Class 12

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 1.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 2 \\
2 & 2 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 3 \\
-1 & 4 \\
-2 & 1
\end{array}\right]\) show that
(i) A + B = B + A
(ii) (A + B) + C = A + (B + C)
Solution:


From (1) and (2), we get
(A + B) + C = A + (B + C).

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & -3 \\
4 & -7
\end{array}\right]\), then find the matrix A – 2B + 6I, where I is the unit matrix of order 2.
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
-3 & 7 & -8 \\
0 & -6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
9 & -1 & 2 \\
-4 & 2 & 5 \\
4 & 0 & -3
\end{array}\right]\), then find the matrix C such that A + B + C is a zero matrix.
Solution:
A + B + C = 0
∴ C = -A – B

Question 4.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
3 & -5 \\
-6 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & -2 \\
4 & 2 \\
1 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]\), find the matrix X such that 3A – 4B + 5X = C.
Solution:
3A – 4B + 5X = C
∴ 5X = C – 3A + 4B

Question 5.
If A = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\), find (A^T)^T.
Solution:
A = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rr}
5 & 3 \\
1 & 2 \\
-4 & 0
\end{array}\right]\)
∴ (A^T)^T = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\) = A

Question 6.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\), find (A^T)^T.
Solution:
A = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rrr}
7 & -2 & 5 \\
3 & -4 & 9 \\
1 & 1 & 1
\end{array}\right]\)
∴ (A^T)^T = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\) = A

Question 7.
Find a, b, c if \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\) is a symetric matrix.
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\)
Since, A is a symmetric matrix, a_ij = a_ji for all i and j

Question 8.
Find x, y, z if \(\left[\begin{array}{ccc}
0 & -5 i & x \\
y & 0 & z \\
\frac{3}{2} & -\sqrt{2} & 0
\end{array}\right]\) is a skew symmetric matrix.
Solution:
Let A = \(\left[\begin{array}{ccc}
0 & -5 i & x \\
y & 0 & z \\
\frac{3}{2} & -\sqrt{2} & 0
\end{array}\right]\)
Since, A is skew-symmetric matrix,

Question 9.
For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither:
(i) \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Then A^T = \(\left[\begin{array}{rrr}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Since, A = A^T, A is a symmetric matrix.

(ii) \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
Solution:
Let B = \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
Then B^T = \(\left(\begin{array}{rrr}
2 & -5 & -1 \\
5 & 4 & -6 \\
1 & 6 & 3
\end{array}\right)\)
∴ B ≠ B^T
Also,
-B^T = \(\left(\begin{array}{rrr}
2 & -5 & -1 \\
5 & 4 & -6 \\
1 & 6 & 3
\end{array}\right)=\left(\begin{array}{rrr}
-2 & 5 & 1 \\
-5 & -4 & 6 \\
-1 & -6 & -3
\end{array}\right)\)
∴ B ≠ -B^T
Hence, B is neither symmetric nor skew-symmetric matrix.

(iii) \(\left[\begin{array}{ccc}
0 & 1+2 i & i-2 \\
-1-2 i & 0 & -7 \\
2-i & 7 & 0
\end{array}\right]\)
Solution:

Hence, C is a skew-symmetric matrix.

Question 10.
Construct the matrix A = [a_ij]_3×3, where a_ij = i – j. State whether A is symmetric or skew-symmetric.
Solution:

Question 11.
Solve the following equations for X and Y, if 3X – Y = \(\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\) and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\)
Solution:

Question 12.
Find matrices A and B, if 2A – B = \(\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\)
Solution:

Question 13.
Find x and y, if \(\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]\)
Solution:

By equality of matrices, we get
2x + y – 1 = 3 ……..(1)
and 4y = 18 ……….(2)
From (2), y = \(\frac{9}{2}\)
Substituting y = \(\frac{9}{2}\) in (1), we get
2x + \(\frac{9}{2}\) – 1 = 3
∴ 2x = 3 – \(\frac{7}{2}\) = \(\frac{-1}{2}\)
∴ x = \(\frac{-1}{4}\)
Hence, x = \(\frac{-1}{4}\) and y = \(\frac{9}{2}\).

Question 14.
If \(\left[\begin{array}{cc}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\), find a, b, c and d.
Solution:
\(\left[\begin{array}{cc}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\)
By equality of matrices,
2a + b = 2 ….. (1)
3a – b = 3 …… (2)
c + 2d = 4 …… (3)
2c – d = -1 …… (4)
Adding (1) and (2), we get
5a = 5
∴ a = 1
Substituting a = 1 in (1), we get
2(1) + b = 2
∴ b = 0
Multiplying equation (4) by 2, we get
4c – 2d = -2 …… (5)
Adding (3) and (5), we get
5c = 2
∴ c = \(\frac{2}{5}\)
Substituting c = \(\frac{2}{5}\) in (4), we get
2(\(\frac{2}{5}\)) – d = -1
∴ d = \(\frac{4}{5}\) + 1 = \(\frac{9}{5}\)
Hence, a = 1, b = 0, c = \(\frac{2}{5}\) and d = \(\frac{9}{5}\).

Question 15.
There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B:
July sales (in Rupees), Physics, Chemistry, Mathematics
A = \(\left[\begin{array}{lll}
5600 & 6750 & 8500 \\
6650 & 7055 & 8905
\end{array}\right]\) First Row Suresh / Second Row Ganesh
August Sales (in Rupees), Physics, Chemistry, Mathematics
B = \(\left[\begin{array}{ccc}
6650 & 7055 & 8905 \\
7000 & 7500 & 10200
\end{array}\right]\) First Row Suresh / Second Row Ganesh
(i) Find the increase in sales in Z from July to August 2017.
(ii) If both book shops get 10% profit in the month of August 2017,
find the profit for each bookseller in each subject in that month.
Solution:
The sales for July and August 2017 for Suresh and Ganesh are given by the matrices A and B as:

(i) The increase in sales (in ₹) from July to August 2017 is obtained by subtracting the matrix A from B.


Hence, the increase in sales (in ₹) from July to August 2017 for:
Suresh book shop: ₹ 1050 in Physics, ₹ 305 in Chemistry, and ₹ 405 in Mathematics.
Ganesh book shop: ₹ 350 in Physics, ₹ 445 in Chemistry, and ₹ 1295 in Mathematics.
(ii) Both the book shops get 10% profit in August 2017,
the profit for each bookseller in each subject in August 2017 is obtained by the scalar multiplication of matrix B by 10%,
i.e. \(\frac{10}{100}=\frac{1}{10}\)

Hence, the profit for Suresh book shop are ₹ 665 in Physics, ₹ 705.50 in Chemistry and ₹ 890.50 in Mathematics and for Ganesh book shop are ₹ 700 in Physics, ₹ 750 in Chemistry and ₹ 1020 in Mathematics.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.3

Question 1.
Write the negation of each of the following statements:
(i) All men are animals.
Solution:
Some men are not animals.

(ii) 3 is a natural number.
Solution:
-3 is not a natural number.

(iii) It is false that Nagpur is the capital of Maharashtra.
Solution:
Nagpur is the capital of Maharashtra.

(iv) 2 + 3 ≠ 5.
Solution:
2 + 3 = 5.

Question 2.
Write the truth value of the negation of each of the following statements:
(i) √5 is an irrational number.
Solution:
Let p : √5 is an irrational number.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(ii) London is in England.
Solution:
Let p : London is in England.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(iii) For every x ∈ N, x + 3 < 8.
Solution:
Let p : For every x ∈ N, x + 3 < 8.
The truth value of p is F.
Therefore, the truth value of ~p is T.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

Question 1.
Express the following statements in symbolic form:
(i) e is a vowel or 2 + 3 = 5.
Solution:
Let p : e is a vowel.
q: 2 + 3 = 5.
Then the symbolic form of the given statement is p ∨ q.

(ii) Mango is a fruit but potato is a vegetable.
Solution:
Let p : Mango is a fruit.
q : Potato is a vegetable.
Then the symbolic form of the given statement is p ∧ q.

(iii) Milk is white or grass is green.
Solution:
Let p : Milk is white.
q : Grass is green.
Then the symbolic form of the given statement is p ∨ q.

(iv) I like playing but not singing.
Solution:
Let p : I like playing.
q : I am not singing.
Then the symbolic form of the given statement is p ∧ q.

(v) Even though it is cloudy, it is still raining.
Solution:
The given statement is equivalent to:
It is cloudy and it is still raining.
Let p : It is cloudy.
q : It is still raining.
Then the symbolic form of the given statement is p ∧ q.

Question 2.
Write the truth values of the following statements:
(I) Earth is a planet and Moon is a star.
Solution:
Let p : Earth is a planet.
q : Moon is a star.
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]

(ii) 16 is an even number and 8 is a perfect square.
Solution:
Let p : 16 is an even number.
q : 8 is a perfect square.
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) A quadratic equation has two distinct roots or 6 has three prime factors.
Solution:
Let p : A quadratic equation has two distinct roots.
q : 6 has three prime factors.
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 F. …..[F ∨ F ≡ F]

(iv) The Himalayas are the highest mountains but they are part of India in the northeast.
Solution:
Let p : the Himalayas are the highest mountains.
q : They are part of India in the northeast.
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]

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.1 Questions and Answers.

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

State which of the following sentences are statements. Justify your answer. In case of statements, write down the truth value:

Question (i).
A triangle has ‘ n’ sides.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (ii).
The sum of interior angles of a triangle is 180°.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (iii).
You are amazing!
Solution:
It is an exclamatory sentence, hence it is not a statement.

Question (iv).
Please grant me a loan.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (v).
√-4 is an irrational number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vi).
x² – 6x + 8 = 0 implies x = -4 or x = -2.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vii).
He is an actor.
Solution:
It is an open sentence, hence it is not a statement.

Question (viii).
Did you eat lunch yet?
Solution:
It is an interrogative sentence, hence it is not a statement.

Question (ix).
Have a cup of cappuccino.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (x).
(x + y)² = x² + 2xy + y² for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xi).
Every real number is a complex number.
Solution:
It is a statement that is true, hence its truth value is ‘T.

Question (xii).
1 is a prime number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xiii).
With the sunset, the day ends.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xiv).
1! = 0.
Solution:
It is a statement that is false, hence its truth value is

Question (xv).
3 + 5 > 11.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xvi).
The number π is an irrational number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xvii).
x² – y² = (x + y)(x – y) for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xviii).
The number 2 is only even a prime number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xix).
Two coplanar lines are either parallel or intersecting.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xx).
The number of arrangements of 7 girls in a row for a photograph is 7!
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxi).
Give me a compass box.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxii).
Bring the motor car here.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxiii).
It may rain today.
Solution:
It is an open sentence, hence it is not a statement.

Question (xxiv).
If a + b < 7, where a ≥ 0 and b ≥ 0, then a < 7 and b < 7.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxv).
Can you speak English?
Solution:
It is an interrogative sentence, hence it is not a statement.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

I. Choose the correct option from the given alternatives:

Question 1.
\(\int_{2}^{3} \frac{d x}{x\left(x^{3}-1\right)}=\)
(a) \(\frac{1}{3} \log \left(\frac{208}{189}\right)\)
(b) \(\frac{1}{3} \log \left(\frac{189}{208}\right)\)
(c) \(\log \left(\frac{208}{189}\right)\)
(d) \(\log \left(\frac{189}{208}\right)\)
Answer:
(a) \(\frac{1}{3} \log \left(\frac{208}{189}\right)\)

Question 2.
\(\int_{0}^{\pi / 2} \frac{\sin ^{2} x \cdot d x}{(1+\cos x)^{2}}=\)
(a) \(\frac{4-\pi}{2}\)
(b) \(\frac{\pi-4}{2}\)
(c) 4 – \(\frac{\pi}{2}\)
(d) \(\frac{4+\pi}{2}\)
Answer:
(a) \(\frac{4-\pi}{2}\)

Question 3.
\(\int_{0}^{\log 5} \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3} \cdot d x=\)
(a) 3 + 2π
(b) 4 – π
(c) 2 + π
(d) 4 + π
Answer:
(b) 4 – π

Question 4.
\(\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{2} x \cdot d x=\)
(a) \(\frac{7 \pi}{256}\)
(b) \(\frac{3 \pi}{256}\)
(c) \(\frac{5 \pi}{256}\)
(d) \(\frac{-5 \pi}{256}\)
Answer:
(c) \(\frac{5 \pi}{256}\)

Question 5.
If \(\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{X}}=\frac{k}{3}\), then k is equal to
(a) √2(2√2 – 2)
(b) \(\frac{\sqrt{2}}{3}\)(2 – 2√2)
(c) \(\frac{2 \sqrt{2}-2}{3}\)
(d) 4√2
Answer:
(d) 4√2

Question 6.
\(\int_{1}^{2} \frac{1}{x^{2}} e^{\frac{1}{x}} \cdot d x=\)
(a) √e + 1
(b) √e − 1
(c) √e(√e − 1)
(d) \(\frac{\sqrt{e}-1}{e}\)
Answer:
(c) √e(√e − 1)

Question 7.
If \(\int_{2}^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] \cdot d x=a+\frac{b}{\log 2}\), then
(a) a = e, b = -2
(b) a = e, b = 2
(c) a = -e, b = 2
(d) a = -e, b = -2
Answer:
(a) a = e, b = -2

Question 8.
Let \(\mathrm{I}_{1}=\int_{e}^{e^{2}} \frac{d x}{\log x}\) and \(\mathrm{I}_{2}=\int_{1}^{2} \frac{e^{x}}{\boldsymbol{X}} \cdot d x\), then
(a) I₁ = \(\frac{1}{3}\) I₂
(b) I₁ + I₂ = 0
(c) I₁ = 2I₂
(d) I₁ = I₂
Answer:
(d) I₁ = I₂

Question 9.
\(\int_{0}^{9} \frac{\sqrt{X}}{\sqrt{X}+\sqrt{9-X}} \cdot d x=\)
(a) 9
(b) \(\frac{9}{2}\)
(c) 0
(d) 1
Answer:
(b) \(\frac{9}{2}\)

Question 10.
The value of \(\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right) \cdot d \theta\) is
(a) 0
(b) 1
(c) 2
(d) π
Answer:
(a) 0

II. Evaluate the following:

Question 1.
\(\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x\)
Solution:
Let I = \(\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x\)
Put Numerator = A(Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ cos x = A(3 cos x + sin x) + B[\(\frac{d}{d x}\)(3 cos x + sin x)]
= A(3 cos x + sin x) + B(-3 sin x + cos x)
∴ cos x + 0 . sin x = (3A + B) cos x + (A – 3B) sin x
Comparing the coefficients of sinx and cos x on both the sides, we get
3A + B = 1 ………. (1)
A – 3B = 0 ………. (2)
Multiplying equation (1) by 3, we get
9A + 3B = 3 ………(3)
Adding (2) and (3), we get
10A = 3
∴ A = \(\frac{3}{10}\)

Question 2.
\(\int_{\pi / 4}^{\pi / 2} \frac{\cos \theta}{\left[\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right]^{3}} d \theta\)
Solution:

Question 3.
\(\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x\)
Solution:
Let I = \(\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x\)
Put √x = t
∴ x = t² and dx = 2t . dt
When x = 0, t = 0
When x = 1, t = 1

Question 4.
\(\int_{0}^{\pi / 4} \frac{\tan ^{3} x}{1+\cos 2 x} d x\)
Solution:

Question 5.
\(\int_{0}^{1} t^{5} \sqrt{1-t^{2}} d t\)
Solution:

Question 6.
\(\int_{0}^{1}\left(\cos ^{-1} x\right)^{2} d x\)
Solution:

Question 7.
\(\int_{-1}^{1} \frac{1+x^{3}}{9-x^{2}} d x\)
Solution:

Question 8.
\(\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{4} x d x\)
Solution:

Question 9.
\(\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x} d x\)
Solution:

Question 10.
\(\int_{1}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x\)
Solution:

III. Evaluate the following:

Question 1.
\(\int_{0}^{1}\left(\frac{1}{1+x^{2}}\right) \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Solution:

Question 2.
\(\int_{0}^{\pi / 2} \frac{1}{6-\cos x} d x\)
Solution:

Question 3.
\(\int_{0}^{a} \frac{1}{a^{2}+a x-x^{2}} d x\)
Solution:

Question 4.
\(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x\)
Solution:

Question 5.
\(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Solution:
Let I = \(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)

Question 6.
\(\int_{0}^{\pi / 4} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x\)
Solution:

Question 7.
\(\int_{0}^{\pi / 2}[2 \log (\sin x)-\log (\sin 2 x)] d x\)
Solution:

Question 8.
\(\int_{0}^{\pi}\left(\sin ^{-1} x+\cos ^{-1} x\right)^{3} \sin ^{3} x d x\)
Solution:

Question 9.
\(\int_{0}^{4}\left[\sqrt{x^{2}+2 x+3}\right]^{-1} d x\)
Solution:

Question 10.
\(\int_{-2}^{3}|x-2| d x\)
Solution:
|x – 2|= 2 – x, if x < 2
= x – 2, if x ≥ 2

IV. Evaluate the following:

Question 1.
If \(\int_{a}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x\), find the value of \(\int_{a}^{a+1} x d x\).
Solution:

Question 2.
If \(\int_{0}^{k} \frac{1}{2+8 x^{2}} \cdot d x=\frac{\pi}{16}\), find k.
Solution:

Question 3.
If f(x) = a + bx + cx², show that \(\int_{0}^{1} f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right]\)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

I. Evaluate:

Question 1.
\(\int_{1}^{9} \frac{x+1}{\sqrt{x}} \cdot d x\)
Solution:

Question 2.
\(\int_{2}^{3} \frac{1}{x^{2}+5 x+6} \cdot d x\)
Solution:

Question 3.
\(\int_{0}^{\pi / 4} \cot ^{2} \cdot d x\)
Solution:

The integral does not exist since cot 0 is not defined.

Question 4.
\(\int_{-\pi / 4}^{\pi / 4} \frac{1}{1-\sin x} \cdot d x\)
Solution:

Question 5.
\(\int_{3}^{5} \frac{1}{\sqrt{2 x+3}-\sqrt{2 x-3}} \cdot d x\)
Solution:

Question 6.
\(\int_{0}^{1} \frac{x^{2}-2}{x^{2}+1} \cdot d x\)
Solution:

Question 7.
\(\int_{0}^{\pi / 4} \sin 4 x \sin 3 x \cdot d x\)
Solution:

Question 8.
\(\int_{0}^{\pi / 4} \sqrt{1+\sin 2 x} \cdot d x\)
Solution:

Question 9.
\(\int_{0}^{\pi / 4} \sin ^{4} x \cdot d x\)
Solution:

Question 10.
\(\int_{-4}^{2} \frac{1}{x^{2}+4 x+13} \cdot d x\)
Solution:

Question 11.
\(\int_{0}^{4} \frac{1}{\sqrt{4 x-x^{2}}} \cdot d x\)
Solution:

Question 12.
\(\int_{0}^{1} \frac{1}{\sqrt{3+2 x-x^{2}}} \cdot d x\)
Solution:

Question 13.
\(\int_{0}^{\pi / 2} x \cdot \sin x \cdot d x\)
Solution:

Question 14.
\(\int_{0}^{1} x \cdot \tan ^{-1} x \cdot d x\)
Solution:

Question 15.
\(\int_{0}^{\infty} x \cdot e^{-x} \cdot d x\)
Solution:

II. Evaluate:

Question 1.
\(\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} \cdot d x\)
Solution:

Question 2.
\(\int_{0}^{\pi / 4} \frac{\sec ^{2} x}{3 \tan ^{2} x+4 \tan x+1} \cdot d x\)
Solution:

Question 3.
\(\int_{0}^{4 \pi} \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} \cdot d x\)
Solution:

Question 4.
\(\int_{0}^{2 \pi} \sqrt{\cos x} \cdot \sin ^{3} x \cdot d x\)
Solution:

Question 5.
\(\int_{0}^{\pi / 2} \frac{1}{5+4 \cos x} \cdot d x\)
Solution:

Question 6.
\(\int_{0}^{\pi / 4} \frac{\cos x}{4-\sin ^{2} x} \cdot d x\)
Solution:

Question 7.
\(\int_{0}^{\pi / 2} \frac{\cos X}{(1+\sin x)(2+\sin x)} \cdot d x\)
Solution:

Question 8.
\(\int_{-1}^{1} \frac{1}{a^{2} e^{x}+b^{2} e^{-x}} \cdot d x\)
Solution:

Question 9.
\(\int_{0}^{\pi} \frac{1}{3+2 \sin x+\cos x} \cdot d x\)
Solution:

Question 10.
\(\int_{0}^{\pi / 4} \sec ^{4} x \cdot d x\)
Solution:

Question 11.
\(\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \cdot d x\)
Solution:

Question 12.
\(\int_{0}^{\pi} \sin ^{3} x(1+2 \cos x)(1+\cos x)^{2} \cdot d x\)
Solution:

Question 13.
\(\int_{0}^{\pi / 2} \sin 2 x \cdot \tan ^{-1}(\sin x) \cdot d x\)
Solution:

Question 14.
\(\int_{\frac{1}{\sqrt{2}}}^{1} \frac{\left(e^{\cos ^{-1} x}\right)\left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}} \cdot d x\)
Solution:

Question 15.
\(\int_{2}^{3} \frac{\cos (\log x)}{x} \cdot d x\)
Solution:

III. Evaluate:

Question 1.
\(\int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} \cdot d x\)
Solution:

Question 2.
\(\int_{0}^{\pi / 2} \log \tan x \cdot d x\)
Solution:

Question 3.
\(\int_{0}^{1} \log \left(\frac{1}{x}-1\right) \cdot d x\)
Solution:

Question 4.
\(\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cdot \cos x} \cdot d x\)
Solution:

Question 5.
\(\int_{0}^{3} x^{2}(3-x)^{\frac{5}{2}} \cdot d x\)
Solution:

Question 6.
\(\int_{-3}^{3} \frac{x^{3}}{9-x^{2}} \cdot d x\)
Solution:

Question 7.
\(\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2+\sin x}{2-\sin x}\right) \cdot d x\)
Solution:

Question 8.
\(\int_{-\pi / 4}^{\pi / 4} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} \cdot d x\)
Solution:

Question 9.
\(\int_{-\pi / 4}^{\pi / 4} x^{3} \cdot \sin ^{4} x \cdot d x\)
Solution:

Question 10.
\(\int_{0}^{1} \frac{\log (x+1)}{x^{2}+1} \cdot d x\)
Solution:

Question 11.
\(\int_{-1}^{1} \frac{x^{3}+2}{\sqrt{x^{2}+4}} \cdot d x\)
Solution:

Question 12.
\(\int_{-a}^{a} \frac{x+x^{3}}{16-x^{2}} \cdot d x\)
Solution:

Question 13.
\(\int_{0}^{1} t^{2} \sqrt{1-t} \cdot d t\)
Solution:

Question 14.
\(\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{2} x \cdot d x\)
Solution:

Question 15.
\(\int_{0}^{1} \frac{\log x}{\sqrt{1-x^{2}}} \cdot d x\)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

I. Evaluate the following integrals as a limit of a sum.

Question 1.
\(\int_{1}^{3}(3 x-4) \cdot d x\)
Solution:
Let f(x) = 3x – 4, for 1 ≤ x ≤ 3
Divide the closed interval [1, 3] into n subintervals each of length h at the points
1, 1 + h, 1 + 2h, 1 + rh, ….., 1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 1
∴ f(a + rh) = f(1 + rh)
= 3(1 + rh) – 4
= 3rh – 1

Question 2.
\(\int_{0}^{4} x^{2} d x\)
Solution:
Let f(x) = x², for 0 ≤ x ≤ 4
Divide the closed interval [0, 4] into n subintervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 4
i.e. 0, h, 2h, ….., rh, ….., nh = 4
∴ h = \(\frac{4}{n}\) as n → ∞, h → 0
Here, a = 0

Question 3.
\(\int_{0}^{2} e^{x} d x\)
Solution:
Let f(x) = e^x, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n equal subntervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 0

Question 4.
\(\int_{0}^{2}\left(3 x^{2}-1\right) d x\)
Solution:
Let f(x) = 3x² – 1, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n subintervals each of length h at the points.
0, 0 + h, 0 + 2h, ….., 0 + rh, ……, 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 0
∴ f(a + rh) = f(0 + rh)
= f(rh)
= 3(rh)² – 1
= 3r²h² – 1

Question 5.
\(\int_{1}^{3} x^{3} d x\)
Solution:
Let f(x) = x³, for 1 ≤ x ≤ 3.
Divide the closed interval [1, 3] into n equal su bintervals each of length h at the points
1, 1 + h, 1 + 2h, ……, 1 + rh, ……, 1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here a = 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Miscellaneous Exercise 3

I. Choose the correct options from the given alternatives:

Question 1.
\(\int \frac{1+x+\sqrt{x+x^{2}}}{\sqrt{x}+\sqrt{1+x}} \cdot d x=\)
(a) \(\frac{1}{2} \sqrt{x+1}+c\)
(b) \(\frac{2}{3}(x+1)^{\frac{3}{2}}+c\)
(c) \(\sqrt{x+1}+c\)
(d) \(2(x+1)^{\frac{3}{2}}+c\)
Answer:
(b) \(\frac{2}{3}(x+1)^{\frac{3}{2}}+c\)

Question 2.
\(\int \frac{1}{x+x^{5}} \cdot d x\) = f(x) + c, then \(\int \frac{x^{4}}{x+x^{5}} \cdot d x=\)
(a) log x – f(x) + c
(b) f(x) + log x + c
(c) f(x) – log x + c
(d) \(\frac{1}{5}\) x⁵ f(x) + c
Answer:
(a) log x – f(x) + c

Question 3.
\(\int \frac{\log (3 x)}{x \log (9 x)} \cdot d x=\)
(a) log(3x) – log(9x) + c
(b) log(x) – (log 3) . log(log 9x) + c
(c) log 9 – (log x) . log(log 3x) + c
(d) log(x) + log(3) . log(log 9x) + c
Answer:
(b) log(x) – (log 3) . log(log 9x) + c

Question 4.
\(\int \frac{\sin ^{m} X}{\cos ^{m+2} X} \cdot d x=\)
(a) \(\frac{\tan ^{m+1} \boldsymbol{X}}{m+1}+c\)
(b) (m + 2) tan^m+1 x + c
(c) \(\frac{\tan ^{m} \boldsymbol{X}}{m}+c\)
(d) (m + 1) tan^m+1 x + c
Answer:
(a) \(\frac{\tan ^{m+1} \boldsymbol{X}}{m+1}+c\)

Question 5.
∫tan(sin^-1 x) . dx =
(a) \(\left(1-x^{2}\right)^{-\frac{1}{2}}+c\)
(b) \(\left(1-x^{2}\right)^{\frac{1}{2}}+c\)
(c) \(\frac{\tan ^{m} \boldsymbol{X}}{\sqrt{1-x^{2}}}+c\)
(d) \(-\sqrt{1-x^{2}}+c\)
Answer:
(d) \(-\sqrt{1-x^{2}}+c\)

Hint: sin^-1 x = \(\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)\)

Question 6.
\(\int \frac{x-\sin x}{1-\cos x} \cdot d x=\)
(a) x cot(\(\frac{x}{2}\)) + c
(b) -x cot(\(\frac{x}{2}\)) + c
(c) cot(\(\frac{x}{2}\)) + c
(d) x tan(\(\frac{x}{2}\)) + c
Answer:
(b) -x cot(\(\frac{x}{2}\)) + c

Question 7.
If f(x) = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\), g(x) = \(e^{\sin ^{-1} x}\), then ∫f(x) . g(x) . dx =
(a) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x-1\right)+c\)
(b) \(e^{\sin ^{-1} x} \cdot\left(1-\sin ^{-1} x\right)+c\)
(c) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x+1\right)+c\)
(d) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} X-1\right)+c\)
Answer:
(a) \(e^{\sin ^{-1} x} \cdot\left(\sin ^{-1} x-1\right)+c\)

Question 8.
If ∫tan³ x . sec³ x . dx = (\(\frac{1}{m}\)) sec^m x – (\(\frac{1}{n}\)) secⁿ x + c, then (m, n) =
(a) (5, 3)
(b) (3, 5)
(c) \(\left(\frac{1}{5}, \frac{1}{3}\right)\)
(d) (4, 4)
Answer:
(a) (5, 3)

Hint: ∫tan³ x . sec³ x dx
= ∫sec² x . tan² x . sec x tan x dx
= ∫sec² x (sec² x – 1) sec x tan x dx
Put sec x = t.

Question 9.
\(\int \frac{1}{\cos x-\cos ^{2} x} \cdot d x=\)
(a) log(cosec x – cot x) + tan(\(\frac{x}{2}\)) + c
(b) sin 2x – cos x + c
(c) log(sec x + tan x) – cot(\(\frac{x}{2}\)) + c
(d) cos 2x – sin x + c
Answer:
(c) log(sec x + tan x) – cot(\(\frac{x}{2}\)) + c

Question 10.
\(\int \frac{\sqrt{\cot x}}{\sin x \cdot \cos x} \cdot d x=\)
(a) \(2 \sqrt{\cot x}+c\)
(b) \(-2 \sqrt{\cot x}+c\)
(c) \(\frac{1}{2} \sqrt{\cot x}+c\)
(d) \(\sqrt{\cot X}+c\)
Answer:
(b) \(-2 \sqrt{\cot x}+c\)

Question 11.
\(\int \frac{e^{x}(x-1)}{x^{2}} \cdot d x=\)
(a) \(\frac{e^{x}}{x}+c\)
(b) \(\frac{e^{x}}{x^{2}}+c\)
(c) \(\left(x-\frac{1}{x}\right) e^{x}+c\)
(d) x e^-x + c
Answer:
(a) \(\frac{e^{x}}{x}+c\)

Question 12.
∫sin(log x) . dx =
(a) \(\frac{x}{2}\) [sin(log x) – cos(log x)] + c
(b) \(\frac{x}{2}\) [sin(log x) + cos(log x)] + c
(c) \(\frac{x}{2}\) [cos(log x) – sin(log x)] + c
(d) \(\frac{x}{4}\) [cos(log x) – sin(log x)] + c
Answer:
(a) \(\frac{x}{2}\) [sin(log x) – cos(log x)] + c

Question 13.
∫x^x (1 + log x) . dx =
(a) \(\frac{1}{2}\) (1 + log x)² + c
(b) x^2x + c
(c) x^x log x + c
(d) x^x + c
Answer:
(d) x^x + c

Hint: \(\frac{d}{d x}\)(xx) = x^x (1 + log x)

Question 14.
\(\int \cos ^{-\frac{3}{7}} x \cdot \sin ^{-\frac{11}{7}} x \cdot d x=\)
(a) \(\log \left(\sin ^{-\frac{4}{7}} x\right)+c\)
(b) \(\frac{4}{7} \tan ^{\frac{4}{7}} x+c\)
(c) \(-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c\)
(d) \(\log \left(\cos ^{\frac{3}{7}} x\right)+c\)
Answer:
(c) \(-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c\)

Hint: \(\int \cos ^{-\frac{3}{7}} x \sin ^{-\frac{11}{7}} x d x\)
= \(\int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{-\frac{11}{7}} x \cdot \cos ^{2} x} d x\)
= \(\int \tan ^{-\frac{11}{7}} x \sec ^{2} x d x\)
Put tan x = t.

Question 15.
\(2 \int \frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x+\sin ^{2} x} \cdot d x=\)
(a) sin 2x + c
(b) cos 2x + c
(c) tan 2x + c
(d) 2 sin 2x + c
Answer:
(a) sin 2x + c

Question 16.
\(\int \frac{d x}{\cos x \sqrt{\sin ^{2} x-\cos ^{2} x}} \cdot d x=\)
(a) log(tan x – \(\sqrt{\tan ^{2} x-1}\)) + c
(b) sin^-1 (tan x) + c
(c) 1 + sin^-1 (cot x) + c
(d) log(tan x + \(\sqrt{\tan ^{2} x-1}\)) + c
Answer:
(d) log(tan x + \(\sqrt{\tan ^{2} x-1}\)) + c

Question 17.
\(\int \frac{\log x}{(\log e x)^{2}} \cdot d x=\)
(a) \(\frac{x}{1+\log x}+c\)
(b) x(1 + log x) + c
(c) \(\frac{1}{1+\log x}+c\)
(d) \(\frac{1}{1-\log x}+c\)
Answer:
(a) \(\frac{x}{1+\log x}+c\)

Question 18.
∫[sin(log x) + cos(log x)] . dx =
(a) x cos(log x) + c
(b) sin(log x) + c
(c) cos(log x) + c
(d) x sin(log x) + c
Answer:
(d) x sin(log x) + c

Question 19.
\(\int \frac{\cos 2 x-1}{\cos 2 x+1} \cdot d x=\)
(a) tan x – x + c
(b) x + tan x + c
(c) x – tan x + c
(d) -x – cot x + c
Answer:
(c) x – tan x + c

Question 20.
\(\int \frac{e^{2 x}+e^{-2 x}}{e^{x}} \cdot d x=\)
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)
(b) \(e^{x}+\frac{1}{3 e^{3 x}}+c\)
(c) \(e^{-x}+\frac{1}{3 e^{3 x}}+c\)
(d) \(e^{-x}-\frac{1}{3 e^{3 x}}+c\)
Answer:
(a) \(e^{x}-\frac{1}{3 e^{3 x}}+c\)

II. Integrate the following with respect to the respective variable:

Question 1.
(x – 2)² √x
Solution:

Question 2.
\(\frac{x^{7}}{x+1}\)
Solution:

Question 3.
\((6 x+5)^{\frac{3}{2}}\)
Solution:

Question 4.
\(\frac{t^{3}}{(t+1)^{2}}\)
Solution:

Question 5.
\(\frac{3-2 \sin x}{\cos ^{2} x}\)
Solution:

Question 6.
\(\frac{\sin ^{6} \theta+\cos ^{6} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}\)
Solution:

Question 7.
cos 3x cos 2x cos x
Solution:

Question 8.
\(\frac{\cos 7 x-\cos 8 x}{1+2 \cos 5 x}\)
Solution:

Question 9.
\(\cot ^{-1}\left(\frac{1+\sin x}{\cos x}\right)\)
Solution:
Let I = \(\int \cot ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x\)

III. Integrate the following w.r.t. x:

Question 1.
\(\frac{(1+\log x)^{3}}{x}\)
Solution:

Question 2.
cot^-1 (1 – x + x²)
Solution:

Question 3.
\(\frac{1}{x \sin ^{2}(\log x)}\)
Solution:

Question 4.
\(\sqrt{x} \sec \left(x^{\frac{3}{2}}\right) \tan \left(x^{\frac{3}{2}}\right)\)
Solution:

Question 5.
log(1 + cos x) – x tan(\(\frac{x}{2}\))
Solution:

Question 6.
\(\frac{x^{2}}{\sqrt{1-x^{6}}}\)
Solution:

Question 7.
\(\frac{1}{(1-\cos 4 x)(3-\cot 2 x)}\)
Solution:

Question 8.
log(log x) + (log x)^-2
Solution:

Question 9.
\(\frac{1}{2 \cos x+3 \sin x}\)
Solution:

Question 10.
\(\frac{1}{x^{3} \sqrt{x^{2}-1}}\)
Solution:

Question 11.
\(\frac{3 x+1}{\sqrt{-2 x^{2}+x+3}}\)
Solution:

Question 12.
log(x² + 1)
Solution:

Question 13.
e^2x sin x cos x
Solution:

Question 14.
\(\frac{x^{2}}{(x-1)(3 x-1)(3 x-2)}\)
Solution:

Question 15.
\(\frac{1}{\sin x+\sin 2 x}\)
Solution:

Question 16.
\(\sec ^{2} x \sqrt{7+2 \tan x-\tan ^{2} x}\)
Solution:

Question 17.
\(\frac{x+5}{x^{3}+3 x^{2}-x-3}\)
Solution:

Question 18.
\(\frac{1}{x\left(x^{5}+1\right)}\)
Solution:

Question 19.
\(\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}\)
Solution:

Question 20.
sec⁴ x cosec² x
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.4

I. Integrate the following w. r. t. x:

Question 1.
\(\frac{x^{2}+2}{(x-1)(x+2)(x+3)}\)
Solution:

Question 2.
\(\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}-2\right)\left(x^{2}+3\right)}\)
Solution:

Question 3.
\(\frac{12 x+3}{6 x^{2}+13 x-63}\)
Solution:

Question 4.
\(\frac{2 x}{4-3 x-x^{2}}\)
Solution:

Question 5.
\(\frac{x^{2}+x-1}{x^{2}+x-6}\)
Solution:

Question 6.
\(\frac{6 x^{3}+5 x^{2}-7}{3 x^{2}-2 x-1}\)
Solution:

Question 7.
\(\frac{12 x^{2}-2 x-9}{\left(4 x^{2}-1\right)(x+3)}\)
Solution:

Question 8.
\(\frac{1}{x\left(x^{5}+1\right)}\)
Solution:

Question 9.
\(\frac{2 x^{2}-1}{x^{4}+9 x^{2}+20}\)
Solution:

Question 10.
\(\frac{x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}-2\right)}\)
Solution:

Question 11.
\(\frac{2 x}{\left(2+x^{2}\right)\left(3+x^{2}\right)}\)
Solution:

Question 12.
\(\frac{2^{x}}{4^{x}-3 \cdot 2^{x}-4}\)
Solution:

Question 13.
\(\frac{3 x-2}{(x+1)^{2}(x+3)}\)
Solution:

Question 14.
\(\frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x}\)
Solution:
Let I = ∫\(\frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x}\) dx

Question 15.
\(\frac{1}{x\left(1+4 x^{3}+3 x^{6}\right)}\)
Solution:

Question 16.
\(\frac{1}{x^{3}-1}\)
Solution:

Question 17.
\(\frac{(3 \sin x-2) \cdot \cos x}{5-4 \sin x-\cos ^{2} x}\)
Solution:

Question 18.
\(\frac{1}{\sin x+\sin 2 x}\)
Solution:

Question 19.
\(\frac{1}{2 \sin x+\sin 2 x}\)
Solution:

Question 20.
\(\frac{1}{\sin 2 x+\cos x}\)
Solution:

Question 21.
\(\frac{1}{\sin x \cdot(3+2 \cos x)}\)
Solution:

Question 22.
\(\frac{5 \cdot e^{x}}{\left(e^{x}+1\right)\left(e^{2 x}+9\right)}\)
Solution:

Question 23.
\(\frac{2 \log x+3}{x(3 \log x+2)\left[(\log x)^{2}+1\right]}\)
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

(I) Choose the correct option from the given alternatives:

Question 1.
Let f(1) = 3, f'(1) = \(-\frac{1}{3}\), g(1) = -4 and g'(1) = \(-\frac{8}{3}\). The derivative of \(\sqrt{[f(x)]^{2}+[g(x)]^{2}}\) w.r.t. x at x = 1 is
(a) \(-\frac{29}{15}\)
(b) \(\frac{7}{3}\)
(c) \(\frac{31}{15}\)
(d) \(\frac{29}{15}\)
Answer:
(d) \(\frac{29}{15}\)

Question 2.
If y = sec(tan^-1 x), then \(\frac{d y}{d x}\) at x = 1, is equal to
(a) \(\frac{1}{2}\)
(b) 1
(c) \(\frac{1}{\sqrt{2}}\)
(d) 2
Answer:
(c) \(\frac{1}{\sqrt{2}}\)

Question 3.
If f(x) = \(\sin ^{-1}\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)\), which of the following is not the derivative of f(x)?
(a) \(\frac{2 \cdot 4^{x} \log 4}{1+4^{2 x}}\)
(b) \(\frac{4^{x+1} \log 2}{1+4^{2 x}}\)
(c) \(\frac{4^{x+1} \log 4}{1+4^{4 x}}\)
(d) \(\frac{2^{2(x+1)} \log 2}{1+2^{4 x}}\)
Answer:
(c) \(\frac{4^{x+1} \log 4}{1+4^{4 x}}\)

Question 4.
If x^y = y^x, then\(\frac{d y}{d x}\) = _______
(a) \(\frac{x(x \log y-y)}{y(y \log x-x)}\)
(b) \(\frac{y(y \log x-x)}{x(x \log y-y)}\)
(c) \(\frac{y^{2}(1-\log x)}{x^{2}(1-\log y)}\)
(d) \(\frac{y(1-\log x)}{x(1-\log y)}\)
Answer:
(b) \(\frac{y(y \log x-x)}{x(x \log y-y)}\)

Question 5.
If y = sin (2 sin^-1 x), then \(\frac{d y}{d x}\) = _______
(a) \(\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}\)
(b) \(\frac{2+4 x^{2}}{\sqrt{1-x^{2}}}\)
(c) \(\frac{4 x^{2}-1}{\sqrt{1-x^{2}}}\)
(d) \(\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\)
Answer:
(a) \(\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}\)

Question 6.
If y = \(\tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)+\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\), then \(\frac{d y}{d x}\) = _______
(a) \(\frac{x}{\sqrt{1-x^{2}}}\)
(b) \(\frac{1-2 x}{\sqrt{1-x^{2}}}\)
(c) \(\frac{1-2 x}{2 \sqrt{1-x^{2}}}\)
(d) \(\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\)
Answer:
(c) \(\frac{1-2 x}{2 \sqrt{1-x^{2}}}\)

Question 7.
If y is a function of x and log(x + y) = 2xy, then the value of y'(0) = _______
(a) 2
(b) 0
(c) -1
(d) 1
Answer:
(d) 1

Question 8.
If g is the inverse of function f and f'(x) = \(\frac{1}{1+x^{7}}\), then the value of g'(x) is equal to:
(a) 1 + x⁷
(b) \(\frac{1}{1+[g(x)]^{7}}\)
(c) 1 + [g(x)]⁷
(d) 7x⁶
Answer:
(c) 1 + [g(x)]⁷

Question 9.
If \(x \sqrt{y+1}+y \sqrt{x+1}=0\) and x ≠ y, then \(\frac{d y}{d x}\) = _______
(a) \(\frac{1}{(1+x)^{2}}\)
(b) \(-\frac{1}{(1+x)^{2}}\)
(c) (1 + x)²
(d) \(-\frac{x}{x+1}\)
Answer:
(b) \(-\frac{1}{(1+x)^{2}}\)

Question 10.
If y = \(\tan ^{-1}\left(\sqrt{\frac{a-x}{a+x}}\right)\), where -a < x < a, then \(\frac{d y}{d x}\) = _______
(a) \(\frac{x}{\sqrt{a^{2}-x^{2}}}\)
(b) \(\frac{a}{\sqrt{a^{2}-x^{2}}}\)
(c) \(-\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
(d) \(\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
Answer:
(c) \(-\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
[Hint: Put x = a cos θ]

Question 11.
If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), then \(\left[\frac{d^{2} y}{d x^{2}}\right]_{\theta=\frac{\pi}{4}}\) = _______
(a) \(\frac{8 \sqrt{2}}{a \pi}\)
(b) \(-\frac{8 \sqrt{2}}{a \pi}\)
(c) \(\frac{a \pi}{8 \sqrt{2}}\)
(d) \(\frac{4 \sqrt{2}}{a \pi}\)
Answer:
(a) \(\frac{8 \sqrt{2}}{a \pi}\)

Question 12.
If y = a cos (log x) and \(A \frac{d^{2} y}{d x^{2}}+B \frac{d y}{d x}+C=0\), then the values of A, B, C are _______
(a) x², -x, -y
(b) x², x, y
(c) x², x, -y
(d) x², -x, y
Answer:
(b) x², x, y

(II) Solve the following:

Question 1.


Let u(x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v'(1) and w'(1). if it doesn’t exist then explain why?
Solution:

Question 2.
The values of f(x), g(x), f'(x) and g'(x) are given in the following table:

Match the following:

Solution:

Question 3.
Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.

(i) The derivative of f[g(x)] w.r.t. x at x = 0 is _______
(ii) The derivative of g[f(x)] w.r.t. x at x = 0 is _______
(iii) The value of \(\left[\frac{d}{d x}\left[x^{10}+f(x)\right]^{-2}\right]_{x=1}\) is _______
(iv) The derivative of f[(x+g(x))] w.r.t. x at x = 0 is _______
Solution:

Question 4.
Differentiate the following w.r.t. x:
(i) \(\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\)
Solution:
Let y = \(\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\)
Put x = cos θ, Then θ = cos^-1 x and

(ii) \(\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]\)
Solution:
Let y = \(\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]\)
Put x = cos θ. Then θ = cos^-1 x and

(iii) \(\tan ^{-1}\left[\frac{\sqrt{x}(3-x)}{1-3 x}\right]\)
Solution:

(iv) \(\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)\)
Solution:
Let y = \(\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)\)
Put x = cos θ. Then θ = cos^-1 x and

(v) \(\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)+\cot ^{-1}\left(\frac{1-10 x^{2}}{7 x}\right)\)
Solution:

(vi) \(\tan ^{-1}\left[\sqrt{\frac{\sqrt{1+x^{2}+x}}{\sqrt{1+x^{2}}-x}}\right]\)
Solution:

Question 5.
(i) If \(\sqrt{y+x}+\sqrt{y-x}=c\), show that \(\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}\)
Solution:
\(\sqrt{y+x}+\sqrt{y-x}=c\)
Differentiating both sides w.r.t. x, we get

(ii) If \(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1\), then show that \(\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
Solution:
\(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1\)
\(y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1\)
Differentiating both sides w.r.t. x, we get

(iii) If x sin(a + y) + sin a cos(a + y) = 0, then show \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)
Solution:
x sin(a + y) + sin a . cos (a + y) = 0 ….. (1)
Differentiating w.r.t. x, we get

(iv) If sin y = x sin(a + y), then show that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)
Solution:

(v) If x = \(e^{\frac{x}{y}}\), then show that \(\frac{d y}{d x}=\frac{x-y}{x \log x}\)
Solution:
x = \(e^{\frac{x}{y}}\)
\(\frac{x}{y}\) = log x …..(1)
y = \(\frac{x}{\log x}\)

(vi) If y = f(x) is a differentiable function of x, then show that \(\frac{d^{2} x}{d y^{2}}=-\left(\frac{d y}{d x}\right)^{-3} \cdot \frac{d^{2} y}{d x^{2}}\)
Solution:
If y = f(x) is a differentiable function of x such that inverse function x = f^-1(y) exists,

Question 6.
(i) Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) w.r.t. \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Solution:

(ii) Differentiate \(\log \left[\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}-x}\right]\) w.r.t. cos(log x)
Solution:

(iii) Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) w.r.t. \(\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^{2}}}{2 \sqrt{1+x^{2}}}}\right)\)
Solution:

Question 7.
(i) If y² = a² cos²x + b² sin²x, show that \(y+\frac{d^{2} y}{d x^{2}}=\frac{a^{2} b^{2}}{y^{3}}\)
Solution:
y² = a² cos²x + b² sin²x …… (1)
Differentiating both sides w.r.t. x, we get

(ii) If log y = log(sin x) – x², show that \(\frac{d^{2} y}{d x^{2}}+4 x \frac{d y}{d x}+\left(4 x^{2}+3\right) y=0\)
Solution:

(iii) If x = a cos θ, y = b sin θ, show that \(a^{2}\left[y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]+b^{2}=0\)
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get

(iv) If y = A cos(log x) + B sin(log x), show that x²y₂ + xy₁ + y = o.
Solution:
y = A cos (log x) + B sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get

(v) If y = A e^mx + B e^nx, show that y₂ – (m + n) y₁ + (mn) y = 0.
Solution:
y = A e^mx + B e^nx
Differentiating w.r.t. x, we get