the woman in it is literate.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If the woman in the family is not literate, then the family does not become literate.
Contrapositive : ~q → ~p is the contrapositive of p → q. i e. If a family does not become literate, then the woman in it is not literate.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) If surface area decreases then pressure increases.
Solution:
Let p : The surface area decreases.
q : The pressure increases.
Then the symbolic form of the given statement is p → q.
Converse : q → p is the converse of p→ q.
i.e. If the pressure increases, then the surface area decreases.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If the surface area does not decrease, then the pressure does not increase.
Contrapositive : ~q → ~p is the contrapositive of p → q.
i.e. If the pressure does not increase, then the surface area does not decrease.

(iv) If voltage increases then current decreases.
Solution:
Let p : Voltage increases.
q : Current decreases.
Then the symbolic form of the given statement is p → q.
Converse : q →p is the converse of p → q.
i.e. If current decreases, then voltage increases.
Inverse : ~p → ~q is the inverse of p → q.
i.e. If voltage does not increase, then current does not decrease.
Contrapositive : ~q → ~p, is the contrapositive of p → q.
i.e. If current does not decrease, then voltage doesnot increase.

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.2 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.2 Questions And Answers Maharashtra Board

Question 1.
Construct the truth table for each of the following statement patterns:
(i) [(p → q) ∧ q] → p
Solution :
Here are two statements and three connectives.
∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 1

(ii) (p ∧ ~q) ↔ (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 2

(iii) (p ∧ q) ↔ (q ∨ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 3

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) p → [~(q ∧ r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 4

(v) ~p ∧ [(p ∨ ~q ) ∧ q]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 5

(vi) (~p → ~q) ∧ (~q → ~p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 6

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vii) (q → p) ∨ (~p ↔ q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 7

(viii) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 8

(ix) p → [~(q ∧ r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 4

(x) (p ∨ ~q) → (r ∧ p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 9

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Using truth tables prove the following logical equivalences.
(i) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 10
The entries in the columns 4 and 6 are identical.
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(ii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 11
The entries in the columns 3 and 7 are identical.
∴ ~(p ∨ q) ∧ (~p ∧ q) = ~p.

(iii) p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 12
The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)].

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) p → (q → p) ≡ ~p → (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 13
The entries in the columns 4 and 7 are identical.
∴ p → (q → p) ≡ ~p → (p → q).

(v) (p ∨ q ) → r ≡ (p → r) ∧ (q → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 14
The entries in the columns 5 and 8 are identical.
∴ (p ∨ q ) → r ≡ (p → r) ∧ (q → r).

(vi) p → (q ∧ r) ≡ (p → q) ∧ (p → r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 15
The entries in the columns 5 and 8 are identical.
∴ p → (q ∧ r) ≡ (p → q) ∧ (p → r).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vii) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 16
The entries in the columns 5 and 8 are identical.
∴ p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r).

(viii) [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 17
The entries in the columns 3 and 7 are identical.
∴ [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r.

(ix) ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 18
The entries in the columns 6 and 9 are identical.
∴ ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p).

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
Examine whether each of the following statement patterns is a tautology or a contradiction or a contingency.
(i) (p ∧ q) → (q ∨ p)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 19
All the entries in the last column of the above truth table are T.
∴ (p ∧ q) → (q ∨ p) is a tautology.

(ii) (p → q) ↔ (~p ∨ q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 20
All the entries in the last column of the above truth table are T.
∴ (p → q) ↔ (~p ∨ q) p is a tautology.

(iii) [~(~p ∧ ~q)] ∨ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 21
The entries in the last column of the above truth table are neither all T nor all F.
∴ [~(~p ∧ ~q)] ∨ q is a contingency.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) [(p → q) ∧ q)] → p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 22
The entries in the last column of the above truth table are neither all T nor all F.
∴ [(p → q) ∧ q)] → p is a contingency

(v) [(p → q) ∧ ~q] → ~p
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 23
All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q] → ~p is a tautology.

(vi) (p ↔ q) ∧ (p → ~q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 24
The entries in the last column of the above truth table are neither all T nor all F.
∴ (p ↔ q) ∧ (p → ~q) is a contingency.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vii) ~(~q ∧ p) ∧ q
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 25
The entries in the last column of the above truth table are neither all T nor all F.
∴ ~(~q ∧ p) ∧ q is a contingency.

(viii) (p ∧ ~q) ↔ (p → q)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 26
All the entries in the last column of the above truth table are F.
∴ (p ∧ ~q) ↔ (p → q) is a contradiction.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ix) (~p → q) ∧ (p ∧ r)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 27
The entries in the last column of the above truth table are neither all T nor all F.
∴ (~p → q) ∧ (p ∧ r) is a contingency.

(x) [p → (~q ∨ r)] ↔ ~[p → (q → r)]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2 28
All the entries in the last column of the above truth table are F.
∴ [p → (~q ∨ r)] ↔ ~[p → (q → r)] is a contradiction

Class 12 Maharashtra State Board Maths Solution 

Mathematical Logic Class 12 Maths 1 Exercise 1.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.1 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.1 Questions And Answers Maharashtra Board

Question 1.
State which of the following sentences are statements. Justify your answer. In case of the statement, write down the truth value :
(i) 5 + 4 = 13.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(ii) x – 3 = 14.
Solution:
It is an open sentence, hence it is not a statement.

(iii) Close the door.
Solution:
It is an imperative sentence, hence it is not a statement.

(iv) Zero is a complex number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) Please get me breakfast.
Solution:
It is an imperative sentence, hence it is not a statement.

(vi) Congruent triangles are also similar.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(vii) x2 = x.
Solution:
It is an open sentence, hence it is not a statement,

(viii) A quadratic equation cannot have more than two roots.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ix) Do you like Mathematics ?
Solution:
It is an interrogative sentence, hence it is not a statement.

(x) The sun sets in the west.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xi) All real numbers are whole numbers.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

(xii) Can you speak in Marathi ?
Solution:
It is an interrogative sentence, hence it is not a statement.

(xiii) x2 – 6x – 7 = 0, when x = 7.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(xiv) The sum of cuberoots of unity is zero.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(xv) It rains heavily.
Solution :
It is an open sentence, hence it is not a statement.

Question 2.
Write the following compound statements symbolically:
(i) Nagpur is in Maharashtra and Chennai is in Tamil Nadu.
Solution:
Let p : Nagpur is in Maharashtra.
q : Chennai is in Tamil Nadu.
Then the symbolic form of the given statement is P∧q.

(ii) Triangle is equilateral or isosceles,
Solution:
Let p : Triangle is equilateral.
q : Triangle is isosceles.
Then the symbolic form of the given statement is P∨q.

(iii) The angle is right angle if and only if it is of measure 90°.
Solution:
Let p : The angle is right angle.
q : It is of measure 90°.
Then the symbolic form of the given statement is p↔q

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) Angle is neither acute nor obtuse.
Solution:
Let p : Angle is acute.
q : Angle is obtuse.
Then the symbolic form of the given statement is
~p ∧ ~q.

(v) If ∆ ABC is right angled at B, then m∠A + m∠C = 90°.
Solution:
Let p : ∆ ABC is right angled at B.
q : m∠A + m∠C = 90°.
Then the symbolic form of the given statement is p → q

(vi) Hima Das wins gold medal if and only if she runs fast.
Solution:
Let p : Hima Das wins gold medal
q : She runs fast.
Then the symbolic form of the given statement is p ↔ q.

(vii) x is not irrational number but it is a square of an integer.
Solution:
Let p : x is not irrational number
q : It is a square of an integer
Then the symbolic form of the given statement is p ∧ q
Note : If p : x is irrational number, then the symbolic form of the given statement is ~p ∧ q.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
Write the truth values of the following :
(i) 4 is odd or 1 is prime.
Solution:
Let p : 4 is odd.
q : 1 is prime.
Then the symbolic form of the given statement is p∨q.
The truth values of both p and q are F.
∴ the truth value of p v q is F. … [F ∨ F = F]

(ii) 64 is a perfect square and 46 is a prime number.
Solution:
Let p : 64 is a perfect square.
q : 46 is a prime number.
Then the symbolic form of the given statement is p∧q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(iii) 5 is a prime number and 7 divides 94.
Solution:
Let p : 5 is a prime number.
q : 7 divides 94.
Then the symbolic form of the given statement is p∧q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(iv) It is not true that 5 – 3i is a real number.
Solution:
Let p : 5 – 3i is a real number.
Then the symbolic form of the given statement is ~ p.
The truth values of p is F.
∴ the truth values of ~ p is T. … [~ F ≡ T]

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) If 3 × 5 = 8, then 3 + 5 = 15.
Solution:
Let p : 3 × 5 = 8.
q : 3 + 5 = 15.
Then the symbolic form of the given statement is p → q.
The truth values of both p and q are F.
∴ the truth value of p → q is T. … [F → F ≡ T]

(vi) Milk is white if and only if sky is blue.
Solution:
Let p : Milk is white.
q : Sky is blue
Then the symbolic form of the given statement is p ↔ q.
The truth values of both p and q are T.
∴ the truth value of p ↔ q is T. … [T ↔ T ≡ T]

(vii) 24 is a composite number or 17 is a prime number.
Solution :
Let p : 24 is a composite number.
q : 17 is a prime number.
Then the symbolic form of the given statement is p ∨ q.
The truth values of both p and q are T.
∴ the truth value of p ∨ q is T. … [T ∨ T ≡ T]

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 4.
If the statements p, q are true statements and r, s are false statements, then determine the truth values of the following:
(i) p ∨ (q ∧ r)
Solution:
Truth values of p and q are T and truth values of r and s are F.
p ∨ (q ∧ r) ≡ T ∨ (T ∧ F)
≡ T ∧ F ≡ T
Hence the truth value of the given statement is true.

(ii) (p → q) ∨ (r → s)
Solution:
(p → q) ∨ (r → s) ≡ (T → T) ∨ (F → F)
≡ T ∨ T ≡ T
Hence the truth value of the given statement is true.

(iii) (q ∧ r) ∨ (~p ∧ s)
Solution:
(q ∧ r) ∨ (~p ∧ s) ≡ (T ∧ F) ∨ (~T ∧ F)
≡ F ∨ (F ∧ F)
≡ F ∨ F ≡ F
Hence the truth value of the given statement is false.

(iv) (p → q) ∧ (~ r)
Solution:
(p → q) ∧ (~ r) ≡ (T → T) ∧ (~ F)
≡ T ∧ T ≡ T
Hence the truth value of the given statement is true.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) (~r ↔ p) → (~q)
Solution:
(~r ↔ p) → (~q) ≡ (~F ↔ T) → (~T)
≡ (T ↔ T) → F
≡ T → F ≡ F
Hence the truth value of the given statement is false.

(vi) [~p ∧ (~q ∧ r) ∨ (q ∧ r) ∨ (p ∧ r)]
Solution:
[~p ∧ (~q ∧ r)∨(q ∧ r)∨(p ∧ r)]
≡ [~T ∧ (~T ∧ F)] ∨ [(T ∧ F) V (T ∧ F)]
≡ [F ∧ (F ∧ F)] ∨ [F V F]
≡ (F ∧ F) ∨ F
≡ F ∨ F ≡ F
Hence the truth value of the given statement is false.

(vii) [(~ p ∧ q) ∧ (~ r)] ∨ [(q → p) → (~ s ∨ r)]
Solution:
[(~ p ∧ q) ∧ (~ r)] ∨ [(q → p) → (~ s ∨ r)]
≡ [(~T ∧ T) ∧ (~F)] ∨ [(T → T) → (~F ∨ F)]
≡ [(F ∧ T) ∧ T] ∨ [T → (T ∨ F)]
≡ (F ∧ T) ∨ (T → T)
≡ F ∨ T ≡ T
Hence the truth value of the given statement is true.

(viii) ~ [(~p ∧ r) ∨ (s → ~q)] ↔ (p ∧ r)
Solution :
~ [(~p ∧ r) ∨ (s → ~q)] ↔ (p ∧ r)
≡ ~ [(~T ∧ F) ∨ (F → ~T)] ↔ (T ∧ F)
≡ ~ [(F ∧ F) ∨ (F → F)] ↔ F
≡ ~ (F ∨ T) ↔ F
≡ ~T ↔ F
≡ F ↔ F ≡ T
Hence the truth value of the given statement is true.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
Write the negations of the following :
(i) Tirupati is in Andhra Pradesh.
Solution:
The negations of the given statements are :
Tirupati is not in Andhra Pradesh.

(ii) 3 is not a root of the equation x2 + 3x – 18 = 0.
Solution:
3 is a root of the equation x2 + 3x – 18 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) \(\sqrt {2}\) is a rational number.
Solution:
\(\sqrt {2}\) is not a rational number.

(iv) Polygon ABCDE is a pentagon.
Solution:
Polygon ABCDE is not a pentagon.

(v) 7 + 3 > 5.
Solution :
7 + 3 > 5.

Class 12 Maharashtra State Board Maths Solution 

Differentiation Class 11 Maths 2 Miscellaneous Exercise 9 Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Miscellaneous Exercise 9 Questions and Answers.

11th Maths Part 2 Differentiation Miscellaneous Exercise 9 Questions And Answers Maharashtra Board

(I) Select the appropriate option from the given alternatives.

Question 1.
If y = \(\frac{x-4}{\sqrt{x+2}}\), then \(\frac{d y}{d x}\) is
(A) \(\frac{1}{x+4}\)
(B) \(\frac{\sqrt{x}}{\left(\sqrt{x+2)^{2}}\right.}\)
(C) \(\frac{1}{2 \sqrt{x}}\)
(D) \(\frac{x}{(\sqrt{x}+2)^{2}}\)
Answer:
(C) \(\frac{1}{2 \sqrt{x}}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q1

Question 2.
If y = \(\frac{a x+b}{c x+d}\),then \(\frac{d y}{d x}\) =
(A) \(\frac{a b-c d}{(c x+d)^{2}}\)
(B) \(\frac{a x-c}{(c x+d)^{2}}\)
(C) \(\frac{a c-b d}{(c x+d)^{2}}\)
(D) \(\frac{a d-b c}{(c x+d)^{2}}\)
Answer:
(D) \(\frac{a d-b c}{(c x+d)^{2}}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q2

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

Question 3.
If y = \(\frac{3 x+5}{4 x+5}\), then \(\frac{d y}{d x}\) =
(A) \(-\frac{15}{(3 x+5)^{2}}\)
(B) \(-\frac{15}{(4 x+5)^{2}}\)
(C) \(-\frac{5}{(4 x+5)^{2}}\)
(D) \(-\frac{13}{(4 x+5)^{2}}\)
Answer:
(C) \(-\frac{5}{(4 x+5)^{2}}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q3

Question 4.
If y = \(\frac{5 \sin x-2}{4 \sin x+3}\), then \(\frac{d y}{d x}\) =
(A) \(\frac{7 \cos x}{(4 \sin x+3)^{2}}\)
(B) \(\frac{23 \cos x}{(4 \sin x+3)^{2}}\)
(C) \(-\frac{7 \cos x}{(4 \sin x+3)^{2}}\)
(D) \(-\frac{15 \cos x}{(4 \sin x+3)^{2}}\)
Answer:
(B) \(\frac{23 \cos x}{(4 \sin x+3)^{2}}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q4

Question 5.
Suppose f(x) is the derivative of g(x) and g(x) is the derivative of h(x).
If h(x) = a sin x + b cos x + c, then f(x) + h(x) =
(A) 0
(B) c
(C) -c
(D) -2(a sin x + b cos x)
Answer:
(B) c
Hint:
h(x) = a sin x + b cos x + c
Differentiating w.r.t. x, we get
h'(x) = a cos x – b sin x = g(x) …..[given]
Differentiating w.r.t. x, we get
g'(x) = -a sin x – b cos x = f(x) …..[given]
∴ f(x) + h(x) = -a sin x – b cos x + a sin x + b cos x + c
∴ f(x) + h(x) = c

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

Question 6.
If f(x) = 2x + 6, for 0 ≤ x ≤ 2
= ax2 + bx, for 2 < x ≤ 4
is differentiable at x = 2, then the values of a and b are
(A) a = \(-\frac{3}{2}\), b = 3
(B) a = \(\frac{3}{2}\), b = 8
(C) a = \(\frac{1}{2}\), b = 8
(D) a = \(-\frac{3}{2}\), b = 8
Answer:
(D) a = \(-\frac{3}{2}\), b = 8
Hint:
f(x) = 2x + 6, 0 ≤ x ≤ 2
= ax2 + bx, 2 < x ≤ 4
Lf'(2) = 2, Rf'(2) = 4a + b
Since f is differentiable at x = 2,
Lf'(2) = Rf'(2)
∴ 2 = 4a + b …..(i)
f is continuous at x = 2.
∴ \(\lim _{x \rightarrow 2^{+}} f(x)=f(2)=\lim _{x \rightarrow 2^{-}} f(x)\)
∴ 4a + 2b = 2(2) + 6
∴ 4a + 2b = 10
∴ 2a + b = 5 …..(ii)
Solving (i) and (ii), we get
a = \(-\frac{3}{2}\), b = 8

Question 7.
If f(x) = x2 + sin x + 1, for x ≤ 0
= x2 – 2x + 1, for x ≤ 0, then
(A) f is continuous at x = 0, but not differentiable at x = 0
(B) f is neither continuous nor differentiable at x = 0
(C) f is not continuous at x = 0, but differentiable at x = 0
(D) f is both continuous and differentiable at x = 0
Answer:
(A) f is continuous at x = 0, but not differentiable at x = 0
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q7

Question 8.
If f(x) = \(\frac{x^{50}}{50}+\frac{x^{49}}{49}+\frac{x^{48}}{48}+\ldots .+\frac{x^{2}}{2}+x+1\), then f'(1) =
(A) 48
(B) 49
(C) 50
(D) 51
Answer:
(C) 50
Hint:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 I Q8

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

(II).

Question 1.
Determine whether the following function is differentiable at x = 3 where,
f(x) = x2 + 2, for x ≥ 3
= 6x – 7, for x < 3.
Solution:
f(x) = x2 + 2, x ≥ 3
= 6x – 7, x < 3
Differentiability at x = 3
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q1.1
Here, Lf'(3) = Rf'(3)
∴ f is differentiable at x = 3.

Question 2.
Find the values of p and q that make function f(x) differentiable everywhere on R.
f(x) = 3 – x, for x < 1
= px2 + qx, for x ≥ 1.
Solution:
f(x) is differentiable everywhere on R.
∴ f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q2
f(x) is differentiable at x = 1.
∴ Lf'(1) = Rf'(1)
∴ -1 = 2p + q …..(ii)
Subtracting (i) from (ii), we get
p = -3
Substituting p = -3 in (i), we get
p + q = 2
∴ -3 + q = 2
∴ q = 5

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

Question 3.
Determine the values of p and q that make the function f(x) differentiable on R where
f(x) = px3, for x < 2
= x2 + q, for x ≥ 2
Solution:
f(x) is differentiable on R.
∴ f(x) is differentiable at x = 2.
∴ f(x) is continuous at x = 2.
Continuity at x = 2:
f(x) is continuous at x = 2.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q3
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q3.1
f(x) is differentiable at x = 2.
∴ Lf'(2) = Rf'(2)
∴ 12p = 4
∴ p = \(\frac{1}{3}\)
Substituting p = \(\frac{1}{3}\) in (i), we get
8(\(\frac{1}{3}\) – q = 4
∴ q = \(\frac{8}{3}\) – 4 = \(\frac{-4}{3}\)

Question 4.
Determine all real values of p and q that ensure the function
f(x) = px + q, for x ≤ 1
= tan(\(\frac{\pi x}{4}\)), for 1 < x < 2
is differentiable at x = 1.
Solution:
f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.
Continuity at x= 1:
f(x) is continuous at x = 1.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q4
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q4.1

Question 5.
Discuss whether the function f(x) = |x + 1| + |x – 1| is differentiable ∀ x ∈ R.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q5
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q5.1
Here, Lf'(1) ≠ Rf'(1)
∴ f is not differentiable at x = 1.
∴ f is not differentiable at x = -1 and x = 1.
∴ f is not differentiable ∀ x ∈ R.

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

Question 6.
Test whether the function
f(x) = 2x – 3, for x ≥ 2
= x – 1, for x < 2
is differentiable at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q6

Question 7.
Test whether the function
f(x) = x2 + 1, for x ≥ 2
= 2x + 1, for x < 2
is differentiable at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q7
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q7.1

Question 8.
Test whether the function
f(x) = 5x – 3x2, for x ≥ 1
= 3 – x, for x < 1
is differentiable at x = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q8
Here, Lf'(1) = Rf'(1)
∴ f(x) is differentiable at x = 1.

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

Question 9.
If f(2) = 4, f'(2) = 1, then find \(\lim _{x \rightarrow 2}\left[\frac{x f(2)-2 f(x)}{x-2}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q9

Question 10.
If y = \(\frac{\mathbf{e}^{x}}{\sqrt{x}}\), find \(\frac{d y}{d x}\) when x = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q10

Class 11 Maharashtra State Board Maths Solution 

Differentiation Class 11 Maths 2 Exercise 9.2 Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.2 Questions and Answers.

11th Maths Part 2 Differentiation Exercise 9.2 Questions And Answers Maharashtra Board

(I) Differentiate the following w.r.t. x

Question 1.
y = \(x^{\frac{4}{3}}+e^{x}-\sin x\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q1

Question 2.
y = √x + tan x – x3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q2

Question 3.
y = log x – cosec x + \(5^{x}-\frac{3}{x^{\frac{3}{2}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q3

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 4.
y = \(x^{\frac{7}{3}}+5 x^{\frac{4}{5}}-\frac{5}{x^{\frac{2}{5}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q4
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q4.1

Question 5.
y = 7x + x7 – \(\frac{2}{3}\) x√x – log x + 77
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q5

Question 6.
y = 3 cot x – 5ex + 3 log x – \(\frac{4}{x^{\frac{3}{4}}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 I Q6

(II) Diffrentiate the following w.r.t. x

Question 1.
y = x5 tan x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q1

Question 2.
y = x3 log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q2

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 3.
y = (x2 + 2)2 sin x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q3

Question 4.
y = ex log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q4

Question 5.
y = \(x^{\frac{3}{2}} e^{x} \log x\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q5

Question 6.
y = \(\log e^{x^{3}} \log x^{3}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 II Q6

(III) Diffrentiate the following w.r.t. x

Question 1.
y = x2√x + x4 log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q1.1

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 2.
y = ex sec x – \(x^{\frac{5}{3}}\) log x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q2

Question 3.
y = x4 + x√x cos x – x2 ex
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q3

Question 4.
y = (x3 – 2) tan x – x cos x + 7x . x7
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q4

Question 5.
y = sin x log x + ex cos x – ex √x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q5

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 6.
y = ex tan x + cos x log x – √x 5x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q6
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 III Q6.1

(IV) Diffrentiate the following w.r.t.x.

Question 1.
y = \(\frac{x^{2}+3}{x^{2}-5}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q1

Question 2.
y = \(\frac{\sqrt{x}+5}{\sqrt{x}-5}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q2
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q2.1

Question 3.
y = \(\frac{x e^{x}}{x+e^{x}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q3

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 4.
y = \(\frac{x \log x}{x+\log x}\)
Solution:
y = \(\frac{x \log x}{x+\log x}\)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q4

Question 5.
y = \(\frac{x^{2} \sin x}{x+\cos x}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q5

Question 6.
y = \(\frac{5 e^{x}-4}{3 e^{x}-2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 IV Q6

(V).

Question 1.
If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12, then find f(x).
Solution:
Let f(x) = ax2 + bx + c …..(i)
∴ f(0) = a(0)2 + b(0) + c
∴ f(0) = c
But, f(0) = 3 …..(given)
∴ c = 3 …..(ii)
Differentiating (i) w.r.t. x, we get
f'(x) = 2ax + b
∴ f'(2) = 2a(2) + b
∴ f'(2) = 4a + b
But, f'(2) = 2 …..(given)
∴ 4a + b = 2 …..(iii)
Also, f'(3) = 2a(3) + b
∴ f'(3) = 6a + b
But, f'(3) = 12 …..(given)
∴ 6a + b = 12 …..(iv)
equation (iv) – equation (iii), we get
2a = 10
∴ a = 5
Substituting a = 5 in (iii), we get
4(5) + b = 2
∴ b = -18
∴ a = 5, b = -18, c = 3
∴ f(x) = 5x2 – 18x + 3

Check:
If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x2 – 18x + 3 and f'(x) = 10x – 18
f(0) = 5(0)2 – 18(0) + 3 = 3
f'(2) = 10(2) – 18 = 2
f'(3) = 10(3) – 18 = 12
Thus, our answer is correct.

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 2.
If f(x) = a sin x – b cos x, f'(\(\frac{\pi}{4}\)) = √2 and f'(\(\frac{\pi}{6}\)) = 2, then find f(x).
Solution:
f(x) = a sin x – b cos x
Differentiating w.r.t. x, we get
f'(x) = a cos x – b (- sin x)
∴ f'(x) = a cos x + b sin x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 V Q2
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 V Q2.1
Now, f(x) = a sin x – b cos x
∴ f(x) = (√3 + 1) sin x + (√3 – 1) cos x

VI. Fill in the blanks. (Activity Problems)

Question 1.
y = ex . tan x
Diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q1.1

Question 2.
y = \(\frac{\sin x}{x^{2}+2}\)
diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q2.1

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

Question 3.
y = (3x2 + 5) cos x
Diff. w.r.t. x
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q3.1

Question 4.
Differentiate tan x and sec x w.r.t. x using the formulae for differentiation of \(\frac{u}{v}\) and \(\frac{1}{v}\) respectively.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q4
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2 VI Q4.1

Class 11 Maharashtra State Board Maths Solution 

Differentiation Class 11 Maths 2 Exercise 9.1 Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.1 Questions and Answers.

11th Maths Part 2 Differentiation Exercise 9.1 Questions And Answers Maharashtra Board

Question 1.
Find the derivatives of the following w.r.t. x by using the method of the first principle.
(a) x2 + 3x – 1
Solution:
Let f(x) = x2 + 3x – 1
∴ f(x + h) = (x + h)2 + 3(x + h) – 1
= x2 + 2xh + h2 + 3x + 3h – 1
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (i)

(b) sin(3x)
Solution:
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (ii)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (ii).1

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

(c) e2x+1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (iii)

(d) 3x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (iv)

(e) log(2x + 5)
Solution:
Let f(x) = log(2x + 5)
∴ f(x + h) = log[2(x + h) + 5] = log (2x + 2h + 5)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (v)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (v).1

(f) tan(2x + 3)
Solution:
Let f(x) = tan(2x + 3)
∴ f(x + h) = tan[2(x + h) + 3] = tan(2x + 2h + 3)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vi)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vi).1

(g) sec(5x – 2)
Solution:
Let f(x) = sec(5x – 2)
f(x + h) = sec[5(x + h) – 2] = sec(5x + 5h – 2)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vii)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (vii).1

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

(h) x√x
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q1 (viii)

Question 2.
Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle.
(i) \(\sqrt{2 x+5}\) at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (i)

(ii) tan x at x = \(\frac{\pi}{4}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (ii)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (ii).1

(iii) 23x+1 at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (iii)

(iv) log(2x + 1) at x = 2
Solution:
Let f(x) = log(2x + 1)
∴ f(2) = log [2(2) + 1] = log 5 and
f(2 + h) = log [2(2 + h) + 1] = log(2h + 5)
By first principle, we get
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (iv)

(v) e3x-4 at x = 2
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (v)

(vi) cos x at x = \(\frac{5 \pi}{4}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (vi)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (vi).1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q2 (vi).2

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 3.
Show that the function f is not differentiable at x = -3,
where f(x) = x2 + 2 for x < -3
= 2 – 3x for x ≥ -3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q3
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q3.1
∴ L f'(-3) ≠ R f'(-3)
∴ f is not differentiable at x = -3.

Question 4.
Show that f(x) = x2 is continuous and differentiable at x = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q4

Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (i)

(ii) f(x) = (2x + 3) |2x + 3| at x = \(-\frac{3}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (ii)
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (ii).1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q5 (ii).2

Question 6.
Discuss the continuity and differentiability of f(x) at x = 2.
f(x) = [x] if x ∈ [0, 4). [where [ ] is a greatest integer (floor) function]
Solution:
Explanation:
x ∈ [0, 4)
∴ 0 ≤ x < 4
We will plot graph for 0 ≤ x < 4
not for x < 0 and upto x = 4 making on X-axis.
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q6
f(x) = [x]
∴ Greatest integer function is discontinuous at all integer values of x and hence not differentiable at all integers.
∴ f is not continuous at x = 2.
∵ f(x) = 1, x < 2
= 2, x ≥ 2
x ∈ neighbourhood of x = 2.
∴ L.H.L. = 1, R.H.L. = 2
∴ f is not continuous at x = 2.
∴ f is not differentiable at x = 2.

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 – x2 if x ≤ 2 at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q7
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q7.1

Question 8.
If f(x) = sin x – cos x if x ≤ \(\frac{\pi}{2}\)
= 2x – π + 1 if x > \(\frac{\pi}{2}\)
Test the continuity and differentiability of f at x = \(\frac{\pi}{2}\).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q8
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q8.1
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q8.2

Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 9.
Examine the function
f(x) = x2 cos(\(\frac{1}{x}\)), for x ≠ 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1 Q9

Class 11 Maharashtra State Board Maths Solution 

Continuity Class 11 Maths 2 Miscellaneous Exercise 8 Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Miscellaneous Exercise 8 Questions and Answers.

11th Maths Part 2 Continuity Miscellaneous Exercise 8 Questions And Answers Maharashtra Board

(I) Select the correct answer from the given alternatives.

Question 1.
f(x) = \(\frac{2^{\cot x}-1}{\pi-2 x}\), for x ≠ \(\frac{\pi}{2}\)
= log √2, for x = \(\frac{\pi}{2}\)
(A) f is continuous at x = \(\frac{\pi}{2}\)
(B) f has a jump discontinuity at x = \(\frac{\pi}{2}\)
(C) f has a removable discontinuity
(D) \(\lim _{x \rightarrow \frac{\pi}{2}} f(x)=2 \log 3\)
Answer:
(A) f is continuous at x = \(\frac{\pi}{2}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q1

Question 2.
If f(x) = \(\frac{1-\sqrt{2} \sin x}{\pi-4 x}\), for x ≠ \(\frac{\pi}{4}\) is continuous at x = \(\frac{\pi}{4}\), then f(\(\frac{\pi}{4}\)) =
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(-\frac{1}{\sqrt{2}}\)
(C) \(-\frac{1}{4}\)
(D) \(\frac{1}{4}\)
Answer:
(D) \(\frac{1}{4}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q2
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q2.1

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 3.
If f(x) = \(\frac{(\sin 2 x) \tan 5 x}{\left(e^{2 x}-1\right)^{2}}\), for x ≠ 0 is continuous at x = 0, then f(0) is
(A) \(\frac{10}{e^{2}}\)
(B) \(\frac{10}{e^{4}}\)
(C) \(\frac{5}{4}\)
(D) \(\frac{5}{2}\)
Answer:
(D) \(\frac{5}{2}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q3

Question 4.
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
(A) f is discontinuous at x = 2
(B) f is discontinuous at x = -4
(C) f is discontinuous at x = 0
(D) f is discontinuous at x = 2 and x = -4
Answer:
(B) f is discontinuous at x = -4
Hint:
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
= \(\frac{x^{2}-7 x+10}{(x+4)(x-2)}\)
Here f(x) is a rational function and is continuous everywhere except at the points Where denominator becomes zero.
Here, denominator becomes zero when x = -4 or x = 2
But x = 2 does not lie in the given interval.
∴ x = -4 is the point of discontinuity.

Question 5.
If f(x) = ax2 + bx + 1, for |x – 1| ≥ 3 and
= 4x + 5, for -2 < x < 4
is continuous everywhere then,
(A) a = \(\frac{1}{2}\), b = 3
(B) a = \(-\frac{1}{2}\), b = -3
(C) a = \(-\frac{1}{2}\), b = 3
(D) a = \(\frac{1}{2}\), b = -3
Answer:
(A) a = \(\frac{1}{2}\), b = 3
Hint:
f(x) = ax2 + bx + 1, |x – 1| ≥ 3
= 4x + 5; -2 < x < 4
The first interval is
|x – 1| ≥ 3
∴ x – 1 ≥ 3 or x – 1 ≤ -3
∴ x ≥ 4 or x ≤ -2
∴ f(x) is same for x ≤ -2 as well as x ≥ 4.
∴ f(x) is defined as:
f(x) = ax2 + bx + 1; x ≤ -2
= 4x + 5; -2 < x < 4
= ax2 + bx + 1; x ≥ 4
f(x) is continuous everywhere.
∴ f(x) is continuous at x = -2 and x = 4.
As f(x) is continuous at x = -2,
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2}\left(a x^{2}+b x+1\right)=\lim _{x \rightarrow-2}(4 x+5)\)
∴ a(-2)2 + b(-2) + 1 = 4(-2) + 5
∴ 4a – 2b + 1 = -3
∴ 4a – 2b = -4
∴ 2a – b = -2 …..(i)
∵ f(x) is continuous at x = 4,
\(\lim _{x \rightarrow 4^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 4^{+}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 4}(4 x+5)=\lim _{x \rightarrow 4}\left(a x^{2}+b x+1\right)\)
4(4) + 5 = a(4)2 + b(4) + 1
16a + 4b + 1 = 21
16a + 4b = 20
4a + b = 5 …..(ii)
Adding (i) and (ii), we get
6a = 3
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (ii), we get
4(\(\frac{1}{2}\)) + b = 5
∴ 2 + b = 5
∴ b = 3
∴ a = \(\frac{1}{2}\), b = 3

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 6.
f(x) = \(\frac{\left(16^{x}-1\right)\left(9^{x}-1\right)}{\left(27^{x}-1\right)\left(32^{x}-1\right)}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then ‘k’ =
(A) \(\frac{8}{3}\)
(B) \(\frac{8}{15}\)
(C) \(-\frac{8}{15}\)
(D) \(\frac{20}{3}\)
Answer:
(B) \(\frac{8}{15}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q6

Question 7.
f(x) = \(\frac{32^{x}-8^{x}-4^{x}+1}{4^{x}-2^{x+1}+1}\), for x ≠ 0
= k, for x = 0,
is continuous at x = 0, then value of ‘k’ is
(A) 6
(B) 4
(C) (log 2) (log 4)
(D) 3 log 4
Answer:
(A) 6
Hint:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q7
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q7.1

Question 8.
If f(x) = \(\frac{12^{x}-4^{x}-3^{x}+1}{1-\cos 2 x}\), for x ≠ 0 is continuous at x = 0 then the value of f(0) is
(A) \(\frac{\log 12}{2}\)
(B) log 2 . log 3
(C) \(\frac{\log 2 \cdot \log 3}{2}\)
(D) None of these
Answer:
(B) log 2 . log 3
Hint:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q8

Question 9.
If f(x) = \(\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}\), for x ≠ 0 and f(0) = k, is continuous at x = 0, then k is
(A) e7
(B) e3
(C) e12
(D) \(e^{\frac{3}{4}}\)
Answer:
(C) e12
Hint:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q9
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 I Q9.1

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 10.
If f(x) = \(\lfloor x\rfloor\) for x ∈ (-1, 2), then f is discontinuous at
(A) x = -1, 0, 1, 2
(B) x = -1, 0, 1
(C) x = 0, 1
(D) x = 2
Answer:
(C) x = 0, 1
Hint:
f(x) = \(\lfloor x\rfloor\), x ∈ (-1, 2)
This function is discontinuous at all integer values of x between -1 and 2.
∴ f(x) is discontinuous at x = 0 and x = 1.

II. Discuss the continuity of the following functions at the point(s) or on the interval indicated against them.

Question 1.
f(x) = \(\frac{x^{2}-3 x-10}{x-5}\), for 3 ≤ x ≤ 6, x ≠ 5
= 10, for x = 5
= \(\frac{x^{2}-3 x-10}{x-5}\), for 6 < x ≤ 9
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q1

Question 2.
f(x) = 2x2 – 2x + 5, for 0 ≤ x ≤ 2
= \(\frac{1-3 x-x^{2}}{1-x}\), for 2 < x < 4
= \(\frac{x^{2}-25}{x-5}\), for 4 ≤ x ≤ 7 and x ≠ 5
= 7, for x = 5
Solution:
The domain of f(x) is [0, 7].
(i) For 0 ≤ x ≤ 2
f(x) = 2x2 – 2x + 5
It is a polynomial function and is Continuous at all point in [0, 2).

(ii) For 2 < x < 4
f(x) = \(\frac{1-3 x-x^{2}}{1-x}\)
It is a rational function and is continuous everwhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 1
But x = 1 does not lie in the interval.
f(x) is continuous at all points in (2, 4).

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

(iii) For 4 ≤ x ≤ 7, x ≤ 5
f(x) = \(\frac{x^{2}-25}{x-5}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 5
But x = 5 does not lie in the interval.
∴ f(x) is continuous at all points in (4, 7] – {5}.

(iv) For continuity at x = 2:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q2 (iv)
∴ f(x) is continuous at x = 2.

(v) For continuity at x = 4:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q2 (v)
∴ f(x) is continuous at x = 4.

(vi) For continuity at x = 5.
f(5) = 7
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q2 (vi)
∴ f(x) is discontinuous at x = 5.
Thus, f(x) is continuous at all points on its domain except at x = 5.

Question 3.
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\), for x ≠ 0
f(0) = \(\frac{68}{15}\), at x = 0 on \(-\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\)
Solution:
The domain of f(x) is [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
(i) For [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}:
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero when cos x = 1,
i.e., x = 0
But x = 0 does not lie in the interval.
∴ f(x) is continuous at all points in [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}

(ii) For continuity at x = 0:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q3 (ii)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q3 (ii).1
∴ \(\lim _{x \rightarrow 0} f(x) \neq f(0)\)
∴ f(x) is discontinuous at x = 0.

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 4.
f(x) = \(\frac{\sin ^{2} \pi x}{3(1-x)^{2}}\), for x ≠ 1
= \(\frac{\pi^{2} \sin ^{2}\left(\frac{\pi x}{2}\right)}{3+4 \cos ^{2}\left(\frac{\pi x}{2}\right)}\), for x = 1, at x = 1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q4

Question 5.
f(x) = \(\frac{|x+1|}{2 x^{2}+x-1}\), for x ≠ -1
= 0, for x = -1, at x = -1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q5
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q5.1

Question 6.
f(x) = [x + 1] for x ∈ [-2, 2)
Where [*] is greatest integer function.
Solution:
f(x) = [x + 1], x ∈ [-2, 2)
∴ f(x) = -1, x ∈ [-2, -1)
= 0, x ∈ [-1, 0)
= 1, x ∈ [0, 1)
= 2, x ∈ [1, 2)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q6
∴ \(\lim _{x \rightarrow-1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow-1^{+}} \mathrm{f}(x)\)
∴ f(x) is discontinuous at x = -1.
Similarly, f(x) is discontinuous at the points x = 0 and x = 1.

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 7.
f(x) = 2x2 + x + 1, for |x – 3| ≥ 2
= x2 + 3, for 1 < x < 5
Solution:
|x – 3| ≥ 2
∴ x – 3 ≥ 2 or x – 3 ≤ -2
∴ x ≥ 5 or x ≤ 1
∴ f(x) = 2x2 + x + 1, x ≤ 1
= x2 + 3, 1 < x < 5
= 2x2 + x + 1, x ≥ 5
Consider the intervals
x < 1 , i.e., (-∞, 1)
1 < x < 5, i.e., (1, 5) x > 5, i.e., (5, ∞)
In all these intervals, f(x) is a polynomial function and hence is continuous at all points.
For continuity at x = 1:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 II Q7
∴ f(x) is discontinuous at x = 5.
∴ f(x) is continuous for all x ∈ R, except at x = 5.

III. Identify discontinuities if any for the following functions as either a jump or a removable discontinuity on their respective domains.

Question 1.
f(x) = x2 + x – 3, for x ∈ [-5, -2)
= x2 – 5, for x ∈ (-2, 5]
Solution:
f(-2) has not been defined.
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{-}}\left(x^{2}+x-3\right)\)
= (-2)2 + (-2) – 3
= 4 – 2 – 3
= -1
\(\lim _{x \rightarrow-2^{+}} f(x)=\lim _{x \rightarrow-2^{+}}\left(x^{2}-5\right)\)
= (-2)2 – 5
= 4 – 5
= -1
∴ \(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2} f(x) \text { exists. }\)
But f(-2) has not been defined.
∴ f(x) has a removable discontinuity at x = -2.

Question 2.
f(x) = x2 + 5x + 1, for 0 ≤ x ≤ 3
= x3 + x + 5, for 3 < x ≤ 6
Solution:
\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}\left(x^{2}+5 x+1\right)\)
= (3)2 + 5(3) + 1
= 9 + 15 + 1
= 25
\(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}\left(x^{3}+x+5\right)\)
= (3)3 + 3 + 5
= 27 + 3 + 5
= 35
∴ \(\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)\)
∴ \(\lim _{x \rightarrow 3} f(x)\) does not exist.
∴ f(x) is discontinuous at x = 3.
∴ f(x) has a jump discontinuity at x = 3.

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 3.
f(x) = \(\frac{x^{2}+x+1}{x+1}\), for x ∈ [0, 3)
= \(\frac{3 x+4}{x^{2}-5}\), for x ∈ [3, 6]
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 III Q3
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 III Q3.1
∴ f(x) is continuous at x = 3.

IV. Discuss the continuity of the following functions at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous.

Question 1.
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
Solution:
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
= \(\frac{(x+3)(x-2)(x-4)}{(x-4)(x+3)}\)
∴ f(x) is not defined at x = 4 and x = -3.
∴ The domain of function f = R – {-3, 4}.
For x ≠ -3 and 4,
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 IV Q1
f(x) is discontinuous at x = 4 and x = -3.
This discontinuity is removable.
∴ f(x) can be redefined as
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\), for x ≠ 4, x ≠ -3
= -5, for x ∈ R – {-3, 4}, x = -3
= 2, for x ∈ R – {-3, 4}, x = 4

Question 2.
f(x) = x2 + 2x + 5, for x ≤ 3
= x3 – 2x2 – 5, for x > 3
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 IV Q2
∴ f(x) is discontinuous at x = 3.
This discontinuity is irremovable.

V. Find k if the following functions are continuous at the points indicated against them.

Question 1.
f(x) = \(\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}\), for x ≠ 2
= k, for x = 2 at x = 2.
Solution:
f(x) is continuous at x = 2. …..(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 V Q1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 V Q1.1

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 2.
f(x) = \(\frac{45^{x}-9^{x}-5^{x}+1}{\left(k^{x}-1\right)\left(3^{x}-1\right)}\), for x ≠ 0
= \(\frac{2}{3}\), for x = 0, at x = 0
Solution:
f(x) is continuous at x = 0 …..(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 V Q2
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 V Q2.1

VI. Find a and b if the following functions are continuous at the points or on the interval indicated against them.

Question 1.
f(x) = \(\frac{4 \tan x+5 \sin x}{a^{x}-1}\), for x < 0
= \(\frac{9}{\log 2}\), for x = 0
= \(\frac{11 x+7 x \cdot \cos x}{b^{x}-1}\), for x < 0
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 VI Q1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 VI Q1.1

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 2.
f(x) = ax2 + bx + 1, for |2x – 3| ≥ 2
= 3x + 2, for \(\frac{1}{2}\) < x < \(\frac{5}{2}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 VI Q2
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 VI Q2.1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 VI Q2.2

VII. Find f(a), if f is continuous at x = a where,

Question 1.
f(x) = \(\frac{1+\cos (\pi x)}{\pi(1-x)^{2}}\), for x ≠ 1 and at a = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 VII Q1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 VII Q1.1

Question 2.
f(x) = \(\frac{1-\cos [7(x-\pi)]}{5(x-\pi)^{2}}\), for x ≠ π and at a = π.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8 VII Q2

VIII. Solve using intermediate value theorem.

Question 1.
Show that 5x – 6x = 0 has a root in [1, 2].
Solution:
Let f(x) = 5x – 6x
5x and 6x are continuous functions for all x ∈ R.
∴ 5x – 6x is also continuous for all x ∈ R.
i.e., f(x) is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = 51 – 6(1) = -1 < 0
f(2) = (5)2 – 6(2) = 13 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 1 and 2 such that f(c) = 0.
∴ There is a root of the given equation in [1, 2].

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

Question 2.
Show that x3 – 5x2 + 3x + 6 = 0 has at least two real roots between x = 1 and x = 5.
Solution:
Let f(x) = x3 – 5x2 + 3x + 6
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
Here, we have been asked to show that f(x) has at least two roots between x = 1 and x = 5.
f(1) = (1)3 – 5(1)2 + 3(1) + 6
= 5 > 0
f(2) = (2)3 – 5(2)2 + 3(2) + 6
= 8 – 20 + 6 + 6
= 0
∴ x = 2 is a root of f(x).
Also, f(3) = (3)3 – 5(3)2 + 3(3) + 6
= 27 – 45 + 9 + 6
= -3 < 0
f(4) = (4)3 – 5(4)2 + 3(4) + 6
= 64 – 80 + 12 + 6
= 2 > 0
∴ f(3) < 0 and f(4) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 3 and 4 such that f(c) = 0.
∴ There are two roots, x = 2 and a root between x = 3 and x = 4.
Thus, there are at least two roots of the given equation between x = 1 and x = 5.

Class 11 Maharashtra State Board Maths Solution 

Continuity Class 11 Maths 2 Exercise 8.1 Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Ex 8.1 Questions and Answers.

11th Maths Part 2 Continuity Exercise 8.1 Questions And Answers Maharashtra Board

Question 1.
Examine the continuity of
(i) f(x) = x3 + 2x2 – x – 2 at x = -2
Solution:
Given, f(x) = x3 + 2x2 – x – 2
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
∴ f(x) is continuous at x = -2.

(ii) f(x) = sin x, for x ≤ \(\frac{\pi}{4}\)
= cos x, for x > \(\frac{\pi}{4}\), at x = \(\frac{\pi}{4}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q1 (ii)

(iii) f(x) = \(\frac{x^{2}-9}{x-3}\), for x ≠ 3
= 8 for x = 3, at x = 3.
Solution:
f(3) = 8 ….(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q1 (iii)
∴ f(x) is discontinuous at x = 3.

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

Question 2.
Examine whether the function is continuous at the points indicated against them.
(i) f(x) = x3 – 2x + 1, if x ≤ 2
= 3x – 2, if x > 2, at x = 2.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q2 (i)

(ii) f(x) = \(\frac{x^{2}+18 x-19}{x-1}\), for x ≠ 1
= 20, for x = 1, at x = 1.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q2 (ii)

(iii) f(x) = \(\frac{x}{\tan 3 x}+2\), for x < 0
= \(\frac{7}{3}\), for x ≥ 0, at x = 0.
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q2 (iii)

Question 3.
Find all the points of discontinuities of f(x) = [x] on the interval (-3, 2).
Solution:
f(x) = [x], x ∈ (-3, 2)
i.e., f(x) = -3, x ∈ (-3, -2)
= -2, x ∈ [-2, -1)
= -1, x ∈ [- 1, 0)
= 0, x ∈ [0, 1)
= 1, x ∈ [1, 2)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q3
Similarly, f(x) is discontinuous at the points x = -1, x = 0, x = 1.
Thus all the integer values of x in the interval (-3, 2),
i.e., the points x = -2, x = -1, x = 0 and x = 1 are the required points of discontinuities.

Question 4.
Discuss the continuity of the function f(x) = |2x + 3|, at x = \(\frac{-3}{2}\).
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q4
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q4.1

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

Question 5.
Test the continuity of the following functions at the points or intervals indicated against them.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5.1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5 (i)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5 (ii)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5 (ii).1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5 (iii)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5 (iv)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5 (iv).1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5 (iv).2
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q5 (v)

Question 6.
Identify discontinuities for the following functions as either a jump or a removable discontinuity.
(i) f(x) = \(\frac{x^{2}-10 x+21}{x-7}\)
Solution:
Given, f(x) = \(\frac{x^{2}-10 x+21}{x-7}\)
It is a rational function and is discontinuous if
x – 7 = 0, i.e., x = 7
∴ f(x) is continuous for all x ∈ R, except at x = 7.
∴ f(7) is not defined.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q6 (i)
Thus, \(\lim _{x \rightarrow 7} \mathrm{f}(x)\) exist but f(7) is not defined.
∴ f(x) has a removable discontinuity.

(ii) f(x) = x2 + 3x – 2, for x ≤ 4
= 5x + 3, for x > 4.
Solution:
f(x) = x2 + 3x – 2, x ≤ 4
= 5x + 3, x > 4
f(x) is a polynomial function for both the intervals.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = 4.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q6 (ii)
∴ f(x) is discontinuous at x = 4.
∴ f(x) has a jump discontinuity at x = 4.

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

(iii) f(x) = x2 – 3x – 2, for x < -3 = 3 + 8x, for x > -3.
Solution:
f(x) = x2 – 3x – 2, x < -3 = 3 + 8x, x > -3
f(x) is a polynomial function for both the intervals.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = -3.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q6 (iii)
∴ f(x) is discontinuous at x = -3.
∴ f(x) has a jump discontinuity at x = -3

(iv) f(x) = 4 + sin x, for x < π = 3 – cos x for x > π.
Solution:
f(x) = 4 + sin x, x < π = 3 – cos x, x > π
sin x and cos x are continuous for all x ∈ R.
4 and 3 are constant functions.
∴ 4 + sin x and 3 – cos x are continuous for all x ∈ R.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = π.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q6 (iv)
But f(π) is not defined.
∴ f(x) has a removable discontinuity at x = π.

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

Question 7.
Show that the following functions have a continuous extension to the point where f(x) is not defined. Also, find the extension.
(i) f(x) = \(\frac{1-\cos 2 x}{\sin x}\), for x ≠ 0.
Solution:
f(x) = \(\frac{1-\cos 2 x}{\sin x}\), for x ≠ 0
Here, f(0) is not defined.
Consider,
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q7 (i)
But f(0) is not defined.
∴ f(x) has a removable discontinuity at x = 0.
∴ The extension of the original function is
f(x) = \(\frac{1-\cos 2 x}{\sin x}\) for x ≠ 0
= 0 for x = 0
∴ f(x) is continuous at x = 0.

(ii) f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), for x ≠ 0.
Solution:
f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), for x ≠ 0
Here, f(0) is not defined.
Consider,
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q7 (ii)
But f(0) is not defined.
∴ f(x) has a removable discontinuity at x = 0.
∴ The extension of the original function is
f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), x ≠ 0
= \(\frac{7}{2}\), x = 0
∴ f(x) is continuous at x = 0.

(iii) f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), for x ≠ -1
Solution:
f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), for x ≠ -1
Here, f(-1) is not defined.
Consider,
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q7 (iii)
But f(-1) is not defined.
∴ f(x) has a removable discontinuity at x = -1.
∴ The extension of the original function is
f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), x ≠ -1
= \(-\frac{2}{3}\), x = -1
∴ f(x) is continuous at x = \(-\frac{2}{3}\)

Question 8.
Discuss the continuity of the following functions at the points indicated against them.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q8
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q8 (i)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q8 (i).1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q8 (i).2
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q8 (ii)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q8 (ii).1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q8 (iii)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q8 (iii).1

Question 9.
Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9.1
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (i)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (i).1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (i).2
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (ii)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (ii).1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (ii).2
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (iii)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (iii).1
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (iv)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (v)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q9 (v).1

Question 10.
(i) If f(x) = \(\frac{\sqrt{2+\sin x}-\sqrt{3}}{\cos ^{2} x}\), for x ≠ \(\frac{\pi}{2}\), is continuous at x = \(\frac{\pi}{2}\) then find f(\(\frac{\pi}{2}\)).
Solution:
f(x) is continuous at x = \(\frac{\pi}{2}\), …..(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q10 (i)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q10 (i).1

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

(ii) If f(x) = \(\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{3 x^{2}+1}-1}\) for x ≠ 0, is continuous at x = 0 then find f(0).
Solution:
f(x) is continuous at x = 0, …..(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q10 (ii)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q10 (ii).1

(iii) If f(x) = \(\frac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^{2}}\) for x ≠ π, is continuous at x = π, then find f(π).
Solution:
f(x) is continuous at x = π, …..(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q10 (iii)

Question 11.
(i) If f(x) = \(\frac{24^{x}-8^{x}-3^{x}+1}{12^{x}-4^{x}-3^{x}+1}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then find k.
Solution:
f(x) is continuous at x = 0 …..(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q11 (i)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q11 (i).1

(ii) If f(x) = \(\frac{5^{x}+5^{-x}-2}{x^{2}}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then find k.
Solution:
f(x) is continuous at x = 0 …..(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q11 (ii)

(iii) If f(x) = \(\frac{\sin 2 x}{5 x}\) – a, for x > 0
= 4 for x = 0
= x2 + b – 3, for x < 0
is continuous at x = 0, find a and b.
Solution:
f(x) is continuous at x = 0 ……(given)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q11 (iii)

(iv) For what values of a and b is the function
f(x) = ax + 2b + 18, for x ≤ 0
= x2 + 3a – b, for 0 < x ≤ 2 = 8x – 2, for x > 2,
continuous for every x?
Solution:
f(x) is continuous for every x …..(given)
∴ f(x) is continuous at x = 0 and x = 2.
As f(x) is continuous at x = 0,
\(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q11 (iv)
∴ (2)2 + 3a – b = 8(2) – 2
∴ 4 + 3a – b = 14
∴ 3a – b = 10 …….(ii)
Subtracting (i) from (ii), we get
2a = 4
∴ a = 2
Substituting a = 2 in (i), we get
2 – b = 6
∴ b = -4
∴ a = 2 and b = -4

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

(v) For what values of a and b is the function
f(x) = \(\frac{x^{2}-4}{x-2}\), for x < 2
= ax2 – bx + 3, for 2 ≤ x < 3
= 2x – a + b, for x ≥ 3
continuous for every x on R?
Solution:
f(x) is continuous for every x on R …..(given)
∴ f(x) is continuous at x = 2 and x = 3.
As f(x) is continuous at x = 2,
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q11 (v)
∴ a(3)2 – b(3) + 3 = 2(3) – a + b
∴ 9a – 3b + 3 = 6 – a + b
∴ 10a – 4b = 3 …..(ii)
Multiplying (i) by 2, we get
8a – 4b = 2 ….(iii)
Subtracting (ii) from (iii), we get
-2a = -1
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (i), we get
4(\(\frac{1}{2}\)) – 2b = 1
∴ 2 – 2b = 1
∴ 1 = 2b
∴ b = \(\frac{1}{2}\)
∴ a = \(\frac{1}{2}\) and b = \(\frac{1}{2}\)

Question 12.
Discuss the continuity of f on its domain, where
f(x) = |x + 1|, for -3 ≤ x ≤ 2
= |x – 5|, for 2 < x ≤ 7
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q12

Question 13.
Discuss the continuity of f(x) at x = \(\frac{\pi}{4}\) where,
f(x) = \(\frac{(\sin x+\cos x)^{3}-2 \sqrt{2}}{\sin 2 x-1}\), for x ≠ \(\frac{\pi}{4}\)
= \(\frac{3}{\sqrt{2}}\), for x = \(\frac{\pi}{4}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q13
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q13.1

Question 14.
Determine the values of p and q such that the following function is continuous on the entire real number line.
f(x) = x + 1, for 1 < x < 3
= x2 + px + q, for |x – 2| ≥ 1.
Solution:
|x – 2| ≥ 1
∴ x – 2 ≥ 1 or x – 2 ≤ -1
∴ x ≥ 3 or x ≤ 1
∴ f(x) = x2 + px + q for x ≥ 3 as well as x ≤ 1
Thus, f(x) = x2 + px + q; x ≤ 1
= x + 1; 1 < x < 3 = x2 + px + q; x > 3
f(x) is continuous for all x ∈ R.
∴ f(x) is continuous at x = 1 and x = 3.
As f(x) is continuous at x = 1,
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q14
Subtracting (i) from (ii), we get
2p = -6
∴ p = -3
Substituting p = -3 in (i), we get
-3 + q = 1
∴ q = 4
∴ p = -3 and q = 4

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

Question 15.
Show that there is a root for the equation 2x3 – x – 16 = 0 between 2 and 3.
Solution:
Let f(x) = 2x3 – x – 16
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(2) = 2(2)3 – 2 – 16 = -2 < 0
f(3) = 2(3)3 – 3 – 16 = 35 > 0
∴ f(2) < 0 and f(3) > 0
∴ By intermediate value theorem,
there has to be point ‘c’ between 2 and 3 such that f(c) = 0.
∴ There is a root of the given equation between 2 and 3.

Question 16.
Show that there is a root for the equation x3 – 3x = 0 between 1 and 2.
Solution:
Let f(x) = x3 – 3x
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = (1)3 – 3(1) = -2 < 0
f(2) = (2)3 – 3(2) = 2 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem,
there has to be point ‘c’ between 1 and 2 such that f(c) = 0.
There is a root of the given equation between 1 and 2.

Question 17.
Let f(x) = ax + b (where a and b are unknown)
= x2 + 5 for x ∈ R
Find the values of a and b, so that f(x) is continuous at x = 1.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q17
Solution:
f(x) = x2 + 5, x ∈ R
∴ f(1) = 1 + 5 = 6
If f(x) = ax + b is continuous at x = 1, then
f(1) = \(\lim _{x \rightarrow 1}(a x+b)\) = a + b
∴ 6 = a + b where, a, b ∈ R
∴ There are infinitely many values of a and b.

Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

Question 18.
Activity: Suppose f(x) = px + 3 for a ≤ x ≤ b
= 5x2 – q for b < x ≤ c
Find the condition on p, q, so that f(x) is continuous on [a, c], by filling in the boxes.
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q18
Solution:
Maharashtra Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1 Q18.1

Class 11 Maharashtra State Board Maths Solution 

Limits Class 11 Maths 2 Miscellaneous Exercise 7 Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Miscellaneous Exercise 7 Questions and Answers.

11th Maths Part 2 Limits Miscellaneous Exercise 7 Questions And Answers Maharashtra Board

I. Select the correct answer from the given alternatives.

Question 1.
\(\lim _{x \rightarrow 2}\left(\frac{x^{4}-16}{x^{2}-5 x+6}\right)=\)
(A) 23
(B) 32
(C) -32
(D) -16
Answer:
(C) -32
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q1

Question 2.
\(\lim _{x \rightarrow-2}\left(\frac{x^{7}+128}{x^{3}+8}\right)=\)
(A) \(\frac{56}{3}\)
(B) \(\frac{112}{3}\)
(C) \(\frac{121}{3}\)
(D) \(\frac{28}{3}\)
Answer:
(B) \(\frac{112}{3}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q2

Question 3.
\(\lim _{x \rightarrow 3}\left(\frac{1}{x^{2}-11 x+24}+\frac{1}{x^{2}-x-6}\right)=\)
(A) \(-\frac{2}{25}\)
(B) \(\frac{2}{25}\)
(C) \(\frac{7}{25}\)
(D) \(-\frac{7}{25}\)
Answer:
(A) \(-\frac{2}{25}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q3

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

Question 4.
\(\lim _{x \rightarrow 5}\left(\frac{\sqrt{x+4}-3}{\sqrt{3 x-11-2}}\right)=\)
(A) \(\frac{-2}{9}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{5}{9}\)
(D) \(\frac{2}{9}\)
Answer:
(D) \(\frac{2}{9}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q4

Question 5.
\(\lim _{x \rightarrow \frac{\pi}{3}}\left(\frac{\tan ^{2} x-3}{\sec ^{3} x-8}\right)=\)
(A) 1
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(\frac{1}{3}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q5

Question 6.
\(\lim _{x \rightarrow 0}\left(\frac{5 \sin x-x \cos x}{2 \tan x-3 x^{2}}\right)=\)
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
(C) 2
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q6

Question 7.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{3 \cos x+\cos 3 x}{(2 x-\pi)^{3}}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(\frac{1}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(-\frac{1}{2}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q7

Question 8.
\(\lim _{x \rightarrow 0}\left(\frac{15^{x}-3^{x}-5^{x}+1}{\sin ^{2} x}\right)=\)
(A) log 15
(B) log 3 + log 5
(C) log 3 . log 5
(D) 3 log 5
Answer:
(C) log 3 . log 5
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q8

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

Question 9.
\(\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}}=\)
(A) e3
(B) e6
(C) e9
(D) e-3
Answer:
(A) e3
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q9

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(-\frac{5}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{2}{5}\)
Answer:
(D) \(\frac{2}{5}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q10

Question 11.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3^{\cos x}-1}{\frac{\pi}{2}-x}\right)=\)
(A) 1
(B) log 3
(C) \(3^{\frac{\pi}{2}}\)
(D) 3 log 3
Answer:
(B) log 3
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q11
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q11.1

Question 12.
\(\lim _{x \rightarrow 0}\left[\frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^{2}}\right]=\)
(A) \(\frac{1}{\mathrm{e}^{9}}\)
(B) \(\frac{1}{\mathrm{e}^{3}}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{3}\)
Answer:
(D) \(\frac{1}{3}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q12

Question 13.
\(\lim _{x \rightarrow 0}\left[\frac{\left(3^{\sin x}-1\right)^{3}}{\left(3^{x}-1\right) \cdot \tan x \cdot \log (1+x)}\right]=\)
(A) 3 log 3
(B) 2 log 3
(C) (log 3)2
(D) (log 3)3
Answer:
(C) (log 3)2
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q13

Question 14.
\(\lim _{x \rightarrow 3}\left[\frac{5^{x-3}-4^{x-3}}{\sin (x-3)}\right]=\)
(A) log 5 – 4
(B) log \(\frac{5}{4}\)
(C) \(\frac{\log 5}{\log 4}\)
(D) \(\frac{\log 5}{4}\)
Answer:
(B) log \(\frac{5}{4}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q14

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+3)^{7}(x-5)^{3}}{(2 x-5)^{10}}\right]=\)
(A) \(\frac{3}{8}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{8}\)
Hint:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q15

(II) Evaluate the following.

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{5}-1}{(1-x)^{3}-1}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q1

Question 2.
\(\lim _{x \rightarrow 0}[x]\) ([*] is a greatest integer function.)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q2
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q2.1

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

Question 3.
If f(r) = πr2 then find \(\lim _{h \rightarrow 0}\left[\frac{f(r+h)-f(r)}{h}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q3

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{x}{|x|+x^{2}}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q4
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q4.1

Question 5.
Find the limit of the function, if it exists, at x = 1
\(f(x)=\left\{\begin{array}{lll}
7-4 x & \text { for } & x<1 \\
x^{2}+2 & \text { for } & x \geq 1
\end{array}\right.\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q5

Question 6.
Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of \(\lim _{x \rightarrow 3} f(x)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q6

Question 7.
\(\lim _{x \rightarrow 0}\left[\frac{\sec x^{2}-1}{x^{4}}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q7
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q7.1

Question 8.
\(\lim _{x \rightarrow 0}\left[\frac{e^{x}+e^{-x}-2}{x \cdot \tan x}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q8

Question 9.
\(\lim _{x \rightarrow 0}\left[\frac{x\left(6^{x}-3^{x}\right)}{\cos (6 x)-\cos (4 x)}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q9
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q9.1

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^{x}+1}{x \cdot \tan x}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q10

Question 11.
\(\lim _{x \rightarrow a}\left[\frac{\sin x-\sin a}{x-a}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q11

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

Question 12.
\(\lim _{x \rightarrow 2}\left[\frac{\log x-\log 2}{x-2}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q12

Question 13.
\(\lim _{x \rightarrow 1}\left[\frac{a b^{x}-a^{x} b}{x^{2}-1}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q13

Question 14.
\(\lim _{x \rightarrow 0}\left[\frac{\left(5^{x}-1\right)^{2}}{\left(2^{x}-1\right) \log (1+x)}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q14
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q14.1

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+1)^{2}(7 x-3)^{3}}{(5 x+2)^{5}}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q15

Question 16.
\(\lim _{x \rightarrow a}\left[\frac{x \cos a-a \cos x}{x-a}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q16
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q16.1

Question 17.
\(\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{(\sin x-\cos x)^{2}}{\sqrt{2}-\sin x-\cos x}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q17
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q17.1

Question 18.
\(\lim _{x \rightarrow 1}\left[\frac{2^{2 x-2}-2^{x}+1}{\sin ^{2}(x-1)}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q18

Question 19.
\(\lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^{2}}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q19
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q19.1

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

Question 20.
\(\lim _{x \rightarrow 1}\left[\frac{\sqrt{x}-1}{\log x}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q20

Question 21.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q21
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q21.1
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q21.2

Question 22.
\(\lim _{x \rightarrow 1}\left(\frac{x+3 x^{2}+5 x^{3}+\cdots \cdots \cdots \cdots \cdots+(2 n-1) x^{n}-n^{2}}{x-1}\right)\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q22
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q22.1
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q22.2

Question 23.
\(\lim _{x \rightarrow 0} \frac{1}{x^{12}}\left[1-\cos \left(\frac{x^{2}}{2}\right)-\cos \left(\frac{x^{4}}{4}\right)+\cos \left(\frac{x^{2}}{2}\right) \cdot \cos \left(\frac{x^{4}}{4}\right)\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q23
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q23.1

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

Question 24.
\(\lim _{x \rightarrow \infty}\left(\frac{8 x^{2}+5 x+3}{2 x^{2}-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q24
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q24.1

Class 11 Maharashtra State Board Maths Solution 

Limits Class 11 Maths 2 Exercise 7.7 Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.7 Questions and Answers.

11th Maths Part 2 Limits Exercise 7.7 Questions And Answers Maharashtra Board

I. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \infty}\left[\frac{a x^{3}+b x^{2}+c x+d}{e x^{3}+f x^{2}+g x+h}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 I Q1

Question 2.
\(\lim _{x \rightarrow \infty}\left[\frac{x^{3}+3 x+2}{(x+4)(x-6)(x-3)}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 I Q2

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7

Question 3.
\(\lim _{x \rightarrow \infty}\left[\frac{7 x^{2}+5 x-3}{8 x^{2}-2 x+7}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 I Q3

II. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \infty}\left[\frac{7 x^{2}+2 x-3}{\sqrt{x^{4}+x+2}}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 II Q1
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 II Q1.1

Question 2.
\(\lim _{x \rightarrow \infty}\left[\sqrt{x^{2}+4 x+16}-\sqrt{x^{2}+16}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 II Q2
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 II Q2.1

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7

Question 3.
\(\lim _{x \rightarrow \infty}\left[\sqrt{x^{4}+4 x^{2}}-x^{2}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 II Q3
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 II Q3.1

III. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \infty}\left[\frac{\left(3 x^{2}+4\right)\left(4 x^{2}-6\right)\left(5 x^{2}+2\right)}{4 x^{6}+2 x^{4}-1}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q1
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q1.1

Question 2.
\(\lim _{x \rightarrow \infty}\left[\frac{(3 x-4)^{3}(4 x+3)^{4}}{(3 x+2)^{7}}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q2

Question 3.
\(\lim _{x \rightarrow \infty}[\sqrt{x}(\sqrt{x+1}-\sqrt{x})]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q3
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q3.1

Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7

Question 4.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x-1)^{20}(3 x-1)^{30}}{(2 x+1)^{50}}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q4

Question 5.
\(\lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^{2}+5}-\sqrt{x^{2}-3}}{\sqrt{x^{2}+3}-\sqrt{x^{2}+1}}\right]\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q5
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q5.1
Maharashtra Board 11th Maths Solutions Chapter 7 Limits Ex 7.7 III Q5.2

Class 11 Maharashtra State Board Maths Solution