11th Commerce Maths 1 Chapter 5 Exercise 5.1 Answers Maharashtra Board

Locus and Straight Line Class 11 Commerce Maths 1 Chapter 5 Exercise 5.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.1 Questions and Answers.

Std 11 Maths 1 Exercise 5.1 Solutions Commerce Maths

Question 1.
If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3) and B(2, 1).
PA = PB
∴ PA2 = PB2
∴ (x – 1)2 + (y – 3)2 = (x – 2)2 + (y – 1)2
∴ x2 – 2x + 1 + y2 – 6y + 9 = x2 – 4x + 4 + y2 – 2y + 1
∴ -2x – 6y + 10 = -4x – 2y + 5
∴ 2x – 4y + 5 = 0
∴ The required equation of locus is 2x – 4y + 5 = 0.

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.1

Question 2.
A(-5, 2) and B(4, 1). Find the equation of the locus of point P, which is equidistant from A and B.
Solution:
Let P(x, y) be any point on the required locus.
P is equidistant from A(-5, 2) and B(4, 1).
∴ PA = PB
∴ PA2 = PB2
∴ (x + 5)2 + (y – 2)2 = (x – 4)2 + (y – 1)2
∴ x2 + 10x + 25 + y2 – 4y + 4 = x2 – 8x + 16 + y2 – 2y + 1
∴ 10x – 4y + 29 = -8x – 2y + 17
∴ 18x – 2y + 12 = 0
∴ 9x – y + 6 = 0
∴ The required equation of locus is 9x – y – 6 = 0

Question 3.
If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 0), B(0, 3) and AP = 2BP
∴ AP2 = 4BP2
∴ (x – 2)2 + (y – 0)2 = 4[(x – 0)2 + (y – 3)2]
∴ x2 – 4x + 4 + y2 = 4(x2 + y2 – 6y + 9)
∴ x2 – 4x + 4 + y2 = 4x2 + 4y2 – 24y + 36
∴ 3x2 + 3y2 + 4x – 24y + 32 = 0
∴ The required equation of locus is 3x2 + 3y2 + 4x – 24y + 32 = 0

Question 4.
If A(4, 1) and B(5, 4), find the equation of the locus of point P if PA2 = 3PB2.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(4, 1), B(5, 4) and PA2 = 3PB2
∴ (x – 4)2 + (y – 1)2 = 3[(x – 5)2 + (y – 4)2]
∴ x2 – 8x + 16 + y2 – 2y + 1 = 3(x2 – 10x + 25 + y2 – 8y + 16)
∴ x2 – 8x + y2 – 2y + 17 = 3x2 – 30x + 75 + 3y2 – 24y + 48
∴ 2x2 + 2y2 – 22x – 22y + 106 = 0
∴ x2 + y2 – 11x – 11y + 53 = 0
∴ The required equation of locus is x2 + y2 – 11x – 11y + 53 = 0.

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.1

Question 5.
A(2, 4) and B(5, 8), find the equation of the locus of point P such that PA2 – PB2 = 13.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 4), B(5, 8) and PA2 – PB2 = 13
∴ [(x – 2)2 + (y – 4)2] – [(x – 5)2 + (y – 8)2] = 13
∴ (x2 – 4x + 4 + y2 – 8y + 16) – (x2 – 10x + 25 + y2 – 16y + 64) = 13
∴ 6x + 8y – 69 = 13
∴ 6x + 8y – 82 = 0
∴ 3x + 4y – 41 = 0
∴ The required equation of locus is 3x + 4y – 41 = 0

Question 6.
A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends a right angle at P. (∠APB = 90°)
Solution:
Let P(x. y) be any point on the required locus.
Given, A(1, 6) and B(3, 5), ∠APB = 90°
∴ ΔAPB is a right-angled triangle.
Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.1 Q6
By Pythagoras theorem,
AP2 + PB2 = AB2
∴ [(x – 1)2 + (y – 6)2] + [(x – 3)2 + (y – 5)2] = (1 – 3)2 + (6 – 5)2
∴ x2 – 2x + 1 + y2 – 12y + 36 + x2 – 6x + 9 + y2 – 10y + 25 = 4 + 1
∴ 2x2 + 2y2 – 8x – 22y + 66 = 0
∴ x2 + y2 – 4x – 11y + 33 = 0
∴ The required equation of locus is x2 + y2 – 4x – 11y + 33 = 0

Question 7.
If the origin is shifted to the point O'(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points (a) A(1, 3) (b) B(2, 5)
Solution:
Origin is shifted to (2, 3) = (h, k)
Let the new co-ordinates be (X, Y).
∴ x = X + h and y = Y + k
∴ x = X + 2 and y = Y + 3 …..(i)
(a) Given, A(x, y) = A(1, 3)
x = X + 2 and y = Y + 3 …..[From (i)]
∴ 1 = X + 2 and 3 = Y + 3
∴ X = -1 and Y = 0
∴ the new co-ordinates of point A are (-1, 0).

(b) Given, B(x, y) = B(2, 5)
x = X + 2 andy = Y + 3 ……[From (i)]
∴ 2 = X + 2 and 5 = Y + 3
∴ X = 0 and Y = 2
∴ the new co-ordinates of point B are (0, 2).

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.1

Question 8.
If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points (a) C(5, 4) (b) D(3, 3)
Solution:
Origin is shifted to (1, 3) = (h, k)
Let the new co-ordinates be (X, Y)
x = X + h and y = Y + k
∴ x = X + 1 and 7 = Y + 3 …..(i)
(a) Given, C(X, Y) = C(5, 4)
∴ x = X + 1 andy = Y + 3 …..[From(i)]
∴ x = 5 + 1 = 6 and y = 4 + 3 = 7
∴ the old co-ordinates of point C are (6, 7).

(b) Given, D(X, Y) = D(3, 3)
∴ x = X + 1 and y = Y + 3 …..[From (i)]
∴ x = 3 + 1 = 4 and y = 3 + 3 = 6
∴ the old co-ordinates of point D are (4, 6).

Question 9.
If the co-ordinates (5, 14) change to (8, 3) by the shift of origin, find the co-ordinates of the point, where the origin is shifted.
Solution:
Let the origin be shifted to (h, k).
Given, (x,y) = (5, 14), (X, Y) = (8, 3)
Since, x = X + h and y = Y + k
∴ 5 = 8 + h and 14 = 3 + k
∴ h = -3 and k = 11
∴ the co-ordinates of the point, where the origin is shifted are (-3, 11).

Question 10.
Obtain the new equations of the following loci if the origin is shifted to the point O'(2, 2), the direction of axes remaining the same:
(a) 3x – y + 2 = 0
(b) x2 + y2 – 3x = 7
(c) xy – 2x – 2y + 4 = 0
Solution:
Given, (h, k) = (2, 2)
Let (X, Y) be the new co-ordinates of the point (x, y).
∴ x = X + h and y = Y + k
∴ x = X + 2 and y = Y + 2
(a) Substituting the values of x and y in the equation 3x – y + 2 = 0, we get
3(X + 2) – (Y + 2) + 2 = 0
∴ 3X + 6 – Y – 2 + 2 = 0
∴ 3X – Y + 6 = 0, which is the new equation of locus.

(b) Substituting the values of x and y in the equation x2 + y2 – 3x = 7, we get
(X + 2)2 + (Y + 2)2 – 3(X + 2) = 7
∴ X2 + 4X + 4 + Y2 + 4Y + 4 – 3X – 6 = 7
∴ X2 + Y2 + X + 4Y – 5 = 0, which is the new equation of locus.

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.1

(c) Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get
(X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0
∴ XY + 2X + 2Y + 4 – 2X – 4 – 2Y – 4 + 4 = 0
∴ XY = 0, which is the new equation of locus.

11th Commerce Maths Digest Pdf

11th Commerce Maths 1 Chapter 6 Exercise 6.3 Answers Maharashtra Board

Determinants Class 11 Commerce Maths 1 Chapter 6 Exercise 6.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Determinants Ex 6.3 Questions and Answers.

Std 11 Maths 1 Exercise 6.3 Solutions Commerce Maths

Question 1.
Solve the following equations using Cramer’s Rule.
(i) x + 2y – z = 5, 2x – y + z = 1, 3x + 3y = 8
Solution:
Given equations are
x + 2y – z = 5
2x – y + z = 1
3x + 3y = 8 i.e. 3x + 3y + 0z = 8
∴ D = \(\left|\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 1 \\
3 & 3 & 0
\end{array}\right|\)
= 1(0 – 3) – 2(0 – 3) – 1(6 + 3)
= -3 + 6 – 9
= -6
Dx = \(\left|\begin{array}{ccc}
5 & 2 & -1 \\
1 & -1 & 1 \\
8 & 3 & 0
\end{array}\right|\)
= 5(0 – 3) – 2(0 – 8) + (-1)(3 + 8)
= -15 + 16 – 11
= -10
Dy = \(\left|\begin{array}{ccc}
1 & 5 & -1 \\
2 & 1 & 1 \\
3 & 8 & 0
\end{array}\right|\)
= 1(0 – 8) – 5(0 – 3) + 1(16 – 3)
= -8 + 15 – 13
= -6
Dz = \(\left|\begin{array}{ccc}
1 & 2 & 5 \\
2 & -1 & 1 \\
3 & 3 & 8
\end{array}\right|\)
= 1(-8 – 3) – 2(16 – 3) + 5(6 + 3)
= -11 – 26 + 45
= 8
By Cramer’s Rule,
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q1(i)
x = \(\frac{5}{3}\), y = 1 and z = \(\frac{-4}{3}\) are the solutions of the given equations.

Check:
We can check if our answer is right or wrong.
In order to do so, substitute the values of x, y and z in the given equations.
x = \(\frac{5}{3}\), y = 1 and z = \(\frac{-4}{3}\) satisfy the given equations.
If either one of the equations is not satisfied, then our answer is wrong.
If x = \(\frac{5}{3}\), y = 1 and z = \(\frac{-4}{3}\) are the solutions of the given equations.
L.H.S. = x + 2y – z
= \(\frac{5}{3}+2-\frac{4}{3}\)
= \(\frac{7}{3}\)
≠ R.H.S.
L.H.S. = 2x – y + z
= \(\frac{10}{3}-1+\frac{4}{3}\)
= \(\frac{11}{3}\)
≠ R.H.S.
L.H.S. = 3x + 3y
= 5 + 3
= 8
= R.H.S.

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3

(ii) 2x – y + 6z = 10, 3x + 4y – 5z = 11, 8x – 7y – 9z = 12
Solution:
Given equations are
2x – y + 6z = 10
3x + 4y – 5z = 11
8x – 7y – 9z = 12
∴ D = \(\left|\begin{array}{ccc}
2 & -1 & 6 \\
3 & 4 & -5 \\
8 & -7 & -9
\end{array}\right|\)
= 2(-36 – 35) – (-1)(-27 + 40) + 6(-21 – 32)
= -142 + 13 – 318
= -447
Dx = \(\left|\begin{array}{ccc}
10 & -1 & 6 \\
11 & 4 & -5 \\
12 & -7 & -9
\end{array}\right|\)
= 10(-36 – 35) – (-1)(-99 + 60) + 6(-77 – 48)
= -710 – 39 – 750
= -1499
Dy = \(\left|\begin{array}{ccc}
2 & 10 & 6 \\
3 & 11 & -5 \\
8 & 12 & -9
\end{array}\right|\)
= 2(-99 + 60) – 10(-27 + 40) + 6(36 – 88)
= -78 – 130 – 312
= -520
Dz = \(\left|\begin{array}{ccc}
2 & -1 & 10 \\
3 & 4 & 11 \\
8 & -7 & 12
\end{array}\right|\)
= 2(48 + 77) – (-1)(36 – 88) + 10(-21 – 32)
= 250 – 52 – 530
= -332
By Cramer’s Rule,
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q1(ii)
∴ x = \(\frac{1499}{447}\), y = \(\frac{520}{447}\) and z = \(\frac{332}{447}\) are the solutions of the given equations.

(iii) 11x – y – z = 31, x – 6y + 2z = -26, x + 2y – 7z = -24
Solution:
Given equations are
11x – y – z = 31
x – 6y + 2z = -26
x + 2y – 7z = -24
D = \(\left|\begin{array}{ccc}
11 & -1 & -1 \\
1 & -6 & 2 \\
1 & 2 & -7
\end{array}\right|\)
= 11(42 – 4) – (-1)(-7 – 2) + (-1)(2 + 6)
= 418 – 9 – 8
= 401
Dx = \(\left|\begin{array}{ccc}
31 & -1 & -1 \\
-26 & -6 & 2 \\
-24 & 2 & -7
\end{array}\right|\)
= 31(42 – 4) – (-1)(182 + 48) + (-1)(-52 – 144)
= 1178 + 230 + 196
= 1604
Dy = \(\left|\begin{array}{ccc}
11 & 31 & -1 \\
1 & -26 & 2 \\
1 & -24 & -7
\end{array}\right|\)
= 11(182 + 48) – 31(-7 – 2) + (-1)(-24 + 26)
= 2530 + 279 – 2
= 2807
Dz = \(\left|\begin{array}{ccc}
11 & -1 & 31 \\
1 & -6 & -26 \\
1 & 2 & -24
\end{array}\right|\)
= 11(144 + 52) – (-1)(-24 + 26) + 31(2 + 6)
= 2156 + 2 + 248
= 2406
By Cramer’s Rule,
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q1(iii)
∴ x = 4, y = 7 and z = 6 are the solutions of the given equations.

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3

(iv) \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-2\), \(\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3\), \(\frac{2}{x}-\frac{1}{y}+\frac{3}{z}=-1\)
Solution:
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
The given equations become
p + q + r = -2
p – 2q + r = 3
2p – q + 3r = -1
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
2 & -1 & 3
\end{array}\right|\)
= 1(-6 + 1) – 1(3 – 2) + 1(-1 + 4)
= -5 – 1 + 3
= -3
Dp = \(\left|\begin{array}{rrr}
-2 & 1 & 1 \\
3 & -2 & 1 \\
-1 & -1 & 3
\end{array}\right|\)
= -2(-6 + 1) – 1(9 + 1) + 1(-3 – 2)
= 10 – 10 – 5
= -5
Dq = \(\left|\begin{array}{ccc}
1 & -2 & 1 \\
1 & 3 & 1 \\
2 & -1 & 3
\end{array}\right|\)
= 1(9 + 1) + 2(3 – 2) + 1(-1 – 6)
= 10 + 2 – 7
= 5
Dr = \(\left|\begin{array}{rrr}
1 & 1 & -2 \\
1 & -2 & 3 \\
2 & -1 & -1
\end{array}\right|\)
= 1(2 + 3) – 1(-1 – 6) – 2(-1 + 4)
= 5 + 7 – 6
= 6
By Cramer’s Rule,
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q1(iv)
∴ x = \(\frac{3}{5}\), y = \(\frac{-3}{5}\), z = \(\frac{-1}{2}\) are the solutions of the given equations.

(v) \(\frac{2}{x}-\frac{1}{y}+\frac{3}{z}=4, \frac{1}{x}-\frac{1}{y}+\frac{1}{z}=2, \frac{3}{x}+\frac{1}{y}-\frac{1}{z}=2\)
Solution:
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
The given equations become
2p – q – 3r = 4
p – q + r = 2
3p + q – r = 2
D = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & -1 & 1 \\
3 & 1 & -1
\end{array}\right|\)
= 2(1 – 1) – (-1)(-1 – 3) + 3(1 + 3)
= 0 – 4 + 12
= 8
Dp = \(\left|\begin{array}{ccc}
4 & -1 & 3 \\
2 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 4(1 – 1) – (-1)(-2 – 2) + 3(2 + 2)
= 0 – 4 + 12
= 8
Dq = \(\left|\begin{array}{ccc}
2 & 4 & 3 \\
1 & 2 & 1 \\
3 & 2 & -1
\end{array}\right|\)
= 2(-2 – 2) – 4(-1 – 3) + 3(2 – 6)
= -8 + 16 – 12
= -4
Dr = \(\left|\begin{array}{ccc}
2 & -1 & 4 \\
1 & -1 & 2 \\
3 & 1 & 2
\end{array}\right|\)
= 2(-2 – 2) – (-1)(2 – 6) + 4(1 + 3)
= -8 – 4 + 16
= 4
By Cramer’s Rule,
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q1(v)
∴ x = 1, y = -2 and z = 2 are the solutions of the given equations.

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3

Question 2.
An amount of ₹ 5,000 is invested in three plans at rates 6%, 7% and 8% per annum respectively. The total annual income from these investments is ₹ 350. If the total annual income from first two investments is ₹ 70 more than the income from the third, find the amount invested in each plan by using Cramer’s Rule.
Solution:
Let the amount of each investment be ₹ x, ₹ y and ₹ z.
According to the given conditions,
x + y + z = 5000
6%x + 7%y + 8%z = 350
∴ \(\frac{6}{100} x+\frac{7}{100} y-\frac{8}{100} z=350\)
∴ 6x + 7y + 8z = 35000
6%x + 7%y = 8%z + 70
∴ \(\frac{6}{100} x+\frac{7}{100} y=\frac{8}{100} z+70\)
∴ 6x + 7y = 8z + 7000
∴ 6x + 7y – 8z = 7000
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q2
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q2.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q2.2
∴ Amounts of investments are ₹ 1750, ₹ 1500, and ₹ 1750.

Check:
First condition:
1750 + 1500 + 1750 = 5000
Second condition:
6% of 1750 + 7% of 1500 + 8% of 1750
= 105 + 105 + 140
= 350
Third condition:
Combined income = 105 + 105
= 210
= 140 + 70
Thus, all the conditions are satisfied.

Question 3.
Show that the following equations are consistent.
2x + 3y + 4 = 0, x + 2y + 3 = 0, 3x + 4y + 5 = 0
Solution:
Given equations are
2x + 3y + 4 = 0
x + 2y + 3 = 0
3x + 4y + 5 = 0
∴ \(\left|\begin{array}{lll}
2 & 3 & 4 \\
1 & 2 & 3 \\
3 & 4 & 5
\end{array}\right|\)
= 2(10 – 12) – 3(5 – 9) + 4(4 – 6)
= 2(-2) – 3(-4) + 4(-2)
= -4 + 12 – 8
= 0
∴ The given equations are consistent.

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3

Question 4.
Find k, if the following equations are consistent.
(i) x + 3y + 2 = 0, 2x + 4y – k = 0, x – 2y – 3k = 0
Solution:
Given equations are
x + 3y + 2 = 0
2x + 4y – k = 0
x – 2y – 3k = 0
Since, these equations are consistent.
∴ \(\left|\begin{array}{ccc}
1 & 3 & 2 \\
2 & 4 & -k \\
1 & -2 & -3 k
\end{array}\right|=0\)
∴ 1(-12k – 2k) – 3(-6k + k) + 2(-4 – 4) = 0
∴ -14k + 15k – 16 = 0
∴ k – 16 = 0
∴ k = 16
Check:
If the value of k satisfies the condition for the given equations to be consistent, then our answer is correct.
Substitute k = 16 in the given equation.
\(\left|\begin{array}{ccc}
1 & 3 & 2 \\
2 & 4 & -16 \\
1 & -2 & -48
\end{array}\right|\)
= 1(-192 – 32) – 3(-96 + 16) + 2(-4 – 4)
= 0
Thus, our answer is correct.

(ii) (k – 2)x + (k – 1)y = 17, (k – 1)x + (k – 2)y = 18, x + y = 5
Solution:
Given equations are
(k – 2)x + (k – 1)y = 17
(k – 1)x + (k – 2)y = 18
x + y = 5
Since, these equations are consistent.
∴ \(\left|\begin{array}{ccc}
k-2 & k-1 & -17 \\
k-1 & k-2 & -18 \\
1 & 1 & -5
\end{array}\right|=0\)
Applying R1 → R1 – R2, we get
\(\left|\begin{array}{ccc}
-1 & 1 & 1 \\
k-1 & k-2 & -18 \\
1 & 1 & -5
\end{array}\right|=0\)
∴ -1(-5k + 10 + 18) – 1(-5k + 5 + 18) + 1(k – 1 – k + 2) = 0
∴ -1(-5k – 28) – 1(- 5k + 23) + 1(1) = 0
∴ 5k – 28 + 5k – 23 – 1 = 0
∴ 10k – 50 = 0
∴ k = 5

Question 5.
Find the area of the triangle whose vertices are:
(i) (4, 5), (0, 7), (-1, 1)
Solution:
Here, A(x1, y1) ≡ A(4, 5), B(x2, y2) ≡ B(0, 7), C(x3, y3) ≡ C(-1, 1)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
4 & 5 & 1 \\
0 & 7 & 1 \\
-1 & 1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [4(7 – 1) – 5(0 + 1) + 1(0 + 7)]
= \(\frac{1}{2}\) (24 – 5 + 7)
= 13 sq.units.

(ii) (3, 2), (-1, 5), (-2, -3)
Solution:
Here, A(x1, y1) ≡ A(3, 2), B(x2, y2) = B(-1, 5), C(x3, y3) ≡ C(-2, -3)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & 2 & 1 \\
-1 & 5 & 1 \\
-2 & -3 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [3(5 + 3) – 2(-1 + 2) + 1(3 + 10)]
= \(\frac{1}{2}\) (24 – 2 + 13)
= \(\frac{35}{2}\) sq. units

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3

(iii) (0, 5), (0, -5), (5, 0)
Solution:
Here, A(x1, y1) ≡ A(0, 5), B(x2, y2) ≡ B(0, -5), C(x3, y3) ≡ C(5,0)
Area of a triangle = \(\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & 5 & 1 \\
0 & -5 & 1 \\
5 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [0(-5 – 0) – 5(0 – 5) + 1(0 + 25)]
= \(\frac{1}{2}\) (0 + 25 + 25)
= \(\frac{50}{2}\)
= 25 sq.units

Question 6.
Find the value of k, if the area of the triangle with vertices at A(k, 3), B(-5, 7), C(-1, 4) is 4 square units.
Solution:
Here, A(x1, y1) ≡ A(k, 3), B(x2, y2) ≡ B(-5, 7), C(x3, y3) ≡ C(-1, 4)
A(ΔABC) = 4 sq.units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{b} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ \(\frac{1}{2}\left|\begin{array}{ccc}
k & 3 & 1 \\
-5 & 7 & 1 \\
-1 & 4 & 1
\end{array}\right|\) = ±4
∴ k(7 – 4) – 3(-5 + 1) + 1(-20 + 7) = ±8
∴ 3k + 12 – 13 = ±8
∴ 3k – 1 = ±8
∴ 3k – 1 = 8 or 3k – 1 = -8
∴ 3k = 9 or 3k = -7
∴ k = 3 or k = \(\frac{-7}{3}\)

Check:
For k = 3,
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q6
Thus, our answer is correct.

Question 7.
Find the area of the quadrilateral whose vertices are A(-3, 1), B(-2, -2), C(4, 1), D(2, 3).
Solution:
A(-3, 1), B(-2, -2), C(4, 1), D(2, 3)
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3 Q7
A(ABCD) = A(ΔABC) + A(ΔACD)
= \(\frac{21}{2}\) + 7
= \(\frac{35}{2}\) sq.units.

Question 8.
By using determinant, show that the following points are collinear.
P(5, 0), Q(10, -3), R(-5, 6)
Solution:
Here, P(x1, y1) ≡ P(5, 0), Q(x2, y2) ≡ Q(10, -3), R(x3, y3) ≡ R(-5, 6)
If A(ΔPQR) = 0, then the points P, Q, R are collinear.
∴ A(ΔPQR) = \(\frac{1}{2}\left|\begin{array}{ccc}
5 & 0 & 1 \\
10 & -3 & 1 \\
-5 & 6 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [5(-3 – 6) – 0(10 + 5) + 1(60 – 15)]
= \(\frac{1}{2}\) (-45 + 0 + 45)
= 0
∴ A(ΔPQR) = 0
∴ Points P, Q and R are collinear.

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.3

Question 9.
The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers, then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.
Solution:
Let the three numbers be x, y and z.
According to the given conditions,
x + y + z = 15
x + z – y = 5 i.e. x – y + z = 5
2x + y – z = 4
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 1(1 – 1) – 1 (-1 – 2) + 1(1 + 2)
= 1(0) – 1(-3) + 1(3)
= 0 + 3 + 3
= 6 ≠ 0
Dx = \(\left|\begin{array}{ccc}
15 & 1 & 1 \\
5 & -1 & 1 \\
4 & 1 & -1
\end{array}\right|\)
= 15(1 – 1) – 1(-5 – 4) + 1(5 + 4)
= 15(0) – 1(-9) + 1(9)
= 0 + 9 + 9
= 18
Dy = \(\left|\begin{array}{ccc}
1 & 15 & 1 \\
1 & 5 & 1 \\
2 & 4 & -1
\end{array}\right|\)
= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10)
= 1(-9) – 15(-3) + 1(-6)
= -9 + 45 – 6
= 30
Dz = \(\left|\begin{array}{ccc}
1 & 1 & 15 \\
1 & -1 & 5 \\
2 & 1 & 4
\end{array}\right|\)
= 1(-4 – 5) – 1(4 – 10) + 15(1 + 2)
= 1(-9) – 1(-6) + 15(3)
= -9 + 6 + 45
= 42
By Cramer’s Rule,
x = \(\frac{D_{x}}{D}=\frac{18}{6}\) = 3
y = \(\frac{D_{y}}{D}=\frac{30}{6}\) = 5
z = \(\frac{D_{z}}{D}=\frac{42}{6}\) = 7
∴ The three numbers are 3, 5 and 7.

11th Commerce Maths Digest Pdf

11th Commerce Maths 1 Chapter 4 Exercise 4.3 Answers Maharashtra Board

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Exercise 4.3 Answers Maharashtra Board

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Std 11 Maths 1 Exercise 4.3 Solutions Commerce Maths

Question 1.
Determine whether the sum to infinity of the following G.P’.s exist. If exists, find it.
(i) \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots\)
(ii) \(2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots\)
(iii) \(-3,1, \frac{-1}{3}, \frac{1}{9}, \ldots\)
(iv) \(\frac{1}{5}, \frac{-2}{5}, \frac{4}{5}, \frac{-8}{5}, \frac{16}{5}, \ldots\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q1
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q1.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q1.2

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3

Question 2.
Express the following recurring decimals as a rational number.
(i) \(0 . \overline{32}\)
(ii) 3.5
(iii) \(4 . \overline{18}\)
(iv) \(0.3 \overline{45}\)
(v) \(3.4 \overline{56}\)
Solution:
(i) \(0 . \overline{32}\) = 0.323232…..
= 0.32 + 0.0032 + 0.000032 + …..
Here, 0.32, 0.0032, 0.000032, … are in G.P. with a = 0.32 and r = 0.01
Since, |r| = |0.01| < 1
∴ Sum to infinity exists.
∴ Sum to infinity = \(\frac{a}{1-r}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q2

(ii) 3.5 = 3.555… = 3 + 0.5 + 0.05 + 0.005 + …
Here, 0.5, 0.05, 0.005, … are in G.P. with a = 0.5 and r = 0.1
Since, |r| = |0.1| < 1
∴ Sum to infinity exists.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q2.1

(iii) \(4 . \overline{18}\) = 4.181818…..
= 4 + 0.18 + 0.0018 + 0.000018 + …..
Here, 0.18, 0.0018, 0.000018, … are in G.P. with a = 0.18 and r = 0.01
Since, |r| = |0.01| < 1
∴ Sum to infinity exists.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q2.2

(iv) 0.345 = 0.3454545…..
= 0.3 + 0.045 + 0.00045 + 0.0000045 + …..
Here, 0.045, 0.00045, 0.0000045, … are in G.P. with a = 0.045, r = 0.01
Since, |r| = |0.01| < 1
∴ Sum to infinity exists.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q2.3
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q2.4

(v) \(3.4 \overline{56}\) = 3.4565656 …..
= 3.4 + 0.056 + 0.00056 + 0.0000056 + ….
Here, 0.056, 0.00056, 0.0000056, … are in G.P. with a = 0.056 and r = 0.01
Since, |r| = |0.01| < 1
∴ Sum to infinity exists.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q2.5
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q2.6

Question 3.
If the common ratio of a G.P. is \(\frac{2}{3}\) and sum of its terms to infinity is 12. Find the first term.
Solution:
r = \(\frac{2}{3}\), sum to infinity = 12 … [Given]
Sum to infinity = \(\frac{a}{1-r}\)
∴ 12 = \(\frac{a}{1-\frac{2}{3}}\)
∴ a = 12 × \(\frac{1}{3}\)
∴ a = 4

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3

Question 4.
If the first term of a G.P. is 16 and sum of its terms to infinity is \(\frac{176}{5}\), find the common ratio.
Solution:
a = 16, sum to infinity = \(\frac{176}{5}\) … [Given]
Sum to infinity = \(\frac{a}{1-r}\)
∴ \(\frac{176}{5}=\frac{16}{1-r}\)
∴ \(\frac{11}{5}=\frac{1}{1-r}\)
∴ 11 – 11r = 5
∴ 11r = 6
∴ r = \(\frac{6}{11}\)

Question 5.
The sum of the terms of an infinite G.P. is 5 and the sum of the squares of those terms is 15. Find the G.P.
Solution:
Let the required G.P. be a, ar, ar2, ar3, …..
Sum to infinity of this G.P. = 5
∴ 5 = \(\frac{a}{1-r}\)
∴ a = 5(1 – r) ……(i)
Also, the sum of the squares of the terms is 15.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.3 Q5

11th Commerce Maths Digest Pdf

11th Commerce Maths 1 Chapter 4 Exercise 4.2 Answers Maharashtra Board

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Exercise 4.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.2 Questions and Answers.

Std 11 Maths 1 Exercise 4.2 Solutions Commerce Maths

Question 1.
For the following G.P.’s, find Sn.
(i) 3, 6, 12, 24, …..
(ii) \(\mathbf{p}, \mathbf{q}, \frac{\mathbf{q}^{2}}{\mathbf{p}}, \frac{\mathbf{q}^{3}}{\mathbf{p}^{2}}, \ldots\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q1
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q1.1

Question 2.
For a G.P., if
(i) a = 2, r = \(-\frac{2}{3}\), find S6.
(ii) S5 = 1023, r = 4, find a.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q2

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2

Question 3.
For a G. P., if
(i) a = 2, r = 3, Sn = 242, find n.
(ii) sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r.
Solution:
(i) a = 2, r = 3, Sn = 242
Sn = \(a\left(\frac{r^{n}-1}{r-1}\right)\), for r > 1
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q3
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q3.1

Question 4.
For a G. P.,
(i) if t3 = 20, t6 = 160, find S7.
(ii) if t4 = 16, t9 = 512, find S10.
Solution:
(i) t3 = 20, t6 = 160
tn = arn-1
∴ t3 = ar3-1 = ar2
∴ ar2 = 20
∴ a = \(\frac{20}{\mathrm{r}^{2}}\) ……(i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q4
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q4.1

Question 5.
Find the sum to n terms:
(i) 3 + 33 + 333 + 3333 + ……
(ii) 8 + 88 + 888 + 8888 + ……..
Solution:
(i) Sn = 3 + 33 + 333 +….. upto n terms
= 3(1 + 11 + 111 +….. upto n terms)
= \(\frac{3}{9}\)(9 + 99 + 999 + … upto n terms)
= \(\frac{3}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]
= \(\frac{3}{9}\)[(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)]
But 10, 100, 1000, … n terms are in G.P.
with a = 10, r = \(\frac{100}{10}\) = 10
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q5

(ii) Sn = 8 + 88 + 888 + … upto n terms
= 8(1 + 11 + 111 + … upto n terms)
= \(\frac{8}{9}\) (9 + 99 + 999 + … upto n terms)
= \(\frac{8}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]
= \(\frac{8}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)]
But 10, 100, 1000, … n terms are in G.P. with
a = 10, r = \(\frac{100}{10}\) = 10
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q5.1

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2

Question 6.
Find the sum to n terms:
(i) 0.4 + 0.44 + 0.444 + ……
(ii) 0.7 + 0.77 + 0.777 + …..
Solution:
(i) Sn = 0.4 + 0.44 + 0.444 + ….. upto n terms
= 4(0.1 + 0.11 + 0.111 + …. upto n terms)
= \(\frac{4}{9}\) (0.9 + 0.99 + 0.999 + … upto n terms)
= \(\frac{4}{9}\) [(i – 0.1) + (1 – 0.01) + (1 – 0.001) … upto n terms]
= \(\frac{4}{9}\) [(1 + 1 + 1 + …n times) – (0.1 + 0.01 + 0.001 +… upto n terms)]
But 0.1, 0.01, 0.001, … n terms are in G.P.
with a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1
∴ Sn = \(\frac{4}{9}\left\{\mathrm{n}-0.1\left[\frac{1-(0.1)^{\mathrm{n}}}{1-0.1}\right]\right\}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q6

(ii) Sn = 0.7 + 0.77 + 0.777 + … upto n terms
= 7(0.1 + 0.11 + 0.111 + … upto n terms)
= \(\frac{7}{9}\) (0.9 + 0.99 + 0.999 + … upto n terms)
= \(\frac{7}{9}\) [(1 – 0.1) + (1 – 0.01) + (1 – 0.001) +… upto n terms]
= \(\frac{7}{9}\) [(1 + 1 + 1 +… n times) – (0.1 + 0.01 + 0.001 +… upto n terms)]
But 0.1, 0.01, 0.001, … n terms are in G.P.
with a = 0.1, r = \(\frac{0.01}{0.1}\) = 0.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q6.1

Question 7.
Find the nth terms of the sequences:
(i) 0.5, 0.55, 0.555,…..
(ii) 0.2, 0.22, 0.222,…..
Solution:
(i) Let t1 = 0.5, t2 = 0.55, t3 = 0.555 and so on.
t1 = 0.5
t2 = 0.55 = 0.5 + 0.05
t3 = 0.555 = 0.5 + 0.05 + 0.005
∴ tn = 0.5 + 0.05 + 0.005 + … upto n terms
But 0.5, 0.05, 0.005, … upto n terms are in G.P. with a = 0.5 and r = 0.1
∴ tn = the sum of first n terms of the G.P.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q7

(ii) Let t1 = 0.2, t2 = 0.22, t3 = 0.222 and so on
t1 = 0.2
t2 = 0.22 = 0.2 + 0.02
t3 = 0.222 = 0.2 + 0.02 + 0.002
∴ tn = 0.2 + 0.02 + 0.002 + … upto n terms
But 0.2, 0.02, 0.002, … upto n terms are in G.P. with a = 0.2 and r = 0.1
∴ tn = the sum of first n terms of the G.P.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q7.1

Question 8.
For a sequence, if Sn = 2(3n-1), find the nth term, hence showing that the sequence is a G.P.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q8

Question 9.
If S, P, R are the sum, product and sum of the reciprocals of n terms of a G.P. respectively, then verify that \(\left(\frac{\mathbf{S}}{\mathbf{R}}\right)^{\mathbf{n}}\) = P2.
Solution:
Let a be the 1st term and r be the common ratio of the G.P.
∴ the G.P. is a, ar, ar2, ar3, …, arn-1
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q9
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q9.1

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2

Question 10.
If Sn, S2n, S3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that Sn (S3n – S2n) = (S2n – Sn)2.
Solution:
Let a and r be the 1st term and common ratio of the G.P. respectively.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q10
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2 Q10.1

11th Commerce Maths Digest Pdf

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Std 11 Maths 1 Miscellaneous Exercise 4 Solutions Commerce Maths

Question 1.
In a G.P., the fourth term is 48 and the eighth term is 768. Find the tenth term.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q1

Question 2.
For a G.P. a = \(\frac{4}{3}\) and t7 = \(\frac{243}{1024}\), find the value of r.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q2

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4

Question 3.
For a sequence, if tn = \(\frac{5^{n-2}}{7^{n-3}}\), verify whether the sequence is a G.P. If it is a G.P., find its first term and the common ratio.
Solution:
The sequence (tn) is a G.P., if \(\frac{5^{n-2}}{7^{n-3}}\) = constant, for all n ∈ N.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q3
∴ the sequence is a G.P. with common ratio = \(\frac{5}{7}\)
∴ first term = t1 = \(\frac{5^{1-2}}{7^{1-3}}=\frac{5^{-1}}{7^{-2}}=\frac{7^{2}}{5}=\frac{49}{5}\)

Question 4.
Find three numbers in G.P., such that their sum is 35 and their product is 1000.
Solution:
Let the three numbers in G.P. be \(\frac{a}{r}\), a, ar.
According to the first condition,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q4
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q4.1
∴ the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.

Question 5.
Find 4 numbers in G. P. such that the sum of the middle 2 numbers is \(\frac{10}{3}\) and their product is 1.
Solution:
Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\).
According to the second condition,
\(\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1\)
∴ a4 = 1
∴ a = 1
According to the first condition,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q5

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4

Question 6.
Find five numbers in G.P. such that their product is 243 and the sum of the second and fourth numbers is 10.
Solution:
Let the five numbers in G.P. be
\(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\)
According to the first condition,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q6
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q6.1

Question 7.
For a sequence, Sn = 4(7n – 1), verify whether the sequence is a G.P.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q7

Question 8.
Find 2 + 22 + 222 + 2222 + …… upto n terms.
Solution:
Sn = 2 + 22 + 222 +….. upto n terms
= 2(1 + 11 + 111 +…… upto n terms)
= \(\frac{2}{9}\) (9 + 99 + 999 + … upto n terms)
= \(\frac{2}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) +…… upto n terms]
= \(\frac{2}{9}\) [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + ….. n times)]
Since, 10, 100, 1000, …… n terms are in G.P.
with a = 10, r = \(\frac{100}{10}\) = 10
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q8

Question 9.
Find the nth term of the sequence 0.6, 0.66, 0.666, 0.6666,…..
Solution:
0.6, 0.66, 0.666, 0.6666, ……
∴ t1 = 0.6
t2 = 0.66 = 0.6 + 0.06
t3 = 0.666 = 0.6 + 0.06 + 0.006
Hence, in general
tn = 0.6 + 0.06 + 0.006 + …… upto n terms.
The terms are in G.P.with
a = 0.6, r = \(\frac{0.06}{0.6}\) = 0.1
∴ tn = the sum of first n terms of the G.P.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q9

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4

Question 10.
Find \(\sum_{r=1}^{n}\left(5 r^{2}+4 r-3\right)\).
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q10

Question 11.
Find \(\sum_{\mathbf{r}=1}^{\mathbf{n}} \mathbf{r}(\mathbf{r}-\mathbf{3})(\mathbf{r}-\mathbf{2})\).
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q11
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q11.1

Question 12.
Find \(\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+r^{2}}{2 r+1}\)
Solution:
We know that,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q12

Question 13.
Find \(\sum_{r=1}^{n} \frac{1^{3}+2^{3}+3^{3}+\ldots+r^{3}}{(r+1)^{2}}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q13
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q13.1

Question 14.
Find 2 × 6 + 4 × 9 + 6 × 12 + …… upto n terms.
Solution:
2, 4, 6, … are in A.P.
∴ rth term = 2 + (r – 1)2 = 2r
6, 9, 12, … are in A.P.
∴ rth term = 6 + (r – 1) (3) = (3r + 3)
∴ 2 × 6 + 4 × 9 + 6 × 12 +…… upto n terms
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q14
= n(n + 1) (2n + 1 + 3)
= 2n(n + 1)(n + 2)

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4

Question 15.
Find 122 + 132 + 142 + 152 + …… + 202.
Solution:
122 + 132 + 142 + 152 + …… + 202
= (12 + 22 + 32 + 42 + ……. + 202) – (12 + 22 + 32 + 42 + …… + 112)
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q15
= 2870 – 506
= 2364

Question 16.
Find (502 – 492) + (482 – 472) + (462 – 452) + …… + (22 – 12).
Solution:
(502 – 492) + (482 – 472) + (462 – 452) + …… + (22 – 12)
= (502 + 482 + 462 + …… + 22) – (492 + 472 + 452 + …… + 12)
= \(\sum_{r=1}^{25}(2 r)^{2}-\sum_{r=1}^{25}(2 r-1)^{2}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q16
= 1300 – 25
= 1275

Question 17.
In a G.P., if t2 = 7, t4 = 1575, find r.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q17

Question 18.
Find k so that k – 1, k, k + 2 are consecutive terms of a G.P.
Solution:
Since k – 1, k, k + 2 are consecutive terms of a G.P.
∴ \(\frac{k}{k-1}=\frac{k+2}{k}\)
∴ k2 = k2 + k – 2
∴ k – 2 = 0
∴ k = 2

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4

Question 19.
If pth, qth and rth terms of a G.P. are x, y, z respectively, find the value of \(x^{q-r} \cdot y^{r-p} \cdot z^{p-q}\).
Solution:
Let a be the first term and R be the common ratio of the G.P.
∴ tn = \(\text { a. } R^{n-1}\)
∴ x = \(\text { a. } R^{p-1}\), y = \(\text { a. } R^{q-1}\), z = \(\text { a. } R^{r-1}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Miscellaneous Exercise 4 Q19

11th Commerce Maths Digest Pdf

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Std 11 Maths 1 Exercise 3.1 Solutions Commerce Maths

Question 1.
Write the conjugates of the following complex numbers:
(i) 3 + i
(ii) 3 – i
(iii) -√5 – √7i
(iv) -√-5
(v) 5i
(vi) √5 – i
(vii) √2 + √3i
Solution:
(i) Conjugate of (3 + i) is (3 – i)
(ii) Conjugate of (3 – i) is (3 + i)
(iii) Conjugate of (-√5 – √7i) is (-√5 + √7i)
(iv) -√-5 = -√5 × √-1 = -√5i
Conjugate of -√-5 is √5i
(v) Conjugate of 5i is -5i
(vi) Conjugate of √5 – i is √5 + i
(vii) Conjugate of √2 + √3i is √2 – √3i

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 2.
Express the following in the form of a + ib, a, b ∈ R, i = √-1. State the values of a and b:
(i) (1 + 2i)(-2 + i)
(ii) \(\frac{\mathrm{i}(4+3 \mathrm{i})}{(1-\mathrm{i})}\)
(iii) \(\frac{(2+i)}{(3-i)(1+2 i)}\)
(iv) \(\frac{3+2 i}{2-5 i}+\frac{3-2 i}{2+5 i}\)
(v) \(\frac{2+\sqrt{-3}}{4+\sqrt{-3}}\)
(vi) (2 + 3i)(2 – 3i)
(vii) \(\frac{4 i^{8}-3 i^{9}+3}{3 i^{11}-4 i^{10}-2}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.2
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.3
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.4
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q2.5

Question 3.
Show that (-1 + √3i)3 is a real number.
Solution:
(-1 + √3i)3
= (-1)3 + 3(-1)2 (√3i) + 3(-1)(√3i)2 +(√3i)3 [∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
= -1 + 3√3i – 3(3i2) + 3√3 i3
= -1 + 3√3i – 3(-3) – 3√3i [∵ i2 = -1, i3 = -1]
= -1 + 9
= 8, which is a real number.

Question 4.
Evaluate the following:
(i) i35
(ii) i888
(iii) i93
(iv) i116
(v) i403
(vi) \(\frac{1}{i^{58}}\)
(vii) i30 + i40 + i50 + i60
Solution:
We know that, i2 = -1, i3 = -i, i4 = 1
(i) i35 = (i4)8 (i2) i = (1)8 (-1) i = -i
(ii) i888 = (i4)222 = (1)222 = 1
(iii) i93 = (i4)23 . i = (1)23 . i = i
(iv) i116 = (i4)29 = (1)29 = 1
(v) i403 = (i4)100 (i2) i = (1)100 (-1) i = -i
(vi) \(\frac{1}{i^{88}}=\frac{1}{\left(i^{4}\right)^{14} \cdot i^{2}}=\frac{1}{(1)^{14}(-1)}=-1\)
(vii) i30 + i40 + i50 + i60
= (i4)7 i2 + (i4)10 + (i4)12 i2 + (i4)15
= (1)7 (-1) + (1)10 + (1)12 (-1) + (1)15
= -1 + 1 – 1 + 1
= 0

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 5.
Show that 1 + i10 + i20 + i30 is a real number.
Solution:
1 + i10 + i20 + i30
= 1 + (i4)2 . i2 + (i4)5 + (i4)7 . i2
= 1 + (1)2 (-1) + (1)5 + (1)7 (-1) [∵ i4 = 1, i2 = -1]
= 1 – 1 + 1 – 1
= 0, which is a real number.

Question 6.
Find the value of
(i) i49 + i68 + i89 + i110
(ii) i + i2 + i3 + i4
Solution:
(i) i49 + i68 + i89 + i110
= (i4)12 . i + (i4)17 + (i4)22 . i + (i4)27 . i2
= (1)12 . i + (1)17 + (1)22 . i + (1)27(-1) ……[∵ i4 = 1, i2 = -1]
= i + 1 + i – 1
= 2i

(ii) i + i2 + i3 + i4
= i + i2 + i2 . i + i4
= i – 1 – i + 1 [∵ i2 = -1, i4 = 1]
= 0

Question 7.
Find the value of 1 + i2 + i4 + i6 + i8 + …… + i20.
Solution:
1 + i2 + i4 + i6 + i8 + ….. + i20
= 1 + (i2 + i4) + (i6 + i8) + (i10 + i12) + (i14 + i16) + (i18 + i20)
= 1 + [i2 + (i2)2] + [(i2)3 + (i2)4] + [(i2)5 + (i2)6] + [(i2)7 + (i2)8] + [(i2)9 + (i2)10]
= 1 + [-1 + (- 1)2] + [(-1)3 + (-1)4] + [(-1)5 + (-1)6] + [(-1)7 + (-1)8] + [(-1)9 + (-1)10] [∵ i2 = -1]
= 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1)
= 1 + 0 + 0 + 0 + 0 + 0
= 1

Question 8.
Find the values of x and y which satisfy the following equations (x, y ∈ R):
(i) (x + 2y) + (2x – 3y)i + 4i = 5
(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)
Solution:
(i) (x + 2y) + (2x – 3y)i + 4i = 5
∴ (x + 2y) + (2x – 3y)i = 5 – 4i
Equating real and imaginary parts, we get
x + 2y = 5 ……..(i)
and 2x – 3y = -4 ………(ii)
Equation (i) × 2 – equation (ii) gives
7y = 14
∴ y = 2
Putting y- 2 in (i), we get
x + 2(2) = 5
∴ x + 4 = 5
∴ x = 1
∴ x = 1 and y = 2
Check:
If x = 1 and y = 2 satisfy the given condition, then our answer is correct.
L.H.S. = (x + 2y) + (2x – 3y)i + 4i
= (1 + 4) + (2 – 6)i + 4i
= 5 – 4i + 4i
= 5
= R.H.S.
Thus, our answer is correct.

(ii) \(\frac{x+1}{1+\mathrm{i}}+\frac{y-1}{1-\mathrm{i}}=\mathrm{i}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q8
(x + y) + (y – x – 2)i = 2i
(x + y) + (y – x – 2)i = 0 + 2i
Equating real and imaginary parts, we get
x + y = 0 and y – x – 2 = 2
∴ x + y = 0 ……(i)
and -x + y = 4 ……..(ii)
Adding (i) and (ii), we get
2y = 4
∴ y = 2
Putting y = 2 in (i), we get
x + 2 = 0
∴ x = -2
∴ x = -2 and y = 2

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1

Question 9.
Find the value of:
(i) x3 – x2 + x + 46, if x = 2 + 3i
(ii) 2x3 – 11x2 + 44x + 27, if x = \(\frac{25}{3-4 i}\)
Solution:
(i) x = 2 + 3i
∴ x – 2 = 3i
∴ (x – 2)2 = 9i2
∴ x2 – 4x + 4 = 9(-1) …..[∵ i2 = -1]
∴ x2 – 4x + 13 = 0 ……(i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q9
∴ x3 – x2 + x + 46 = (x2 – 4x + 13)(x + 3) + 7
= 0(x + 3) + 7 ……[From (i)]
= 7

(ii) x = \(\frac{25}{3-4 i}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q9.1
∴ x = 3 + 4i
∴ x – 3 = 4i
∴ (x – 3)2 = 16i2
∴ x2 – 6x + 9 = 16(-1) …….[∵ i2 = -1]
∴ x2 – 6x + 25 = 0 …….(i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.1 Q9.2
∴ 2x3 – 11x2 + 44x + 27
= (x2 – 6x + 25) (2x + 1) + 2
= 0 . (2x + 1) + 2 ……[From (i)]
= 0 + 2
= 2

11th Commerce Maths Digest Pdf 

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Std 11 Maths 1 Miscellaneous Exercise 7 Solutions Commerce Maths

I.

Question 1.
If \(\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=80\) then find the value of n.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 I Q1

II. Evaluate the following Limits:

Question 1.
\(\lim _{x \rightarrow a} \frac{(x+2)^{\frac{5}{3}}-(a+2)^{\frac{5}{3}}}{x-a}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q1

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

Question 2.
\(\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q2
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q2.1

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{(x-2)}{2 x^{2}-7 x+6}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q3

Question 4.
\(\lim _{x \rightarrow 1}\left[\frac{x^{3}-1}{x^{2}+5 x-6}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q4

Question 5.
\(\lim _{x \rightarrow 3}\left[\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q5
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q5.1

Question 6.
\(\lim _{x \rightarrow 4}\left[\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q6

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

Question 7.
\(\lim _{x \rightarrow 0}\left[\frac{5^{x}-1}{x}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q7

Question 8.
\(\lim _{x \rightarrow 0}\left(1+\frac{x}{5}\right)^{\frac{1}{x}}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q8

Question 9.
\(\lim _{x \rightarrow 0}\left[\frac{\log (1+9 x)}{x}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q9

Question 10.
\(\lim _{x \rightarrow 0} \frac{(1-x)^{5}-1}{(1-x)^{3}-1}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q10

Question 11.
\(\lim _{x \rightarrow 0}\left[\frac{a^{x}+b^{x}+c^{x}-3}{x}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q11

Question 12.
\(\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{x^{2}}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q12

Question 13.
\(\lim _{x \rightarrow 0}\left[\frac{x\left(6^{x}-3^{x}\right)}{\left(2^{x}-1\right) \cdot \log (1+x)}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q13
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q13.1

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

Question 14.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^{x}+1}{x^{2}}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q14

Question 15.
\(\lim _{x \rightarrow 0}\left[\frac{\left(5^{x}-1\right)^{2}}{x \cdot \log (1+x)}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q15
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q15.1

Question 16.
\(\lim _{x \rightarrow 0}\left[\frac{a^{4 x}-1}{b^{2 x}-1}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q16

Question 17.
\(\lim _{x \rightarrow 0}\left[\frac{\log 100+\log (0.01+x)}{x}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q17

Question 18.
\(\lim _{x \rightarrow 0}\left[\frac{\log (4-x)-\log (4+x)}{x}\right]\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q18
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q18.1

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

Question 19.
Evaluate the limit of the function if exist at x = 1 where,
\(f(x)= \begin{cases}7-4 x & x<1 \\ x^{2}+2 & x \geq 1\end{cases}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7 II Q19

11th Commerce Maths Digest Pdf

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Std 11 Maths 1 Miscellaneous Exercise 9 Solutions Commerce Maths

I. Differentiate the following functions w.r.t.x.

Question 1.
x5
Solution:
Let y = x5
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} x^{5}=5 x^{4}\)

Question 2.
x-2
Solution:
Let y = x-2
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{-2}\right)=-2 x^{-3}=\frac{-2}{x^{3}}\)

Question 3.
√x
Solution:
Let y = √x
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} \sqrt{x}=\frac{1}{2 \sqrt{x}}\)

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

Question 4.
x√x
Solution:
Let y = x√x
∴ y = \(x^{\frac{3}{2}}\)
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{3}{2}}=\frac{3}{2} x^{\frac{1}{2}}\)

Question 5.
\(\frac{1}{\sqrt{x}}\)
Solution:
Let y = \(\frac{1}{\sqrt{x}}\)
∴ y = \(x^{\frac{-1}{2}}\)
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-1}{2} x^{\frac{-3}{2}}=\frac{-1}{2 x^{\frac{3}{2}}}\)

Question 6.
7x
Solution:
Let y = 7x
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} 7^{x}=7^{x} \log 7\)

II. Find \(\frac{d y}{d x}\) if

Question 1.
y = x2 + \(\frac{1}{x^{2}}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q1

Question 2.
y = (√x + 1)2
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q2

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

Question 3.
y = \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q3
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q3.1

Question 4.
y = x3 – 2x2 + √x + 1
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q4

Question 5.
y = x2 + 2x – 1
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q5

Question 6.
y = (1 – x)(2 – x)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q6

Question 7.
y = \(\frac{1+x}{2+x}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q7
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q7.1

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

Question 8.
y = \(\frac{(\log x+1)}{x}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q8

Question 9.
y = \(\frac{e^{x}}{\log x}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q9

Question 10.
y = x log x (x2 + 1)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 II Q10

III. Solve the following:

Question 1.
The relation between price (P) and demand (D) of a cup of Tea is given by D = \(\frac{12}{P}\). Find
the rate at which the demand changes when the price is ₹ 2/-. Interpret the result.
Solution:
Demand, D = \(\frac{12}{P}\)
Rate of change of demand
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 III Q1
When price P = 2,
Rate of change of demand,
\(\left(\frac{\mathrm{dD}}{\mathrm{dP}}\right)_{\mathrm{P}=2}=\frac{-12}{(2)^{2}}=-3\)
∴ When the price is 2, the rate of change of demand is -3.
∴ Here, the rate of change of demand is negative demand would fall when the price becomes ₹ 2.

Question 2.
The demand (D) of biscuits at price P is given by D = \(\frac{64}{P^{3}}\), find the marginal demand
when the price is ₹ 4/-.
Solution:
Given demand D = \(\frac{64}{P^{3}}\)
Now, marginal demand
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 III Q2
When P = 4
Marginal demand
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 III Q2.1

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

Question 3.
The supply S of electric bulbs at price P is given by S = 2p3 + 5. Find the marginal supply when the price is ₹ 5/-. Interpret the result.
Solution:
Given, supply S = 2p3 + 5
Now, marginal supply
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 III Q3
∴ When p = 5
Marginal supply = \(\left(\frac{\mathrm{dS}}{\mathrm{dp}}\right)_{\mathrm{p}=5}\)
= 6(5)2
= 150
Here, the rate of change of supply with respect to the price is positive which indicates that the supply increases.

Question 4.
The total cost of producing x items is given by C = x2 + 4x + 4. Find the average cost and the marginal cost. What is the marginal cost when x = 7?
Solution:
Total cost C = x2 + 4x + 4
Now. Average cost = \(\frac{C}{x}=\frac{x^{2}+4 x+4}{x}\)
= x + 4 + \(\frac{4}{x}\)
and Marginal cost = \(\frac{\mathrm{dC}}{\mathrm{d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\)(x2 + 4x + 4)
= \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x2) + 4\(\frac{\mathrm{d}}{\mathrm{d} x}\) (x) + \(\frac{\mathrm{d}}{\mathrm{d} x}\) (4)
= 2x + 4(1) + 0
= 2x + 4
∴ When x = 7,
Marginal cost = \(\left(\frac{\mathrm{d} \mathrm{C}}{\mathrm{d} x}\right)_{x=7}\)
= 2(7) + 4
= 14 + 4
= 18

Question 5.
The demand D for a price P is given as D = \(\frac{27}{P}\), find the rate of change of demand when the price is ₹ 3/-.
Solution:
Demand, D = \(\frac{27}{P}\)
Rate of change of demand = \(\frac{dD}{dP}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 III Q5
When price P = 3,
Rate of change of demand,
\(\left(\frac{\mathrm{dD}}{\mathrm{dP}}\right)_{\mathrm{P}=3}=\frac{-27}{(3)^{2}}=-3\)
∴ When price is 3, Rate of change of demand is -3.

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

Question 6.
If for a commodity; the price demand relation is given as D = \(\left(\frac{P+5}{P-1}\right)\). Find the marginal demand when price is ₹ 2/-
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 III Q6

Question 7.
The price function P of a commodity is given as P = 20 + D – D2 where D is demand. Find the rate at which price (P) is changing when demand D = 3.
Solution:
Given, P = 20 + D – D2
Rate of change of price = \(\frac{dP}{dD}\)
= \(\frac{d}{dD}\)(20 + D – D2)
= 0 + 1 – 2D
= 1 – 2D
Rate of change of price at D = 3 is
\(\left(\frac{\mathrm{dP}}{\mathrm{dD}}\right)_{\mathrm{D}=3}\) = 1 – 2(3) = -5
∴ Price is changing at a rate of -5, when demand is 3.

Question 8.
If the total cost function is given by C = 5x3 + 2x2 + 1; find the average cost and the marginal cost when x = 4.
Solution:
Total cost function C = 5x3 + 2x2 + 1
Average cost = \(\frac{C}{x}\)
= \(\frac{5 x^{3}+2 x^{2}+1}{x}\)
= 5x2 + 2x + \(\frac{1}{x}\)
When x = 4,
Average cost = 5(4)2 + 2(4) + \(\frac{1}{4}\)
= 80 + 8 + \(\frac{1}{4}\)
= \(\frac{320+32+1}{4}\)
= \(\frac{353}{4}\)
Marginal cost = \(\frac{\mathrm{dC}}{\mathrm{d} x}\)
= \(\frac{d}{dx}\) (5x3 + 2x2 + 1)
= 5\(\frac{d}{dx}\) (x3) + 2 \(\frac{d}{dx}\) (x2) + \(\frac{d}{dx}\) (1)
= 5(3x2) + 2(2x) + 0
= 15x2 + 4x
When x = 4, marginal cost = \(\left(\frac{\mathrm{dC}}{\mathrm{d} x}\right)_{x=4}\)
= 15(4)2 + 4(4)
= 240 + 16
= 256
∴ The average cost and marginal cost at x = 4 are \(\frac{353}{4}\) and 256 respectively.

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

Question 9.
The supply S for a commodity at price P is given by S = P2 + 9P – 2. Find the marginal supply when the price is 7/-.
Solution:
Given, S = P2 + 9P – 2
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 III Q9
∴ The marginal supply is 23, at P = 7.

Question 10.
The cost of producing x articles is given by C = x2 + 15x + 81. Find the average cost and marginal cost functions. Find the marginal cost when x = 10. Find x for which the marginal cost equals the average cost.
Solution:
Given, cost C = x2 + 15x + 81
Maharashtra Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9 III Q10
If marginal cost = average cost, then
2x + 15 = x + 15 + \(\frac{81}{x}\)
∴ x = \(\frac{81}{x}\)
∴ x2 = 81
∴ x = 9 …..[∵ x > 0]

11th Commerce Maths Digest Pdf

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Std 11 Maths 1 Exercise 4.1 Solutions Commerce Maths

Question 1.
Verify whether the following sequences are G.P. If so, write tn.
(i) 2, 6, 18, 54, ……
(ii) 1, -5, 25, -125, …….
(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots\)
(iv) 3, 4, 5, 6, ……
(v) 7, 14, 21, 28, …..
Solution:
(i) 2, 6, 18, 54, …….
t1 = 2, t2 = 6, t3 = 18, t4 = 54, …..
Here, \(\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}=\frac{t_{4}}{t_{3}}=3\)
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, a = 2, r = 3
tn= arn-1
∴ tn = 2(3n-1)

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

(ii) 1, -5, 25, -125, ……
t1 = 1, t2 = -5, t3 = 25, t4 = -125, …..
Here, \(\frac{t_{2}}{t_{1}}=\frac{t_{3}}{t_{2}}=\frac{t_{4}}{t_{3}}=-5\)
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Here, a = 1, r = -5
tn = arn-1
∴ tn = (-5)n-1

(iii) \(\sqrt{5}, \frac{1}{\sqrt{5}}, \frac{1}{5 \sqrt{5}}, \frac{1}{25 \sqrt{5}}, \ldots\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q1
Since, the ratio of any two consecutive terms is a constant, the given sequence is a geometric progression.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q1.1

(iv) 3, 4, 5, 6,……
t1 = 3, t2 = 4, t3 = 5, t4 = 6, …..
Here, \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}=\frac{4}{3}, \frac{\mathrm{t}_{3}}{\mathrm{t}_{2}}=\frac{5}{4}, \frac{\mathrm{t}_{4}}{\mathrm{t}_{3}}=\frac{6}{5}\)
Since, \(\frac{t_{2}}{t_{1}} \neq \frac{t_{3}}{t_{2}} \neq \frac{t_{4}}{t_{3}}\)
∴ the given sequence is not a geometric progression.

(v) 7, 14, 21, 28, …..
t1 = 7, t2 = 14, t3 = 21, t4 = 28, …..
Here, \(\frac{t_{2}}{t_{1}}=2, \frac{t_{3}}{t_{2}}=\frac{3}{2}, \frac{t_{4}}{t_{3}}=\frac{4}{3}\)
Since, \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}} \neq \frac{\mathrm{t}_{3}}{\mathrm{t}_{2}} \neq \frac{\mathrm{t}_{4}}{\mathrm{t}_{3}}\)
∴ the given sequence is not a geometric progression.

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Question 2.
For the G.P.,
(i) if r = \(\frac{1}{3}\), a = 9, find t7.
(ii) if a = \(\frac{7}{243}\), r = \(\frac{1}{3}\), find t3.
(iii) if a = 7, r = -3, find t6.
(iv) if a = \(\frac{2}{3}\), t6 = 162, find r.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q2
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q2.1

Question 3.
Which term of the G. P. 5, 25, 125, 625, ….. is 510?
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q3

Question 4.
For what values of x, \(\frac{4}{3}\), x, \(\frac{4}{27}\) are in G. P.?
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q4

Question 5.
If for a sequence, \(t_{n}=\frac{5^{n-3}}{2^{n-3}}\), show that the sequence is a G. P. Find its first term and the common ratio.
Solution:
The sequence (tn) is a G.P., if \(\frac{t_{n}}{t_{n-1}}\) = constant, for all n ∈ N
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q5
∴ the sequence is a G. P. with common ratio \(\frac{5}{2}\)
First term, t1 = \(\frac{5^{\mathrm{l}-3}}{2^{1-3}}=\frac{2^{2}}{5^{2}}=\frac{4}{25}\)

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Question 6.
Find three numbers in G. P. such that their sum is 21 and sum of their squares is 189.
Solution:
Let the three numbers in G. P. be \(\frac{a}{\mathrm{r}}\), a, ar.
According to the first condition,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q6
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q6.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q6.2
∴ the three numbers are 12, 6, 3 or 3, 6, 12.
Check:
First condition:
12, 6, 3 are in G.P. with r = \(\frac{1}{2}\)
12 + 6 + 3 = 21
Second condition:
122 + 62 + 32 = 144 + 36 + 9 = 189
Thus, both the conditions are satisfied.

Question 7.
Find four numbers in G. P. such that sum of the middle two numbers is \(\frac{10}{3}\) and their product is 1.
Solution:
Let the four numbers in G.P. be \(\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}\).
According to the second condition,
\(\frac{\mathrm{a}}{\mathrm{r}^{3}}\left(\frac{\mathrm{a}}{\mathrm{r}}\right)(\mathrm{ar})\left(\mathrm{ar}^{3}\right)=1\)
∴ a4 = 1
∴ a = 1
According to the first condition,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q7

Question 8.
Find five numbers in G. P. such that their product is 1024 and the fifth term is square of the third term.
Solution:
Let the five numbers in G. P. be
\(\frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2}\)
According to the given conditions,
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q8
When a = 4, r = -2
\(\frac{a}{r^{2}}\) = 1, \(\frac{a}{r}\) = -2, a = 4, ar = -8, ar2 = 16
∴ the five numbers in G.P. are 1, 2, 4, 8, 16 or 1, -2, 4, -8, 16.

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1

Question 9.
The fifth term of a G. P. is x, eighth term of the G. P. is y and eleventh term of the G. P. is z. Verify whether y2 = xz.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q9

Question 10.
If p, q, r, s are in G. P., show that p + q, q + r, r + s are also in G.P.
Solution:
p, q, r, s are in G.P.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.1 Q10
∴ p + q, q + r, r + s are in G.P.

11th Commerce Maths Digest Pdf

11th Commerce Maths 1 Chapter 2 Miscellaneous Exercise 2 Answers Maharashtra Board

Functions Class 11 Commerce Maths 1 Chapter 2 Miscellaneous Exercise 2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Functions Miscellaneous Exercise 2 Questions and Answers.

Std 11 Maths 1 Miscellaneous Exercise 2 Solutions Commerce Maths

Question 1.
Which of the following relations are functions? If it is a function determine its domain and range.
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}
(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)}
(iii) {(1, 1), (3, 1), (5, 2)}
Solution:
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2 Q1
Every element of set A has been assigned a unique element in set B.
∴ Given relation is a function.
Domain = {2, 4, 6, 8, 10, 12, 14},
Range = {1, 2, 3, 4, 5, 6, 7}

(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)}
∴ (1, 1), (1, -1) ∈ the relation
∴ Given relation is not a function.
As element 1 of the domain has not been assigned a unique element of co-domain.

(iii) {(1, 1), (3, 1), (5, 2)}
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2 Q1.1
Every element of set A has been assigned a unique element in set B.
∴ Given relation is a function.
Domain = {1, 3, 5}, Range = {1, 2}

Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2

Question 2.
A function f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2, x ∈ R. Show that f is one-one and onto. Hence, find f-1.
Solution:
f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2
First we have to prove that f is one-one function for that we have to prove if
f(x1) = f(x2) then x1 = x2
Here f(x) = \(\frac{3 x}{5}\) + 2
Let f(x1) = f(x2)
∴ \(\frac{3 x_{1}}{5}+2=\frac{3 x_{2}}{5}+2\)
∴ \(\frac{3 x_{1}}{5}=\frac{3 x_{2}}{5}\)
∴ x1 = x2
∴ f is a one-one function.
Now, we have to prove that f is an onto function.
Let y ∈ R be such that
y = f(x)
∴ y = \(\frac{3 x}{5}\) + 2
∴ y – 2 = \(\frac{3 x}{5}\)
∴ x = \(\frac{5(y-2)}{3}\) ∈ R
∴ for any y ∈ co-domain R, there exist an element x = \(\frac{5(y-2)}{3}\) ∈ domain R such that f(x) = y
∴ f is an onto function.
∴ f is one-one onto function.
∴ f-1 exists.
∴ \(\mathrm{f}^{-1}(y)=\frac{5(y-2)}{3}\)
∴ \(f^{-1}(x)=\frac{5(x-2)}{3}\)

Question 3.
A function f is defined as follows:
f(x) = 4x + 5, for -4 ≤ x < 0. Find the values of f(-1), f(-2), f(0), if they exist.
Solution:
f(x) = 4x + 5, -4 ≤ x < 0
f(-1) = 4(-1) + 5 = -4 + 5 = 1
f(-2) = 4(-2) + 5 = -8 + 5 = -3
x = 0 ∉ domain of f
∴ f(0) does not exist.

Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2

Question 4.
A function f is defined as follows:
f(x) = 5 – x for 0 ≤ x ≤ 4. Find the value of x such that f(x) = 3.
Solution:
f(x) = 5 – x
f(x) = 3
∴ 5 – x = 3
∴ x = 5 – 3 = 2

Question 5.
If f(x) = 3x2 – 5x + 7, find f(x – 1).
Solution:
f(x) = 3x2 – 5x + 7
∴ f(x – 1) = 3(x – 1)2 – 5(x – 1) + 7
= 3(x2 – 2x + 1) – 5(x – 1) + 7
= 3x2 – 6x + 3 – 5x + 5 + 7
= 3x2 – 11x + 15

Question 6.
If f(x) = 3x + a and f(1) = 7, find a and f(4).
Solution:
f(x) = 3x + a,
f(1) = 7
∴ 3(1) + a = 7
∴ a = 7 – 3 = 4
∴ f(x) = 3x + 4
∴ f(4) = 3(4) + 4
= 12 + 4
= 16

Question 7.
If f(x) = ax2 + bx + 2 and f(1) = 3, f(4) = 42, find a and b.
Solution:
f(x) = ax2 + bx + 2
f(1) = 3
∴ a(1)2 + b(1) + 2 = 3
∴ a + b = 1 …….(i)
f(4) = 42
∴ a(4)2 + b(4) + 2 = 42
∴ 16a + 4b = 40
Dividing by 4, we get
4a + b = 10 ……….(ii)
Solving (i) and (ii), we get
a = 3, b = -2

Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2

Question 8.
If f(x) = \(\frac{2 x-1}{5 x-2}, x \neq \frac{2}{5}\), verify whether (fof)(x) = x
Solution:
(fof)(x) = f(f(x))
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2 Q8

Question 9.
If f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\), then verify that (fog)(x) = x.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2 Q9

11th Commerce Maths Digest Pdf