11th Commerce Maths 1 Chapter 3 Miscellaneous Exercise 3 Answers Maharashtra Board

Complex Numbers Class 11 Commerce Maths 1 Chapter 3 Miscellaneous Exercise 3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Miscellaneous Exercise 3 Questions and Answers.

Std 11 Maths 1 Miscellaneous Exercise 3 Solutions Commerce Maths

Question 1.
Find the value of \(\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3 Q1

Question 2.
Find the value of √-3 × √-6.
Solution:
√-3 × √-6 = √3 × √-1 + √6 × √-1
= √3i × √6i
= √18i2
= -3√2 ……[∵ i2 = -1]

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3

Question 3.
Simplify the following and express in the form a + ib.
(i) 3 + √-64
(ii) (2i3)2
(iii) (2 + 3i) (1 – 4i)
(iv) \(\frac{5}{2}\) i(-4 – 3i)
(v) (1 + 3i)2 (3 + i)
(vi) \(\frac{4+3 i}{1-i}\)
(vii) \(\left(1+\frac{2}{i}\right)\left(3+\frac{4}{i}\right)(5+i)^{-1}\)
(viii) \(\frac{\sqrt{5}+\sqrt{3} i}{\sqrt{5}-\sqrt{3} i}\)
(ix) \(\frac{3 i^{5}+2 i^{7}+i^{9}}{i^{6}+2 i^{8}+3 i^{18}}\)
(x) \(\frac{5+7 i}{4+3 i}+\frac{5+7 i}{4-3 i}\)
Solution:
(i) 3 + √-64
= 3 + √64 . √-1
= 3 + 8i

(ii) (2i3)2
= 4i6
= 4(i2)3
= 4(-1)3 …..[∵ i2 = -1]
= -4
= -4 + 0i

(iii) (2 + 3i)(1 – 4i) = 2 – 8i + 3i – 12i2
= 2 – 5i – 12(-1) ……[∵ i2 = -1]
= 14 – 5i

(iv) \(\frac{5}{2}\) i(-4 – 3i)
= \(\frac{5}{2}\) (-4i – 3i2)
= \(\frac{5}{2}\) [-4i – 3(-1)] ……[∵ i2 = -1]
= \(\frac{5}{2}\) (3 – 4i)
= \(\frac{15}{2}\) – 10i

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3

(v) (1 + 3i)2 (3 + i)
= (1 + 6i + 9i2) (3 + i)
= (1 + 6i – 9)(3 + i) ……[∵ i2 = -1]
= (-8 + 6i)(3 + i)
= -24 – 8i + 18i + 6i2
= -24 + 10i + 6(-1)
= -24 + 10i – 6
= -30 + 10i

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3 Q3
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3 Q3.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3 Q3.2

Question 4.
Solve the following equations for x, y ∈ R:
(i) (4 – 5i) x + (2 + 3i) y = 10 – 7i
(ii) (1 – 3i) x + (2 + 5i) y = 1 + i
(iii) \(\frac{x+i y}{2+3 i}\) = 7 – i
(iv) (x + iy) (5 + 6i) = 2 + 3i
(v) 2x + i9 y (2 + i) = x i7 + 10 i16
Solution:
(i) (4 – 5i) x + (2 + 3i)y = 10 – 7i
∴ (4x + 2y) + (3y – 5x) i = 10 – 7i
Equating real and imaginary parts, we get
4x + 2y = 10
i.e., 2x + y = 5 …….(i)
and 3y – 5x = -7 ……..(ii)
Equation (i) × 3 – equation (ii) gives
11x = 22
∴ x = 2
Putting x = 2 in (i), we get
2(2) + y = 5
∴ y = 1
∴ x = 2 and y = 1

(ii) (1 – 3i) x + (2 + 5i) y = 7 + i
∴ (x + 2y) + (-3x + 5y)i = 7 + i
Equating real and imaginary parts, we get
x + 2y = 7 ……..(i)
and -3x + 5y = 1 ……..(ii)
Equation (i) × 3 + equation (ii) gives
11y = 22
∴ y = 2
Putting y = 2 in (i), we get
x + 2(2) = 7
∴ x = 3
∴ x = 3 and y = 2

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3

(iii) \(\frac{x+i y}{2+3 i}\) = 7 – i
∴ x + iy = (7 – i)(2 + 3i)
∴ x + iy = 14 + 21i – 2i – 3i2
∴ x + iy = 14 + 19i – 3(-1) …..[∵ i2 = -1]
∴ x + iy = 17 + 19i
Equating real and imaginary parts, we get
x = 17 and y = 19

(iv) (x + iy)(5 + 6i) = 2 + 3i
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3 Q4
Equating real and imaginary parts, we get
x = \(\frac{28}{61}\) and y = \(\frac{3}{61}\)

(v) 2x + i9 y (2 + i) = x i7 + 10 i16
∴ 2x + (i4)2 . i . y (2 + i) = x (i2)3 . i + 10 . (i4)4
∴ 2x + (1)2 . iy (2 + i) = x (-1)3 . i + 10 (1)4 ……[∵ i2 = -1, i4 = 1]
∴ 2x + 2yi + yi2 = -xi + 10
∴ 2x + 2yi – y + xi = 10
∴ (2x – y) + (x + 2y)i = 10 + 0.i
Equating real and imaginary parts, we get
2x – y = 10 ……(i)
and x + 2y = 0 ……..(ii)
Equation (i) × 2 + equation (ii) gives
5x = 20
∴ x = 4
Putting x = 4 in (i), we get
2(4) – y = 10
∴ y = 8 – 10
∴ y = -2
∴ x = 4 and y = -2

Question 5.
Find the value of:
(i) x3 + 2x2 – 3x + 21, if x = 1 + 2i
(ii) x3 – 5x2 + 4x + 8, if x = \(\frac{10}{3-i}\)
(iii) x3 – 3x2 + 19x – 20, if x = 1 – 4i
Solution:
(i) x = 1 + 2i
∴ x – 1 = 2i
∴ (x – 1)2 = 4i2
∴ x2 – 2x + 1 = -4 ……[∵ i2 = -1]
∴ x2 – 2x + 5 = 0 ……(i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3 Q5
∴ x3 + 2x2 – 3x + 21
= (x2 – 2x + 5)(x + 4) + 1
= 0.(x + 4) + 1 ……[From (i)]
= 0 + 1
= 1
∴ x3 + 2x2 – 3x + 21 = 1

(ii) x = \(\frac{10}{3-i}\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3 Q5.1
x3 – 5x2 + 4x + 8
= (x2 – 6x + 10)(x + 1) – 2
= 0 . (x + 1) – 2 ……[From (i)]
= 0 – 2
∴ x3 – 5x2 + 4x + 8 = -2

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3

(iii) x = 1 – 4i
∴ x – 1 = -4i
∴ (x – 1)2 = 16i2
∴ x2 – 2x + 1 = -16 ……[∵ i2 = -1]
∴ x2 – 2x + 17 = 0 ……(i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3 Q5.2
∴ x3 – 3x2 + 19x – 20
= (x2 – 2x + 17) (x – 1) – 3
= 0 . (x – 1) – 3 ….[From (i)]
= 0 – 3
= -3
∴ x3 – 3x2 + 19x – 20 = -3

Question 6.
Find the square roots of:
(i) -16 + 30i
(ii) 15 – 8i
(iii) 2 + 2√3i
(iv) 18i
(v) 3 – 4i
(vi) 6 + 8i
Solution:
(i) Let \(\sqrt{-16+30 \mathrm{i}}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
-16 + 30i = a2 + b2i2 + 2abi
∴ -16 + 30i = (a2 – b2) + 2abi …..[∵ i2 = -1]
Equating real and imaginary parts, we get
a2 – b2 = -16 and 2ab = 30
∴ a2 – b2 = -16 and b = \(\frac{15}{a}\)
∴ a2 – \(\left(\frac{15}{a}\right)^{2}\) = -16
∴ a2 – \(\frac{225}{a^{2}}\) = -16
∴ a4 – 225 = – 16a2
∴ a4 + 16a2 – 225 = 0
∴ (a2 + 25)(a2 – 9) = 0
∴ a2 = -25 or a2 = 9
But a ∈ R, a2 ≠ -25
∴ a2 = 9
∴ a = ±3
When a = 3, b = \(\frac{15}{3}\) = 5
When a = -3, b = \(\frac{15}{-3}\) = -5
∴ \(\sqrt{-16+30 \mathrm{i}}\) = ±(3 + 5i)

(ii) Let \(\sqrt{15-8 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
15 – 8i = a2 + b2i2 + 2abi
∴ 15 – 8i = (a2 – b2) + 2abi ……[∵ i2 = -1]
Equating real and imaginary parts, we get
a2 – b2 = 15 and 2ab = -8
∴ a2 – b2 = 15 and b = \(\frac{-4}{a}\)
∴ a2 – (\(\left(\frac{-4}{a}\right)^{2}\)) = 15
∴ a2 – \(\frac{16}{a^{2}}\) = 15
∴ a4 – 16 = 15a2
∴ a4 – 15a2 – 16 = 0
∴ (a2 – 16)(a2 + 1) = 0
∴ a2 = 16 or a2 = -1
But a ∈ R, a2 ≠ -1
∴ a2 = 16
∴ a = ±4
When a = 4, b = \(\frac{-4}{4}\) = -1
When a = -4, b = \(\frac{-4}{-4}\) = 1
\(\sqrt{15-8 i}\) = ±(4 – i)

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3

(iii) Let \(\sqrt{2+2 \sqrt{3} i}\) = a + bi, where a, b ∈ R.
Squaring on both sides, we get
2 – 2√3i = a2 + b2i2 + 2abi
∴ 2 – 2√3i = a2 – b2 + 2abi …..[∵ i2 = -1]
Equating real and imaginary parts, we get
a2 – b2 = 2 and 2ab = 2√3
∴ a2 – b2 = 2 and b = \(\frac{\sqrt{3}}{\mathrm{a}}\)
∴ a2 – \(\left(\frac{\sqrt{3}}{a}\right)^{2}\) = 2
∴ a2 – \(\frac{3}{a^{2}}\) = 2
∴ a4 – 3 = 2a2
∴ a4 – 2a2 – 3 = 0
∴ (a2 – 3)(a2 + 1) = 0
∴ a2 = 3 or a2 = -1
But a ∈ R, a2 ≠ -1
∴ a2 = 3
∴ a = ±√3
When a = √3 , b = \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1
When a = -√3 , b = \(\frac{\sqrt{3}}{-\sqrt{3}}\) = -1
∴ \(\sqrt{2+2 \sqrt{3} i}\) = ±(√3 + i)

(iv) Let \(\sqrt{18 \mathrm{i}}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
18i = a2 + b2i2 + 2abi
∴ 0 + 18i = a2 – b2 + 2abi …..[∵ i2 = -1]
Equating real and imaginary parts, we get
a2 – b2 = 0 and 2ab = 18
∴ a2 – b2 = 0 and b = \(\frac{9}{\mathrm{a}}\)
∴ a2 – \(\left(\frac{9}{a}\right)^{2}\) = 0
∴ a2 – \(\frac{81}{a^{2}}\) = 0
∴ a4 – 81 = 0
∴ (a2 – 9) (a2 + 9) = 0
∴ a2 = 9 or a2 = -9
But a ∈ R, a2 ≠ -9
∴ a2 = 9
∴ a = ±3
When a = 3, b = \(\frac{9}{3}\) = 3
When a = 3, b = \(\frac{9}{-3}\) = -3
∴ \(\sqrt{18 \mathrm{i}}\) = ±3(1 + i)

(v) Let \(\sqrt{3-4 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
3 – 4i = a2 + b2i2 + 2abi
∴ 3 – 4i = a2 – b2 + 2abi ……[∵ i2 = -1]
Equating real and imaginary parts, we get
a2 – b2 = 3 and 2ab = -4
∴ a2 – b2 = 3 and b = \(\frac{-2}{a}\)
∴ a2 – \(\left(-\frac{2}{a}\right)^{2}\) = 3
∴ a2 – \(\frac{4}{a^{2}}\) = 3
∴ a4 – 4 = 3a2
∴ a4 – 3a2 – 4 = 0
∴ (a2 – 4)(a2 + 1) = 0
∴ a2 = 4 or a2 = -1
But, a ∈ R, a2 ≠ -1
∴ a2 = 4
∴ a = ±2
When a = 2, b = \(\frac{-2}{2}\) = -1
When a = -2, b = \(\frac{-2}{-2}\) = 1
∴ \(\sqrt{3-4 i}\) = ±(2 – i)

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Miscellaneous Exercise 3

(vi) Let \(\sqrt{6+8 i}\) = a + bi, where a, b ∈ R
Squaring on both sides, we get
6 + 8i = a2 + b2i2 + 2abi
∴ 6 + 8i = a2 – b2 + 2abi ……[∵ i2 = -1]
Equating real and imaginary parts, we get
a2 – b2 = 6 and 2ab = 8
∴ a2 – b2 = 6 and b = \(\frac{4}{\mathrm{a}}\)
∴ a2 – \(\left(\frac{4}{a}\right)^{2}\) = 6
∴ a2 – \(\frac{16}{a^{2}}\) = 6
∴ a4 – 16 = 6a2
∴ a4 – 6a2 – 16 = 0
∴ (a2 – 8)(a2 + 2) = 0
∴ a2 = 8 or a2 = -2
But a ∈ R, a2 ≠ -2
∴ a2 = 8
∴ a = ±2√2
When a = 2√2, b = \(\frac{4}{2 \sqrt{2}}\) = √2
When a = -2√2, b = \(\frac{4}{-2 \sqrt{2}}\) = -√2
∴ \(\sqrt{6+8 i}\) = ±(2√2 + √2i) = ±√2(2 + i)

11th Commerce Maths Digest Pdf 

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

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

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

Std 11 Maths 1 Miscellaneous Exercise 6 Solutions Commerce Maths

Question 1.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q1(i)
= 2(4 – 0) + 5(10 – 9) + 7(0 – 18)
= 2(4) + 5(1) + 7(-18)
= 8 + 5 – 126
= -113

(ii) \(\left|\begin{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
\end{array}\right|\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q1(ii)
= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)
= 1(32) + 3(36) + 12(-18)
= 32 + 108 – 216
= -76

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6

Question 2.
Find the value(s) of x, if
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
Solution:
\(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
∴ 1(-10x2 + 10x) – 4(5x2 + 5) + 20(2x + 2) = 0
∴ -10x2 + 10x – 20x2 – 20 + 40x + 40 = 0
∴ -30x2 + 50x + 20 = 0
∴ 3x2 – 5x – 2 = 0 ……[Dividing throughout by (-10)]
∴ 3x2 – 6x + x – 2 = 0
∴ 3x(x – 2) + 1(x – 2) = 0
∴ (x – 2) (3x + 1) = 0
∴ x – 2 = 0 or 3x + 1 = 0
∴ x = 2 or x = \(-\frac{1}{3}\)

(ii) \(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
Solution:
\(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
∴ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0
∴ 1(-12) – 2x(-15) + 4x(-3) = 0
∴ -12 + 30x – 12x = 0
∴ 18x = 12
∴ x = \(\frac{2}{3}\)

Question 3.
By using properties of determinants, prove that \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|=0\).
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q3

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6

Question 4.
Without expanding the determinants, show that
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q4
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q4.1
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q4.2
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q4.3
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q4.4
Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6 Q4.5

Question 5.
Solve the following linear equations by Cramer’s Rule.
(i) 2x – y + z = 1, x + 2y + 3z = 8, 3x + y – 4z = 1
Solution:
Given equations are
2x – y + z = 1
x + 2y + 3z = 8
3x + y – 4z = 1
D = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-8 – 3) – (-1)(-4 – 9) + 1(1 – 6)
= 2(-11) + 1(-13) + 1(-5)
= -22 – 13 – 5
= -40 ≠ 0
Dx = \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
8 & 2 & 3 \\
1 & 1 & -4
\end{array}\right|\)
= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)
= 1(-11) + 1(-35) + 1(6)
= -11 – 35 + 6
= -40
Dy = \(\left|\begin{array}{ccc}
2 & 1 & 1 \\
1 & 8 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-32 – 3) – 1(-4 – 9) + 1(1 – 24)
= 2(-35) – 1(-13) + 1(-23)
= -70 + 13 – 23
= -80
Dz = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 8 \\
3 & 1 & 1
\end{array}\right|\)
= 2(2 – 8) – (-1)(1 – 24) + 1(1 – 6)
= 2(-6) + 1(-23) + 1(-5)
= -12 – 23 – 5
= -40
By Cramer’s Rule,
x = \(\frac{D_{x}}{D}=\frac{-40}{-40}\) = 1
y = \(\frac{\mathrm{D}_{y}}{\mathrm{D}}=\frac{-80}{-40}\) = 2
z = \(\frac{\mathrm{D}_{z}}{\mathrm{D}}=\frac{-40}{-40}\) = 1
∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6

(ii) \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-2\), \(\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3\), \(\frac{2}{x}-\frac{1}{y}+\frac{3}{z}=-1\)
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,
p = \(\frac{\mathrm{D}_{\mathrm{p}}}{\mathrm{D}}=\frac{-5}{-3}=\frac{5}{3}\)
q = \(\frac{\mathrm{D}_{y}}{\mathrm{D}}=\frac{-80}{-40}\) = 2
r = \(\frac{D_{2}}{D}=\frac{-40}{-40}\) = 1
∴ x = \(\frac{3}{5}\), y = \(\frac{-3}{5}\), z = \(\frac{-1}{2}\) are the solutions of the given equations.

(iii) x – y + 2z = 7, 3x + 4y – 5z = 5, 2x – y + 3z = 12
Solution:
Given equations are
x – y + 2z = 1
3x + 4y – 5z = 5
2x – y + 3z = 12
D = \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right|\)
= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8)
= 1(7) + 1(19) + 2(-11)
= 7 + 19 – 22
= 4 ≠ 0
Dx = \(\left|\begin{array}{ccc}
7 & -1 & 2 \\
5 & 4 & -5 \\
12 & -1 & 3
\end{array}\right|\)
= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)
= 7(7)+ 1(75) +2(-53)
= 49 + 75 – 106
= 18
Dy = \(\left|\begin{array}{ccc}
1 & 7 & 2 \\
3 & 5 & -5 \\
2 & 12 & 3
\end{array}\right|\)
= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= -6
Dz = \(\left|\begin{array}{ccc}
1 & -1 & 7 \\
3 & 4 & 5 \\
2 & -1 & 12
\end{array}\right|\)
= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8)
= 1(53)+ 1(26) + 7(-11)
= 53 + 26 – 77
= 2
By Cramer’s Rule,
x = \(\frac{\mathrm{D}_{x}}{\mathrm{D}}=\frac{18}{4}=\frac{9}{2}\)
y = \(\frac{D_{y}}{D}=\frac{-6}{4}=\frac{-3}{2}\)
z = \(\frac{D_{z}}{D}=\frac{2}{4}=\frac{1}{2}\)
∴ x = \(\frac{9}{2}\), y = \(\frac{-3}{2}\) and z = \(\frac{1}{2}\) are the solutions of the given equations.

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6

Question 6.
Find the value(s) of k, if the following equations are consistent.
(i) 3x + y – 2 = 0, kx + 2y – 3 = 0 and 2x – y = 3
Solution:
Given equations are
3x + y – 2 = 0
kx + 2y – 3 = 0
2x – y = 3 i.e. 2x – y – 3 = 0
Since, these equations are consistent.
∴ \(\left|\begin{array}{rrr}
3 & 1 & -2 \\
k & 2 & -3 \\
2 & -1 & -3
\end{array}\right|=0\)
∴ 3(-6 – 3) – 1(-3k + 6) – 2(-k – 4) = 0
∴ 3(-9) – 1 (-3k + 6) – 2(-k – 4) = 0
∴ -27 + 3k – 6 + 2k + 8 = 0
∴ 5k – 25 = 0
∴ k = 5

(ii) kx + 3y + 4 = 0, x + ky + 3 = 0, 3x + 4y + 5 = 0
Solution:
Given equations are
kx + 3y + 4 = 0
x + ky + 3 = 0
3x + 4y + 5 = 0
Since, these equations are consistent.
∴ \(\left|\begin{array}{lll}
\mathrm{k} & 3 & 4 \\
1 & \mathrm{k} & 3 \\
3 & 4 & 5
\end{array}\right|=0\)
∴ k(5k – 12) – 3(5 – 9) + 4(4 – 3k) = 0
∴ 5k2 – 12k + 12 + 16 – 12k = 0
∴ 5k2 – 24k + 28 = 0
∴ 5k2 – 10k – 14k + 28 = 0
∴ 5k(k – 2) – 14(k – 2) = 0
∴ (k – 2) (5k – 14) = 0
∴ k – 2 = 0 or 5k – 14 = 0
∴ k = 2 or k = \(\frac{14}{5}\)

Question 7.
Find the area of triangles whose vertices are
(i) A(-1, 2), B(2, 4), C(0, 0)
Solution:
Here, A(x1, y1) ≡ A(-1, 2), B(x2, y2) ≡ B(2, 4), C(x3, y3) ≡ C(0, 0)
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(∆ABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
-1 & 2 & 1 \\
2 & 4 & 1 \\
0 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-1(4 – 0) – 2(2 – 0) + 1(0 – 0)]
= \(\frac{1}{2}\) (-4 – 4)
= \(\frac{1}{2}\) (-8)
= -4
Since, area cannot be negative.
∴ A(∆ABC) = 4 sq.units

(ii) P(3, 6), Q(-1, 3), R(2, -1)
Solution:
Here, P(x1, y1) ≡ P(3, 6), Q(x2, y2) ≡ Q(-1, 3), R(x3, y3) ≡ R(2, -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(∆PQR) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & 6 & 1 \\
-1 & 3 & 1 \\
2 & -1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [3(3 + 1) – 6(-1 – 2) + 1(1 – 6)]
= \(\frac{1}{2}\) [3(4) – 6(-3) + 1(-5)]
= \(\frac{1}{2}\) (12 + 18 – 5)
∴ A(∆PQR) = \(\frac{25}{2}\) sq.units

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6

(iii) L(1, 1), M(-2, 2), N(5, 4)
Solution:
Here, L(x1, y1) ≡ L(1, 1), M(x2, y2) ≡ M(-2, 2), N(x3, y3) ≡ N(5, 4)
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(∆LMN) = \(\frac{1}{2}\left|\begin{array}{rrr}
1 & 1 & 1 \\
-2 & 2 & 1 \\
5 & 4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [1(2 – 4) -1(-2 – 5) + 1(-8 – 10)]
= \(\frac{1}{2}\) [1(-2) – 1(-7) + 1(-18)]
= \(\frac{1}{2}\) (-2 + 7 – 18)
= \(\frac{-13}{2}\)
Since, area cannot be negative.
∴ A(∆LMN) = \(\frac{13}{2}\) sq.units

Question 8.
Find the value of k,
(i) if the area of ∆PQR is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).
Solution:
Here, P(x1, y1) ≡ P(k, 0), Q(x2, y2) ≡ Q(4, 0), R(x3, y3) ≡ R(0, 2)
A(∆PQR) = 4 sq.units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ ±4 = \(\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
4 & 0 & 1 \\
0 & 2 & 1
\end{array}\right|\)
∴ ±4 = \(\frac{1}{2}\) [k(0 – 2) – 0 + 1(8 – 0)]
∴ ±8 = -2k + 8
∴ 8 = -2k + 8 or -8 = -2k + 8
∴ -2k = 0 or 2k = 16
∴ k = 0 or k = 8

Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6

(ii) if area of ∆LMN is \(\frac{33}{2}\) square units and vertices are L(3, -5), M(-2, k), N(1, 4).
Solution:
Here, L(x1, y1) ≡ L(3, -5), M(x2, y2) ≡ M(-2, k), N(x3, y3) ≡ N(1, 4)
A(∆LMN) = \(\frac{33}{2}\) sq.units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ \(\pm \frac{33}{2}=\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
-2 & \mathrm{k} & 1 \\
1 & 4 & 1
\end{array}\right|\)
∴ ±\(\frac{33}{2}\) = \(\frac{1}{2}\) [3(k – 4) – (-5) (-2 – 1) + 1(-8 – k)]
∴ ±33 = 3k – 12 – 15 – 8 – k
∴ 33 = 2k – 35
∴ 2k – 35 = 33 or 2k – 35 = -33
∴ 2k = 68 or 2k = 2
∴ k = 34 or k = 1

11th Commerce Maths Digest Pdf

11th Commerce Maths 1 Chapter 3 Exercise 3.3 Answers Maharashtra Board

Complex Numbers Class 11 Commerce Maths 1 Chapter 3 Exercise 3.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Complex Numbers Ex 3.3 Questions and Answers.

Std 11 Maths 1 Exercise 3.3 Solutions Commerce Maths

Question 1.
If ω is a complex cube root of unity, show that
(i) (2 – ω)(2 – ω2) = 7
(ii) (2 + ω + ω2)3 – (1 – 3ω + ω2)3 = 65
(iii) \(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\) = ω2
Solution:
ω is the complex cube root of unity.
∴ ω3 = 1 and 1 + ω + ω2 = 0
Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1
(i) L.H.S. = (2 – ω)(2 – ω2)
= 4 – 2ω2 – 2ω + ω3
= 4 – 2(ω2 + ω) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1
= 7
= R.H.S.

(ii) L.H.S. = (2 + ω + ω2)3 – (1 – 3ω + ω2)3
= [2 + (ω + ω2)]3 – [-3ω + (1 + ω2)]3
= (2 – 1)3 – (-3ω – ω)3
= 13 – (-4ω)3
= 1 + 64ω3
= 1 + 64(1)
= 65
= R.H.S.

(iii) L.H.S. =\(\frac{\left(\mathbf{a}+\mathbf{b} \omega+\mathbf{c} \omega^{2}\right)}{\mathbf{c}+\mathbf{a} \omega+\mathbf{b} \omega^{2}}\)
= \(\frac{a \omega^{3}+b \omega^{4}+c \omega^{2}}{c+a \omega+b \omega^{2}}\) ……[∵ ω3 = 1, ω4 = ω]
= \(\frac{\omega^{2}\left(c+a \omega+b \omega^{2}\right)}{c+a \omega+b \omega^{2}}\)
= ω2
= R.H.S.

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3

Question 2.
If ω is a complex cube root of unity, find the value of
(i) ω + \(\frac{1}{\omega}\)
(ii) ω2 + ω3 + ω4
(iii) (1 + ω2)3
(iv) (1 – ω – ω2)3 + (1 – ω + ω2)3
(v) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)
Solution:
ω is the complex cube root of unity.
∴ ω3 = 1 and 1 + ω + ω2 = 0
Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1
(i) ω + \(\frac{1}{\omega}\)
= \(\frac{\omega^{2}+1}{\omega}\)
= \(\frac{-\omega}{\omega}\)
= -1

(ii) ω2 + ω3 + ω4
= ω2 (1 + ω + ω2)
= ω2 (0)
= 0

(iii) (1 + ω2)3
= (-ω)3
= -ω3
= -1

(iv) (1 – ω – ω2)3 + (1 – ω + ω2)3
= [1 – (ω + ω2)]3 + [(1 + ω2) – ω]3
= [1 – (-1)]3 + (-ω – ω)3
= 23 + (-2ω)3
= 8 – 8ω3
= 8 – 8(1)
= 0

(v) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)
= (1 + ω)(1 + ω2)(1 + ω)(1 + ω2) …..[∵ ω3 = 1, ω4 = ω]
= (-ω2)(-ω)(-ω2)(-ω)
= ω6
= (ω3)2
= (1)2
= 1

Question 3.
If α and β are the complex cube roots of unity, show that α2 + β2 + αβ = 0.
Solution:
α and β are the complex cube roots of unity.
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3 Q3
∴ α – β = -1
L.H.S. = α2 + β2 + αβ
= α2 + 2αβ + β2 + αβ – 2αβ ……[Adding and subtracting 2αβ]
= (α2 + 2αβ + β2) – αβ
= (α + β)2 – αβ
= (-1)2 – 1
= 1 – 1
= 0
= R.H.S.

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3

Question 4.
If x = a + b, y = αa + βb and z = aβ + bα, where α and β are the complex cube roots of unity, show that xyz = a3 + b3.
Solution:
x = a + b, y = αa + βb, z = aβ + bα
α and β are the complex cube roots of unity.
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3 Q4
Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3 Q4.1

Question 5.
If ω is a complex cube root of unity, then prove the following:
(i) (ω2 + ω – 1)3 = -8
(ii) (a + b) + (aω + bω2) + (aω2 + bω) = 0
Solution:
ω is the complex cube root of unity.
∴ ω3 = 1 and 1 + ω + ω2 = 0
Also, 1 + ω2 = -ω, 1 + ω = -ω2 and ω + ω2 = -1
(i) L.H.S. = (ω2 + ω – 1)3
= (-1 – 1)3
= (-2)3
= -8
= R.H.S.

Maharashtra Board 11th Commerce Maths Solutions Chapter 3 Complex Numbers Ex 3.3

(ii) L.H.S. = (a + b) + (aω + bω2) + (aω2 + bω)
= (a + aω + aω2) + (b + bω + bω2)
= a(1 + ω + ω2) + b(1 + ω + ω2)
= a(0) + b(0)
= 0
= R.H.S.

11th Commerce Maths Digest Pdf 

11th Commerce Maths 1 Chapter 4 Exercise 4.4 Answers Maharashtra Board

Sequences and Series Class 11 Commerce Maths 1 Chapter 4 Exercise 4.4 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.4 Questions and Answers.

Std 11 Maths 1 Exercise 4.4 Solutions Commerce Maths

Question 1.
Verify whether the following sequences are H.P.
(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \ldots \ldots \ldots \ldots\)
(iii) \(\frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \ldots\)
Solution:
(i) \(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \ldots\)
Here, the reciprocal sequence is 3, 5, 7, 9, …
∴ t1 = 3, t2 = 5, t3 = 7, …..
∵ t2 – t1 = t3 – t2 = t4 – t3 = 2, constant
∴ The reciprocal sequence is an A.P.
∴ the given sequence is H.P.

(ii) \(\frac{1}{3}, \frac{1}{6}, \frac{1}{9}, \frac{1}{12}, \ldots \ldots \ldots \ldots\)
Here, the reciprocal sequence is 3, 6, 9, 12 …
∴ t1 = 3, t2 = 6, t3 = 9, t4 = 12, …
∵ t2 – t1 = t3 – t2 = t4 – t3 = 3, constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is H.P.

(iii) \(\frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \ldots\)
Here, the reciprocal sequence is 7, 9, 11, 13, 15, ……
∴ t1 = 7, t2 = 9, t3 = 11, t4 = 13, …..
∵ t2 – t1 = t3 – t2 = t4 – t3 = 2, constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is H.P.

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4

Question 2.
Find the nth term and hence find the 8th term of the following H.P.s:
(i) \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \ldots \ldots \ldots\)
(ii) \(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \ldots \ldots \ldots \ldots\)
(iii) \(\frac{1}{5}, \frac{1}{10}, \frac{1}{15}, \frac{1}{20}, \cdots \cdots \cdots\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4 Q2
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4 Q2.1

Question 3.
Find A.M. of two positive numbers whose G.M. and H.M. are 4 and \(\frac{16}{5}\).
Solution:
G.M. = 4, H.M. = \(\frac{16}{5}\)
∵ (G.M.)2 = (A.M.) (H.M.)
∴ 16 = A.M. × \(\frac{16}{5}\)
∴ A.M. = 5

Question 4.
Find H.M. of two positive numbers whose A.M. and G.M. are \(\frac{15}{2}\) and 6.
Solution:
A.M. = \(\frac{15}{2}\), G.M. = 6
Now, (G.M.)2 = (A.M.) (H.M.)
∴ 62 = \(\frac{15}{2}\) × H.M.
∴ H.M. = 36 × \(\frac{2}{15}\)
∴ H.M. = \(\frac{24}{5}\)

Question 5.
Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48.
Solution:
A.M. = 75, H.M. = 48
(G.M.)2 = (A.M.) (H.M.)
∵ (G.M.)2 = 75 × 48
∵ (G.M.)2 = 25 × 3 × 16 × 3
∵ (G.M.)2 = 52 × 42 × 32
∴ G.M. = 5 × 4 × 3
∴ G.M. = 60

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4

Question 6.
Insert two numbers between \(\frac{1}{7}\) and \(\frac{1}{13}\) so that the resulting sequence is a H.P.
Solution:
Let the required numbers be \(\frac{1}{\mathrm{H}_{1}}\) and \(\frac{1}{\mathrm{H}_{2}}\).
∴ \(\frac{1}{7}, \frac{1}{\mathrm{H}_{1}}, \frac{1}{\mathrm{H}_{2}}, \frac{1}{13}\) are in H.P.
∴ 7, H1, H2 and 13 are in A.P.
∴ t1 = a = 7 and t4 = a + 3d = 13
∴ 7 + 3d = 13
∴ 3d = 6
∴ d = 2
∴ H1 = t2 = a + d = 7 + 2 = 9
and H2 = t3 = a + 2d = 7 + 2(2) = 11
∴ \(\frac{1}{9}\) and \(\frac{1}{11}\) are the required numbers to be inserted between \(\frac{1}{7}\) and \(\frac{1}{13}\) so that the resulting sequence is a H.P.

Question 7.
Insert two numbers between 1 and -27 so that the resulting sequence is a G.P.
Solution:
Let the required numbers be G1 and G2.
∴ 1, G1, G2, -27 are in G.P.
∴ t1 = 1, t2 = G1, t3 = G2, t4 = -27
∴ t1 = a = 1
tn = arn-1
∴ t4 = (1) r4-1
∴ -27 = r3
∴ r3 = (-3)3
∴ r = -3
∴ G1 = t2 = ar = 1(-3) = -3
∴ G2 = t3 = ar = 1(-3)2 = 9
∴ -3 and 9 are the required numbers to be inserted between 1 and -27 so that the resulting sequence is a G.P.

Question 8.
Find two numbers whose A.M. exceeds their G.M. by \(\frac{1}{2}\) and their H.M. by \(\frac{25}{26}\).
Solution:
Let a, b be the two numbers.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4 Q8
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4 Q8.1
∴ a + b = 13
∴ b = 13 – a …….(iii)
and ab = 36
∴ a(13 – a) = 36 …… [From (iii)]
∴ a2 – 13a + 36 = 0
∴ (a – 4)(a – 9) = 0
∴ a = 4 or a = 9
When a = 4, b = 13 – 4 = 9
When a = 9, b = 13 – 9 = 4
∴ the two numbers are 4 and 9.

Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4

Question 9.
Find two numbers whose A.M. exceeds G.M. bv 7 and their H.M. by \(\frac{63}{5}\).
Solution:
Let a, b be the two numbers.
Maharashtra Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4 Q9
∴ a + b = 70
∴ b = 70 – a …..(ii)
∴ G = A – 7 = 35 – 7 = 28 …….[From (i)]
∴ √ab = 28
∴ ab = 282 = 784
∴ a(70 – a) = 784 ……[From (ii)]
∴ 70a – a2 = 784
∴ a2 – 70a + 784 = 0
∴ a2 – 56a – 14a + 784 = 0
∴ (a – 56) (a – 14) = 0
∴ a = 14 or a = 56
When a = 14, b = 70 – 14 = 56
When a = 56, b = 70 – 56 = 14
∴ the two numbers are 14 and 56.

11th Commerce Maths Digest Pdf

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

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

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

Std 11 Maths 1 Miscellaneous Exercise 5 Solutions Commerce Maths

Question 1.
Find the slopes of the lines passing through the following points:
(i) (1, 2), (3, -5)
(ii) (1, 3), (5, 2)
(iii) (-1, 3), (3, -1)
(iv) (2, -5), (3, -1)
Solution:
(i) Let A = (1, 2) = (x1, y1) and B = (3, -5) = (x2, y2) say.
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-5-2}{3-1}=\frac{-7}{2}\)

(ii) Let C = (1, 3) = (x1, y1) and D = (5, 2) = (x2, y2) say.
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-3}{5-1}=\frac{-1}{4}\)

(iii) Let E = (-1, 3) = (x1, y1) and F = (3, -1) = (x2, y2) say.
Slope of line EF = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{3-(-1)}=\frac{-4}{4}\) = -1

(iv) Let P = (2, -5) = (x1, y1) and Q = (3, -1) = (x2, y2) say.
Slope of line PQ = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-(-5)}{3-2}\) = \(\frac{-1+5}{1}\) = 4

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5

Question 2.
Find the slope of the line which
(i) makes an angle of 120° with the positive X-axis.
(ii) makes intercepts 3 and -4 on the axes.
(iii) passes through the points A(-2, 1) and the origin.
Solution:
(i) θ = 120°
Slope of the line = tan 120°
= tan (180° – 60°)
= -tan 60° …..[tan(180° – θ) = -tan θ]
= -√3

(ii) Given, x-intercept of line is 3 and y-intercept of line is -4
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, -4).
∴ The line passes through (3, 0) = (x1, y1) and (0, -4) = (x2, y2) say.
∴ Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-4-0}{0-3}=\frac{-4}{-3}=\frac{4}{3}\)

(iii) Required line passes through O(0, 0) = (x1, y1) and A(-2, 1) = (x2, y2) say.
Slope of line OA = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-0}{-2-0}=\frac{1}{-2}\) = \(\frac{-1}{2}\)

Question 3.
Find the value of k:
(i) if the slope of the line passing through the points (3, 4), (5, k) is 9.
(ii) the points (1, 3), (4, 1), (3, k) are collinear.
(iii) the point P(1, k) lies on the line passing through the points A(2, 2) and B(3, 3).
Solution:
(i) Let P(3, 4), Q(5, k).
Slope of PQ = 9 …….[Given]
∴ \(\frac{\mathrm{k}-4}{5-3}\) = 9
∴ \(\frac{\mathrm{k}-4}{2}\) = 9
∴ k – 4 = 18
∴ k = 22

(ii) The points A(1, 3), B(4, 1) and C(3, k) are collinear.
∴ Slope of AB = Slope of BC
∴ \(\frac{1-3}{4-1}=\frac{k-1}{3-4}\)
∴ \(\frac{-2}{3}=\frac{\mathrm{k}-1}{-1}\)
∴ 2 = 3k – 3
∴ k = \(\frac{5}{3}\)

(iii) Given, point P(1, k) lies on the line joining A(2, 2) and B(3, 3).
∴ Slope of AB = Slope of BP
∴ \(\frac{3-2}{3-2}=\frac{3-k}{3-1}\)
∴ 1 = \(\frac{3-k}{2}\)
∴ 2 = 3 – k
∴ k = 1

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5

Question 4.
Reduce the equation 6x + 3y + 8 = 0 into slope-intercept form. Hence, find its slope.
Solution:
Given equation is 6x + 3y + 8 = 0, which can be written as
3y = -6x – 8
∴ y = \(\frac{-6 x}{3}-\frac{8}{3}\)
∴ y = -2x – \(\frac{8}{3}\)
This is of the form y = mx + c with m = -2
∴ y = -2x – \(\frac{8}{3}\) is in slope-intercept form with slope = -2

Question 5.
Verify that A(2, 7) is not a point on the line x + 2y + 2 = 0.
Solution:
Given equation is x + 2y + 2 = 0.
Substituting x = 2 and y = 7 in L.H.S. of given equation, we get
L.H.S. = x + 2y + 2
= 2 + 2(7) + 2
= 2 + 14 + 2
= 18
≠ R.H.S.
∴ Point A does not lie on the given line.

Question 6.
Find the X-intercept of the line x + 2y – 1 = 0.
Solution:
Given equation of the line is x + 2y – 1 = 0
To find the x-intercept, put y = 0 in given equation of the line
∴ x + 2(0) – 1 = 0
∴ x + 0 – 1 = 0
∴ x = 1
∴ X-intercept of the given line is 1.
Alternate method:
Given equation of the line is x + 2y – 1 = 0
i.e. x + 2y = 1
∴ \(\frac{x}{1}+\frac{y}{\frac{1}{2}}=1\)
Comparing with \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}=1\), we get a = 1
X-intercept of the line is 1.

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5

Question 7.
Find the slope of the line y – x + 3 = 0.
Solution:
Equation of given line is y – x + 3 = 0
i.e. y = x – 3
Comparing with y = mx + c, we get
m = Slope = 1

Question 8.
Does point A(2, 3) lie on the line 3x + 2y – 6 = 0? Give reason.
Solution:
Given equation is 3x + 2y – 6 = 0.
Substituting x = 2 and y = 3 in L.H.S. of given equation, we get
L.H.S. = 3x + 2y – 6
= 3(2)+ 2(3) – 6
= 6
≠ R.H.S.
∴ Point A does not lie on the given line.

Question 9.
Which of the following lines passes through the origin?
(a) x = 2
(b) y = 3
(c) y = x + 2
(d) 2x – y = 0
Solution:
Any line passing through origin is of the form y = mx or ax + by = 0.
Here in the given option, 2x – y = 0 is in the form ax + by = 0.

Question 10.
Obtain the equation of the line which is:
(i) parallel to the X-axis and 3 units below it.
(ii) parallel to the Y-axis and 2 units to the left of it.
(iii) parallel to the X-axis and making an intercept of 5 on the Y-axis.
(iv) parallel to the Y-axis and making an intercept of 3 on the X-axis.
Solution:
(i) Equation of a line parallel to X-axis is y = k.
Since, the line is at a distance of 3 units below X-axis.
∴ k = -3
∴ the equation of the required line is y = -3
i.e., y + 3 = 0.

(ii) Equation of a line parallel to Y-axis is x = h.
Since, the line is at a distance of 2 units to the left of Y-axis.
∴ h = -2
∴ the equation of the required line is x = -2
i.e., x + 2 = 0.

(iii) Equation of a line parallel to X-axis with y-intercept ‘k’ is y = k.
Here, y-intercept = 5
∴ the equation of the required line is y = 5.

(iv) Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 3
∴ the equation of the required line is x = 3.

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5

Question 11.
Obtain the equation of the line containing the point:
(i) (2, 3) and parallel to the X-axis.
(ii) (2, 4) and perpendicular to the Y-axis.
(iii) (2, 5) and perpendicular to the X-axis.
Solution:
(i) Equation of a line parallel to X-axis is of the form y = k.
Since, the line passes through (2, 3).
∴ k = 3
∴ the equation of the required line is y = 3.

(ii) Equation of a line perpendicular to Y-axis
i.e., parallel to X-axis, is of the form y = k.
Since, the line passes through (2, 4).
∴ k = 4
∴ the equation of the required line is y = 4.

(iii) Equation of a line perpendicular to X-axis
i.e., parallel to Y-axis, is of the form x = h.
Since, the line passes through (2, 5).
∴ h = 2
∴ the equation of the required line is x = 2.

Question 12.
Find the equation of the line:
(i) having slope 5 and containing point A(-1, 2).
(ii) containing the point (2, 1) and having slope 13.
(iii) containing the point T(7, 3) and having inclination 90°.
(iv) containing the origin and having inclination 90°.
(v) through the origin which bisects the portion of the line 3x + 2y = 2 intercepted between the co-ordinate axes.
Solution:
(i) Given, slope (m) = 5 and the line passes through A(-1, 2).
Equation of the line in slope point form is y – y1 = m(x – x1)
∴ the equation of the required line is y – 2 = 5(x + 1)
∴ y – 2 = 5x + 5
∴ 5x – y + 7 = 0

(ii) Given, slope (m) = 13 and the line passes through (2, 1).
Equation of the line in slope point form is y – y1 = m(x – x1)
∴ the equation of the required line is y – 1 = 13(x – 2)
∴ y – 1 = 13x – 26
∴ 13x – y = 25.

(iii) Given, Inclination of line = θ = 90°
∴ the required line is parallel to Y-axis (or lies on the Y-axis.)
Equation of a line parallel to Y-axis is of the form x = h.
Since, the line passes through (7, 3).
∴ h = 7
∴ the equation of the required line is x = 7.

(iv) Given, Inclination of line = θ = 90°
∴ the required line is parallel to Y-axis (or lies on the Y-axis.)
Equation of a line parallel to Y-axis is of the form x = h.
Since, the line passes through origin (0, 0).
∴ h = 0
∴ the equation of the required line is x = 0.

(v) Given equation of the line is 3x + 2y = 2.
∴ \(\frac{3 x}{2}+\frac{2 y}{2}=1\)
∴ \(\frac{x}{\frac{2}{3}}+\frac{y}{1}=1\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5 Q12(v)
This equation is of the form \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}=1\), with
a = \(\frac{2}{3}\), b = 1
∴ the line 3x + 2y = 2 intersects the X-axis at A(\(\frac{2}{3}\), 0) and Y-axis at B(0, 1).
Required line is passing through the midpoint of AB.
∴ Midpoint of AB = \(\left(\frac{\frac{2}{3}+0}{2}, \frac{0+1}{2}\right)=\left(\frac{1}{3}, \frac{1}{2}\right)\)
∴ Required line passes through (0, 0) and \(\left(\frac{1}{3}, \frac{1}{2}\right)\).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ the equation of the required line is
\(\frac{y-0}{\frac{1}{2}-0}=\frac{x-0}{\frac{1}{3}-0}\)
∴ 2y = 3x
∴ 3x – 2y = 0

Question 13.
Find the equation of the line passing through the points A(-3, 0) and B(0, 4).
Solution:
Since, the required line passes through the points A(-3, 0) and B(0, 4).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
Here, (x1, y1) = (-3, 0) and (x2, y2) = (0, 4)
∴ the equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-(-3)}{0-(-3)}\)
∴ \(\frac{y}{4}=\frac{x+3}{3}\)
∴ 4x + 12 = 3y
∴ 4x – 3y + 12 = 0

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5

Question 14.
Find the equation of the line:
(i) having slope 5 and making intercept 5 on the X-axis.
(ii) having an inclination 60° and making intercept 4 on the Y-axis.
Solution:
(i) Since, the x-intercept of the required line is 5.
∴ it passes through (5, 0).
Also, slope(m) of the line is 5
Equation of the line in slope point form is y – y1 = m(x – x1)
∴ the equation of the required line is y – 0 = 5(x – 5)
∴ y = 5x – 25
∴ 5x – y – 25 = 0

(ii) Given, Inclination of line = θ = 60°
∴ Slope of the line (m) = tan θ
= tan 60°
= √3
and the y-intercept of the required line is 4.
∴ it passes through (0, 4).
Equation of the line in slope point form is y – y1 = m(x – x1)
∴ the equation of the required line is y – 4 = √3(x – 0)
∴ y – 4 = √3x
∴ √3x – y + 4 = 0

Question 15.
The vertices of a triangle are A(1, 4), B(2, 3), and C(1, 6). Find equations of
(i) the sides
(ii) the medians
(iii) Perpendicular bisectors of sides
(iv) altitudes of ∆ABC
Solution:
Vertices of ∆ABC are A(1, 4), B(2, 3), and C(1, 6)
(i) Equation of the line in two-point form is
Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5 Q15(i)
Since, both the points A and C have same x co-ordinates i.e. 1
∴ the points A and C lie on a line parallel to Y-axis.
∴ the equation of side AC is x = 1.

(ii) Let D, E, and F be the midpoints of sides BC, AC, and AB respectively of ∆ABC.
Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5 Q15(ii)
Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5 Q15(ii).1

(iii) Slope of side BC = \(\left(\frac{6-3}{1-2}\right)=\left(\frac{3}{-1}\right)\) = -3
∴ Slope of perpendicular bisector of BC is \(\frac{1}{3}\) and the line passes through \(\left(\frac{3}{2}, \frac{9}{2}\right)\)
∴ Equation of the perpendicular bisector of side BC is \(\left(y-\frac{9}{2}\right)=\frac{1}{3}\left(x-\frac{3}{2}\right)\)
∴ \(\frac{2 y-9}{2}=\frac{1}{3}\left(\frac{2 x-3}{2}\right)\)
∴ 3(2y – 9) = (2x – 3)
∴ 2x – 6y + 24 = 0
∴ x – 3y + 12 = 0
Since, both the points A and C have the same x co-ordinates i.e. 1
∴ the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, the perpendicular bisector of side AC is parallel to X-axis.
Since, the perpendicular bisector of side AC passes through E(1, 5).
∴ the equation of the perpendicular bisector of side AC is y = 5.
Slope of side AB = \(\left(\frac{3-4}{2-1}\right)\) = -1
∴ Slope of perpendicular bisector of AB is 1 and the line passes through \(\left(\frac{3}{2}, \frac{7}{2}\right)\).
∴ Equation of the perpendicular bisector of side AB is \(\left(y-\frac{7}{2}\right)=1\left(x-\frac{3}{2}\right)\)
∴ \(\frac{2 y-7}{2}=\frac{2 x-3}{2}\)
∴ 2y – 7 = 2x – 3
∴ 2x – 2y + 4 = 0
∴ x – y + 2 = 0

Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5

(iv) Let AX, BY and CZ be the altitudes through the vertices A, B, and C respectively of ∆ABC.
Maharashtra Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5 Q15(iv)
Slope of BC = -3
∴ Slope of AX = \(\frac{1}{3}\) …..[∵ AX ⊥ BC]
Since, altitude AX passes through (1, 4) and has slope \(\frac{1}{3}\)
∴ equation of altitude AX is y – 4 = \(\frac{1}{3}\)(x – 1)
∴ 3y – 12 = x – 1
∴ x – 3y + 11 = 0
Since, both the points A and C have the same x co-ordinates i.e. 1
∴ the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, altitude BY is parallel to X-axis.
Since, the altitude BY passes through B(2, 3).
∴ the equation of altitude BY is y = 3.
Also, slope of AB = -1
∴ Slope of CZ = 1 …..[∵ CZ ⊥ AB]
Since, altitude CZ passes through (1, 6) and has slope 1
∴ equation of altitude CZ is y – 6 = 1(x – 1)
∴ y – 6 = x – 1
∴ x – y + 5 = 0

11th Commerce Maths Digest Pdf

11th Commerce Maths 1 Chapter 8 Exercise 8.1 Answers Maharashtra Board

Continuity Class 11 Commerce Maths 1 Chapter 8 Exercise 8.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Continuity Ex 8.1 Questions and Answers.

Std 11 Maths 1 Exercise 8.1 Solutions Commerce Maths

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

(ii) f(x) = \(\frac{x^{2}-9}{x-3}\) on R
Solution:
f(x) = \(\frac{x^{2}-9}{x-3}\); x ∈ R
f(x) is a rational function and is continuous for all x ∈ R, except at the points where denominator becomes zero.
Here, denominator x – 3 = 0 when x = 3.
∴ Function f is continuous for all x ∈ R, except at x = 3, where it is not defined.

Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1

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

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

Question 3.
Test the continuity of the following functions at the points indicated against them.
(i) f(x) = \(\frac{\sqrt{x-1}-(x-1)^{\frac{1}{3}}}{x-2}\) for x ≠ 2
= \(\frac{1}{5}\) for x = 2, at x = 2
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q3(i)
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q3(i).1

(ii) f(x) = \(\frac{x^{3}-8}{\sqrt{x+2}-\sqrt{3 x-2}}\) for x ≠ 2
= -24 for x = 2, at x = 2
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q3(ii)
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q3(ii).1

(iii) f(x) = 4x + 1 for x ≤ \(\frac{8}{3}\)
= \(\frac{59-9 x}{3}\), for x > \(\frac{8}{3}\), at x = \(\frac{8}{3}\)
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q3(iii)
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q3(iii).1

Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1

(iv) f(x) = \(\frac{x^{3}-27}{x^{2}-9}\) for 0 ≤ x < 3
= \(\frac{9}{2}\), for 3 ≤ x ≤ 6, at x = 3
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q3(iv)

Question 4.
(i) If f(x) = \(\frac{24^{x}-8^{x}-3^{x}+1}{12^{x}-4^{x}-3^{x}+1}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, find k.
Solution:
Function f is continuous at x = 0
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q4(i)

(ii) If f(x) = \(\frac{5^{x}+5^{-x}-2}{x^{2}}\), for x ≠ 0
= k for x = 0
is continuous at x = 0, find k.
Solution:
Function f is continuous at x = 0
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q4(ii)

(iii) For what values of a and b is the function
f(x) = ax + 2b + 18 for x ≤ 0
= x2 + 3a – b for 0 < x ≤ 2 = 8x – 2 for x > 2,
continuous for every x?
Solution:
Function f is continuous for every x.
∴ Function f is continuous at x = 0 and x = 2
As f is continuous at x = 0.
∴ \(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 0^{-}}(a x+2 b+18)=\lim _{x \rightarrow 0^{+}}\left(x^{2}+3 a-b\right)\)
∴ a(0) + 2b + 18 = (0)2 + 3a – b
∴ 3a – 3b = 18
∴ a – b = 6 …..(i)
Also, Function f is continous at x = 2
∴ \(\lim _{x \rightarrow 2^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 2^{-}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 2^{-}}\left(x^{2}+3 a-b\right)=\lim _{x \rightarrow 2^{-}}(8 x-2)\)
∴ (2)2 + 3a – b = 8(2) – 2
∴ 4 + 3a – b = 14
∴ 3a – b = 10 …..(ii)
Subtracting (i) from (ii), we get
2a = 4
∴ a = 2
Substituting a = 2 in (i), we get
2 – b = 6
∴ b = -4
∴ a = 2 and b = -4

Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1

(iv) For what values of a and b is the function
f(x) = \(\frac{x^{2}-4}{x-2}\) for x < 2
= ax2 – bx + 3 for 2 ≤ x < 3
= 2x – a + b for x ≥ 3
continuous in its domain.
Solution:
Function f is continuous for every x on R.
∴ Function f is continuous at x = 2 and x = 3.
As f is continuous at x = 2.
∴ \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1 Q4(iv)
∴ 2 + 2 = a(2)2 – b(2) + 3
∴ 4 = 4a – 2b + 3
∴ 4a – 2b = 1 …..(i)
Also function f is continuous at x = 3
∴ \(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)\)
∴ \(\lim _{x \rightarrow 3^{-}}\left(a x^{2}-b x+3\right)=\lim _{x \rightarrow 3^{+}}(2 x-a+b)\)
∴ a(3)2 – b(3) + 3 = 2(3) – a + b
∴ 9a – 3b + 3 = 6 – a + b
∴ 10a – 4b = 3 …..(ii)
Multiplying (i) by 2, we get
8a – 4b = 2 …..(iii)
Subtracting (iii) from (ii), we get
2a = 1
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (i), we get
4(\(\frac{1}{2}\)) – 2b = 1
∴ 2 – 2b = 1
∴ 1 = 2b
∴ b = \(\frac{1}{2}\)
∴ a = \(\frac{1}{2}\) and b = \(\frac{1}{2}\)

Read More

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

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Sequences and Series Ex 4.3 Questions and Answers.

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