Category: Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.6

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{9^{x}-5^{x}}{4^{x}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{5^{x}+3^{x}-2^{x}-1}{x}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left(\frac{a^{x}+b^{x}+c^{x}-3}{\sin x}\right)\)
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left(\frac{6^{x}+5^{x}+4^{x}-3^{x+1}}{\sin x}\right)\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left(\frac{8^{\sin x}-2^{\tan x}}{e^{2 x}-1}\right)\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{3^{x}+3^{-x}-2}{x \cdot \tan x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{3+x}{3-x}\right]^{\frac{1}{x}}\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{5 x+3}{3-2 x}\right]^{\frac{2}{x}}\)
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{\log (3-x)-\log (3+x)}{x}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left[\frac{4 x+1}{1-4 x}\right]^{\frac{1}{x}}\)
Solution:

Question 6.
\(\lim _{x \rightarrow 0}\left[\frac{5+7 x}{5-3 x}\right]^{\frac{1}{3 x}}\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{a^{x}-b^{x}}{\sin (4 x)-\sin (2 x)}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{\left(2^{x}-1\right)^{3}}{\left(3^{x}-1\right) \cdot \sin x \cdot \log (1+x)}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{15^{x}-5^{x}-3^{x}+1}{x \cdot \sin x}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{(25)^{x}-2(5)^{x}+1}{x \cdot \sin x}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left[\frac{(49)^{x}-2(35)^{x}+(25)^{x}}{\sin x \cdot \log (1+2 x)}\right]\)
Solution:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Miscellaneous Exercise 7 Questions and Answers.

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

I.

Question 1.
If \(\lim _{x \rightarrow 2} \frac{x^{n}-2^{n}}{x-2}=80\) then find the value of n.
Solution:

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:

Question 2.
\(\lim _{x \rightarrow 0} \frac{(1+x)^{n}-1}{x}\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{(x-2)}{2 x^{2}-7 x+6}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 1}\left[\frac{x^{3}-1}{x^{2}+5 x-6}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 3}\left[\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right]\)
Solution:

Question 6.
\(\lim _{x \rightarrow 4}\left[\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right]\)
Solution:

Question 7.
\(\lim _{x \rightarrow 0}\left[\frac{5^{x}-1}{x}\right]\)
Solution:

Question 8.
\(\lim _{x \rightarrow 0}\left(1+\frac{x}{5}\right)^{\frac{1}{x}}\)
Solution:

Question 9.
\(\lim _{x \rightarrow 0}\left[\frac{\log (1+9 x)}{x}\right]\)
Solution:

Question 10.
\(\lim _{x \rightarrow 0} \frac{(1-x)^{5}-1}{(1-x)^{3}-1}\)
Solution:

Question 11.
\(\lim _{x \rightarrow 0}\left[\frac{a^{x}+b^{x}+c^{x}-3}{x}\right]\)
Solution:

Question 12.
\(\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{x^{2}}\)
Solution:

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:

Question 14.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^{x}+1}{x^{2}}\right]\)
Solution:

Question 15.
\(\lim _{x \rightarrow 0}\left[\frac{\left(5^{x}-1\right)^{2}}{x \cdot \log (1+x)}\right]\)
Solution:

Question 16.
\(\lim _{x \rightarrow 0}\left[\frac{a^{4 x}-1}{b^{2 x}-1}\right]\)
Solution:

Question 17.
\(\lim _{x \rightarrow 0}\left[\frac{\log 100+\log (0.01+x)}{x}\right]\)
Solution:

Question 18.
\(\lim _{x \rightarrow 0}\left[\frac{\log (4-x)-\log (4+x)}{x}\right]\)
Solution:

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:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Limits Ex 7.4

I. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{9^{x}-5^{x}}{4^{x}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{5^{x}+3^{x}-2^{x}-1}{x}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{\log (2+x)-\log (2-x)}{x}\right]\)
Solution:

II. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{3^{x}+3^{-x}-2}{x^{2}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{3+x}{3-x}\right]^{\frac{1}{x}}\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{\log (3-x)-\log (3+x)}{x}\right]\)
Solution:

III. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-b^{2 x}}{\log (1+4 x)}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{\left(2^{x}-1\right)^{2}}{\left(3^{x}-1\right) \cdot \log (1+x)}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{15^{x}-5^{x}-3^{x}+1}{x^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 2}\left[\frac{3^{\frac{x}{2}}-3}{3^{x}-9}\right]\)
Solution:

IV. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{(25)^{x}-2(5)^{x}+1}{x^{2}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{(49)^{x}-2(35)^{x}+(25)^{x}}{x^{2}}\right]\)
Solution:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Limits Ex 7.3

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^{2}}-\sqrt{6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow 0}\left[\frac{\sqrt{1-y^{2}}-\sqrt{1+y^{2}}}{y^{2}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow a}\left[\frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{2}-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 1}\left[\frac{x^{2}+x \sqrt{x}-2}{x-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt{1+x^{2}}-\sqrt{1+x}}{\sqrt{1+x^{3}}-\sqrt{1+x}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 4}\left[\frac{x^{2}+x-20}{\sqrt{3 x+4}-4}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-8}{\sqrt{x+2}-\sqrt{3 x-2}}\right]\)
Solution:

IV. Evaluate the following limits:

Question 1.
\(\lim _{y \rightarrow 2}\left[\frac{2-y}{\sqrt{3-y}-1}\right]\)
Solution:

Question 2.
\(\lim _{z \rightarrow 4}\left[\frac{3-\sqrt{5+z}}{1-\sqrt{5-z}}\right]\)
Solution:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Limits Ex 7.2

I. Evaluate the following limits:

Question 1.
\(\lim _{z \rightarrow 2}\left[\frac{z^{2}-5 z+6}{z^{2}-4}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow-3}\left[\frac{x+3}{x^{2}+4 x+3}\right]\)
Solution:

Question 3.
\(\lim _{y \rightarrow 0}\left[\frac{5 y^{3}+8 y^{2}}{3 y^{4}-16 y^{2}}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow-2}\left[\frac{-2 x-4}{x^{3}+2 x^{2}}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{u \rightarrow 1}\left[\frac{u^{4}-1}{u^{3}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{x^{3}-27}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-1}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow-2}\left[\frac{x^{7}+x^{5}+160}{x^{3}+8}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow \frac{1}{2}}\left[\frac{1-8 y^{3}}{y-4 y^{3}}\right]\)
Solution:

Question 3.
\(\lim _{v \rightarrow \sqrt{2}}\left[\frac{v^{2}+v \sqrt{2}-4}{v^{2}-3 v \sqrt{2}+4}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 3}\left[\frac{x^{2}+2 x-15}{x^{2}-5 x+6}\right]\)
Solution:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Limits Ex 7.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Limits Ex 7.1

I. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{x+6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{-3}-2^{-3}}{x-2}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 5}\left[\frac{x^{3}-125}{x^{5}-3125}\right]\)
Solution:

Question 4.
If \(\lim _{x \rightarrow 1}\left[\frac{x^{4}-1}{x-1}\right]=\lim _{x \rightarrow a}\left[\frac{x^{3}-a^{3}}{x-a}\right]\), find all possible values of a.
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 7}\left[\frac{(\sqrt[3]{x}-\sqrt[3]{7})(\sqrt[3]{x}+\sqrt[3]{7})}{x-7}\right]\)
Solution:

Question 2.
If \(\lim _{x \rightarrow 5}\left[\frac{x^{k}-5^{k}}{x-5}\right]=500\), find all possible values of k.
Solution:

Question 3.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{8}-1}{(1-x)^{2}-1}\right]\)
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt[3]{1+x}-\sqrt{1+x}}{x}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow 1}\left[\frac{2 y-2}{\sqrt[3]{7+y}-2}\right]\)
Solution:

Question 3.
\(\lim _{z \rightarrow a}\left[\frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}\right]\)
Solution:

Question 4.
\(\lim _{x \rightarrow 5}\left[\frac{x^{3}-125}{x^{2}-25}\right]\)
Solution:

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

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

Question 1.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)
Solution:

= 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:

= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)
= 1(32) + 3(36) + 12(-18)
= 32 + 108 – 216
= -76

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:

Question 4.
Without expanding the determinants, show that

Solution:

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
D_x = \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
8 & 2 & 3 \\
1 & 1 & -4
\end{array}\right|\)
= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)
= 1(-11) + 1(-35) + 1(6)
= -11 – 35 + 6
= -40
D_y = \(\left|\begin{array}{ccc}
2 & 1 & 1 \\
1 & 8 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-32 – 3) – 1(-4 – 9) + 1(1 – 24)
= 2(-35) – 1(-13) + 1(-23)
= -70 + 13 – 23
= -80
D_z = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 8 \\
3 & 1 & 1
\end{array}\right|\)
= 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.

(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
D_p = \(\left|\begin{array}{rrr}
-2 & 1 & 1 \\
3 & -2 & 1 \\
-1 & -1 & 3
\end{array}\right|\)
= -2(-6 + 1) – 1(9 + 1) + 1(-3 – 2)
= 10 – 10 – 5
= -5
D_q = \(\left|\begin{array}{ccc}
1 & -2 & 1 \\
1 & 3 & 1 \\
2 & -1 & 3
\end{array}\right|\)
= 1(9 + 1) + 2(3 – 2) + 1(-1 – 6)
= 10 + 2 – 7
= 5
D_r = \(\left|\begin{array}{rrr}
1 & 1 & -2 \\
1 & -2 & 3 \\
2 & -1 & -1
\end{array}\right|\)
= 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
D_x = \(\left|\begin{array}{ccc}
7 & -1 & 2 \\
5 & 4 & -5 \\
12 & -1 & 3
\end{array}\right|\)
= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)
= 7(7)+ 1(75) +2(-53)
= 49 + 75 – 106
= 18
D_y = \(\left|\begin{array}{ccc}
1 & 7 & 2 \\
3 & 5 & -5 \\
2 & 12 & 3
\end{array}\right|\)
= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= -6
D_z = \(\left|\begin{array}{ccc}
1 & -1 & 7 \\
3 & 4 & 5 \\
2 & -1 & 12
\end{array}\right|\)
= 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.

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

(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

(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

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

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

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
D_x = \(\left|\begin{array}{ccc}
5 & 2 & -1 \\
1 & -1 & 1 \\
8 & 3 & 0
\end{array}\right|\)
= 5(0 – 3) – 2(0 – 8) + (-1)(3 + 8)
= -15 + 16 – 11
= -10
D_y = \(\left|\begin{array}{ccc}
1 & 5 & -1 \\
2 & 1 & 1 \\
3 & 8 & 0
\end{array}\right|\)
= 1(0 – 8) – 5(0 – 3) + 1(16 – 3)
= -8 + 15 – 13
= -6
D_z = \(\left|\begin{array}{ccc}
1 & 2 & 5 \\
2 & -1 & 1 \\
3 & 3 & 8
\end{array}\right|\)
= 1(-8 – 3) – 2(16 – 3) + 5(6 + 3)
= -11 – 26 + 45
= 8
By Cramer’s Rule,

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.

(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
D_x = \(\left|\begin{array}{ccc}
10 & -1 & 6 \\
11 & 4 & -5 \\
12 & -7 & -9
\end{array}\right|\)
= 10(-36 – 35) – (-1)(-99 + 60) + 6(-77 – 48)
= -710 – 39 – 750
= -1499
D_y = \(\left|\begin{array}{ccc}
2 & 10 & 6 \\
3 & 11 & -5 \\
8 & 12 & -9
\end{array}\right|\)
= 2(-99 + 60) – 10(-27 + 40) + 6(36 – 88)
= -78 – 130 – 312
= -520
D_z = \(\left|\begin{array}{ccc}
2 & -1 & 10 \\
3 & 4 & 11 \\
8 & -7 & 12
\end{array}\right|\)
= 2(48 + 77) – (-1)(36 – 88) + 10(-21 – 32)
= 250 – 52 – 530
= -332
By Cramer’s Rule,

∴ 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
D_x = \(\left|\begin{array}{ccc}
31 & -1 & -1 \\
-26 & -6 & 2 \\
-24 & 2 & -7
\end{array}\right|\)
= 31(42 – 4) – (-1)(182 + 48) + (-1)(-52 – 144)
= 1178 + 230 + 196
= 1604
D_y = \(\left|\begin{array}{ccc}
11 & 31 & -1 \\
1 & -26 & 2 \\
1 & -24 & -7
\end{array}\right|\)
= 11(182 + 48) – 31(-7 – 2) + (-1)(-24 + 26)
= 2530 + 279 – 2
= 2807
D_z = \(\left|\begin{array}{ccc}
11 & -1 & 31 \\
1 & -6 & -26 \\
1 & 2 & -24
\end{array}\right|\)
= 11(144 + 52) – (-1)(-24 + 26) + 31(2 + 6)
= 2156 + 2 + 248
= 2406
By Cramer’s Rule,

∴ x = 4, y = 7 and z = 6 are the solutions of the given equations.

(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
D_p = \(\left|\begin{array}{rrr}
-2 & 1 & 1 \\
3 & -2 & 1 \\
-1 & -1 & 3
\end{array}\right|\)
= -2(-6 + 1) – 1(9 + 1) + 1(-3 – 2)
= 10 – 10 – 5
= -5
D_q = \(\left|\begin{array}{ccc}
1 & -2 & 1 \\
1 & 3 & 1 \\
2 & -1 & 3
\end{array}\right|\)
= 1(9 + 1) + 2(3 – 2) + 1(-1 – 6)
= 10 + 2 – 7
= 5
D_r = \(\left|\begin{array}{rrr}
1 & 1 & -2 \\
1 & -2 & 3 \\
2 & -1 & -1
\end{array}\right|\)
= 1(2 + 3) – 1(-1 – 6) – 2(-1 + 4)
= 5 + 7 – 6
= 6
By Cramer’s Rule,

∴ 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
D_p = \(\left|\begin{array}{ccc}
4 & -1 & 3 \\
2 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 4(1 – 1) – (-1)(-2 – 2) + 3(2 + 2)
= 0 – 4 + 12
= 8
D_q = \(\left|\begin{array}{ccc}
2 & 4 & 3 \\
1 & 2 & 1 \\
3 & 2 & -1
\end{array}\right|\)
= 2(-2 – 2) – 4(-1 – 3) + 3(2 – 6)
= -8 + 16 – 12
= -4
D_r = \(\left|\begin{array}{ccc}
2 & -1 & 4 \\
1 & -1 & 2 \\
3 & 1 & 2
\end{array}\right|\)
= 2(-2 – 2) – (-1)(2 – 6) + 4(1 + 3)
= -8 – 4 + 16
= 4
By Cramer’s Rule,

∴ x = 1, y = -2 and z = 2 are the solutions of the given equations.

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



∴ 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.

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

(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,

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)

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.

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
D_x = \(\left|\begin{array}{ccc}
15 & 1 & 1 \\
5 & -1 & 1 \\
4 & 1 & -1
\end{array}\right|\)
= 15(1 – 1) – 1(-5 – 4) + 1(5 + 4)
= 15(0) – 1(-9) + 1(9)
= 0 + 9 + 9
= 18
D_y = \(\left|\begin{array}{ccc}
1 & 15 & 1 \\
1 & 5 & 1 \\
2 & 4 & -1
\end{array}\right|\)
= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10)
= 1(-9) – 15(-3) + 1(-6)
= -9 + 45 – 6
= 30
D_z = \(\left|\begin{array}{ccc}
1 & 1 & 15 \\
1 & -1 & 5 \\
2 & 1 & 4
\end{array}\right|\)
= 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.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Determinants Ex 6.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Determinants Ex 6.2

Question 1.
Without expanding, evaluate the following determinants.
(i) \(\left|\begin{array}{lll}
1 & a & b+c \\
1 & b & c+a \\
1 & c & a+b
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 x & 9 x & 12 x
\end{array}\right|\)
Solution:

(iii) \(\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
Solution:

Question 2.
Using properties of determinants, show that \(\left|\begin{array}{ccc}
a+b & a & b \\
a & a+c & c \\
b & c & b+c
\end{array}\right|\) = 4abc
Solution:

Question 3.
Solve the following equation.
\(\left|\begin{array}{ccc}
x+2 & x+6 & x-1 \\
x+6 & x-1 & x+2 \\
x-1 & x+2 & x+6
\end{array}\right|=0\)
Solution:
\(\left|\begin{array}{ccc}
x+2 & x+6 & x-1 \\
x+6 & x-1 & x+2 \\
x-1 & x+2 & x+6
\end{array}\right|=0\)
Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₁, we get
\(\left|\begin{array}{ccc}
x+2 & x+6 & x-1 \\
4 & -7 & 3 \\
-3 & -4 & 7
\end{array}\right|=0\)
∴ (x + 2)(-49 + 12) – (x + 6)(28 + 9) + (x – 1)(-16 – 21) = 0
∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0
∴ -37(x + 2 + x + 6 + x – 1) = 0
∴ 3x + 7 = 0
∴ x = \(\frac{-7}{3}\)

Question 4.
If \(\left|\begin{array}{lll}
4+x & 4-x & 4-x \\
4-x & 4+x & 4-x \\
4-x & 4-x & 4+x
\end{array}\right|=0\), then find the values of x.
Solution:

∴ (12 – x)[1(4x² – 0) – (4 – x)(0 – 0) + (4 – x)(0 – 0)] = 0
∴ (12 – x)(4x²) = 0
∴ x²(12 – x) = 0
∴ x = 0 or 12 – x = 0
∴ x = 0 or x = 12

Question 5.
Without expanding determinants, show that
\(\left|\begin{array}{ccc}
1 & 3 & 6 \\
6 & 1 & 4 \\
3 & 7 & 12
\end{array}\right|+4\left|\begin{array}{lll}
2 & 3 & 3 \\
2 & 1 & 2 \\
1 & 7 & 6
\end{array}\right|=10\left|\begin{array}{lll}
1 & 2 & 1 \\
3 & 1 & 7 \\
3 & 2 & 6
\end{array}\right|\)
Solution:

Question 6.
Without expanding determinants, find the value of
(i) \(\left|\begin{array}{lll}
10 & 57 & 107 \\
12 & 64 & 124 \\
15 & 78 & 153
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{lll}
2014 & 2017 & 1 \\
2020 & 2023 & 1 \\
2023 & 2026 & 1
\end{array}\right|\)
Solution:

Question 7.
Without expanding determinants, prove that
(i) \(\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|=\left|\begin{array}{lll}
b_{1} & c_{1} & a_{1} \\
b_{2} & c_{2} & a_{2} \\
b_{3} & c_{3} & a_{3}
\end{array}\right|=\left|\begin{array}{lll}
c_{1} & a_{1} & b_{1} \\
c_{2} & a_{2} & b_{2} \\
c_{3} & a_{3} & b_{3}
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{lll}
1 & y z & y+z \\
1 & z x & z+x \\
1 & x y & x+y
\end{array}\right|=\left|\begin{array}{lll}
1 & x & x^{2} \\
1 & y & y^{2} \\
1 & z & z^{2}
\end{array}\right|\)
Solution:

In 1st determinant, taking (x + y + z) common from C₃ and in 2nd determinant, taking \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) common from R₁, R₂, R₃ respectively, we get

Balbharti Maharashtra State Board Class 11 Secretarial Practice Solutions Chapter 9 Business Communication Skills of a Secretary Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Secretarial Practice Solutions Chapter 9 Business Communication Skills of a Secretary

1A. Select the correct answer from the options given below and rewrite the statements.

Question 1.
Business communication is concerned with ____________ activities.
(a) economic
(b) business
(c) social
Answer:
(b) business

Question 2.
Written communication is a ____________ record.
(a) permanent
(b) temporary
(c) unauthorised
Answer:
(a) permanent

Question 3.
E-mail is ____________ mode of communication.
(a) fastest
(b) slowest
(c) costliest
Answer:
(a) fastest

Question 4.
A unique internet address of website is known as ____________
(a) World Wide Web
(b) Uniform Resource Locater
(c) .com
Answer:
(b) Uniform Resource Locater

Question 5.
____________ is an organised statement of facts.
(a) Report
(b) Notice
(c) Heading
Answer:
(a) Report

Question 6.
There should be proper ____________ between words, lines and between paragraphs.
(a) margin
(b) typing
(c) spacing
Answer:
(c) spacing

Question 7.
____________ refers to use of minimum words.
(a) Courtesy
(b) Conciseness
(c) Correctness
Answer:
(b) Conciseness

Question 8.
A letter without ____________ is invalid.
(a) ‘You’ attitude
(b) signature
(c) clarity
Answer:
(b) signature

1B. Match the pairs.

Question 1.

Group ‘A’	Group ‘B’
(a) Twitter	(1) Hearing and understanding
(b) Consideration	(2) Personable
(c) Active listening	(3) Harsh, rude words
(d) Body language	(4) Social Media
(e) Courtesy	(5) Non-verbal communication
	(6) Blog
	(7) ‘You’ attitude
	(8) Empathy
	(9) SMS
	(10) Politeness

Answer:

Group ‘A’	Group ‘B’
(a) Twitter	(4) Social Media
(b) Consideration	(7) ‘You’ attitude
(c) Active listening	(1) Hearing and understanding
(d) Body language	(5) Non-verbal communication
(e) Courtesy	(10) Politeness

1C. Write a word or a term or a phrase that can substitute each of the following statements.

Question 1.
Process of communication, conveying a message in spoken form.
Answer:
Verbal communication

Question 2.
A set of interconnected web pages located on a single web domain.
Answer:
Website

Question 3.
Part of a business letter that introduces the sender to the receiver.
Answer:
Heading

Question 4.
A written summary of the business transacted at the meeting.
Answer:
Minutes

Question 5.
Part of a letter that contains the name and address of the sender.
Answer:
Heading or Letterhead

Question 6.
Audio-Visual means of electronic communication.
Answer:
Video conference

1D. State whether the following statements are True or False.

Question 1.
Notice is a written summary of business transacted at a meeting.
Answer:
False

Question 2.
Written communication provides permanent records.
Answer:
True

Question 3.
Active listening is essential for effective communication.
Answer:
True

Question 4.
The inside address gives the name and address of the sender.
Answer:
False

Question 5.
A letter without a date is incomplete and invalid.
Answer:
True

Question 6.
The reference number shows the purpose of the letter.
Answer:
False

Question 7.
Coherence refers to the logical arrangement of the contents of a letter.
Answer:
True

Question 8.
A letter should have minimum folds.
Answer:
True

1E. Find the odd one.

Question 1.
Paper, Margin, Typing, Courtesy
Answer:
Courtesy

Question 2.
Clarity, Courtesy, Spacing, Correctness
Answer:
Spacing

Question 3.
Date, Inside Address, Conciseness, Subject
Answer:
Conciseness

1F. Complete the sentences.

Question 1.
When communication is done through Reports, Letters, Circulars, etc it is called as ____________
Answer:
written communication

Question 2.
Proper arrangement of different parts of business letter is called as ____________
Answer:
layout

Question 3.
The part of the letter which contains the name and address of the receiver of the letter is called as ____________
Answer:
inside address

1G. Select the correct option from the bracket.

Question 1.

Group ‘A’	Group ‘B’
(1) You Attitude	…………………………
(2) Conciseness	………………………..
(3) ………………….	Complete information
(4) …………………	Polite language

(Minimum words, Completeness, Courtesy, Consideration)
Answer:

Group ‘A’	Group ‘B’
(1) You Attitude	Consideration
(2) Conciseness	Minimum words
(3) Completeness	Complete Information
(4) Courtesy	Polite language

1H. Answer in one sentence.

Question 1.
Name the type of communication in which words are not used.
Answer:
Non-verbal communication is the type of communication in which words are not used.

Question 2.
Name the type of communication in which communication is done in spoken form.
Answer:
Verbal communication is the type of communication in which communication is done in spoken form.

Question 3.
Name the type of communication which can be re-read.
Answer:
Written communication is the type of communication that can be re-read.

1I. Correct the underlined word and rewrite the following sentences.

Question 1.
Consideration means the letter should be in a logical sequence.
Answer:
Coherence means the letter should be in a logical sequence.

Question 2.
Completeness means the use of minimum words.
Answer:
Conciseness means the use of minimum words.

Question 3.
Complimentary close contains greetings to the reader of the letter.
Answer:
Salutation contains greetings to the reader of the letter.

1J. Arrange in proper order.

Question 1.
(a) Heading
(b) Complimentary close
(c) Subject
Answer:
(a) Heading
(b) Subject
(c) Complimentary close

Question 2.
(a) Enclosure
(b) Body of letter
(c) Date
Answer:
(a) Date
(b) Body of letter
(c) Enclosure

2. Explain the following terms/concepts.

Question 1.
Communication
Answer:

• Communication is derived from the Latin term ‘communis’ that means ‘common’- ‘Shared by all’.
• Communication is an exchange of facts, ideas, opinions, or emotions by two or more persons.
• Communication is a two-way process where the thoughts feelings and opinion is transmitted.
• Effective communication is when a message is conveyed by the sender and received by the receiver exactly the same it was intended.
• It is giving or exchanging information, signals, or messages by talk, gestures, or writing.

Question 2.
Business Communication
Answer:

• Business communication is the branch of general communication especially concerned with business activities.
• It is a process through which information, facts, ideas, orders, decisions, etc. are exchanged between the person associated with the business.
• The success of the business enterprise depends largely upon good communication.
• Ineffective communication may cause loss of money and even goodwill of a business.
• Thus, business communication relating to trade, law, management, finance, etc. of a business enterprise is termed as business communication.

Question 3.
Written Communication
Answer:

• The exchange of information or ideas in a written form is known as written communication.
• Written communication includes reports, letters, circulars, etc.
• Written communication is the most important and the most effective mode of business communication.
• The words written should convey specific meaning and should not confuse the reader.
• Letters, memos, notices, circulars, reports, minutes are some common types of written communication.

Question 4.
Business Correspondence
Answer:

• Communication through the exchange of letters is known as correspondence.
• A businessman who writes letters in his day-to-day transactions is called business correspondence.
• Business correspondence is a written communication between two parties.
• Business correspondence takes place because the place of production and place of consumption is not the same.

Question 5.
Report
Answer:

• A report is an organized statement of facts or opinions leading to some conclusion with or without some recommendations.
• It is a systematic presentation of facts on a specific topic.
• Some reports are made as per the Companies Act and some are prepared as per the requirement of the company.
• A report may be prepared by an individual or by a committee.

Question 6.
Minutes
Answer:

• It is a written summary of the business transacted at the meeting.
• It is a concise and accurate official record of the discussion and decision at company meetings.
• It can be used for future reference.
• Minutes is the official record of the meeting so it is necessary to draft minutes in a proper format.
• Minutes should be prepared by the secretary within 15 days of a meeting.
• It is always written in the past tense.
• Minutes are prepared by the secretary, confirmed by a member, signed by a chairman, and countersigned by the secretary.

3. Study the following case/situation and express your opinion.

1. Mr. Rahul is the secretary who has been asked by the Managing Director to inform a director about a decision taken in a board meeting in which he was absent. Which aspect of essentials of a good business letter he follows:
(Clarity, conciseness, coherence, courtesy, completeness, correctness)

Question (a).
When he is giving the required information in a very short and brief manner?
Answer:
Clarity and conciseness

Question (b).
When he is using courteous words so as to be polite?
Answer:
Courtesy

Question (c).
When he is giving the entire information about the meeting in a proper manner?
Answer:
Coherence, completeness, and correctness.

4. Answer in brief.

Question 1.
Explain any four essentials of effective communication.
Answer:

• Communication is an exchange of facts, ideas, opinions, or emotions by two or more persons.
• Effective communication is when a message is conveyed by the sender and received by the receiver exactly the same it was intended.
• Being able to communicate effectively is an essential skill.

The following are the essentials skills for effective communication:
(a) Listening:

• One of the most important aspects of business communication is being a good listener.
• Effective communication requires active listening.
• Active listening involves hearing and understanding a person.

(b) Body language:

• Body language is an important communication tool.
• Body language should convey words.
• Tone, hand gestures, and ensuring eye contact are involved in body language.

(c) Clear and concise:

• The message should be conveyed by using as few words as possible, whether in person or through telephone, or e-mail.
• The message should be clear concise and direct.
• Excessive words should be avoided.
• Thought should be given to the message before being conveyed in order to avoid confusion.

(d) Confident:

• For effective communication, confidence is needed.
• Making eye contact but having a friendly tone always shows confidence.

Question 2.
State any four essentials of a good business letter.
Answer:

• A business letter is a type of written communication written by the secretary.
• Good letter writing is important for maintaining the image of the business.
• It helps in understanding the objective of the content and helps to make the correct decision.

A business letter should possess the following qualities:
(a) Clarity:

• The message of the letter must be clear.
• Simple and common words are to be used.
• Technical and short forms should be avoided.
• Names and figures should be correct and clear.

(b) Conciseness:

• The letter should be brief.
• Minimum words are to be used.
• Unnecessary and irrelevant information should be avoided.
• A brief letter saves time for the reader.

(c) Completeness:

• A letter must give complete information to the reader.
• The letter should cover all possible facts related to the subject matter.
• An incomplete letter does not achieve the desired results.

(d) Courtesy:

• Courtesy means the language of the letter must be polite and kind.
• A courteous letter gets a favorable response from the reader.
• Harsh, rude words, insulting remarks must be avoided.

5. Justify the following statements.

Question 1.
Written communication is very useful to the organization.
Answer:

• The exchange of information or ideas in a written form is known as written communication.
• Written communication is the most important and effective mode of business communication.
• It provides us with records, references, etc. on which important decisions are taken.
• It provides legal defense to the organization through records, letters, instructions, etc.
• It provides uniformity of policy and procedures and builds proper guidelines for the working of the organization.
• It builds an image of the company.
• It leads to accuracy and dependability.
• Responsibility can be easily assigned through written communication.
• It is permanent in nature.
• Thus, written communication is very useful for the organization.

Question 2.
Social media network is very useful to the business.
Answer:

• Social media are online interactive groups created using advanced mobile and web-based technologies.
• From the business point of view, it provides a great opportunity to interact with the public and communicate about their product and services.
• It helps in developing loyalties.
• It builds a strong relationships with the audience and consumers.
• Social networking makes relationships more personal.
• Business can be promoted more effectively through advertising.
• Thus, social media network is very useful to the business.

Question 3.
Listening is the most important aspect of effective communication.
Answer:

• The most important aspect of effective communication is being a good listener.
• Effective communication requires active listening.
• Active listening involves hearing and understanding what a person is saying to you.
• Without the ability to listen effectively, messages are generally misunderstood.
• Good listening skills can lead to better customer satisfaction.
• It can increase productivity with fewer mistakes.
• Increased sharing of information will lead to more creative and innovative work.
• Thus listening is the most important aspect of communication.

6. Answer the following questions.

Question 1.
State the merits of written communication.
Answer:

• When the exchange of information or ideas is in a written form is known as written communication.
• Written communication includes reports, letters, circulars, etc.
• Written communication is the most important and the most effective of any mode of business communication.
• The words written should contain specific meaning and should not confuse the reader.
• Letter, memos, notices, circulars, minutes are some common types of written communication.

The following are the merits of written communication:
(a) Accurate and precise:

• Written communication is drafted with great care.
• The communicator has to be accurate and factual as it is open to verification.
• Therefore written communication focuses greater on accuracy and precision.

(b) Re-read many times:

• The receiver of written communication can read the message any time again in the future.
• He can re-read it till it is properly understood by him.

(c) Permanent record:
Written communication becomes a permanent record of the organization and can prove very useful for future reference.

(d) Documentary evidence:
Written communication is acceptable as legal documents and as legal evidence also.

(e) Wide access:
Written communication is the best channel of communication for conveying information to people living in different places.

(f) No need for personal contact:

• It is not necessary for both parties to be available at the time of communication.
• Messages can be sent to the concerned person who can read when receives and gets spare time.

(g) Completeness:

• Written messages are prepared with perfect knowledge related to the matter.
• So there is completeness in the message.

(h) Economical:

• This method is economical when the receiver is far away from the business place.
• E-mails are the most popular method of written communication.

Question 2.
Explain different parts of a business letter.
Answer:

• A business letter is a type of written communication written by a secretary.
• Good letter writing is important for maintaining the image of the business.
• The business letters are written with the objective of understanding and take the correct decisions.

The following are the different parts of the business letter:
(a) Heading:

• The heading contains the name, address, telephone number, email id, website CIN of the company.
• It is that part of the business letter which introduces the sender to the receiver.

Eg. TATA MOTORS LTD. 176, S.B. Road. MIDC, Pune – 411015 CIN – L28920MH1945PLC004520 Telephone – (022)4756823 Email – tatamotors@gmail.com Website – www.tatamotor.com

(b) Date:

• The date is written on the right-hand side of the letter just below the heading.
• The date includes the date, month, and year.
• A letter without a date is incomplete.
• The date is very important as the letter acts as legal evidence.
• Eg. British style – 1st April 2019
  American style – April 1st, 2019

(c) Reference number:

• It is written on the left-hand side below the heading.
• The reference number is given to have a quick reference to the matter concerned.
• The outgoing letter is given a reference number.

(d) Inside address:

• It contains the name, address of the receiver of the letter.
• It is written on the left-hand side of the letter.
• For names, Mr, Shri, Mrs, or Smt are used and for firms, Messrs is used.

(e) Subject:

• The reader gets the idea of the matter of the letter without reading the letter completely.
• It helps to send it to the concerned section.
• It is written in brief as the subject.

(f) Salutation:

• Salutation is a greeting from the writer.
• It creates a favorable impression on the reader’s mind.
• It appears on the left margin below the inside address.

(g) Body of the letter:

• It is the most important part of the business letter.
• It contains the actual message for the receiver of the letter.
• The message should be divided into paragraphs. The first paragraph, Main paragraph, Closing paragraph.

(h) Complimentary close:

• This is the concluding part of the letter.
• It is written below the body of the letter on the right-hand side.
• It shows the polite end of the letter.
• It should match salutation.

(i) Signature:

• It is the final part of the letter.
• A letter without a signature is incomplete and invalid.
• Below the signature, the name and his or her designation are written.
• The person who signs is responsible for the matter written in the letter.

(j) Enclosure:

• It includes documents, cheques, etc. which are attached with the letter.
• It is shown by word enclosure which is written on the left-hand side.

Activity (Textbook Page No. 144)

How can a company use Twitter and Blogs to reach out to outsiders?
Answer:

• Twitter is considered to be an effective tool to grow your business and its brand name.
• Twitter uses attractive hashtags that attract the attention of the targeted audience or customers.
• It is an excellent platform to approach and get connected to new audiences gaining their positive opinions.
• The cost of Twitter is very nominal, so small business organizations can use its benefits easily.
• Twitter helps to connect a huge number of new customers as well as keep connected to old customers.
• It also helps to build the brand name and recognition with the help of advertisements.
• It helps to create a customer support channel and recognition who tweets positively as well as negatively about your product. Thus company or firm can improve its brand name.
• Similarly, blogging also helps businesses to compete with competitors in the market.
• Blogging helps to post in detail the qualities of your product which reaches a number of customers.
• Needed customers can easily approach your business or profession by viewing your blog.