Category: Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.3 Questions and Answers.

Maharashtra State Board 11th 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.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^{2}-9}\right]\)
Solution:

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

Question 4.
\(\lim _{x \rightarrow 2}\left[\frac{\sqrt{2+x}-\sqrt{6-x}}{\sqrt{x}-\sqrt{2}}\right]\)
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:

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

Question 4.
\(\lim _{y \rightarrow 0}\left[\frac{\sqrt{a+y}-\sqrt{a}}{y \sqrt{a+y}}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{x^{2}+9}-\sqrt{2 x^{2}+9}}{\sqrt{3 x^{2}+4}-\sqrt{2 x^{2}+4}}\right)\)
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.
\(\lim _{z \rightarrow 4}\left[\frac{3-\sqrt{5+z}}{1-\sqrt{5-z}}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x}\right)\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.2 Questions and Answers.

Maharashtra State Board 11th 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:

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

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

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

Question 6.
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-7 x+6}{x^{3}-7 x^{2}+16 x-12}\right]\)
Solution:
\(\lim _{x \rightarrow 2}\left[\frac{x^{3}-7 x+6}{x^{3}-7 x^{2}+16 x-12}\right]\)
As x → 2, numerator and denominator both tend to zero
∴ x – 2 is a factor of both.
To find the other factor for both of them, by synthetic division
Consider, Numerator = x³ + 0x² – 7x + 6


∴ The limit does not exist

III. Evaluate the following limits:

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

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

Question 3.
\(\lim _{x \rightarrow 1}\left[\frac{x^{4}-3 x^{2}+2}{x^{3}-5 x^{2}+3 x+1}\right]\)
Solution:
\(\lim _{x \rightarrow 1}\left[\frac{x^{4}-3 x^{2}+2}{x^{3}-5 x^{2}+3 x+1}\right]\)
As x → 1, numerator and denominator both tend to zero
∴ x – 1 is a factor of both.
To find the factor of numerator and denominator by synthetic division
Consider, numerator = x⁴ + 0x³ – 3x² + 0x + 2

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

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

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.1

Question 1.
Find the equation of a circle with
(i) centre at origin and radius 4.
(ii) centre at (-3, -2) and radius 6.
(iii) centre at (2, -3) and radius 5.
(iv) centre at (-3, -3) passing through point (-3, -6).
Solution:
(i) The equation of a circle with centre at origin and radius ‘r’ is given by
x² + y² = r²
Here, r = 4
∴ The required equation of the circle is x² + y² = 4² i.e., x² + y² = 16.

(ii) The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = -3, k = -2 and r = 6
∴ The required equation of the circle is
[x – (-3)]² + [y – (-2)]² = 6²
⇒ (x + 3)² + (y + 2)² = 36
⇒ x² + 6x + 9 + y² + 4y + 4 – 36 = 0
⇒ x² + y² + 6x + 4y – 23 = 0

(iii) The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = 2, k = -3 and r = 5
The required equation of the circle is
(x – 2)² + [y – (-3)]² = 5²
⇒ (x – 2)² + (y + 3)² = 25
⇒ x² – 4x + 4 + y² + 6y + 9 – 25 = 0
⇒ x² + y² – 4x + 6y – 12 = 0

(iv) Centre of the circle is C (-3, -3) and it passes through the point P (-3, -6).

The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = -3, k = -3, r = 3
The required equation of the circle is
[x – (-3)]² + [y – (-3)]² = 3²
⇒ (x + 3)² + (y + 3)² = 9
⇒ x² + 6x + 9 + y² + 6y + 9 – 9 = 0
⇒ x² + y² + 6x + 6y + 9 = 0

Check:
If the point (-3, -6) satisfies x² + y² + 6x + 6y + 9 = 0, then our answer is correct.
L.H.S. = x² + y² + 6x + 6y + 9
= (-3)² + (-6)² + 6(-3) – 6(-6) + 9
= 9 + 36 – 18 – 36 + 9
= 0
= R.H.S.
Thus, our answer is correct.

Question 2.
Find the centre and radius of the following circles:
(i) x² + y² = 25
(ii) (x – 5)² + (y – 3)² = 20
(iii) \(\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{3}\right)^{2}=\frac{1}{36}\)
Solution:
(i) Given equation of the circle is
x² + y² = 25
⇒ x² + y² = (5)²
Comparing this equation with x² + y² = r², we get r = 5
Centre of the circle is (0, 0) and radius of the circle is 5.

(ii) Given equation of the circle is
(x – 5)² + (y – 3)² = 20
⇒ (x – 5)² + (y – 3)² = (√20)²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 5, k = 3 and r = √20 = 2√5
Centre of the circle = (h, k) = (5, 3)
and radius of the circle = 2√5.

(iii) Given the equation of the circle is

Comparing this equation with (x – h)² + (y – k)² = r², we get
h = \(\frac{1}{2}\), k = \(\frac{-1}{3}\) and r = \(\frac{1}{6}\)
Centre of the circle = (h, k) = (\(\frac{1}{2}\), \(\frac{-1}{3}\)) and radius of the circle = \(\frac{1}{6}\)

Question 3.
Find the equation of the circle with centre
(i) at (a, b) and touching the Y-axis.
(ii) at (-2, 3) and touching the X-axis.
(iii) on the X-axis and passing through the origin having radius 4.
(iv) at (3, 1) and touching the line 8x – 15y + 25 = 0.
Solution:
(i) Since the circle is touching the Y-axis, the radius of the circle is X-co-ordinate of the centre.

∴ r = a
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = a, k = b
The required equation of the circle is
⇒ (x – a)² + (y – b)² = a²
⇒ x² – 2ax + a² + y² – 2by + b² = a²
⇒ x² + y² – 2ax – 2by + b² = 0

(ii) Since the circle is touching the X-axis, the radius of the circle is the Y co-ordinate of the centre.

∴ r = 3
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = -2, k = 3
The required equation of the circle is
⇒ (x + 2)² + (y – 3)² = 3²
⇒ x² + 4x + 4 + y² – 6y + 9 = 9
⇒ x² + y² + 4x – 6y + 4 = 0

(iii) Let the co-ordinates of the centre of the required circle be C (h, 0).
Since the circle passes through the origin i.e., O(0, 0)
OC = radius
⇒ \(\sqrt{(h-0)^{2}+(0-0)^{2}}=4\)
⇒ h² = 16
⇒ h = ±4

the co-ordinates of the centre are (4, 0) or (-4, 0).
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = ± 4, k = 0, r = 4
The required equation of the circle is
⇒ (x – 4)² + (y – 0)² = 4² or (x + 4)² + (y – 0)² = 4²
⇒ x² – 8x + 16 + y² = 16 or x² + 8x + 16 + y² = 16
⇒ x² + y² – 8x = 0 or x² + y² + 8x = 0

(iv) Centre of the circle is C (3, 1).
Let the circle touch the line 8x – 15y + 25 = 0 at point M.

CM = radius (r)
CM = Length of perpendicular from centre C(3, 1) on the line 8x – 15y + 25 = 0

The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = 3, k = 1 and r = 2
The required equation of the circle is
⇒ (x – 3)² + (y – 1)² = 2²
⇒ x² – 6x + 9 + y² – 2y + 1 = 4
⇒ x² + y² – 6x – 2y + 10 – 4 = 0
⇒ x² + y² – 6x – 2y + 6 = 0

Question 4.
Find the equation of the circle, if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4 and radius is 9.
Solution:

Given equations of diameters are 2x + y = 6 and 3x + 2y = 4.
Let C (h, k) be the centre of the required circle.
Since point of intersection of diameters is the centre of the circle,
x = h, y = k
Equations of diameters become
2h + k = 6 …..(i)
and 3h + 2k = 4 ……..(ii)
By (ii) – 2 × (i), we get
-h = -8
⇒ h = 8
Substituting h = 8 in (i), we get
2(8) + k = 6
⇒ k = 6 – 16
⇒ k = -10
Centre of the circle is C (8, -10) and radius, r = 9
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = 8, k = -10
The required equation of the circle is
⇒ (x – 8)² + (y + 10)² = 9²
⇒ x² – 16x + 64 + y² + 20y + 100 = 81
⇒ x² + y² – 16x + 20y + 100 + 64 – 81 = 0
⇒ x² + y² – 16x + 20y + 83 = 0

Question 5.
If y = 2x is a chord of the circle x² + y² – 10x = 0, find the equation of the circle with this chord as diameter.
Solution:

y = 2x is the chord of the given circle.
It satisfies the equation of a given circle.
Substituting y = 2x in x² + y² – 10x = 0, we get
⇒ x² + (2x)² – 10x = 0
⇒ x² + 4x² – 10x = 0
⇒ 5x² – 10x = 0
⇒ 5x(x – 2) = 0
⇒ x = 0 or x = 2
When x = 0, y = 2x = 2(0) = 0
∴ A = (0, 0)
When x = 2, y = 2x = 2 (2) = 4
∴ B = (2, 4)
End points of chord AB are A(0, 0) and B(2, 4).
Chord AB is the diameter of the required circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as end points of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 0, y₁ = 0, x₂ = 2, y₂ = 4
The required equation of the circle is
⇒ (x – 0) (x – 2) + (y – 0) (y – 4 ) = 0
⇒ x² – 2x + y² – 4y = 0
⇒ x² + y² – 2x – 4y = 0

Question 6.
Find the equation of a circle with a radius of 4 units and touch both the co-ordinate axes having centre in the third quadrant. Solution:
The radius of the circle = 4 units
Since the circle touches both the co-ordinate axes and its centre is in the third quadrant,
the centre of the circle is C(-4, -4).

The equation of a circle with centre at (h, k) and radius r is given by (x – h)² + (y – k)² = r²
Here, h = -4, k = -4, r = 4
the required equation of the circle is
⇒ [x – (-4)]² + [y – (-4)]² = 4²
⇒ (x + 4)² + (y + 4)² = 16
⇒ x² + 8x + 16 + y² + 8y + 16 – 16 = 0
⇒ x² + y² + 8x + 8y + 16 = 0

Question 7.
Find the equation of the circle passing through the origin and having intercepts 4 and -5 on the co-ordinate axes.
Solution:
Let the circle intersect X-axis at point A and intersect Y-axis at point B.
the co-ordinates of point A are (4, 0) and the co-ordinates of point B are (0, -5).

Since ∠AOB is a right angle,
AB represents the diameter of the circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as end points of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 4, y₁ = 0, x₂ = 0, y₂ = -5
The required equation of the circle is
⇒ (x – 4) (x – 0) + (y – 0) [y – (-5)] = 0
⇒ x(x – 4) + y(y + 5) = 0
⇒ x² – 4x + y² + 5y = 0
⇒ x² + y² – 4x + 5y = 0

Question 8.
Find the equation of a circle passing through the points (1, -4), (5, 2) and having its centre on line x – 2y + 9 = 0.
Solution:

Let C(h, k) be the centre of the required circle which lies on the line x – 2y + 9 = 0.
Equation of line becomes
h – 2k + 9 = 0 …..(i)
Also, the required circle passes through points A(1, -4) and B(5, 2).
CA = CB = radius
CA = CB
By distance formula,
\(\sqrt{(\mathrm{h}-1)^{2}+[\mathrm{k}-(-4)]^{2}}=\sqrt{(\mathrm{h}-5)^{2}+(\mathrm{k}-2)^{2}}\)
Squaring both the sides, we get
⇒ (h – 1)² + (k + 4)² = (h – 5)² + (k – 2)²
⇒ h² – 2h + 1 + k² + 8k + 16 = h² – 10h + 25 + k² – 4k + 4
⇒ -2h + 8k + 17 = -10h – 4k + 29
⇒ 8h + 12k – 12 = 0
⇒ 2h + 3k – 3 = 0 ……(ii)
By (ii) – (i) × 2, we get
7k = 21
⇒ k = 3
Substituting k = 3 in (i), we get
h – 2(3) + 9 = 0
⇒ h – 6 + 9 = 0
⇒ h = -3
Centre of the circle is C(-3, 3).
radius (r) = CA

The equation of a circle with centre at (h, k) and radius r is given by (x – h)² + (y – k)² = r²
Here, h = -3, k = 3, r = √65
The required equation of the circle is
⇒ [x – (-3)]² + (y – 3)² = (√65)²
⇒ (x + 3)² + (y – 3)² = 65
⇒ x² + 6x + 9 + y² – 6y + 9 – 65 = 0
⇒ x² + y² + 6x – 6y – 47 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Miscellaneous Exercise 5

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

Question 1.
If A is (5, -3) and B is a point on the X-axis such that the slope of line AB is -2, then B ≡
(a) (7, 2)
(b) (\(\frac{7}{2}\), 0)
(c) (0, \(\frac{7}{2}\))
(d) (\(\frac{2}{7}\), 0)
Answer:
(b) (\(\frac{7}{2}\), 0)
Hint:
Let B(x, 0) be the point on X-axis.
We have A = (5, -3)
slope of AB = -2
⇒ \(\frac{0-(-3)}{x-5}\) = -2
⇒ 3 = -2(x – 5)
⇒ 3 = -2x + 10
⇒ x = \(\frac{7}{2}\)
Co-ordinates of point B = (\(\frac{7}{2}\), 0)

Question 2.
If the point (1, 1) lies on the line passing through the points (a, 0) and (0, b), then \(\frac{1}{a}+\frac{1}{b}=\)
(a) -1
(b) 0
(c) 1
(d) \(\frac{1}{a b}\)
Answer:
(c) 1
Hint:
Line passes through (a, 0), (0, b).
x-intercept = a, y-intercept = b
∴ Equation of line is \(\frac{x}{a}+\frac{y}{b}=1\) …….(i)
Since line (i) passes through (1, 1), (1, 1) satisfies (i)
∴ \(\frac{1}{a}+\frac{1}{b}=1\)

Question 3.
If A(1, -2), B(-2, 3) and C(2, -5) are the vertices of ΔABC, then the equation of median BE is
(a) 7x + 13y + 47 = 0
(b) 13x + 7y + 5 = 0
(c) 7x – 13y + 5 = 0
(d) 13x – 7y – 5 = 0
Answer:
(b) 13x + 7y + 5 = 0
Hint:

Question 4.
The equation of the line through (1, 2), which makes equal intercepts on the axes, is
(a) x + y = 1
(b) x + y = 2
(c) x + y = 4
(d) x + y = 3
Answer:
(d) x + y = 3
Hint:
Let the equation of required line be
\(\frac{x}{a}+\frac{y}{b}=1\) ……..(i)
Since the line makes equal intercepts on the axes, a = b
\(\frac{x}{a}+\frac{y}{a}=1\)
∴ x + y = a ……(ii)
But, equation (ii) passes through (1, 2).
1 + 2 = a
∴ a = 3
Substituting a = 3 in equation (ii), we get
x + y = 3

Question 5.
If the line kx + 4y = 6 passes through the point of intersection of the two lines 2x + 3y = 4 and 3x + 4y = 5, then k =
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
Hint:
Given two lines are
2x + 3y = 4 ……(i)
3x + 4y = 5 …….(ii)
Multiplying (i) by 3 and (ii) by 2 and then subtracting, we get
y = 2
Substituting y = 2 in (i), we get
x = -1
∴ Point of intersection of lines (i) and (ii) is (-1, 2).
Given that the line kx + 4y = 6 passes through (-1, 2).
k(-1) + 4(2) = 6
∴ k = 2

Question 6.
The equation of a line, having inclination 120° with positive direction of X-axis, which is at a distance of 3 units from the origin is
(a) √3x ± y + 6 = 0
(b) √3x + y ± 6 = 0
(c) x + y = 6
(d) x + y = -6
Answer:
(b) √3x + y ± 6 = 0
Hint:

Here, α = 30° and p = 3 units
Equation of line with inclination a and distance from origin as p is
x cos α + y sin α = p
∴ x cos 30° + y sin 30° = ±3
∴ \(\frac{\sqrt{3} x}{2}+\frac{y}{2}=\pm 3\)
∴ √3x + y ± 6 = 0

Question 7.
A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) 1
(d) \(\frac{4}{3}\)
Answer:
(d) \(\frac{4}{3}\)
Hint:
Slope of line 3x + y = 3 is -3
∴ Slope of line perpendicular to given line = \(\frac{1}{3}\)
Equation of required line passing through (2, 2) and having slope \(\frac{1}{3}\) is
y – 2 = \(\frac{1}{3}\)(x – 2)
3y – 6 = x – 2
∴ x – 3y + 4 = 0
∴ y-intercept = \(\frac{-4}{-3}=\frac{4}{3}\)

Question 8.
The angle between the line √3x – y – 2 = 0 and x – √3y + 1 = 0 is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Answer:
(b) 30°
Hint:

Question 9.
If kx + 2y – 1 = 0 and 6x – 4y + 2 = 0 are identical lines, then determine k.
(a) -3
(b) \(-\frac{1}{3}\)
(c) \(\frac{1}{3}\)
(d) 3
Answer:
(a) -3
Hint:
Lines kx + 2y – 1 = 0 and 6x – 4y + 2 = 0 are identical.
∴ \(\frac{k}{6}=\frac{2}{-4}=\frac{-1}{2}\)
∴ k = -3

Question 10.
Distance between the two parallel lines y = 2x + 7 and y = 2x + 5 is
(a) \(\frac{\sqrt{2}}{\sqrt{5}}\)
(b) \(\frac{1}{\sqrt{5}}\)
(c) \(\frac{\sqrt{5}}{2}\)
(d) \(\frac{2}{\sqrt{5}}\)
Answer:
(d) \(\frac{2}{\sqrt{5}}\)
Hint:
Here, c₁ = 7, c₂ = 5, a = 2 and b = -1
Distance between parallel lines

II. Answer the following questions.

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

(b) Given, 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 = \(\frac{5}{3}\)

(c) 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 = \(\frac{3-\mathrm{k}}{2}\)
2 = 3 – k
k = 1

Question 2.
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 3.
Find the distance of the origin from the line x = -2.
Solution:
Given equation of line is x = -2
This equation represents a line parallel to Y-axis and at a distance of 2 units to the left of Y-axis.
∴ Distance of the origin from the line is 2 units.

Question 4.
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 5.
Which of the following lines passes through the origin?
(a) x = 2
(b) y = 3
(c) y = x + 2
(d) 2x – y = 0
Answer:
(d) 2x – y = 0
Hint:
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.
∴ Option (d) is the correct answer.

Question 6.
Obtain the equation of the line which is:
(a) parallel to the X-axis and 3 units below it.
(b) parallel to the Y-axis and 2 units to the left of it.
(c) parallel to the X-axis and making an intercept of 5 on the Y-axis.
(d) parallel to the Y-axis and making an intercept of 3 on the X-axis.
Solution:
(a) 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.

(b) 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.

(c) 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.

(d) 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.

Question 7.
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.
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.

Question 8.
Find the equation of the line:
(a) having slope 5 and containing point A(-1, 2).
(b) containing the point T(7, 3) and having inclination 90°.
(c) through the origin which bisects the portion of the line 3x + 2y = 2 intercepted between the co-ordinate axes.
Solution:
(a) 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

(b) Given, Inclination of line = θ = 90°
the required line is parallel to 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.

(c) Given equation of the line is 3x + 2y = 2.

\(\frac{3 x}{2}+\frac{2 y}{2}=1\)
\(\frac{x}{\frac{2}{3}}+\frac{y}{1}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), 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 9.
Find the equation of the line passing through the points S(2, 1) and T(2, 3).
Solution:
The required line passes through the points S(2, 1) and T(2, 3).
Since both the given points have same x co-ordinates i.e. 2
the given points lie on a line parallel to Y-axis.
∴ The equation of the required line is x = 2.

Question 10.
Find the distance of the origin from the line 12x + 5y + 78 = 0.
Solution:
Let p be the perpendicular distance of origin from the line 12x + 5y + 78 = 0.
Here, a = 12, b = 5, c = 78

Question 11.
Find the distance between the parallel lines 3x + 4y + 3 = 0 and 3x + 4y + 15 = 0.
Solution:
Equations of the given parallel lines are 3x + 4y + 3 = 0 and 3x + 4y + 15 = 0
Here, a = 3, b = 4, c₁ = 3 and c₂ = 15
∴ Distance between the parallel lines

Question 12.
Find the equation of the line which contains the point A(3, 5) and makes equal intercepts on the co-ordinates axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …….(i)
This line passes through A(3, 5).
∴ \(\frac{3}{a}+\frac{5}{b}=1\) ……..(ii)
Since the required line makes equal intercepts on the co-ordinates axes,
a = b …….(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{5}{a}=1\)
∴ a = 8
∴ b = 8 …… [From (iii)]
Substituting the values of a and b in equation (i), the equation of the required line is
\(\frac{x}{8}+\frac{y}{8}=1\)
∴ x + y = 8

Case II: Line passing through origin.
Slope of line passing through origin and A(3, 5) is
m = \(\frac{5-0}{3-0}=\frac{5}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is
y = \(\frac{5}{3}\)x
∴ 5x – 3y = 0

Question 13.
The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6). Find equations of
(a) the sides
(b) the medians
(c) perpendicular bisectors of sides
(d) altitudes of ?ABC
Solution:
Vertices of ∆ABC are A(1, 4), B(2, 3) and C(1, 6)
(a) Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}\) = \(\frac{x-x_{1}}{x_{2}-x_{1}}\)
Equation of side AB is
\(\frac{y-4}{3-4}=\frac{x-1}{2-1}\)
y – 4 = -1(x – 1)
y – 4 = -x + 1
x + y = 5
Equation of side BC is
\(\frac{y-3}{6-3}=\frac{x-2}{1-2}\)
-1(y – 3) = 3(x – 2)
-y + 3 = 3x – 6
∴ 3x + y = 9
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.

(b) Let D, E and F be the midpoints of sides AC and AB respectively of ∆ABC.

(c) 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
\(\left(y-\frac{9}{2}\right)=\frac{1}{3}\left(x-\frac{3}{2}\right)\)
3(2y – 9) = (2x – 3)
6y – 27 = 2x – 3
2x – 6y + 24 = 0
∴ x – 3y + 12 = 0
Since both the points A and C have 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, 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 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
\(\left(y-\frac{7}{2}\right)=1\left(x-\frac{3}{2}\right)\)
2y – 7 = 2x – 3
2x – 2y + 4 = 0
∴ x – y + 2 = 0

(d) Let AX, BY, and CZ be the altitudes through the vertices A, B and C respectively of ∆ABC.

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 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
Since altitude CZ passes through (1, 6) and has slope 1,
equation of altitude CZ is
y – 6 = 1(x – 1)
∴ x – y + 5 = 0

Question 14.
Find the equation of the line which passes through the point of intersection of lines x + y – 3 = 0, 2x – y + 1 = 0 and which is parallel to X-axis.
Solution:
Let u ≡ x + y – 3 = 0 and v ≡ 2x – y + 1 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
(x + y – 3) + k(2x – y + 1) = 0 …..(i)
x + y – 3 + 2kx – ky + k = 0
x + 2kx + y – ky – 3 + k = 0
(1 + 2k)x + (1 – k)y – 3 + k = 0
But, this line is parallel to X-axis
Its slope = 0
⇒ \(\frac{-(1+2 k)}{1-k}=0\)
⇒ 1 + 2k = 0
⇒ k = \(\frac{-1}{2}\)
Substituting the value of k in (i), we get
(x + y – 3) + \(\frac{-1}{2}\) (2x – y + 1) = 0
⇒ 2(x + y – 3) – (2x – y + 1 ) = 0
⇒ 2x + 2y – 6 – 2x + y – 1 = 0
⇒ 3y – 7 = 0, which is the equation of the required line.

Question 15.
Find the equation of the line which passes through the point of intersection of lines x + y + 9 = 0, 2x + 3y + 1 = 0 and which makes x-intercept 1.
Solution:
Let u ≡ x + y + 9 = 0 and v ≡ 2x + 3y + 1 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
(x + y + 9) + k(2x + 3y + 1) = 0 ……(i)
⇒ x + y + 9 + 2kx + 3ky + k = 0
⇒ (1 + 2k)x + (1 + 3k)y + 9 + k = 0
But, x-intercept of this line is 1.
⇒ \(\frac{-(9+\mathrm{k})}{1+2 \mathrm{k}}\)
⇒ -9 – k = 1 + 2k
⇒ k = \(\frac{-10}{3}\)
Substituting the value of k in (i), we get
(x + y + 9) + (\(\frac{-10}{3}\)) (2x + 3y + 1) = 0
⇒ 3(x + y + 9) – 10(2x + 3y + 1) = 0
⇒ 3x + 3y + 27 – 20x – 30y – 10 = 0
⇒ -17x – 27y+ 17 = 0
⇒ 17x + 27y – 17 = 0, which is the equation of the required line.

Question 16.
Find the equation of the line through A(-2, 3) and perpendicular to the line through S(1, 2) and T(2, 5).
Solution:
Slope of ST = \(\frac{5-2}{2-1}\) = 3
Since the required line is perpendicular to ST,
slope of required line = \(\frac{-1}{3}\) and line passes through A(-2, 3)
Equation of the line in slope point form is y – y₁ = m(x – x₁)
The equation of the required line is
y – 3 = \(\frac{-1}{3}\)(x + 2)
⇒ 3(y – 3) = -(x + 2)
⇒ 3y – 9 = -x – 2
⇒ x + 3y = 7

Question 17.
Find the x-intercept of the line whose slope is 3 and which makes intercept 4 on the Y-axis.
Solution:
Equation of a line having slope ‘m’ and y-intercept ‘c’ is y = mx + c
Given, m = 3, c = 4
The equation of the line is y = 3x + 4
3x – y = -4
\(\frac{3 x}{(-4)}-\frac{y}{(-4)}=1\)
\(\frac{x}{\left(\frac{-4}{3}\right)}+\frac{y}{4}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), where
x-intercept = a
x-intercept = \(\frac{-4}{3}\)

Alternate Method:

Let θ be the inclination of the line.
Then tan θ = 3 …..[∵ slope = 3 (given)]
\(\frac{\mathrm{OB}}{\mathrm{OA}}=3\)
\(\frac{4}{\mathrm{OA}}=3\)
OA = \(\frac{4}{3}\)
x-intercept = –\(\frac{4}{3}\) as point A is to the left side of Y-axis.

Question 18.
Find the distance of P(-1, 1) from the line 12(x + 6) = 5(y – 2).
Solution:
Given equation of the line is
12(x + 6) = 5(y – 2)
12x + 72 = 5y – 10
12x – 5y + 82 = 0
Let p be the perpendicular distance of the point (-1, 1) from the line 12x – 5y + 82 = 0.

Question 19.
Line through A(h, 3) and B(4,1) intersect the line lx – 9y -19 = 0 at right angle. Find the value of h.
Solution:
Given, A(h, 3) and B(4, 1)
Slope of AB (m₁) = \(\frac{1-3}{4-h}\)
m₁ = \(\frac{2}{h-4}\)
Slope of line 7x – 9y – 19 = 0 is m₂ = \(\frac{7}{9}\)
Since line AB and line 7x – 9y – 19 = 0 are perpendicular to each other,
m₁ × m₂ = -1
\(\frac{2}{h-4} \times \frac{7}{9}=-1\)
14 = 9(4 – h)
14 = 36 – 9h
9h = 22
h = \(\frac{22}{9}\)

Question 20.
Two lines passing through M(2, 3) intersect each other at an angle of 45°. If slope of one line is 2, find the equation of the other line.
Solution:
Let m be the slope of the required line which make an angle of 45° with the other line.
Slope of one of the lines is 2.
tan 45° = \(\left|\frac{\mathrm{m}-2}{1+\mathrm{m}(2)}\right|\)
1 = \(\left|\frac{m-2}{1+2 m}\right|\)
\(\frac{m-2}{1+2 m}=\pm 1\)
\(\frac{\mathrm{m}-2}{1+2 \mathrm{~m}}\) = 1 or \(\frac{\mathrm{m}-2}{1+2 \mathrm{~m}}\) = -1
m – 2 = 1 + 2m or m – 2 = -1 – 2m
m = -3 or 3m = 1
m = -3 or m = \(\frac{1}{3}\)
Required line passes through M(2, 3)
When m = -3, equation of the line is
y – 3 = -3(x – 2)
y – 3 = -3x + 6
∴ 3x + y = 9
When m = \(\frac{1}{3}\), equation of the line is
y – 3 = \(\frac{1}{3}\)(x – 2)
3y – 9 = x – 2
∴ x – 3y + 7 = 0

Question 21.
Find the y-intercept of the line whose slope is 4 and which has x-intercept 5.
Solution:
Given, slope = 4, x-intercept = 5
Since the x-intercept of the line is 5, it passes through (5, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of the required line is
y – 0 = 4(x – 5)
y = 4x – 20
4x – y = 20
\(\frac{4 x}{20}-\frac{y}{20}=1\)
\(\frac{x}{5}+\frac{y}{(-20)}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), where
x-intercept = b, y-intercept = -20

Question 22.
Find the equations of the diagonals of the rectangle whose sides are contained in the lines x = 8, x = 10, y = 11 and y = 12.
Solution:
Given, equations of sides of rectangle are x = 8, x = 10, y = 11 and y = 12

From the above diagram,
Vertices of rectangle are A(8, 11), B(10, 11), C(10, 12) and D(8, 12).
Equation of diagonal AC is
\(\frac{y-11}{12-11}=\frac{x-8}{10-8}\)
\(\frac{y-11}{1}=\frac{x-8}{2}\)
2y – 22 = x – 8
x – 2y + 14 = 0
Equation of diagonal BD is
\(\frac{y-11}{12-11}=\frac{x-10}{8-10}\)
\(\frac{y-11}{1}=\frac{x-10}{-2}\)
-2y + 22 = x – 10
x + 2y = 32

Question 23.
A(1, 4), B(2, 3) and C(1, 6) are vertices of AABC. Find the equation of the altitude through B and hence find the co-ordinates of the point where this altitude cuts the side AC of ∆ABC.
Solution:
Vertices of triangle are A(1, 4), B(2, 3) and C(1, 6).
Let BD be the altitude through the vertex B.

Since both the points A and C have same x co-ordinates i.e. 1
the given points lie on a line parallel to Y-axis.
The equation of the line AC is x = 1 …..(i)
AC is parallel to Y-axis and therefore, altitude BD is parallel to X-axis.
Since the altitude BD passes through B(2, 3), the equation of altitude BD is y = 3 ……(ii)
From (i) and (ii),
Point of intersection of AC and altitude BD is (1, 3).

Question 24.
The vertices of ∆PQR are P(2, 1), Q(-2, 3) and R(4, 5). Find the equation of the median through R.
Solution:

Let S be the midpoint of side PQ.
Then RS is the median through R.
S = \(\left(\frac{2-2}{2}, \frac{3+1}{2}\right)\) = (0, 2)
The median RS passes through the points R(4, 5) and S(0, 2).
∴ Equation of median RS is
\(\frac{y-5}{2-5}=\frac{x-4}{0-4}\)
⇒ \(\frac{y-5}{-3}=\frac{x-4}{-4}\)
⇒ 4(y – 5) = 3(x – 4)
⇒ 4y – 20 = 3x – 12
∴ 3x – 4y + 8 = 0

Question 25.
A line perpendicular to segment joining A(1, 0) and B(2, 3) divides it internally in the ratio 1 : 2. Find the equation of the line. Solution:
Given, A(1, 0), B(2, 3)
Slope of AB = \(\frac{3-0}{2-1}\) = 3
Required line is perpendicular to AB.
Slope of required line = \(\frac{-1}{3}\)
Let point C divide AB in the ratio 1 : 2.

Required line passes through \(\left(\frac{4}{3}, 1\right)\) and has slope = \(\frac{-1}{3}\)
Equation of the line in slope point form is y – y₁ = m(x – x₁)
The equation of the required line is
y – 1 = \(\frac{-1}{3}\left(x-\frac{4}{3}\right)\)
⇒ 3(y – 1) = \(-1\left(x-\frac{4}{3}\right)\)
⇒ 3y – 3 = -x + \(\frac{4}{3}\)
⇒ 9y – 9 = -3x + 4
⇒ 3x + 9y = 13

Question 26.
Find the co-ordinates of the foot of the perpendicular drawn from the point P(-1, 3) to the line 3x – 4y – 16 = 0.
Solution:

Let M be the foot of perpendicular drawn from P(-1, 3) to the line 3x – 4y – 16 = 0
Slope of the line 3x – 4y – 16 = 0 is \(\frac{-3}{-4}=\frac{3}{4}\)
Since PM ⊥ to line (i),
slope of PM = \(\frac{-4}{3}\)
Equation of PM is
y – 3 = \(\frac{-4}{3}\) (x + 1)
⇒ 3(y – 3) = -4(x + 1)
⇒ 3y – 9 = -4x – 4
∴ 4x + 3y – 5 = 0 ……(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equation (i) and (ii).
By (i) × 3 + (ii) × 4, we get
25x = 68
x = \(\frac{68}{25}\)
Substituting x = \(\frac{68}{25}\) in (ii), we get

The co-ordinates of the foot of perpendicular M are \(\left(\frac{68}{25}, \frac{-49}{25}\right)\)

Question 27.
Find points on the X-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}=1\) is 4 units.
Solution:
The equation of line is \(\frac{x}{3}+\frac{y}{4}=1\)
i.e. 4x + 3y – 12 = 0 …..(i)
Let (h, 0) be a point on the X-axis.
The distance of this point from line (i) is 4.
⇒ \(\frac{|4 h+3(0)-12|}{\sqrt{4^{2}+3^{2}}}=4\)
⇒ \(\frac{|4 \mathrm{~h}-12|}{5}=4\)
⇒ |4h – 12| = 20
⇒ 4h – 12 = 20 or 4h – 12 = -20
⇒ 4h = 32 or 4h = -8
⇒ h = 8 or h = -2
∴ The required points are (8, 0) and (-2, 0).

Question 28.
The perpendicular from the origin to a line meets it at (-2, 9). Find the equation of the line.
Solution:

Slope of ON = \(\frac{9-0}{-2-0}=\frac{-9}{2}\)
Since line AB ⊥ ON,
slope of the line AB perpendicular to ON is \(\frac{2}{9}\) and it passes through point N(-2, 9).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line AB is
y – 9 = \(\frac{2}{9}\)(x + 2)
⇒ 9(y – 9) = 2(x + 2)
⇒ 9y – 81 = 2x + 4
⇒ 2x – 9y + 85 = 0

Question 29.
P(a, b) is the midpoint of a line segment intercepted between the axes. Show that the equation of the line is \(\frac{x}{a}+\frac{y}{b}=2\).
Solution:
Let the intercepts of a line AB be x₁ and y₁ on the X and Y-axes respectively.
A ≡ (x₁, 0), B = (0, y₁)
P(a, b) is the midpoint of a line segment AB intercepted between the axes.

Question 30.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle of 135° with the positive X-axis.
Solution:

Let a line L make angle 135° with positive X-axis.
Required distance = PQ, where PQ || line L
Slope of PQ = tan 135°
= tan (180° – 45°)
= -tan 45°
= -1
Equation of PQ is
y – 1 = (-1)(x – 4)
y – 1 = -x + 4
x + y = 5 …..(i)
To get point Q we solve the equation 4x – y = 0 with (i)
Substituting y = 4x in (i), we get
5x = 5
x = 1
Substituting x = 1 in (i), we get
1 + y = 5
y = 4
∴ Q = (1, 4)
PQ = \(\sqrt{(4-1)^{2}+(1-4)^{2}}\)
= \(\sqrt{9+9}\)
= 3√2

Question 31.
Show that there are two lines which pass through A(3, 4) and the sum of whose intercepts is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) ……(1)
This line passes through (3, 4)
\(\frac{3}{a}+\frac{4}{b}=1\) …..(ii)
Since the sum of the intercepts of the line is zero,
a + b = 0
a = -b ……(iii)
Substituting the value of a in (ii), we get
\(\frac{3}{-b}+\frac{4}{b}=1\)
\(\frac{1}{b}\) = 1
b = 1
a = -1 ……[From (iii)]
Substituting the values of a and b in (i),
the equation of the required line is
\(\frac{x}{-1}+\frac{y}{1}=1\)
x – y = -1
∴ x – y + 1 = 0

Case II: Line passing through origin.
Slope of line passing through origin and A(3, 4) is
m = \(\frac{4-0}{3-0}=\frac{4}{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
The equation of the required line is
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0
∴ There are two lines which pass through A(3, 4) and the sum of whose intercepts is zero.

Question 32.
Show that there is only one line which passes through B(5, 5) and the sum of whose intercepts is zero.
Solution:
When line is passing through origin, the sum of intercepts made by the line is zero.
Slope of line passing through origin and B(5, 5) is
m = \(\frac{5-0}{5-0}\) = 1
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
The equation of the required line is y = x
∴ x – y = 0
∴ There is only one line which passes through B(5, 5) and the sum of whose intercepts is zero.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.4

Question 1.
Find the slope, x-intercept, y-intercept of each of the following lines, i. 2x + 3y-6 = 0 ii. 3x-y-9 = 0 iii. x + 2y = 0
Solution:
i. Given equation of the line is 2x + 3y – 6 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 2, b = 3, c = -6

ii. Given equation of the line is 3x – y – 9 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 3, b = – 1, c = – 9

iii. Given equation of the line is x + 2y = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 1, b = 2, c = 0

Question 2.
Write each of the following equations in ax + by + c = 0 form.
i. y = 2x – 4
ii. y = 4
iii. \(\frac{x}{2}+\frac{y}{4}=1\)
iv. \(\frac{x}{3}-\frac{y}{2}=0\)
i. y = 2x – 4
∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

ii. y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

iii. \(\frac{x}{2}+\frac{y}{4}=1\)
∴ \(\frac{2 x+y}{4}\)
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

iv. \(\frac{x}{3}-\frac{y}{2}=0\)
∴ 2x – 3y = 0
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.
[Note: Answer given in the textbook is ‘2x – 3y – 6 = 0’. However, as per our calculation it is ‘2x-3y + 0 = 0’.]

Question 3.
Show that the lines x – 2y – 7 = 0 and 2x – 4y + 15 = 0 are parallel to each other.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x – 4y + 15 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{-4}=\frac{1}{2}\)
Since m₁ = m₂
the given lines are parallel to each other.

Question 4.
Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x + y + 1 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{1}=-2\)
Since m₁ x m₁ = \(\frac{1}{2}\) x (- 2) = -1,
the given lines are perpendicular to each other. Consider,
x – 2y – 7 = 0 …(i)
2x + y + 1 =0 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y + 2 = 0 …(iii)
Adding equations (i) and (iii), we get
5x – 5 = 0
∴ x = 1
Substituting x = 1 in equation (ii), we get
2 + y + 1 = 0
∴ y = – 3
∴ The point of intersection of the given lines is (1,-3).

Question 5.
If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Solution:
Let the line 3x + 4y = p cuts the X and Y axes at points A and B respectively.
3x + 4y = p

This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
∴ A (a, 0) ≡ (\(\frac{p}{3}\), 0) and B ≡ (0, b) = (0, \(\frac{p}{4}\))
∴ OA = \(\frac{p}{3}\) and OB = \(\frac{p}{4}\)
Given, A (∆OAB) = 24 sq. units
∴ \(\left|\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\right|=24\)
∴ \(\left|\frac{1}{2} \times \frac{\mathrm{p}}{3} \times \frac{\mathrm{p}}{4}\right|=24\)
∴ p² = 576
∴ p = ± 24

Question 6.
Find the co-ordinates of the foot of the perpendicular drawn from the point A(- 2,3) to the line 3x-y -1 = 0.
Solution:
Let M be the foot of perpendicular drawn from
point A(- 2,3) to the line
3x-y- 1 = 0 …(i)
Slope of the line 3x-y – 1 = 0 is \(\frac{-3}{-1}\) =3.
Since AM ⊥ to line (i),
slope of AM = \(\frac{-1}{3}\)
∴ Equation of AM is
y – 3 = \(\frac{-1}{3}\)(x + 2)

∴ 3(y – 3) = – 1(x + 2)
∴ 3y – 9 = -x – 2
∴ x + 3y – 7 = 0 …………(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).
By (i) x 3 + (ii), we get 10x -10 = 0
∴ x = 1
Substituting x = 1 in (ii), we get
1 + 3y – 7 = 0
∴ 3y = 6
∴ y = 2
∴ The co-ordinates of the foot of the perpendicular Mare (1,2).

Question 7.
Find the co-ordinates of the circumcentre of the triangle whose vertices are A(- 2, 3), B(6, -1), C(4,3),e
Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.
Let F be the circumcentre of AABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.
∴ D and E are the midpoints of side BC and AC respectively.

Since FD passes through (5, 1) and has slope 1/2 equation of FD is
y – 1 = \(\frac{1}{2}\)(x-5)
∴ 2 (y – 1) = x – 5
∴ 2y – 2 = x – 5
∴ x – 2y – 3 = 0 …(i)
Since both the points A and C have same y co-ordinates i.e. 3,
the given points lie on the line y = 3.
Since the equation FE passes through E(1, 3),
the equation of FE is x = 1. .. .(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y -3 = 0
∴ y = -1
∴ Co-ordinates of circumcentre F ≡ (1, – 1).

Question 8.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(3, – 2), B(7,6), C (-1,2).
Solution:
Let O be the orthocentre of ∆ABC.
Let AD and BE be the altitudes on the sides BC and AC respectively.

Slope of side BC = \(\frac{2-6}{-1-7}=\frac{-4}{-8}=\frac{1}{2}\)
∴ Slope of AD = – 2 [∵ AD ⊥ BC]
∴ Equation of line AD is
y – (-2) = (- 2) (x – 3)
∴ y + 2 = -2x + 6
∴ 2x + y -4 = 0 …(i)
Slope of side AC = \(\frac{-2-2}{3-(-1)}=\frac{-4}{4}\) = -1
∴ Slope of BE = 1 …[ ∵ BE ⊥ AC]
∴ Equation of line BE is
y – 6 = 1(x – 7)
∴ y – 6 = x – 1
∴ x = y + 1 …(ii)
Substituting x = y + 1 in (i), we get
2(y + 1) + y – 4 = 0
∴ 2y + 2 + y – 4 = 0
∴ 3y – 2 = 0
∴ y = \(\frac{2}{3} in (ii), we get
Substituting y = [latex]\frac{2}{3}\) in (ii), we get
x = \(\frac{2}{3}+1=\frac{5}{3}\)
∴ Co-ordinates of orthocentre, O = \(\left(\frac{5}{3}, \frac{2}{3}\right)\)

Question 9.
Show that the lines 3 – 4y + 5 = 0, lx – 8y + 5 = 0 and 4JC + 5y – 45 = 0 are concurrent. Find their point of concurrence.
Solution:
The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence. Equations of the given lines are
3x – 4y + 5 = 0 …(i)
7x-8y + 5 = 0 …(ii)
4x + 5y – 45 = 0 …(iii)
By (i) x 2 – (ii), we get
– x + 5 = 0
∴ x = 5
Substituting x = 5 in (i), we get
3(5) – 4y + 5 = 0
∴ -4y = – 20
∴ y = 5
∴ The point of intersection of lines (i) and (ii) is given by (5, 5).
Substituting x = 5 and y = 5 in L.H.S. of (iii), we get
L.H.S. = 4(5) + 5(5) – 45
= 20 + 25 – 45
= 0
= R.H.S.
∴ Line (iii) also passes through (5, 5).
Hence, the given three lines are concurrent and the point of concurrence is (5, 5).

Question 10.
Find the equation of the line whose x-intercept is 3 and which ¡s perpendicular to the line 3x – y + 23 = 0.
Solution:
Slope of the line 3x – y + 23 = 0 is 3.
∴ Slope of the required line perpendicular to
3x – y + 23 = 0 is \(\frac{-1}{3}\)
Since the x-intercept of the required line is 3, it passes through (3, 0).
∴ The equation of the required line is ‘
y – 0 = \(\frac{-1}{3}\)(x – 3)
∴ 3y = x + 3
∴ x + 3y = 3

Question 11.
Find the distance of the origin from the line 7x + 24y – 50 = 0.
Solution:
Let p be the perpendicular distance of origin
fromtheline7x + 24y – 50 = 0
Here, a = 7, b = 24, c = -50

Question 12.
Find the distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0.
Solution:
Let p be the perpendicular distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0
Here, a = 12, b = – 5, c = – 13, x₁ = -2, y₁ = 3

Question 13.
Find the distance between parallel lines 4x – 3y + 5 = 0 and 4xr – 3y + 7 = 0.
Solution:
Equations of the given parallel lines are 4x – 3y + 5 = 0 and 4x – 3y + 1 = 0
Here, a = 4, b = – 3, c₁ = 5 and c₂ = 7
∴ Distance between the parallel lines

Question 14.
Find the distance between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution:
Equations of the given parallel lines are 3x + 2y + 6 = 0 and
9x + 6y – 1 = 0 i.e., 3x + 2y – \(\frac{7}{3}\) =0
Here, a = 3, b = 2, c₁ = 6 and c₂ = \(\frac{-7}{3}\)
∴ Distance between the parallel lines

Question 15.
Find the points on the line x + y – 4 = 0 which are at a unit distance from the line 4JC + 3y = 10.
Solution:
Let P(x₁, y₁) be a point on the line x + y – 4 = o.
∴ x₁ + y₁ – 4 = 0
∴ y₁ = 4 – x₁ …(i)
Also, distance of P from the line 4x + 3y- 10 = 0 is 1

∴ 5 = | x₁ + 2 |
∴ x₁ + 2 = ± 5
∴ x₁ + 2 = 5 or x₁ + 2 = – 5
∴ x₁ = 3 or x₁ = – 7
From (i), when x₁ = 3, y₁ = 1
and when x₁ = -7, y₁ = 11
∴ The required points are (3, 1) and (-7, 11).
[Note: The question has been modified]

Question 16.
Find the equation of the line parallel to the X-axis and passing through the point of intersection of lines x + y – 2 = 0 and 4x + 3y = 10.
Solution:
Let u = x + y – 2 = 0 and v = 4x + 3y – 10 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x + y – 2) + k(4x + 3y – 10) = 0 …(i)
∴ x + y – 2 + 4kx + 3ky – 10k = 0
∴ x + 4kx + y + 3ky – 2 – 10k = 0
∴ (1+ 4k)x + (1 + 3k)y – 2 – 10k = 0
But, this line is parallel to X-axis.
∴ Its slope = 0
∴ \(\frac{-(1+4 k)}{1+3 k}\) = 0
∴ 1 + 4k = 0
∴ k = \(\frac{-1}{4}\)
Substituting the value of k in (i), we get
(x + y – 2) + (4x + 3y – 10) = 0
∴ 4(x +y – 2) – (4x + 3y -10 ) = 0
∴ 4x + 4y – 8 – 4x – 3y + 10 = 0
∴ y + 2 = 0, which is the equation of the required line.
[Note: Answer given in the textbook is 5y – 8= 0. However, as per our calculation it is y + 2 = 0.]

Question 17.
Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2xr – 3y + 4 = 0 and making intercept 3 on the X-axis.
Solution:
Let u ≡ x + y – 2 = 0 and v ≡ 2x – 3y + 4 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x +y – 2) + k(2x – 3y + 4) = 0 …(i)
But, x-intercept of line is 3.
∴ It passes through (3, 0).
Substituting x = 3 and y = 0 in (i), we get
(3 + 0 – 2) + k(6 – 0 + 4) = 0
∴ 1 + 10k = 0
k = \(\frac{-1}{10}\)
Substituting the value of k in (i), we get (x + y – 2) + \(\left(\frac{-1}{10}\right)\) (2x – 3y + 4) = 0
∴ 10(x + y – 2) – (2x – 3y + 4) = 0
∴ 10x + 10y -20 — 2x + 3y-4 = 0
∴ 8x + 13y – 24 = 0, which is the equation of the required line.

Question 18.
If A(4, 3), B(0, 0) and C(2, 3) are the vertices of ΔABC, then find the equation of bisector of angle BAC.
Solution:
Let the bisector of ∠ BAC meets BC at point D.
∴ Point D divides seg BC in the ratio l(AB) : l(AC)

∴ 18 (y – 3) = 6 (x – 4)
∴ 3(y – 3) = x – 4
∴ 3y – 9 = x – 4
∴ x – 3y + 5 = 0

Question 19.
D(- 1, 8), E(4, – 2), F(- 5, – 3) are midpoints of sides BC, CA and AB of AABC. Find
i. equations of sides of ΔABC.
ii. co-ordinates of the circumcentre of ΔABC.
Solution:
Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ΔABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.

∴ x₁ + x₂ = -10 …………. (v)
and y₁ + y₂ = – 6 …………(vi)
For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = – 4
∴ x₁ + x₂ + x₃ = -2 …………..(vii)
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = – 10
Solving (v) and (vii), we get x₃ = 8

For y-coordinates:
Adding (ii), (iv) and (vi), we get 2y₁ + 2y₂ + 2y₃ = 6
y₁ + y₂ + y₃ = 3 …….(viii)
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
∴ Vertices of AABC are A(0, – 13), B(- 10, 7), C(8, 9)

∴ 8(y + 13) = 22x
∴ 4(y + 13) = 11x
∴ 11x – 4y – 52 = 0

ii. Here, A(0, – 13), B(- 10, 7), C(8, 9) are the vertices of ΔABC.
Let F be the circumcentre of AABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.
D and E are the midpoints of side BC and AC.

∴ Slope of FD = -9 … [ ∵ FD ⊥ BC]
Since FD passes through (-1, 8) and has slope -9, equation of FD is
y – 8 = -9 (x +1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1
Also, slope of AC = \(\frac{-13-9}{0-8}=\frac{11}{4}\)
∴ Slope of FE = \(\frac{-4}{11}\) [ ∵ FE ⊥ AC]
Since FE passes through (4, -2) and has slope -4
\(\frac{-4}{11}\), equation of FE is
(y + 2) = \(\frac{-4}{11}\) (x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = – 4x + 16
∴ 4x + 11y = -6 …………(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value ofy in (ii), we get
4x + 11(-9x- 1) = – 6
∴ 4x – 99x -11 = – 6
∴ -95x = 5
∴ x = \(\frac{-1}{19}\)
Substituting the value of x in (i), we get
y = -9(\(\frac{-1}{19}\)) – 1 = \(\frac{-10}{19}\)
∴ Co-ordinates of circumcentre F ≡ \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)

Question 20.
0(0, 0), A(6, 0) and B(0, 8) are vertices of a triangle. Find the co-ordinates of the incentre of ∆OAB.
Solution:
Let bisector of ∠O meet AB at point D and bisector of ∠A meet BO at point E
∴ Point D divides seg AB in the ratio l(OA): l(OB)
and point E divides seg BO in the ratio l(AB): l(AO)
Let I be the incentre of ∠OAB.

∴ Point D divides AB internally in 6 : 8
i.e. 3 :4

∴ y = x …(i)
Now, by distance formula,
l(AB) = \(\begin{aligned}
&=\sqrt{(6-0)^{2}+(0-8)^{2}} \\
&=\sqrt{36+64}=10
\end{aligned}\)
l(AO) = \(\sqrt{(6-0)^{2}+(0-0)^{2}}\) = 6
∴ Point E divides BO internally in 10 : 6 i.e. 5:3

∴ -2y = x – 6
∴ x + 2y = 6 …(ii)
To find co-ordinates of incentre, we have to solve equations (i) and (ii).
Substituting y = x in (ii), we get
x + 2x = 6
∴ x = 2
Substituting the value of x in (i), we get
y = 2
∴ Co-ordinates of incentre I ≡ (2, 2).

Alternate Method:
Let I be the incentre.
I lies in the 1st quadrant.
OPIR is a square having side length r.
Since OA = 6, OP = r,
PA = 6 – r
Since PA = AQ,
AQ = 6 – r …(i)
Since OB = 8, OR = r,
BR = 8 – r
∴ BR = BQ
∴ BQ = 8 – r …(ii)

AB = BQ + AQ
Also, AB = \(\begin{aligned}
&=\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}} \\
&=\sqrt{6^{2}+8^{2}} \\
&=\sqrt{100}=10
\end{aligned}\)
∴ BQ + AQ= 10
∴ (8 – r) + (6 – r) = 10
∴ 2r = 14- 10 = 4
∴ r = 2
∴ I = (2,2)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Sets and Relations Ex 1.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Sets and Relations Ex 1.2

Question 1.
If (x – 1, y + 4) = (1, 2), find the values of x and y.
Solution:
(x – 1, y + 4) = (1, 2)
By the definition of equality of ordered pairs, we have
x – 1 = 1 and y + 4 = 2
∴ x = 2 and y = -2

Question 2.
If \(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{3}, \frac{3}{2}\right)\), find x and y.
Solution:
\(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{3}, \frac{3}{2}\right)\)
By the definition of equality of ordered pairs, we have
\(x+\frac{1}{3}=\frac{1}{3}\) and \(\frac{y}{3}-1=\frac{3}{2}\)
\(x=\frac{1}{3}-\frac{1}{3}\) and \(\frac{y}{3}=\frac{3}{2}+1=\frac{5}{2}\)
x = 0 and y = \(\frac{15}{2}\)

Question 3.
If A = {a, b, c}, B = {x, y}, find A × B, B × A, A × A, B × B.
Solution:
A = {a, b, c}, B = {x, y}
A × B = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)}
B × A = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}
B × B = {(x, x), (x, y), (y, x), (y, y)}

Question 4.
If P = {1, 2, 3} and Q = {6, 4}, find the sets P × Q and Q × P.
Solution:
P = {1, 2, 3}, Q = {6, 4}
P × Q = {(1, 6), (1, 4), (2, 6), (2, 4), (3, 6), (3, 4)}
Q × P = {(6, 1), (6, 2), (6, 3), (4, 1), (4, 2), (4, 3)}

Question 5.
Let A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}. Find
(i) A × (B ∩ C)
(ii) (A × B) ∩ (A × C)
(iii) A × (B ∪ C)
(iv) (A × B) ∪ (A × C)
Solution:
A= {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}
(i) B ∩ C = {5, 6}
∴ A × (B ∩ C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

(ii) A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∩ (A × C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}

(iii) B ∪ C = {4, 5, 6}
∴ A × (B ∪ C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

(iv) A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∪ (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}

Question 6.
Express {(x, y) / x² + y² = 100, where x, y ∈ W} as a set of ordered pairs.
Solution:
{(x, y) / x² + y² = 100, where x, y ∈ W}
We have, x² + y² = 100
When x = 0 and y = 10,
x² + y² = 0² + 10² = 100
When x = 6 andy = 8,
x² + y² = 6² + 8² = 100
When x = 8 and y = 6,
x² + y² = 8² + 6² = 100
When x = 10 and y = 0,
x² + y² = 10² + 0² = 100
∴ Set of ordered pairs = {(0, 10), (6, 8), (8, 6), (10, 0)}

Question 7.
Write the domain and range of the following relations.
(i) {(a, b) / a ∈ N, a < 6 and b = 4}
(ii) {(a, b) / a, b ∈ N, a + b = 12}
(iii) {(2, 4), (2, 5), (2, 6), (2, 7)}
Solution:
(i) Let R₁ = {(a, b)/ a ∈ N, a < 6 and b = 4}
Set of values of ‘a’ are domain and set of values of ‘b’ are range.
a ∈ N and a < 6
∴ a = 1, 2, 3, 4, 5 and b = 4
Domain (R₁) = {1, 2, 3, 4, 5}
Range (R₁) = {4}

(ii) Let R₂ = {(a, b)/a, b ∈ N and a + b = 12}
Now, a, b ∈ N and a + b = 12
When a = 1, b = 11
When a = 2, b = 10
When a = 3, b = 9
When a = 4, b = 8
When a = 5, b = 7
When a = 6, b = 6
When a = 7, b = 5
When a = 8, b = 4
When a = 9, b = 3
When a = 10, b = 2
When a = 11, b = 1
∴ Domain (R₂) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
Range (R₂) = {11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1}

(iii) Let R₃ = {(2, 4), (2, 5), (2, 6), (2, 7)}
Domain (R₃) = {2}
Range (R₃) = {4, 5, 6, 7}

Question 8.
Let A = {6, 8} and B = {1, 3, 5}.
Let R = {(a, b) / a ∈ A, b ∈ B, a – b is an even number}.
Show that R is an empty relation from A to B.
Solution:
A= {6, 8}, B = {1, 3, 5}
R = {(a, b)/ a ∈ A, b ∈ B, a – b is an even number}
a ∈ A
∴ a = 6, 8
b ∈ B
∴ b = 1, 3, 5
When a = 6 and b = 1, a – b = 5 which is odd
When a = 6 and b = 3, a – b = 3 which is odd
When a = 6 and b = 5, a – b = 1 which is odd
When a = 8 and b = 1, a – b = 7 which is odd
When a = 8 and b = 3, a – b = 5 which is odd
When a = 8 and b = 5, a – b = 3 which is odd
Thus, no set of values of a and b gives a – b even.
∴ R is an empty relation from A to B.

Question 9.
Write the relation in the Roster form and hence find its domain and range.
(i) R₁ = {(a, a²) / a is a prime number less than 15}
(ii) R₂ = {(a, \(\frac{1}{a}\)) / 0 < a ≤ 5, a ∈ N}
Solution:
(i) R₁ = {(a, a²) / a is a prime number less than 15}
∴ a = 2, 3, 5, 7, 11, 13
∴ a2 = 4, 9, 25, 49, 121, 169
∴ R₁ = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)}
∴ Domain (R₁) = {a/a is a prime number less than 15} = {2, 3, 5, 7, 11, 13}
Range (R₁) = {a²/a is a prime number less than 15} = {4, 9, 25, 49, 121, 169}

Question 10.
R = {(a, b) / b = a + 1, a ∈ Z, 0 < a < 5}. Find the range of R.
Solution:
R = {(a, b) / b = a + 1, a ∈ Z, 0 < a < 5}
∴ a = 1, 2, 3, 4
∴ b = 2, 3, 4, 5
∴ Range (R) = {2, 3, 4, 5}

Question 11.
Find the following relations as sets of ordered pairs.
(i) {(x, y) / y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
(ii) {(x,y) / y > x + 1, x ∈ {1, 2} and y ∈ {2, 4, 6}}
(iii) {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}
Solution:
(i) {(x, y) / y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
Here y = 3x
When x = 1, y = 3(1) = 3
When x = 2, y = 3(2) = 6
When x = 3, y = 3(3) = 9
∴ Ordered pairs are {(1, 3), (2, 6), (3, 9)}

(ii) {(x, y) / y > x + 1, x ∈ {1, 2} and y ∈ {2, 4, 6}}
Here, y > x + 1
When x = 1 and y = 2, 2 ≯ 1 + 1
When x = 1 and y = 4, 4 > 1 + 1
When x = 1 and y = 6, 6 > 1 + 1
When x = 2 and y = 2, 2 ≯ 2 + 1
When x = 2 and y = 4, 4 > 2 + 1
When x = 2 and y = 6, 6 > 2 + 1
∴ Ordered pairs are {(1, 4), (1, 6), (2, 4), (2, 6)}

(iii) {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}
Here, x + y = 3
When x = 0, y = 3
When x = 1, y = 2
When x = 2, y = 1
When x = 3, y = 0
∴ Ordered pairs are {(0, 3), (1, 2), (2, 1), (3, 0)}

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Sets and Relations Ex 1.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Sets and Relations Ex 1.1

Question 1.
Describe the following sets in Roster form:
(i) {x / x is a letter of the word ‘MARRIAGE’}
(ii) {x / x is an integer, –\(\frac{1}{2}\) < x < \(\frac{9}{2}\)}
(iii) {x / x = 2n, n ∈ N}
Solution:
(i) Let A = {x / x is a letter of the word ‘MARRIAGE’}
∴ A = {M, A, R, I, G, E}

(ii) Let B = {x / x is an integer, –\(\frac{1}{2}\) < x < \(\frac{9}{2}\)}
∴ B = {0, 1, 2, 3, 4}

(iii) Let C = {x / x = 2n, n ∈ N}
∴ C = {2, 4, 6, 8, ….}

Question 2.
Describe the following sets in Set-Builder form:
(i) {0}
(ii) {0, ±1, ±2, ±3}
(iii) \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
Solution:
(i) Let A = {0}
0 is a whole number but it is not a natural number.
∴ A = {x / x ∈ W, x ∉ N}

(ii) Let B = {0, ±1, ±2, ±3}
B is the set of elements which belongs to Z from -3 to 3.
∴ B = {x / x ∈ Z, -3 ≤ x ≤ 3}

(iii) Let C = \(\left\{\frac{1}{2}, \frac{2}{5}, \frac{3}{10}, \frac{4}{17}, \frac{5}{26}, \frac{6}{37}, \frac{7}{50}\right\}\)
∴ C = {x / x = \(\frac{n}{n^{2}+1}\), n ∈ N, n ≤ 7}

Question 3.
If A = {x / 6x² + x – 15 = 0}, B = {x / 2x² – 5x – 3 = 0}, C = {x / 2x² – x – 3 = 0}, then find (i) (A ∪ B ∪ C) (ii) (A ∩ B ∩ C)
Solution:
A = {x / 6x² + x – 15 = o}
∴ 6x² + x – 15 = 0
∴ 6x² + 10x – 9x – 15 = 0
∴ 2x(3x + 5) – 3(3x + 5) = 0
∴ (3x + 5) (2x – 3) = 0
∴ 3x + 5 = 0 or 2x – 3 = 0
∴ x = \(\frac{-5}{3}\) or x = \(\frac{3}{2}\)
∴ A = \(\left\{\frac{-5}{3}, \frac{3}{2}\right\}\)

B = {x / 2x² – 5x – 3 = 0}
∴ 2x² – 5x – 3 = 0
∴ 2x² – 6x + x – 3 = 0
∴ 2x(x – 3) + 1(x – 3) = 0
∴ (x – 3)(2x + 1) = 0
∴ x – 3 = 0 or 2x + 1 = 0
∴ x = 3 or x = \(\frac{-1}{2}\)
∴ B = {\(\frac{-1}{2}\), 3}

C = {x / 2x² – x – 3 = 0}
∴ 2x² – x – 3 = 0
∴ 2x² – 3x + 2x – 3 = 0
∴ x(2x – 3) + 1(2x – 3) = 0
∴ (2x – 3) (x + 1) = 0
∴ 2x – 3 = 0 or x + 1 = 0
∴ x = \(\frac{3}{2}\) or x = -1
∴ C = {-1, \(\frac{3}{2}\)}

(i) A ∪ B ∪ C = \(\left\{-\frac{5}{3}, \frac{3}{2}\right\} \cup\left\{\frac{-1}{2}, 3\right\} \cup\left\{-1, \frac{3}{2}\right\}\) = \(\left\{\frac{-5}{3},-1, \frac{-1}{2}, \frac{3}{2}, 3\right\}\)

(ii) A ∩ B ∩ C = { }

Question 4.
If A, B, C are the sets for the letters in the words ‘college’, ‘marriage’ and ‘luggage’ respectively, then verify that [A – (B ∪ C)] = [(A – B) ∩ (A – C)].
Solution:
A = {c, o, l, g, e}
B = {m, a, r, i, g, e}
C = {l, u, g, a, e}
B ∪ C = {m, a, r, i, g, e, l, u}
A – (B ∪ C) = {c, o}
A – B = {c, o, l}
A – C = {c, o}
∴ [(A – B) ∩ (A – C)] = {c, o} = A – (B ∪ C)
∴ [A – (B ∪ C)] = [(A – B) ∩ (A – C)]

Question 5.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8} and universal set X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then verify the following:
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
(iii) (A ∪ B)’ = A’ ∩ B’
(iv) (A ∩ B)’ = A’ ∪ B’
(v) A = (A ∩ B) ∪ (A ∩ B’)
(vi) B = (A ∩ B) ∪ (A’ ∩ B)
(vii) n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Solution:
A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8}, X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(i) B ∩ C = {4, 5, 6}
∴ A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6} ……(i)
A ∪ B = {1, 2, 3, 4, 5, 6}
A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
∴ (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} ……(ii)
From (i) and (ii), we get
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) B ∪ C = {3, 4, 5, 6, 7, 8}
∴ A ∩ (B ∪ C) = {3, 4} …..(i)
A ∩ B = {3, 4}
A ∩ C = {4}
∴ (A ∩ B) ∪ (A ∩ C) = {3, 4} …..(ii)
From (i) and (ii), we get
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(iii) A ∪ B = {1, 2, 3, 4, 5, 6}
∴ (A ∪ B)’ = {7, 8, 9, 10} …….(i)
A’ = {5, 6, 7, 8, 9, 10}, B’ = {1, 2, 7, 8, 9, 10}
∴ A’ ∩ B’ = {7, 8, 9, 10} ……(ii)
From (i) and (ii), we get
(A ∪ B)’ = A’ ∩ B’

(iv) A ∩ B = {3, 4}
∴ (A ∩ B)’ = {1, 2, 5, 6, 7, 8, 9, 10} ……(i)
A’ = {5, 6, 7, 8, 9, 10}
B’ = {1, 2, 7, 8, 9, 10}
∴ A’ ∪ B’ = {1, 2, 5, 6, 7, 8, 9, 10} ……(ii)
From (i) and (ii), we get
(A ∩ B)’ = A’ ∪ B’

(v) A = {1, 2, 3, 4} …..(i)
A ∩ B = {3, 4}
B’ = {1, 2, 7, 8, 9, 10}
A ∩ B’ = {1, 2}
∴ (A ∩ B) ∪ (A ∩ B’) = {1, 2, 3, 4} ……(ii)
From (i) and (ii), we get
A = (A ∩ B) ∪ (A ∩ B’)

(vi) B = {3, 4, 5, 6} …..(i)
A ∩ B = {3, 4}
A’ = {5, 6, 7, 8, 9, 10}
A’ ∩ B = {5, 6}
∴ (A ∩ B) ∪ (A’ ∩ B) = {3, 4, 5, 6} …..(ii)
From (i) and (ii), we get
B = (A ∩ B) ∪ (A’ ∩ B)

(vii) A = {1, 2, 3, 4}, B = {3, 4, 5, 6},
A ∩ B = {3, 4}, A ∪ B = {1, 2, 3, 4, 5, 6}
∴ n(A) = 4, n(B) = 4,
n(A ∩ B) = 2,
n(A ∪ B) = 6 …..(i)
∴ n(A) + n(B) – n(A ∩ B) = 4 + 4 – 2
∴ n(A) + n(B) – n(A ∩ B) = 6 …..(ii)
From (i) and (ii), we get
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Question 6.
If A and B are subsets of the universal set X and n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5, find
(i) n(A ∪ B)
(ii) n(A ∩ B)
(iii) n(A’ ∩ B)
(iv) n(A ∩ B’)
Solution:
n(X) = 50, n(A) = 35, n(B) = 20, n(A’ ∩ B’) = 5
(i) n(A ∪ B) = n(X) – [n(A ∪ B)’]
= n(X) – n(A’ ∩ B’)
= 50 – 5
= 45

(ii) n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 35 + 20 – 45
= 10

(iii) n(A’ ∩ B) = n(B) – n(A ∩ B)
= 20 – 10
= 10

(iv) n(A ∩ B’) = n(A) – n(A ∩ B)
= 35 – 10
= 25

Question 7.
Out of 200 students, 35 students failed in MHT-CET, 40 in AIEEE and 40 in IIT entrance, 20 failed in MHT-CET and AIEEE, 17 in AIEEE and IIT entrance, 15 in MHT-CET and IIT entrance, and 5 failed in all three examinations. Find how many students
(i) did not fail in any examination.
(ii) failed in AIEEE or IIT entrance.
Solution:
Let A = set of students who failed in MHT-CET
B = set of students who failed in AIEEE
C = set of students who failed in IIT entrance
X = set of all students
∴ n(X) = 200, n(A) = 35, n(B) = 40, n(C) = 40,
n(A ∩ B) = 20, n(B ∩ C) = 17, n(A ∩ C) = 15, n(A ∩ B ∩ C) = 5

(i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
= 35 + 40 + 40 – 20 – 17 – 15 + 5
= 68
∴ No. of students who did not fail in any exam = n(X) – n(A ∪ B ∪ C)
= 200 – 68
= 132

(ii) No. of students who failed in AIEEE or IIT entrance = n(B ∪ C)
= n(B) + n(C) – n(B ∩ C)
= 40 + 40 – 17
= 63

Question 8.
From amongst 2000 literate individuals of a town, 70% read Marathi newspapers, 50% read English newspapers and 32.5% read both Marathi and English newspapers. Find the number of individuals who read
(i) at least one of the newspapers.
(ii) neither Marathi nor English newspaper.
(iii) only one of the newspapers.
Solution:
Let M = set of individuals who read Marathi newspapers
E = set of individuals who read English newspapers
X = set of all literate individuals
∴ n(X) = 2000,
n(M) = \(\frac{70}{100}\) × 2000 = 1400
n(E) = \(\frac{50}{100}\) × 2000 = 1000
n(M ∩ E) = \(\frac{32.5}{2}\) × 2000 = 650
n(M ∪ E) = n(M) + n(E) – n(M ∩ E)
= 1400 + 1000 – 650
= 1750

(i) No. of individuals who read at least one of the newspapers = n(M ∪ E) = 1750.
(ii) No. of individuals who read neither Marathi nor English newspaper = n(M’ ∩ E’)
= n(M ∪ E)’
= n(X) – n(M ∪ E)
= 2000 – 1750
= 250
(iii) No. of individuals who read only one of the newspapers = n(M ∩ E’) + n(M’ ∩ E)
= n(M ∪ E) – n(M ∩ E)
= 1750 – 650
= 1100

Question 9.
In a hostel, 25 students take tea, 20 students take coffee, 15 students take milk, 10 students take both tea and coffee, 8 students take both milk and coffee. None of them take tea and milk both and everyone takes atleast one beverage, find the number of students in the hostel.
Solution:
Let T = set of students who take tea
C = set of students who take coffee
M = set of students who take milk
∴ n(T) = 25, n(C) = 20, n(M) = 15,
n(T ∩ C) = 10, n(M ∩ C) = 8, n(T ∩ M) = 0, n(T ∩ M ∩ C) = 0

∴ Number of students in the hostel = n(T ∪ C ∪ M)
= n(T) + n(C) + n(M) – n(T ∩ C) – n(M ∩ C) – n(T ∩ M) + n(T ∩ M ∩ C)
= 25 + 20 + 15 – 10 – 8 – 0 + 0
= 42

Question 10.
There are 260 persons with skin disorders. If 150 had been exposed to the chemical A, 74 to the chemical B, and 36 to both chemicals A and B, find the number of persons exposed to
(i) Chemical A but not Chemical B
(ii) Chemical B but not Chemical A
(iii) Chemical A or Chemical B.
Solution:
Let A = set of persons exposed to chemical A
B = set of persons exposed to chemical B
X = set of all persons
∴ n(X) = 260, n(A) = 150, n(B) = 74, n(A ∩ B) = 36

(i) No. of persons exposed to chemical A but not to chemical B = n(A ∩ B’)
= n(A) – n(A ∩ B)
= 150 – 36
= 114

(ii) No. of persons exposed to chemical B but not to chemical A = n(A’ ∩ B)
= n(B) – n(A ∩ B)
= 74 – 36
= 38

(iii) No. of persons exposed to chemical A or chemical B = n(A ∪ B)
= n(A) + n(B) – n(A ∩ B)
= 150 + 74 – 36
= 188

Question 11.
If A = {1, 2, 3}, write the set of all possible subsets of A.
Solution:
A = {1, 2, 3}
∴ { }, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3} and {1, 2, 3} are all the possible subsets of A.

Question 12.
Write the following intervals in set-builder form:
(i) (-3, 0)
(ii) [6, 12]
(iii) (6, 12)
(iv) (-23, 5)
Solution:
(i) (-3, 0) = {x / x ∈ R, -3 < x < 0}
(ii) [6, 12] = {x / x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12) = {x / x ∈ R, 6 < x < 12}
(iv) (-23, 5) = {x / x ∈ R, -23 < x < 5}

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.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.1 Questions and Answers.

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

I. Evaluate the following limits:

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

Question 2.
\(\lim _{y \rightarrow-3}\left[\frac{y^{5}+243}{y^{3}+27}\right]\)
Solution:

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

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+6}}{x}\right]\)
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:

III. Evaluate the following limits:

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

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

Question 3.
If \(\lim _{x \rightarrow 5}\left[\frac{x^{k}-5^{k}}{x-5}\right]\) = 500, find all possible values of k.
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{8}-1}{(1-x)^{2}-1}\right]\)
Solution:

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

Question 6.
\(\lim _{y \rightarrow 1}\left[\frac{2 y-2}{\sqrt[3]{7+y}-2}\right]\)
Solution:

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

Question 8.
\(\lim _{x \rightarrow 7}\left[\frac{x^{3}-343}{\sqrt{x}-\sqrt{7}}\right]\)
Solution:

Question 9.
\(\lim _{x \rightarrow 1}\left(\frac{x+x^{3}+x^{5}+\ldots+x^{2 n-1}-n}{x-1}\right)\)
Solution:

IV. In the following examples, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

Question 1.
\(\lim _{x \rightarrow 2}(2 x+3)=7\)
Solution:
We have to find some δ so that
\(\lim _{x \rightarrow 2}(2 x+3)=7\)
Here a = 2, l = 1 and f(x) = 2x + 3
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |(2x + 3) – 7| < ∈
∴ |2x + 4| < ∈
∴ 2(x – 2)|< ∈
∴ |x – 2| < \(\frac{\epsilon}{2}\)
∴ δ ≤ \(\frac{\epsilon}{2}\) such that
|2x + 4| < δ ⇒ |f(x) – 7| < ∈

Question 2.
\(\lim _{x \rightarrow-3}(3 x+2)=-7\)
Solution:
We have to find some δ so that
\(\lim _{x \rightarrow-3}(3 x+2)=-7\)
Here a = -3, l = -7 and f(x) = 3x + 2
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |3x + 2 – (-7)| < ∈
∴ |3x + 9| < ∈
∴ |3(x + 3)| < ∈
∴ |x + 3| < \(\frac{\epsilon}{3}\)
∴ δ < \(\frac{\epsilon}{3}\) such that
|x + 3| ≤ δ ⇒ |f(x) + 7| < ∈

Question 3.
\(\lim _{x \rightarrow 2}\left(x^{2}-1\right)=3\)
Solution:
We have to find some δ > 0 such that
\(\lim _{x \rightarrow 2}\left(x^{2}-1\right)=3\)
Here, a = 2, l = 3 and f(x) = x² – 1
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |(x² – 1) – 3| < ∈
∴ |x² – 4| < ∈
∴ |(x + 2)(x – 2)| < ∈ …..(i)
We have to get rid of the factor |x + 2|
As |x – 2| < δ
-δ < x – 2 < δ
∴ 2 – δ < x < 2 + δ
Since δ can be assumed as very small, let us choose δ < 1
∴ 1 < x < 3
∴ 3 < x + 2 < 5 …..(Adding 2 throughout)
∴ |x + 2| < 5
∴ |(x + 2)(x – 2)| < 5|x – 2| ……(ii)
From (i) and (ii), we get
5|x – 2|< ∈
∴ x – 2 < \(\frac{\epsilon}{5}\)
If δ = \(\frac{\epsilon}{5}\), |x – 2| < δ ⇒ |x² – 4| < ∈
∴ We choose δ = min{\(\frac{\epsilon}{5}\), 1} then
|x – 2| < δ ⇒ |f(x) – 3| < ∈

Question 4.
\(\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)=3\)
Solution:
We have to find some δ > 0 such that
\(\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)=3\)
Here a = 1, l = 3 and f(x) = x² + x + 1
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |x² + x + 1 – 3| < ∈
∴ |x² + x – 2| < ∈
∴ |(x + 2)(x – 1)| < ∈ …..(i)
We have to get rid of the factor |x + 2|
As |x – 1| < δ
-δ < x – 1 < δ
∴ 1 – δ < x < 1 + δ
Since δ can be assumed as very small, let us choose δ < 1
∴ 0 < x < 2
∴ 2 < x + 2 < 4
∴ |x + 2| < 4
∴ |(x + 2)(x – 1)|< 4 |x – 1| …..(ii)
From (i) and (ii), we get
4|x – 1| < ∈
∴ |x – 1| < \(\frac{\epsilon}{4}\)
If δ = \(\frac{\epsilon}{4}\),
|x – 1| < δ ⇒ x² + x – 2 < ∈
∴ We choose δ = min{\(\frac{\epsilon}{4}\), 1} then
|x – 1| < δ ⇒ |f(x) – 3| < ∈

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Ex 6.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Functions Ex 6.2

Question 1.
If f(x) = 3x + 5, g(x) = 6x – 1, then find
(i) (f + g) (x)
(ii) (f – g) (2)
(iii) (fg) (3)
(iv) (f/g) (x) and its domain
Solution:
f(x) = 3x + 5, g (x) = 6x – 1
(i) (f + g) (x) = f (x) + g (x)
= 3x + 5 + 6x – 1
= 9x + 4

(ii) (f – g) (2) = f(2) – g(2)
= [3(2) + 5] – [6(2) – 1]
= 6 + 5 – 12 + 1
= 0

(iii) (fg) (3) = f (3) g(3)
= [3(3) + 5] [6(3) – 1]
= (14) (17)
= 238

(iv) \(\left(\frac{\mathrm{f}}{\mathrm{g}}\right)(x)=\frac{\mathrm{f}(x)}{\mathrm{g}(x)}=\frac{3 x+5}{6 x-1}, x \neq \frac{1}{6}\)
Domain = R – {\(\frac{1}{6}\)}

Question 2.
Let f: (2, 4, 5} → {2, 3, 6} and g: {2, 3, 6} → {2, 4} be given by f = {(2, 3), (4, 6), (5, 2)} and g = {(2, 4), (3, 4), (6, 2)}. Write down gof.
Solution:
f = {(2, 3), (4, 6), (5, 2)}
∴ f(2) = 3, f(4) = 6, f(5) = 2
g ={(2, 4), (3, 4), (6, 2)}
∴ g(2) = 4, g(3) = 4, g(6) = 2
gof: {2, 4, 5} → {2, 4}
(gof) (2) = g(f(2)) = g(3) = 4
(gof) (4) = g(f(4)) = g(6) = 2
(gof) (5) = g(f(5)) = g(2) = 4
∴ gof = {(2, 4), (4, 2), (5, 4)}

Question 3.
If f(x) = 2x² + 3, g(x) = 5x – 2, then find
(i) fog
(ii) gof
(iii) fof
(iv) gog
Solution:
f(x) = 2x² + 3, g(x) = 5x – 2
(i) (fog) (x) = f(g(x))
= f(5x – 2)
= 2(5x – 2)² + 3
= 2(25x² – 20x + 4) + 3
= 50x² – 40x + 11

(ii) (gof) (x) = g(f(x))
= g(2x² + 3)
= 5(2x + 3) – 2
= 10x² + 13

(iii) (fof) (x) = f(f(x))
= f(2x² + 3)
= 2(2x² + 3)² + 3
= 2(4x⁴ + 12x² + 9) + 3
= 8x⁴ + 24x² + 21

(iv) (gog) (x) = g(g(x))
= g(5x – 2)
= 5(5x – 2) – 2
= 25x – 12

Question 4.
Verify that f and g are inverse functions of each other, where
(i) f(x) = \(\frac{x-7}{4}\), g(x) = 4x + 7
(ii) f(x) = x³ + 4, g(x) = \(\sqrt[3]{x-4}\)
(iii) f(x) = \(\frac{x+3}{x-2}\), g(x) = \(\frac{2 x+3}{x-1}\)
Solution:
(i) f(x) = \(\frac{x-7}{4}\)
Replacing x by g(x), we get
f[g(x)] = \(\frac{g(x)-7}{4}=\frac{4 x+7-7}{4}\) = x
g(x) = 4x + 7
Replacing x by f(x), we get
g[f(x)] = 4f(x) + 7 = 4(\(\frac{x-7}{4}\)) + 7 = x
Here, f[g(x)] = x and g[f(x)] = x.
∴ f and g are inverse functions of each other.

(ii) f(x) = x³ + 4
Replacing x by g(x), we get
f[g(x)] = [g(x)]³ + 4
= \((\sqrt[3]{x-4})^{3}+4\)
= x – 4 + 4
= x
g(x) = \(\sqrt[3]{x-4}\)
Replacing x by f(x), we get
g[f(x)] = \(\sqrt[3]{f(x)-4}=\sqrt[3]{x^{3}+4-4}=\sqrt[3]{x^{3}}\) = x
Here, f[g(x)] = x and g[f(x)] = x
∴ f and g are inverse functions of each other.

(iii) f(x) = \(\frac{x+3}{x-2}\)
Replacing x by g(x), we get

Here, f[g(x)] = x and g[f(x)] = x.
∴ f and g are inverse functions of each other.

Question 5.
Check if the following functions have an inverse function. If yes, find the inverse function.
(i) f(x) = 5x²
(ii) f(x) = 8
(iii) f(x) = \(\frac{6 x-7}{3}\)
(iv) f(x) = \(\sqrt{4 x+5}\)
(v) f(x) = 9x³ + 8
(vi) f(x) =
Solution:
(i) f(x) = 5x² = y (say)

For two values (x₁ and x₂) of x, values of the function are equal.
∴ f is not one-one.
∴ f does not have an inverse.

(ii) f(x) = 8 = y (say)
For every value of x, the value of the function f is the same.
∴ f is not one-one i.e. (many-one) function.
∴ f does not have the inverse.

(iii) f(x) = \(\frac{6 x-7}{3}\)
Let f(x₁) = f(x₂)
∴ \(\frac{6 x_{1}-7}{3}=\frac{6 x_{2}-7}{3}\)
∴ x₁ = x₂
∴ f is a one-one function.
f(x) = \(\frac{6 x-7}{3}\) = y (say)
∴ x = \(\frac{3y+7}{6}\)
∴ For every y, we can get x
∴ f is an onto function.
∴ x = \(\frac{3y+7}{6}\) = f^-1 (y)
Replacing y by x, we get
f-1 (x) = \(\frac{3x+7}{6}\)

(iv) f(x) = \(\sqrt{4 x+5}, x \geq \frac{-5}{4}\)
Let f(x₁) = f(x₂)
∴ \(\sqrt{4 x_{1}+5}=\sqrt{4 x_{2}+5}\)
∴ x₁ = x₂
∴ f is a one-one function.
f(x) = \(\sqrt{4 x+5}\) = y, (say) y ≥ 0
Squaring on both sides, we get
y² = 4x + 5
∴ x = \(\frac{y^{2}-5}{4}\)
∴ For every y we can get x.
∴ f is an onto function.
∴ x = \(\frac{y^{2}-5}{4}\) = f^-1 (y)
Replacing y by x, we get
f^-1 (x) = \(\frac{x^{2}-5}{4}\)

(v) f(x) 9x³ + 8
Let f(x₁) = f(x₂)
∴ \(9 x_{1}^{3}+8=9 x_{2}^{3}+8\)
∴ x₁ = x₂
∴ f is a one-one function.
∴ f(x) = 9x³ + 8 = y, (say)
∴ x = \(\sqrt[3]{\frac{y-8}{9}}\)
∴ For every y we can get x.
∴ f is an onto function.
∴ x = \(\sqrt[3]{\frac{y-8}{9}}\) = f^-1 (y)
Replacing y by x, we get
f^-1 (x) = \(\sqrt[3]{\frac{x-8}{9}}\)

(vi) f(x) = x + 7, x < 0
= 8 – x, x ≥ 0

We observe from the graph that for two values of x, say x₁, x₂ the values of the function are equal.
i.e. f(x₁) = f(x₂)
∴ f is not one-one (i.e. many-one) function.
∴ f does not have inverse.

Question 6.
If f(x) = , then find
(i) f(3)
(ii) f(2)
(iii) f(0)
Solution:
f(x) = x² + 3, x ≤ 2
= 5x + 7, x > 2
(i) f(3) = 5(3) + 7
= 15 + 7
= 22

(ii) f(2) = 2² + 3
= 4 + 3
= 7

(iii) f(0) = 0² + 3 = 3

Question 7.
If f(x) = , then find
(i) f(-4)
(ii) f(-3)
(iii) f(1)
(iv) f(5)
Solution:
f(x) = 4x – 2, x ≤ -3
= 5, -3 < x < 3
= x², x ≥ 3
(i) f(-4) = 4(-4) – 2
= -16 – 2
= -18

(ii) f(-3) = 4(-3) – 2
= -12 – 2
= -14

(iii) f(1) = 5

(iv) f(5) = 5² = 25

Question 8.
If f(x) = 2 |x| + 3x, then find
(i) f(2)
(ii) f(-5)
Solution:
f(x) = 2 |x| + 3x
(i) f(2) = 2|2| + 3(2)
= 2 (2) + 6 ….. [∵ |x| = x, x > 0]
= 10

(ii) f(-5) = 2 |-5| + 3(-5)
= 2(5) – 15 …..[∵ |x| = -x, x < 0]
= 10 – 15
= -5

Question 9.
If f(x) = 4[x] – 3, where [x] is greatest integer function of x, then find
(i) f(7.2)
(ii) f(0.5)
(iii) \(f\left(-\frac{5}{2}\right)\)
(iv) f(2π), where π = 3.14
Solution:
f(x) = 4[x] – 3
(i) f(7.2) = 4 [7.2] – 3
= 4(7) – 3 ………[∵ 7 ≤ 7.2 < 8, [7.2] = 7]
= 25

(ii) f(0.5) = 4[0.5] – 3
= 4(0) – 3 ………[∵ 0 ≤ 0.5 < 1, [0.5] = 0]
= -3

(iii) \(f\left(-\frac{5}{2}\right)\) = f(-2.5)
= 4[-2.5] – 3
= 4(-3) – 3 …….[∵-3 ≤ -2.5 ≤ -2, [-2.5] = -3]
= -15

(iv) f(2π) = 4[2π] – 3
= 4[6.28] – 3 …..[∵ π = 3.14]
= 4(6) – 3 …….[∵ 6 ≤ 6.28 < 7, [6.28] = 6]
= 21

Question 10.
If f(x) = 2{x} + 5x, where {x} is fractional part function of x, then find
(i) f(-1)
(ii) f(\(\frac{1}{4}\))
(iii) f(-1.2)
(iv) f(-6)
Solution:
f(x) = 2{x} + 5x
(i) {-1} = -1 – [-1] = -1 + 1 = 0
∴ f(-1) = 2 {-1} + 5(-1)
= 2(0) – 5
= -5

(ii) {\(\frac{1}{4}\)} = \(\frac{1}{4}\) – [latex]\frac{1}{4}[/latex] = \(\frac{1}{4}\) – 0 = \(\frac{1}{4}\)
f(\(\frac{1}{4}\)) = 2{\(\frac{1}{4}\)} + 5(\(\frac{1}{4}\))
= 2(\(\frac{1}{4}\)) + \(\frac{5}{4}\)
= \(\frac{7}{4}\)
= 1.75

(iii) {-1.2} = -1.2 – [-1.2] = -1.2 + 2 = 0.8
f(-1.2) = 2{-1.2} + 5(-1.2)
= 2(0.8) + (-6)
= -4.4

(iv) {-6} = -6 – [-6] = -6 + 6 = 0
f(-6) = 2{-6} + 5(-6)
= 2(0) – 30
= -30

Question 11.
Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function.
(i) |x + 4| ≥ 5
(ii) |x – 4| + |x – 2| = 3
(iii) x² + 7|x| + 12 = 0
(iv) |x| ≤ 3
(v) 2|x| = 5
(vi) [x + [x + [x]]] = 9
(vii) {x} > 4
(viii) {x} = o
(ix) {x} = 0.5
(x) 2{x} = x + [x]
Solution:
(i) |x + 4| ≥ 5
The solution of |x| ≥ a is x ≤ -a or x ≥ a
∴ |x + 4| ≥ 5 gives
∴ x + 4 ≤ -5 or x + 4 ≥ 5
∴ x ≤ -5 – 4 or x ≥ 5 – 4
∴ x ≤ -9 or x ≥ 1
∴ The solution set = (-∞, – 9] ∪ [1, ∞)

(ii) |x – 4| + |x – 2| = 3 …..(i)
Case I: x < 2
Equation (i) reduces to
4 – x + 2 – x = 3 …….[x < 2 < 4, x – 4 < 0, x – 2 < 0]
∴ 6 – 3 = 2x
∴ x = \(\frac{3}{2}\)

Case II: 2 ≤ x < 4
Equation (i) reduces to
4 – x + x – 2 = 3
∴ 2 = 3 (absurd)
There is no solution in [2, 4)

Case III: x ≥ 4
Equation (i) reduces to
x – 4 + x – 2 = 3
∴ 2x = 6 + 3 = 9
∴ x = \(\frac{9}{2}\)
∴ x = \(\frac{3}{2}\), \(\frac{9}{2}\) are solutions.
The solution set = {\(\frac{3}{2}\), \(\frac{9}{2}\)}

(iii) x² + 7|x| + 12 = 0
∴ (|x|)² + 7|x| + 12 = 0
∴ (|x| + 3) (|x| + 4) = 0
∴ There is no x that satisfies the equation.
The solution set = { } or Φ

(iv) |x| ≤ 3 The solution set of |x| ≤ a is -a ≤ x ≤ a
∴ The required solution is -3 ≤ x ≤ 3
∴ The solution set is [-3, 3]

(v) 2|x| = 5
∴ |x| = \(\frac{5}{2}\)
∴ x = ±\(\frac{5}{2}\)

(vi) [x + [x + [x]]] = 9
∴ [x + [x] + [x] ] = 9 …….[[x + n] = [x] + n, if n is an integer]
∴ [x + 2[x]] = 9
∴ [x] + 2[x] = 9 …..[[2[x] is an integer]]
∴ [x] = 3
∴ x ∈ [3, 4)

(vii) {x} > 4
This is a meaningless statement as 0 ≤ {x} < 1
∴ The solution set = { } or Φ

(viii) {x} = 0
∴ x is an integer
∴ The solution set is Z.

(ix) {x} = 0.5
∴ x = ….., -2.5, -1.5, -0.5, 0.5, 1.5, …..
∴ The solution set = {x : x = n + 0.5, n ∈ Z}

(x) 2{x} = x + [x]
= [x] + {x} + [x] ……[x = [x] + {x}]
∴ {x} = 2[x]
R.H.S. is an integer
∴ L.H.S. is an integer
∴ {x} = 0
∴ [x] = 0
∴ x = 0