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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Ex 2.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

Question 1.
Find the equations of tangents and normals to the curve at the point on it.
(i) y = x² + 2e^x + 2 at (0, 4)
Solution:

(ii) x³ + y³ – 9xy = 0 at (2, 4)
Solution:
x³ + y³ – 9xy = 0
Differentiating both sides w.r.t. x, we get


Hence, the equations of tangent and normal are 4x – 5y + 12 = 0 and 5x + 4y – 26 = 0 respectively.

(iii) x² – √3xy + 2y² = 5 at (√3, 2)
Solution:
x² – √3xy + 2y² = 5
Differentiating both sides w.r.t. x, we get

the slope of normal at (√3, 2) does not exist.
normal is parallel to Y-axis.
equation of the normal is of the form x = k
Since, it passes through the point (√3, 2), k = √3
equation of the normal is x = √3.
Hence, the equations of tangent and normal are y = 2 and x = √3 respectively.

(iv) 2xy + π sin y = 2π at (1, \(\frac{\pi}{2}\))
Solution:
2xy + π sin y = 2π
Differentiating both sides w.r.t. x, we get


Hence, the equations of tangent and normal are πx + 2y – 2π = 0 and 4x – 2πy + π² – 4 = 0 respectively.

(v) x sin 2y = y cos 2x at (\(\frac{\pi}{4}\), \(\frac{\pi}{2}\))
Solution:
x sin 2y = y cos 2x
Differentiating both sides w.r.t. x, we get



Hence, the equations of the tangent and normal are 2x – y = 0 and 4x + 8y – 5π = 0 respectively.

(vi) x = sin θ and y = cos 2θ at θ = \(\frac{\pi}{6}\)
Solution:
When θ = \(\frac{\pi}{6}\), x = sin\(\frac{\pi}{6}\) and y = cos\(\frac{\pi}{3}\)
∴ x = \(\frac{1}{2}\) and y = \(\frac{1}{2}\)
Hence, the point at which we want to find the equations of tangent and normal is (\(\frac{1}{2}\), \(\frac{1}{2}\))
Now, x = sin θ, y = cos 2θ
Differentiating x and y w.r.t. θ, we get


2y – 1 = x – \(\frac{1}{2}\)
4y – 2 = 2x – 1
2x – 4y + 1 = 0
Hence, equations of the tangent and normal are 4x + 2y – 3 = 0 and 2x – 4y + 1 = 0 respectively.

(vii) x = √t, y = t – \(\frac{1}{\sqrt{t}}\), at t = 4.
Solution:
When t = 4, x = √4 and y = 4 – \(\frac{1}{\sqrt{4}}\)
∴ x = 2 and y = 4 – \(\frac{1}{2}\) = \(\frac{7}{2}\)
Hence, the point at which we want to find the equations of tangent and normal is (2, \(\frac{7}{2}\)).
Now, x = √t, y = t – \(\frac{1}{\sqrt{t}}\)
Differentiating x and y w.r.t. t, we get



Hence, the equations of tangent and normal are 17x – 4y – 20 = 0 and 8x + 34y – 135 = 0 respectively.

Question 2.
Find the point of the curve y = \(\sqrt{x-3}\) where the tangent is perpendicular to the line 6x + 3y – 5 = 0.
Solution:
Let the required point on the curve y = \(\sqrt{x-3}\) be P(x₁, y₁).
Differentiating y = \(\sqrt{x-3}\) w.r.t. x, we get

Hence, the required points are (4, 1) and (4, -1).

Question 3.
Find the points on the curve y = x³ – 2x² – x where the tangents are parallel to 3x – y + 1 = 0.
Solution:
Let the required point on the curve y = x³ – 2x² – x be P(x₁, y₁).

Question 4.
Find the equations of the tangents to the curve x² + y² – 2x – 4y + 1 = 0 which are parallel to the X-axis.
Solution:
Let P (x₁, y₁) be the point on the curve x² + y² – 2x – 4y + 1 = 0 where the tangent is parallel to X-axis.
Differentiating x² + y² – 2x – 4y + 1 = 0 w.r.t. x, we get


the coordinates of the points are (1, 0) or (1, 4)
Since the tangents are parallel to X-axis, their equations are of the form y = k
If it passes through the point (1, 0), k = 0, and if it passes through the point (1, 4), k = 4
Hence, the equations of the tangents are y = 0 and y = 4.

Question 5.
Find the equations of the normals to the curve 3x² – y² = 8, which are parallel to the line x + 3y = 4.
Solution:
Let P(x₁, y₁) be the foot of the required normal to the curve 3x² – y² = 8.
Differentiating 3x² – y² = 8 w.r.t. x, we get


Hence, the equations of the normals are x + 3y – 8 = 0 and x + 3y + 8 = 0.

Question 6.
If the line y = 4x – 5 touches the curve y² = ax³ + b at the point (2, 3), find a and b.
Solution:
y² = ax³ + b
Differentiating both sides w.r.t. x, we get

= slope of the tangent at (2, 3)
Since, the line y = 4x – 5 touches the curve at the point (2, 3), slope of the tangent at (2, 3) is 4.
2a = 4 ⇒ a = 2
Since (2, 3) lies on the curve y² = ax³ + b
(3)² = a(2)³ + b
9 = 8a + b
9 = 8(2) + b …… [∵ a = 2]
b = -7
Hence, a = 2 and b = -7.

Question 7.
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which y-coordinate is changing 8 times as fast as the x-coordinate.
Solution:
Let P(x₁, y₁) be the point on the curve 6y = x³ + 2 whose y-coordinate is changing 8 times as fast as the x-coordinate.

Question 8.
A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area increasing, when its radius is 5 cm?
Solution:
Let r be the radius and S be the surface area of the soap bubble at any time t.
Then S = 4πr²
Differentiating w.r.t. t, we get

Hence, the surface area of the soap bubble is increasing at the rate of 0.87c cm² / sec.

Question 9.
The surface area of a spherical balloon is increasing at the rate of 2 cm²/sec. At what rate is the volume of the balloon is increasing, when the radius of the balloon is 6 cm?
Solution:
Let r be the radius, S be the surface area and V be the volume of the spherical balloon at any time t.
Then S = 4πr² and V = \(\frac{4}{3} \pi r^{3}\)
Differentiating w.r.t. t, we get

Hence, the volume of the spherical balloon is increasing at the rate of 6 cm³ / sec.

Question 10.
If each side of an equilateral triangle increases at the rate of √2 cm/sec, find the rate of increase of its area when its side of length is 3 cm.
Solution:
If x cm is the side of the equilateral triangle and A is its area, then \(A=\frac{\sqrt{3}}{4} x^{2}\)
Differentiating w.r.t. f, we get

Hence, rate of increase of the area of equilateral triangle = \(\frac{3 \sqrt{6}}{2}\) cm² / sec.

Question 11.
The volume of a sphere increases at the rate of 20 cm³/sec. Find the rate of change of its surface area, when its radius is 5 cm.
Solution:
Let r be the radius, S be the surface area and V be the volume of the sphere at any time t.
Then S = 4πr² and V = \(\frac{4}{3} \pi r^{3}\)
Differentiating w.r.t. t, we get

Hence, the surface area of the sphere is changing at the rate of 8 cm²/sec.

Question 12.
The edge of a cube is decreasing at the rate of 0.6 cm/sec. Find the rate at which its volume is decreasing, when the edge of the cube is 2 cm.
Solution:
Let x be the edge of the cube and V be its volume at any time t.
Then V = x³
Differentiating both sides w.r.t. t, we get

Hence, the volume of the cube is decreasing at the rate of 7.2 cm³/sec.

Question 13.
A man of height 2 meters walks at a uniform speed of 6 km/hr away from a lamp post of 6 meters high. Find the rate at which the length of the shadow is increasing.
Solution:
Let OA be the lamp post, MN the man, MB = x, his shadow, and OM = y, the distance of the man from the lamp post at time t.

Then \(\frac{d y}{d t}\) = 6 km/hr is the rate at which the man is moving at away from the lamp post.
\(\frac{d x}{d t}\) is the rate at which his shadow is increasing.
From the figure,
\(\frac{x}{2}=\frac{x+y}{6}\)
6x = 2x + 2y
4x = 2y
x = \(\frac{1}{2}\) y
\(\frac{d x}{d t}=\frac{1}{2} \frac{d y}{d t}=\frac{1}{2} \times 6=3 \mathrm{~km} / \mathrm{hr}\)
Hence, the length of the shadow is increasing at the rate of 3 km/hr.

Question 14.
A man of height 1.5 meters walks towards a lamp post of height 4.5 meters, at the rate of (\(\frac{3}{4}\)) meter/sec.
Find the rate at which
(i) his shadow is shortening
(ii) the tip of the shadow is moving.
Solution:

Let OA be the lamp post, MN the man, MB = x his shadow and OM = y the distance of the man from lamp post at time t.
Then \(\frac{d y}{d t}=\frac{3}{4}\) is the rate at which the man is moving towards the lamp post.
\(\frac{d x}{d t}\) is the rate at which his shadow is shortening.
B is the tip of the shadow and it is at a distance of x + y from the post.
\(\frac{d}{d t}(x+y)=\frac{d x}{d t}+\frac{d y}{d t}\) is the rate at which the tip of the shadow is moving.
From the figure,
\(\frac{x}{1.5}=\frac{x+y}{4.5}\)
45x = 15x + 15y
30x = 15y
x = \(\frac{1}{2}\)y
\(\frac{d x}{d t}=\frac{1}{2} \cdot \frac{d y}{d t}=\frac{1}{2}\left(\frac{3}{4}\right)=\left(\frac{3}{8}\right) \text { metre/sec }\)
and \(\frac{d x}{d t}+\frac{d y}{d t}=\frac{3}{8}+\frac{3}{4}=\left(\frac{9}{8}\right) \text { metres } / \mathrm{sec}\)
Hence (i) the shadow is shortening at the rate of (\(\frac{3}{8}\)) metre/sec, and
(ii) the tip of shadow is moving at the rate of (\(\frac{9}{8}\)) metres/sec.

Question 15.
A ladder 10 metres long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 metres per second, find how fast the top of the ladder is sliding down the wall, when the bottom is 6 metres away from the wall.
Solution:

Let AB be the ladder, where AB = 10 metres.
Let at time t seconds, the end A of the ladder be x metres from the wall and the end B be y metres from the ground.
Since, OAB is a right angled triangle, by Pythagoras’ theorem
x² + y² = 10² i.e. y² = 100 – x²
Differentiating w.r.t. t, we get
2y \(\frac{d y}{d t}\) = 0 – 2x \(\frac{d x}{d t}\)
∴ \(\frac{d y}{d t}=-\frac{x}{y} \cdot \frac{d x}{d t}\) ……..(1)
Now, \(\frac{d x}{d t}\) = 1.2 metres/sec is the rate at which the bottom at of the ladder is pulled horizontally and \(\frac{d y}{d t}\) is the rate at which the top of ladder B is sliding.
When x = 6, y² = 100 – 36 = 64
y = 8
(1) gives \(\frac{d y}{d t}=-\frac{6}{8}(1.2)=-\frac{6}{8} \times \frac{12}{10}\)
\(=-\frac{9}{10}=-0.9\)
Hence, the top of the ladder is sliding down the wall, at the rate of 0.9 metre/sec.

Question 16.
If water is poured into an inverted hollow cone whose semi-vertical angle is 30° so that its depth (measured along the axis) increases at the rate of 1 cm/sec. Find the rate at which the volume of water increases when the depth is 2 cm.
Solution:

Let r be the radius, h be the height, θ be the semi-vertical angle and V be the volume of the water at any time t.

Hence, the volume of water is increasing at the rate of \(\left(\frac{4 \pi}{3}\right)\) cm3/sec.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.3

Question 1.
Find two unit vectors each of which is perpendicular to both
\(\bar{u}\) and \(\bar{v}\), where \(\bar{u}\) = \(2 \hat{i}+\hat{j}-2 \hat{k}\), \(\bar{v}\) = \(\hat{i}+2 \hat{j}-2 \hat{k}\)
Solution:
Let \(\bar{u}\) = \(2 \hat{i}+\hat{j}-2 \hat{k}\)
\(\bar{v}\) = \(\hat{i}+2 \hat{j}-2 \hat{k}\)

Question 2.
If \(\bar{a}\) and \(\bar{b}\) are two vectors perpendicular to each other, prove that (\(\bar{a}\) + \(\bar{b}\))² = (\(\bar{a}\) – \(\bar{b}\))²
Solution:
\(\bar{a}\) and \(\bar{b}\) are perpendicular to each other.

∴ LHS = RHS
Hence, (\(\bar{a}\) + \(\bar{b}\))² = (\(\bar{a}\) – \(\bar{b}\))².

Question 3.
Find the values of c so that for all real x the vectors \(x c \hat{i}-6 \hat{j}+3 \hat{k}\) and \(x \hat{i}+2 \hat{j}+2 c x \hat{k}\) make an obtuse angle.
Solution:
Let \(\bar{a}\) = \(x c \hat{i}-6 \hat{j}+3 \hat{k}\) and \(\bar{b}\) = \(x \hat{i}+2 \hat{j}+2 c x \hat{k}\)
Consider \(\bar{a}\)∙\(\bar{b}\) = \((x c \hat{i}-6 \hat{j}+3 \hat{k}) \cdot(x \hat{i}+2 \hat{j}+2 c x \hat{k})\)
= (xc)(x) + (-6)(2) + (3)(2cx)
= cx² – 12 + 6cx
= cx² + 6cx – 12
If the angle between \(\bar{a}\) and \(\bar{b}\) is obtuse, \(\bar{a}\)∙\(\bar{b}\) < 0.
∴ cx² + 6cx – 12 < 0
∴ cx² + 6cx < 12

∴ c < 0.
Hence, the angle between a and b is obtuse if c < 0.

Question 4.
Show that the sum of the length of projections of \(\hat{p} \hat{i}+q \hat{j}+r \hat{k}\) on the coordinate axes, where p = 2, q = 3 and r = 4, is 9.
Solution:
Let \(\bar{a}\) = \(\hat{p} \hat{i}+q \hat{j}+r \hat{k}\)
Projection of \(\bar{a}\) on X-axis

Similarly, projections of \(\bar{a}\) on Y- and Z-axes are 3 and 4 respectively.
∴ sum of these projections = 2 + 3 + 4 = 9.

Question 5.
Suppose that all sides of a quadrilateral are equal in length and opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular.
Solution:


∵ \(\overline{\mathrm{AC}}\), \(\overline{\mathrm{BD}}\) are non-zero vectors
∴ \(\overline{\mathrm{AC}}\) is perpendicular to \(\overline{\mathrm{BD}}\)
Hence, the diagonals are perpendicular.

Question 6.
Determine whether \(\bar{a}\) and \(\bar{b}\) are orthogonal, parallel or neither.
(i) \(\bar{a}\) = \(-9 \hat{i}+6 \hat{j}+15 \hat{k}\), \(\bar{b}\) = \(6 \hat{i}-4 \hat{j}-10 \hat{k}\)
Solution:
\(\bar{a}\) = \(-9 \hat{i}+6 \hat{j}+15 \hat{k}\) = -3\((3 \hat{i}-2 \hat{j}-5 \hat{k})\)
= \(-\frac{3}{2}(6 \hat{i}-4 \widehat{j}-19 \hat{k})\)
∴ \(\bar{a}\) = \(-\frac{3}{2} \bar{b}\)
i.e. \(\bar{a}\) is a non-zero scalar multiple of \(\bar{b}\)
Hence, \(\bar{a}\) is parallel to \(\bar{b}\).

(ii) \(\bar{a}\) = \(2 \hat{i}+3 \hat{j}-\hat{k}\), \(\bar{b}\) = \(5 \hat{i}-2 \hat{j}+4 \hat{k}\)
Solution:
\(\bar{a} \cdot \bar{b}\) = \((2 \hat{i}+3 \hat{j}-\hat{k}) \cdot(5 \hat{i}-2 \hat{j}+4 \hat{k})\)
= (2)(5) + (3)(-2) + (-1)(4)
= 10 – 6 – 4 = 0
Since, \(\bar{a}\), \(\bar{b}\) are non-zero vectors and \(\bar{a} \cdot \bar{b}\) = 0, \(\bar{a}\) is orthogonal to \(\bar{b}\).

(iii) \(\bar{a}\) = \(-\frac{3}{5} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{3} \hat{k}\), \(\bar{b}\) = \(5 \hat{i}+4 \hat{j}+3 \hat{k}\).
Solution:

= -3 + 2 + 1
= 0
Since, \(\bar{a}\), \(\bar{b}\) are non-zero vectors and \(\bar{a} \cdot \bar{b}\) = 0
\(\bar{a}\) is orthogonal to \(\bar{b}\).

(iv) \(\bar{a}\) = \(4 \hat{i}-\hat{j}+6 \hat{k}\), \(\bar{a}\) = \(5 \hat{i}-2 \hat{j}+4 \hat{k}\)
Solution:
\(\bar{a} \cdot \bar{b}\) = \((4 \hat{i}-\hat{j}+6 \hat{k}) \cdot(5 \hat{i}-2 \hat{j}+4 \hat{k})\)
= (4)(5) + (-1)(-2) + (6)(4)
= 20 + 2 + 24
= 46 ≠ 0
∴ \(\bar{a}\) is not orthogonal to \(\bar{b}\).
It is clear that \(\bar{a}\) is not a scalar multiple of \(\bar{b}\).
∴ \(\bar{a}\) is not parallel to \(\bar{b}\).
Hence, \(\bar{a}\) is neither parallel nor orthogonal to \(\bar{b}\).

Question 7.
Find the angle P of the triangle whose vertices are P(0, -1, -2), Q(3, 1, 4) and R(5, 7, 1).
Solution:
The position vectors \(\bar{p}\), \(\bar{q}\) and \(\bar{r}\) of the points P(0, -1, -2), Q(3, 1, 4) and R(5, 7, 1) are


∴ P = 45°

Question 8.
If\(\hat{p}\), \(\hat{q}\), and \(\hat{r}\) are unit vectors, find

(i) \(\hat{p} \cdot \hat{q}\)
Solution:

Let the triangle be denoted by ABC,
where \(\overline{\mathrm{AB}}\) = \(\bar{p}\), \(\overline{\mathrm{AC}}\) = \(\bar{q}\) and \(\overline{\mathrm{BC}}\) = \(\bar{r}\)
∵ \(\bar{p}\), \(\bar{r}\), \(\bar{r}\) are unit vectors.
∴ l(AB) = l(BC) = l(CA) = 1
∴ the triangle is equilateral
∴ ∠A = ∠B = ∠C = 60°
(i) Using the formula for angle between two vectors,

(ii) \(\hat{p} \cdot \hat{r}\)
Solution:

Question 9.
Prove by vector method that the angle subtended on semicircle is a right angle.
Solution:
Let seg AB be a diameter of a circle with centre C and P be any point on the circle other than A and B. Then ∠APB is an angle subtended on a semicircle.
Let \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{CB}}\) = \(\bar{a}\) and \(\overline{\mathrm{CP}}\) = \(\bar{r}\).
Then|\(\bar{a}\)| = |\(\bar{r}\)| …(1)

Hence, the angle subtended on a semicircle is the right angle.

Question 10.
If a vector has direction angles 45ºand 60º find the third direction angle.
Solution:
Let α = 45°, β = 60°
We have to find γ.
∴ cos²α + cos²β + cos²γ = 1
∴ cos²45° + cos²60° + cos²γ = 1

Hence, the third direction angle is \(\frac{\pi}{3}\) or \(\frac{2 \pi}{3}\).

Question 11.
If a line makes angles 90º, 135º, 45º with the X, Y and Z axes respectively, then find its direction cosines.
Solution:
Let l, m, n be the direction cosines of the line.
Then l = cos α, m = cos β, n = cos γ
Here, α = 90°, β = 135° and γ = 45°
∴ l = cos 90° = 0
m = cos 135° = cos (180° – 45°) = -cos 45° = \(-\frac{1}{\sqrt{2}}\) and n = cos 45° = \(\frac{1}{\sqrt{2}}\)
∴ the direction cosines of the line are 0, \(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\).

Question 12.
If a line has the direction ratios, 4, -12, 18 then find its direction cosines.
Solution:
The direction ratios of the line are a = 4, b = -12, c = 18.
Let l, m, n be the direction cosines of the line.

Question 13.
The direction ratios of \(\overline{A B}\) are -2, 2, 1. If A = (4, 1, 5) and l(AB) = 6 units, find B.
Solution:
The direction ratio of \(\overline{A B}\) are -2, 2, 1.
∴ the direction cosines of \(\overline{A B}\) are

The coordinates of the points which are at a distance of d units from the point (x₁, y₁, z₁) are given by (x₁ ± ld, y₁ ± md, z₁ ± nd)
Here x₁ = 4, y₁ = 1, z₁ = 5, d = 6, l = \(-\frac{2}{3}\), m = \(\frac{2}{3}\), n = \(\frac{1}{3}\)
∴ the coordinates of the requited points are
(4 ± \(\left(-\frac{2}{3}\right)\)6, 1 ± \(\frac{2}{3}\)(6), 5 ± \(\frac{1}{3}\)(6))
i.e. (4 – 4, 1 + 4, 5 + 2) and (4 + 4, 1 – 4, 5 – 2)
i.e. (0, 5, 7) and (8, -3, 3).

Question 14.
Find the angle between the lines whose direction cosines l, m, n satisfy the equations 5l + m + 3n = 0 and 5mn – 2nl + 6lm = 0.
Solution:
Given, 5l + m + 3n = 0 …(1)
and 5mn – 2nl + 6lm = 0 …(2)
From (1), m = -(51 + 3n)
Putting the value of m in equation (2), we get,
-5(5l + 3n)n – 2nl – 6l(5l + 3n) = 0
∴ -25ln – 15n² – 2nl – 30l² – 18ln = 0
∴ – 30l² – 45ln – 15n² = 0
∴ 2l² + 3ln + n² = 0
∴ 2l² + 2ln + ln + n² = 0
∴ 2l(l + n) + n(l + n) = 0
∴ (l + n)(2l + n) = 0
∴ l + n = 0 or 2l + n = 0
l = -n or n = -2l
Now, m = -(5l + 3n), therefore, if l = -n,
m = -(-5n + 3n) = 2n
∴ -l = \(\frac{m}{2}\) = n
∴ \(\frac{l}{-1}=\frac{m}{2}=\frac{n}{1}\)
∴ the direction ratios of the first line are
a₁ = -1, b₁ = 2, c₁ = 1
If n = -2l, m = -(5l – 6l) – l
∴ l = m = \(\frac{n}{-2}\)
∴ \(\frac{l}{1}=\frac{m}{1}=\frac{n}{-2}\)
∴ the direction ratios of the second line are
a₂ = -1, b₂ = 1, c₂ = -2
Let θ be the angle between the lines.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.2

Question 1.
Find the position vector of point R which divides the line joining the points P and Q whose position vectors are \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(-5 \hat{i}+2 \hat{j}-5 \hat{k}\) in the ratio 3 : 2
(i) internally
Solution:
It is given that the points P and Q have position vectors \(\bar{p}\) = \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\bar{p}\) = \(-5 \hat{i}+2 \hat{j}-5 \hat{k}\) respectively.
(i) If R(\(\bar{r}\)) divides the line segment PQ internally in the ratio 3 : 2, by section formula for internal division,

(ii) externally.
Solution:
If R(\(\bar{r}\)) divides the line segment joining P and Q externally in the ratio 3 : 2, by section formula for external division,

∴ coordinates of R = (-19, 8, -21).
Hence, the position vector of R is \(-19 \hat{i}+8 \hat{j}-21 \hat{k}\) and coordinates of R are (-19, 8, -21).

Question 2.
Find the position vector of midpoint M joining the points L (7, -6, 12) and N (5, 4, -2).
Solution:
The position vectors \(\bar{l}\) and \(\bar{n}\) of the points L(7, -6, 12) and N (5, 4, -2) are given by

∴ coordinates of M = (6, -1, 5).
Hence, position vector of M is \(6 \hat{i}-\hat{j}+5 \hat{k}\) and the coordinates of M are (6, -1, 5).

Question 3.
If the points A(3, 0, p), B (-1, q, 3) and C(-3, 3, 0) are collinear, then find
(i) The ratio in which the point C divides the line segment AB.
Solution:
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be the position vectors of A, B and C respectively.
Then \(\bar{a}\) = \(3 \hat{i}+0 \cdot \hat{j}+p \hat{k}\), \(\bar{b}\) = \(-\hat{i}+q \hat{j}+3 \hat{k}\) and \(\bar{c}\) = \(-3 \hat{i}+3 \hat{j}+0 \hat{k}\).
As the points A, B, C are collinear, suppose the point C divides line segment AB in the ratio λ : 1.
∴ by the section formula,

By equality of vectors, we have,
-3(λ + 1) = -λ + 3 … (1)
3(λ + 1 ) = λ q … (2)
0 = 3λ + p … (3)
From equation (1), -3λ – 3 = -λ + 3
∴ -2λ = 6 ∴ λ = -3
∴ C divides segment AB externally in the ratio 3 : 1.

(ii) The values of p and q.
Solution:
Putting λ = -3 in equation (2), we get
3(-3 + 1) = -3q
∴ -6 = -3q ∴ q = 2
Also, putting λ = -3 in equation (3), we get
0 = -9 + p ∴ p = 9
Hence p = 9 and q = 2.

Question 4.
The position vector of points A and B are 6\(\bar{a}\) + 2\(\bar{b}\) and \(\bar{a}\) – 3\(\bar{b}\). If the point C divides AB in the ratio 3 : 2 then show that the position vector of C is 3\(\bar{a}\) – \(\bar{b}\).
Solution:
Let \(\bar{c}\) be the position vector of C.
Since C divides AB in the ratio 3 : 2,

Hence, the position vector of C is 3\(\bar{a}\) – \(\bar{b}\).

Question 5.
Prove that the line segments joining mid-point of adjacent sides of a quadrilateral form a parallelogram.
Solution:
Let ABCD be a quadrilateral and P, Q, R, S be the midpoints of the sides AB, BC, CD and DA respectively.
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\), \(\bar{p}\), \(\bar{q}\), \(\bar{r}\) and s be the position vectors of the points A, B, C, D, P, Q, R and S respectively.
Since P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively,


∴ □PQRS is a parallelogram.

Question 6.
D and E divide sides BC and CA of a triangle ABC in the ratio 2 : 3 respectively. Find the position vector of the point of intersection of AD and BE and the ratio in which this point divides AD and BE.
Solution:

Let AD and BE intersect at P.
Let A, B, C, D, E, P have position vectos \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\), \(\bar{e}\), \(\bar{p}\) respectively.
D and E divide segments BC and CA internally in the ratio 2 : 3.
By the section formula for internal division,

LHS is the position vector of the point which divides segment AD internally in the ratio 15 : 4.
RHS is the position vector of the point which divides segment BE internally in the ratio 10 : 9.
But P is the point of intersection of AD and BE.
∴ P divides AD internally in the ratio 15 : 4 and P divides BE internally in the ratio 10 : 9.
Hence, the position vector of the point of interaction of
AD and BE is \(\bar{p}\) = \(\frac{15 \bar{d}+4 \bar{a}}{19}=\frac{10 \bar{e}+9 \bar{b}}{19}\) and it divides AD internally in the ratio 15 : 4 and BE internally in the ratio 10 : 9.

Question 7.
Prove that a quadrilateral is a parallelogram if and only if its diagonals bisect each other.
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) be respectively the position vectors of the vertices A, B, C and D of the parallelogram ABCD. Then AB = DC and side AB || side DC.

The position vectors of the midpoints of the diagonals AC and BD are (\(\bar{a}\) + \(\bar{c}\))/2 and (\(\bar{b}\) + \(\bar{d}\))/2. By (1), they are equal.
∴ the midpoints of the diagonals AC and BD are the same.
This shows that the diagonals AC and BD bisect each other.

(ii) Conversely, suppose that the diagonals AC and BD
of □ ABCD bisect each other,
i. e. they have the same midpoint.
∴ the position vectors of these midpoints are equal.
∴ \(\frac{\bar{a}+\bar{c}}{2}=\frac{\bar{b}+\bar{d}}{2}\) ∴ \(\bar{a}+\bar{c}=\bar{b}+\bar{d}\)
∴ \(\bar{b}\) – \(\bar{a}\) = \(\bar{c}\) – \(\bar{d}\) ∴ \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{DC}}\)
∴ \(\overline{\mathrm{AB}}\) ||\(\overline{\mathrm{DC}}\) and \(|\overline{\mathrm{AB}}|\) = \(|\overline{\mathrm{DC}}|\)
∴ side AB || side DC and AB = DC.
∴ □ ABCD is a parallelogram.

Question 8.
Prove that the median of a trapezium is parallel to the parallel sides of the trapezium and its length is half the sum of parallel sides.
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) be respectively the position vectors of the vertices A, B, C and D of the trapezium ABCD, with side AD || side BC.
Then the vectors \(\overline{\mathrm{AD}}\) and \(\overline{\mathrm{BC}}\) are parallel.
∴ there exists a scalar k,
such that \(\overline{\mathrm{AD}}\) = k∙\(\overline{\mathrm{BC}}\)
∴ \(\overline{\mathrm{AD}}\) + \(\overline{\mathrm{BC}}\) = k∙\(\overline{\mathrm{BC}}\) + \(\overline{\mathrm{BC}}\)
= (k + 1)BC …(1)

Let \(\bar{m}\) and \(\bar{n}\) be the position vectors of the midpoints M and N of the non-parallel sides AB and DC respectively.
Then seg MN is the median of the trapezium.
By the midpoint formula,

Thus \(\overline{\mathrm{MN}}\) is a scalar multiple of \(\overline{\mathrm{BC}}\)
∴ \(\overline{\mathrm{MN}}\) and \(\overline{\mathrm{BC}}\) are parallel vectors
∴ \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{BC}}\) where \(\overline{\mathrm{BC}}\) || \(\overline{\mathrm{AD}}\)
∴ the median MN is parallel to the parallel sides AD and BC of the trapezium.
Now \(\overline{\mathrm{AD}}\) and \(\overline{\mathrm{BC}}\) are collinear

Question 9.
If two of the vertices of the triangle are A(3, 1, 4) and B(-4, 5, -3) and the centroid of a triangle is G(-1, 2, 1), then find the coordinates of the third vertex C of the triangle.
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{g}\) be the position vectors of A, B, C and G respectively.
Then \(\bar{a}\) = \(3 \hat{i}+\hat{j}+4 \hat{k}\), \(\bar{b}\) = \(-4 \hat{i}+5 \hat{j}-3 \hat{k}\) and \(\bar{g}\) = \(-\hat{i}+2 \hat{j}+\hat{k}\).
Since G is the centroid of the ∆ABC, by the centroid formula,

∴ the coordinates of third vertex C are (-2, 0, 2).

Question 10.
In∆OAB, E is the mid-point of OB and D is the point on AB such that AD : DB = 2 : 1.
If OD and AE intersect at P, then determine the ratio OP : PD using vector methods.
Solution:

Let A, B, D, E, P have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{d}\), \(\bar{e}\), \(\bar{p}\) respectively w.r.t. O.
∵ AD : DB = 2 : 1.
∴ D divides AB internally in the ratio 2 : 1.
Using section formula for internal division, we get

LHS is the position vector of the point which divides OD internally in the ratio 3 : 2.
RHS is the position vector of the point which divides AE internally in the ratio 4 : 1.
But OD and AE intersect at P
∴ P divides OD internally in the ratio 3 : 2.
Hence, OP : PD = 3 : 2.

Question 11.
If the centroid of a tetrahedron OABC is (1, 2, -1) where A = (a, 2, 3), B = (1, b, 2), C = (2, 1, c) respectively, find the distance of P (a, b, c) from the origin.
Solution:
Let G = (1, 2, -1) be the centroid of the tetrahedron OABC.
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{g}\) be the position vectors of the points A, B, C, G respectively w.r.t. O.

By equality of vectors
a + 3 = 4, b + 3 = 8, c + 5= -4
∴ a = 1, b = 5, c = -9
∴ P = (a, b, c) = (1, 5, -9)
Distance of P from origion = \(\sqrt{1^{2}+5^{2}+(-9)^{2}}\)
= \(\sqrt{1+25+81}\)
= \(\sqrt{107}\)

Question 12.
Find the centroid of tetrahedron with vertices K(5, -7, 0), L(1, 5, 3), M(4, -6, 3), N(6, -4, 2) ?
Solution:
Let \(\bar{p}\), \(\bar{l}\), \(\bar{m}\), \(\bar{n}\) be the position vectors of the points K, L, M, N respectively w.r.t. the origin O.

Hence, the centroid of the tetrahedron is G = (4, -3, 2).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.1

Question 1.
The vector \(\bar{a}\) is directed due north and \(|\bar{a}|\) = 24. The vector \(\bar{b}\) is directed due west and \(|\bar{b}|\) = 7. Find \(|\bar{a}+\bar{b}|\).
Solution:

Let \(\overline{\mathrm{AB}}\) = \(\bar{a}\), \(\overline{\mathrm{BC}}\) = \(\bar{b}\)
Then \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BC}}\) = a + b
Given : \(|\bar{a}|\) = \(|\overline{\mathrm{AB}}|\) = l(AB) = 24 and
\(|\bar{b}|\) = \(|\overline{\mathrm{BC}}|\) = l(BC) = 7
∴ ∠ABC = 90°
∴ [l(AC)]² = [l(AB)]² + [l(BC)]²
= (24)² + (7)² = 625
∴ l(AC) = 25 ∴ \(|\overline{\mathrm{AC}}|\) = 25
∴ \(|\bar{a}+\bar{b}|\) = \(|\overline{\mathrm{AC}}|\) = 25.

Question 2.
In the triangle PQR, \(\overline{\mathrm{PQ}}\) = 2\(\bar{a}\) and \(\overline{\mathrm{QR}}\) = 2\(\bar{b}\). The mid-point of PR is M. Find following vectors in terms of \(\bar{a}\) and \(\bar{b}\).
(i) \(\overline{\mathrm{PR}}\)
Solution:

Given : \(\overline{\mathrm{PQ}}\) = 2\(\bar{a}\), \(\overline{\mathrm{QR}}\) = 2\(\bar{b}\)
(i) \(\overline{\mathrm{PR}}\) = \(\overline{\mathrm{PQ}}\) + \(\overline{\mathrm{QR}}\)
= 2\(\bar{a}\) + 2\(\bar{a}\).

(ii) \(\overline{\mathrm{PM}}\)
Solution:
∵ M is the midpoint of PR
∴ \(\overline{\mathrm{PM}}\) = \(\frac{1}{2} \overline{\mathrm{PR}}\) = \(\frac{1}{2}\)[2\(\bar{a}\) + 2\(\bar{b}\)]
= \(\bar{a}\) + \(\bar{b}\).

(iii) \(\overline{\mathrm{QM}}\)
Solution:
\(\overline{\mathrm{RM}}\) = \(\frac{1}{2}(\overline{\mathrm{RP}})\) = \(-\frac{1}{2} \overline{\mathrm{PR}}\) = \(-\frac{1}{2}\)(2\(\bar{a}\) + 2\(\bar{b}\))
= –\(\bar{a}\) – \(\bar{b}\)
∴ \(\overline{\mathrm{QM}}\) = \(\overline{\mathrm{QR}}\) + \(\overline{\mathrm{RM}}\)
= 2\(\bar{b}\) – \(\bar{a}\) – \(\bar{b}\)
= \(\bar{b}\) – \(\bar{a}\).

Question 3.
OABCDE is a regular hexagon. The points A and B have position vectors \(\bar{a}\) and \(\bar{b}\) respectively, referred to the origin O. Find, in terms of \(\bar{a}\) and \(\bar{b}\) the position vectors of C, D and E.
Solution:

Given : \(\overline{\mathrm{OA}}\) = \(\bar{a}\), \(\overline{\mathrm{OB}}\) = \(\bar{a}\) Let AD, BE, OC meet at M.
Then M bisects AD, BE, OC.
\(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{AO}}\) + \(\overline{\mathrm{OB}}\) = –\(\overline{\mathrm{OA}}\) + \(\overline{\mathrm{OB}}\) = –\(\bar{a}\) + \(\bar{b}\) = \(\bar{b}\) – \(\bar{a}\)
∵ OABM is a parallelogram

Hence, the position vectors of C, D and E are 2\(\bar{b}\) – 2\(\bar{a}\), 2\(\bar{b}\) – 3\(\bar{a}\) and \(\bar{b}\) – 2\(\bar{a}\) respectively.

Question 4.
If ABCDEF is a regular hexagon, show that \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{AC}}\) + \(\overline{\mathrm{AD}}\) + \(\overline{\mathrm{AE}}\) + \(\overline{\mathrm{AF}}\) = 6\(\overline{\mathrm{AO}}\), where O is the center of the hexagon.
Solution:

ABCDEF is a regular hexagon.
∴ \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{ED}}\) and \(\overline{\mathrm{AF}}\) = \(\overline{\mathrm{CD}}\)
∴ by the triangle law of addition of vectors,

Question 5.
Check whether the vectors \(2 \hat{i}+2 \hat{j}+3 \hat{k}\), + \(-3 \hat{i}+3 \hat{j}+2 \hat{k}\), + \(3 \hat{i}+4 \hat{k}\) form a triangle or not.
Solution:
Let, if possible, the three vectors form a triangle ABC
with \(\overline{A B}\) = \(2 \hat{i}+2 \hat{j}+3 \hat{k}\), \(\overline{B C}\) = \(3 \hat{i}+3 \hat{j}+2 \hat{k}\), \(\overline{A C}\) = \(3 \hat{i}+4 \hat{k}\)
Now, \(\overline{A B}\) + \(\overline{B C}\)
= \((2 \hat{i}+2 \hat{j}+3 \hat{k})\) + \((-3 \hat{i}+3 \hat{j}+2 \hat{k})\)
= \(-\hat{i}+5 \hat{j}+5 \hat{k} \neq 3 \hat{i}+4 \hat{k}\) = \(\overline{\mathrm{AC}}\)
Hence, the three vectors do not form a triangle.

Question 6.
In the figure 5.34 express \(\bar{c}\) and \(\bar{d}\) in terms of \(\bar{a}\) and \(\bar{b}\). Find a vector in the direction of \(\bar{a}\) = \(\hat{i}-2 \hat{j}\) that has magnitude 7 units.

Solution:
\(\overline{\mathrm{PQ}}\) = \(\overline{\mathrm{PS}}\) + \(\overline{\mathrm{SQ}}\)
∴ \(\bar{a}\) = \(\bar{c}\) – \(\bar{d}\) … (1)
\(\overline{\mathrm{PR}}\) = \(\overline{\mathrm{PS}}\) + \(\overline{\mathrm{SR}}\)
∴ \(\bar{b}\) = \(\bar{c}\) + \(\bar{d}\) … (2)
Adding equations (1) and (2), we get
\(\bar{a}\) + \(\bar{b}\) = (\(\bar{c}\) – \(\bar{d}\)) + (\(\bar{c}\) + \(\bar{d}\)) = 2\(\bar{c}\)

Question 7.
Find the distance from (4, -2, 6) to each of the following :
(a) The XY-plane
Solution:
Let the point A be (4, -2, 6).
Then,
The distance of A from XY-plane = |z| = 6

(b) The YZ-plane
Solution:
The distance of A from YZ-plane = |x| = 4

(c) The XZ-plane
Solution:
The distance of A from ZX-plane = |y| = 2

(d) The X-axis
Solution:
The distance of A from X-axis
= \(\sqrt{y^{2}+z^{2}}\) = \(\sqrt{(-2)^{2}+6^{2}}\) = \(\sqrt{40}\) = \(2 \sqrt{10}\)

(e) The Y-axis
Solution:
The distance of A from Y-axis
= \(\sqrt{z^{2}+x^{2}}\) = \(\sqrt{6^{2}+4^{2}}\) = \(\sqrt{52}\) = \(2 \sqrt{13}\)

(f) The Z-axis
Solution:
The distance of A from Z-axis
= \(\sqrt{x^{2}+y^{2}}\) = \(\sqrt{4^{2}+(-2)^{2}}\) = \(\sqrt{20}\) = \(2 \sqrt{5}\)

Question 8.
Find the coordinates of the point which is located :
(a) Three units behind the YZ-plane, four units to the right of the XZ-plane and five units above the XY-plane.
Solution:
Let the coordinates of the point be (x, y, z).
Since the point is located 3 units behind the YZ- j plane, 4 units to the right of XZ-plane and 5 units , above the XY-plane,
x = -3, y = 4 and z = 5
Hence, coordinates of the required point are (-3, 4, 5)

(b) In the YZ-plane, one unit to the right of the XZ-plane and six units above the XY-plane.
Solution:
Let the coordinates of the point be (x, y, z).
Since the point is located in the YZ plane, x = 0. Also, the point is one unit to the right of XZ-plane and six units above the XY-plane.
∴ y = 1, z = 6.
Hence, coordinates of the required point are (0, 1, 6).

Question 9.
Find the area of the triangle with vertices (1, 1, 0), (1, 0, 1) and (0, 1, 1).
Solution:
Let A = (1, 1, 0), B = (1, 0, 1), C = (0, 1, 1)

Question 10.
If \(\overline{\mathrm{AB}}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\) and initial point A ≡ (1, 5, ,0). Find the terminal point B.
Solution:
Let \(\bar{a}\) and \(\bar{b}\) be the position vectors of A and B.
Given : A = (1, 5, 0) .’. \(\bar{a}\) = \(\hat{i}+5 \hat{j}\)
Now, \(\overline{\mathrm{AB}}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\)
∴ \(\bar{b}\) – \(\bar{a}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\)
∴ \(\bar{b}\) = \((2 \hat{i}-4 \hat{j}+7 \hat{k})\) + \(\bar{a}\)
= \((2 \hat{i}-4 \hat{j}+7 \hat{k})\) + \((\hat{i}+5 \hat{j})\)
= \(3 \hat{i}+\hat{j}+7 \hat{k}\)
Hence, the terminal point B = (3, 1, 7).

Question 11.
Show that the following points are collinear :
(i) A (3, 2, -4), B (9, 8, -10), C (-2, -3, 1).
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) be the position vectors of the points.
A = (3, 2, -4), B = (9, 8, -10) and C = (-2, -3, 1) respectively.

∴ \(\overline{\mathrm{BC}}\) is a non-zero scalar multiple of \(\overline{\mathrm{AB}}\)
∴ they are parallel to each other.
But they have the point B in common.
∴ \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AB}}\) are collinear vectors.
Hence, the points A, B and C are collinear.

(ii) P (4, 5, 2), Q (3, 2, 4), R (5, 8, 0).
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) be the position vectors of the points.
P = (4, 5, 2), Q = (3, 2, 4), R = (5, 8, 0) respectively.

= 2.\(\overline{\mathrm{AB}}\) …[By (1)]
∴ \(\overline{\mathrm{BC}}\) is a non-zero scalar multiple of \(\overline{\mathrm{AB}}\)
∴ they are parallel to each other.
But they have the point B in common.
∴ \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AB}}\) are collinear vectors.
Hence, the points A, B and C are collinear.

Question 12.
If the vectors \(2 \hat{i}-q \hat{j}+3 \hat{k}\) and \(4 \hat{i}-5 \hat{j}+6 \hat{k}\) are collinear, then find the value of q.
Solution:
The vectors \(2 \hat{i}-q \hat{j}+3 \hat{k}\) and \(4 \hat{i}-5 \hat{j}+6 \hat{k}\) are collinear
∴ the coefficients of \(\hat{i}, \hat{j}, \hat{k}\) are proportional

Question 13.
Are the four points A(1, -1, 1), B(-1, 1, 1), C(1, 1, 1) and D(2, -3, 4) coplanar? Justify your answer.
Solution:
The position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\) of the points A, B, C, D are

By equality of vectors,
y = -2 ….(1)
2x – 2y = 2 … (2)
3y = 0 … (3)
From (1), y = -2
From (3), y = 0 This is not possible.
Hence, the points A, B, C, D are not coplanar.

Question 14.
Express \(-\hat{i}-3 \hat{j}+4 \hat{k}\) as linear combination of the vectors \(2 \hat{i}+\hat{j}-4 \hat{k}\), \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(3 \hat{i}+\hat{j}-2 \hat{k}\).
Solution:

By equality of vectors,
2x + 2y + 3 = -1
x – y + z = -3
-4x + 3y – 2z = 4
We have to solve these equations by using Cramer’s Rule
D = \(\left|\begin{array}{rrr}
2 & 2 & 3 \\
1 & -1 & 1 \\
-4 & 3 & -2
\end{array}\right|\)
= 2(2 – 3) – 2(-2 + 4) + 3(3 – 4)
= -2 – 4 – 3 = -9 ≠ 0

= 2(-4 + 9) – 2(4 – 12) – 1(3 – 4)
= 10 + 16 + 1 = 27

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4

I : Choose correct alternatives.
Question 1.
If the equation 4x² + hxy + y² = 0 represents two coincident lines, then h = _________.
(A) ± 2
(B) ± 3
(C) ± 4
(D) ± 5
Solution:
(C) ± 4

Question 2.
If the lines represented by kx² – 3xy + 6y² = 0 are perpendicular to each other then _________.
(A) k = 6
(B) k = -6
(C) k = 3
(D) k = -3
Solution:
(B) k = -6

Question 3.
Auxiliary equation of 2x² + 3xy – 9y² = 0 is _________.
(A) 2m² + 3m – 9 = 0
(B) 9m² – 3m – 2 = 0
(C) 2m² – 3m + 9 = 0
(D) -9m² – 3m + 2 = 0
Solution:
(B) 9m² – 3m – 2 = 0

Question 4.
The difference between the slopes of the lines represented by 3x² – 4xy + y² = 0 is _________.
(A) 2
(B) 1
(C) 3
(D) 4
Solution:
(A) 2

Question 5.
If the two lines ax² +2hxy+ by² = 0 make angles α and β with X-axis, then tan (α + β) = _____.
(A) \(\frac{h}{a+b}\)
(B) \(\frac{h}{a-b}\)
(C) \(\frac{2 h}{a+b}\)
(D) \(\frac{2 h}{a-b}\)
Solution:
(D) \(\frac{2 h}{a-b}\)

Question 6.
If the slope of one of the two lines \(\frac{x^{2}}{a}+\frac{2 x y}{h}+\frac{y^{2}}{b}\) = 0 is twice that of the other, then ab:h² = ___.
(A) 1 : 2
(B) 2 : 1
(C) 8 : 9
(D) 9 : 8
Solution:
(D) 9 : 8

Question 7.
The joint equation of the lines through the origin and perpendicular to the pair of lines 3x² + 4xy – 5y² = 0 is _________.
(A) 5x² + 4xy – 3y² = 0
(B) 3x² + 4xy – 5y² = 0
(C) 3x² – 4xy + 5y² = 0
(D) 5x² + 4xy + 3y² = 0
Solution:
(A) 5x² + 4xy – 3y² = 0

Question 8.
If acute angle between lines ax² + 2hxy + by² = 0 is, \(\frac{\pi}{4}\) then 4h² = _________.
(A) a² + 4ab + b²
(B) a² + 6ab + b²
(C) (a + 2b)(a + 3b)
(D) (a – 2b)(2a + b)
Solution:
(B) a² + 6ab + b²

Question 9.
If the equation 3x² – 8xy + qy² + 2x + 14y + p = 1 represents a pair of perpendicular lines then
the values of p and q are respectively _________.
(A) -3 and -7
(B) -7 and -3
(C) 3 and 7
(D) -7 and 3
Solution:
(B) -7 and -3

Question 10.
The area of triangle formed by the lines x² + 4xy + y² = 0 and x – y – 4 = 0 is _________.
(A) \(\frac{4}{\sqrt{3}}\) Sq. units
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
(C) \(\frac{16}{\sqrt{3}}\) Sq. units
(D)\(\frac{15}{\sqrt{3}}\) Sq. units
Solution:
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
[Hint : Area = \(\frac{p^{2}}{\sqrt{3}}\), where p is the length of perpendicular from the origin to x – y – 4 = 0]

Question 11.
The combined equation of the co-ordinate axes is _________.
(A) x + y = 0
(B) x y = k
(C) xy = 0
(D) x – y = k
Solution:
(C) xy = 0

Question 12.
If h² = ab, then slope of lines ax² + 2hxy + by² = 0 are in the ratio _________.
(A) 1 : 2
(B) 2 : 1
(C) 2 : 3
(D) 1 : 1
Solution:
(D) 1 : 1
[Hint: If h² = ab, then lines are coincident. Therefore slopes of the lines are equal.]

Question 13.
If slope of one of the lines ax² + 2hxy + by² = 0 is 5 times the slope of the other, then 5h² = _________.
(A) ab
(B) 2 ab
(C) 7 ab
(D) 9 ab
Solution:
(D) 9 ab

Question 14.
If distance between lines (x – 2y)² + k(x – 2y) = 0 is 3 units, then k =
(A) ± 3
(B) ± 5\(\sqrt {5}\)
(C) 0
(D) ± 3\(\sqrt {5}\)
Solution:
(D) ± 3\(\sqrt {5}\)
[Hint: (x – 2y)² + k(x – 2y) = 0
∴ (x – 2y)(x – 2y + k) = 0
∴ equations of the lines are x – 2y = 0 and x – 2y + k = 0 which are parallel to each other.
∴ \(\left|\frac{k-0}{\sqrt{1+4}}\right|\) = 3
∴ k = ± 3\(\sqrt {5}\)

II. Solve the following.
Question 1.
Find the joint equation of lines:
(i) x – y = 0 and x + y = 0
Solution:
The joint equation of the lines x – y = 0 and
x + y = 0 is
(x – y)(x + y) = 0
∴ x² – y² = 0.

(ii) x + y – 3 = 0 and 2x + y – 1 = 0
Solution:
The joint equation of the lines x + y – 3 = 0 and 2x + y – 1 = 0 is
(x + y – 3)(2x + y – 1) = 0
∴ 2x² + xy – x + 2xy + y² – y – 6x – 3y + 3 = 0
∴ 2x² + 3xy + y² – 7x – 4y + 3 = 0.

(iii) Passing through the origin and having slopes 2 and 3.
Solution:
We know that the equation of the line passing through the origin and having slope m is y = mx. Equations of the lines passing through the origin and having slopes 2 and 3 are y = 2x and y = 3x respectively.
i.e. their equations are
2x – y = 0 and 3x – y = 0 respectively.
∴ their joint equation is (2x – y)(3x – y) = 0
∴ 6x² – 2xy – 3xy + y² = 0
∴ 6x² – 5xy + y² = 0.

(iv) Passing through the origin and having inclinations 60° and 120°.
Solution:
Slope of the line having inclination θ is tan θ .
Inclinations of the given lines are 60° and 120°
∴ their slopes are m₁ = tan60° = \(\sqrt {3}\) and
m₂ = tan 120° = tan (180° – 60°)
= -tan 60° = –\(\sqrt {3}\)
Since the lines pass through the origin, their equa-tions are
y = \(\sqrt {3}\)x and y= –\(\sqrt {3}\)x
i.e., \(\sqrt {3}\)x – y = 0 and \(\sqrt {3}\)x + y = 0
∴ the joint equation of these lines is
(\(\sqrt {3}\)x – y)(\(\sqrt {3}\)x + y) = 0
∴ 3x² – y² = 0.

(v) Passing through (1, 2) amd parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0
∴ the equations of the lines passing through (1, 2) and parallel to the coordinate axes are x = 1 and y =1
i.e. x – 1 = 0 and y – 2 0
∴ their combined equation is
(x – 1)(y – 2) = 0
∴ x(y – 2) – 1(y – 2) = 0
∴ xy – 2x – y + 2 = 0

(vi) Passing through (3, 2) and parallel to the line x = 2 and y = 3.
Solution:
Equations of the lines passing through (3, 2) and parallel to the lines x = 2 and y = 3 are x = 3 and y = 2.
i.e. x – 3 = 0 and y – 2 = 0
∴ their joint equation is
(x – 3)(y – 2) = 0
∴ xy – 2x – 3y + 6 = 0.

(vii) Passing through (-1, 2) and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0.
Solution:
Let L₁ and L₂ be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 respectively.
Slopes of the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 are \(-\frac{1}{2}\) and \(-\frac{3}{-4}=\frac{3}{4}\) respectively.
∴ slopes of the lines L₁and L₂ are 2 and \(\frac{-4}{3}\) respectively.
Since the lines L₁ and L₂ pass through the point (-1, 2), their equations are
∴ (y – y₁) = m(x – x₁)
∴ (y – 2) = 2(x + 1)
⇒ y – 1 = 2x + 2
⇒ 2x – y + 4 = 0 and
∴ (y – 2) = \(\left(\frac{-4}{3}\right)\)(x + 1)
⇒ 3y – 6 = (-4)(x + 1)
⇒ 3y – 6 = -4x + 4
⇒ 4x + 3y – 6 + 4 = 0
⇒ 4x + 3y – 2 = 0
their combined equation is
∴ (2x – y + 4)(4x + 3y – 2) = 0
∴ 8x² + 6xy – 4x – 4xy – 3y² + 2y + 16x + 12y – 8 = 0
∴ 8x² + 2xy + 12x – 3y² + 14y – 8 = 0

(viii) Passing through the origin and having slopes 1 + \(\sqrt {3}\) and 1 – \(\sqrt {3}\)
Solution:
Let l₁ and l₂ be the two lines. Slopes of l₁ is 1 + \(\sqrt {3}\) and that of l₂ is 1 – \(\sqrt {3}\)
Therefore the equation of a line (l₁) passing through the origin and having slope is
y = (1 + \(\sqrt {3}\))x
∴ (1 + \(\sqrt {3}\))x – y = 0 ..(1)
Similarly, the equation of the line (l₂) passing through the origin and having slope is
y = (1 – \(\sqrt {3}\))x
∴ (1 – \(\sqrt {3}\))x – y = 0 …(2)
From (1) and (2) the required combined equation is

∴ (1 – 3)x² – 2xy + y² = 0
∴ -2x² – 2xy + y² = 0
∴ 2x² + 2xy – y² = 0
This is the required combined equation.

(ix) Which are at a distance of 9 units from the Y – axis.
Solution:
Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it, are x = 9 and x = -9
i.e. x – 9 = 0 and x + 9 = 0

∴ their combined equation is
(x – 9)(x + 9) = 0
∴ x² – 81 = 0.

(x) Passing through the point (3, 2), one of which is parallel to the line x – 2y = 2 and other is perpendicular to the line y = 3.
Solution:
Let L₁ be the line passes through (3, 2) and parallel to the line x – 2y = 2 whose slope is \(\frac{-1}{-2}=\frac{1}{2}\)
∴ slope of the line L₁ is \(\frac{1}{2}\).
∴ equation of the line L₁ is
y – 2 = \(\frac{1}{2}\)(x – 3)
∴ 2y – 4 = x – 3 ∴ x – 2y + 1 = 0
Let L₂ be the line passes through (3, 2) and perpendicular to the line y = 3.
∴ equation of the line L₂ is of the form x = a.
Since L₂ passes through (3, 2), 3 = a
∴ equation of the line L₂ is x = 3, i.e. x – 3 = 0
Hence, the equations of the required lines are
x – 2y + 1 = 0 and x – 3 = 0
∴ their joint equation is
(x – 2y + 1)(x – 3) = 0
∴ x² – 2xy + x – 3x + 6y – 3 = 0
∴ x² – 2xy – 2x + 6y – 3 = 0.

(xi) Passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18.
Solution:
Let L₁ and L₂ be the lines passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18 respectively.
Slopes of the lines x + 2y = 19 and 3x + y = 18 are \(-\frac{1}{2}\) and \(-\frac{3}{1}\) = -3 respectively.
Since the lines L₁ and L₂ pass through the origin, their equations are
y = 2x and y = \(\frac{1}{3}\)x
i.e. 2x – y = 0 and x – 3y = 0
∴ their combined equation is
(2x – y)(x – 3y) = 0
∴ 2x² – 6xy – xy + 3y² = 0
∴ 2x² – 7xy + 3y² = 0.

Question 2.
Show that each of the following equation represents a pair of lines.
(i) x² + 2xy – y² = 0
Solution:
Comparing the equation x² + 2xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 2, i.e. h = 1 and b = -1
∴ h² – ab = (1)² – 1(-1) = 1 + 1=2 > 0
Since the equation x² + 2xy – y² = 0 is a homogeneous equation of second degree and h² – ab > 0, the given equation represents a pair of lines which are real and distinct.

(ii) 4x² + 4xy + y² = 0
Solution:
Comparing the equation 4x² + 4xy + y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 4, 2h = 4, i.e. h = 2 and b = 1
∴ h² – ab = (2)² – 4(1) = 4 – 4 = 0
Since the equation 4x² + 4xy + y² = 0 is a homogeneous equation of second degree and h² – ab = 0, the given equation represents a pair of lines which are real and coincident.

(iii) x² – y² = 0
Solution:
Comparing the equation x² – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 0, i.e. h = 0 and b = -1
∴ h² – ab = (0)² – 1(-1) = 0 + 1 = 1 > 0
Since the equation x² – y² = 0 is a homogeneous equation of second degree and h² – ab > 0, the given equation represents a pair of lines which are real and distinct.

(iv) x² + 7xy – 2y² = 0
Solution:
Comparing the equation x² + 7xy – 2y² = 0
a = 1, 2h = 7 i.e., h = \(\frac{7}{2}\) and b = -2
∴ h² – ab = \(\left(\frac{7}{2}\right)^{2}\) – 1(-2)
= \(\frac{49}{4}\) + 2
= \(\frac{57}{4}\) i.e. 14.25 = 14 > 0
Since the equation x² + 7xy – 2y² = 0 is a homogeneous equation of second degree and h² – ab > 0, the given equation represents a pair of lines which are real and distinct.

(v) x² – 2\(\sqrt {3}\) xy – y² = 0
Solution:
Comparing the equation x² – 2\(\sqrt {3}\) xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h= -2\(\sqrt {3}\), i.e. h = –\(\sqrt {3}\) and b = 1
∴ h² – ab = (-\(\sqrt {3}\))² – 1(1) = 3 – 1 = 2 > 0
Since the equation x² – 2\(\sqrt {3}\)xy – y² = 0 is a homo¬geneous equation of second degree and h² – ab > 0, the given equation represents a pair of lines which are real and distinct.

Question 3.
Find the separate equations of lines represented by the following equations:
(i) 6x² – 5xy – 6y² = 0
Solution:
6x² – 5xy – 6y² = 0
∴ 6x² – 9xy + 4xy – 6y² = 0
∴ 3x(2x – 3y) + 2y(2x – 3y) = 0
∴ (2x – 3y)(3x + 2y) = 0
∴ the separate equations of the lines are
2x – 3y = 0 and 3x + 2y = 0.

(ii) x² – 4y² = 0
Solution:
x² – 4y² = 0
∴ x² – (2y)² = 0
∴(x – 2y)(x + 2y) = 0
∴ the separate equations of the lines are
x – 2y = 0 and x + 2y = 0.

(iii) 3x² – y² = 0
Solution:
3x² – y² = 0
∴ (\(\sqrt {3}\) x)² – y² = 0
∴ (\(\sqrt {3}\)x – y)(\(\sqrt {3}\)x + y) = 0
∴ the separate equations of the lines are
\(\sqrt {3}\)x – y = 0 and \(\sqrt {3}\)x + y = 0.

(iv) 2x² + 2xy – y² = 0
Solution:
2x² + 2xy – y² = 0
∴ The auxiliary equation is -m² + 2m + 2 = 0
∴ m² – 2m – 2 = 0

m₁ = 1 + \(\sqrt {3}\) and m₂ = 1 – \(\sqrt {3}\) are the slopes of the lines.
∴ their separate equations are
y = m₁x and y = m₂x
i.e. y = (1 + \(\sqrt {3}\))x and y = (1 – \(\sqrt {3}\))x
i.e. (\(\sqrt {3}\) + 1)x – y = 0 and (\(\sqrt {3}\) – 1)x + y = 0.

Question 4.
Find the joint equation of the pair of lines through the origin and perpendicular to the lines
given by :
(i) x² + 4xy – 5y² = 0
Solution:
Comparing the equation x² + 4xy – 5y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 4, b= -5
Let m₁ and m₂ be the slopes of the lines represented by x² + 4xy – 5y² = 0.

Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{4}{5}\)xy – \(\frac{1}{5}\)y² = 0 …[By (1)]
∴ 5x² + 4xy – y² = 0

(ii) 2x² – 3xy – 9y² = 0
Solution:
Comparing the equation 2x² – 3xy – 9y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 2, 2h = -3, b = -9
Let m₁ and m₂ be the slopes of the lines represented by 2x² – 3xy – 9y² = 0
∴ m₁ + m₂ =\(\frac{-2 h}{b}=-\frac{3}{9}\) and m₁m₂ = \(\frac{a}{b}=-\frac{2}{9}\) …(1)
Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y)(x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\left(-\frac{3}{9}\right)\)xy + \(\left(-\frac{2}{9}\right)\)y² = 0 …[By (1)]
∴ 9x² – 3xy – 2y² = 0

(iii) x² + xy – y² = 0
Solution:
Comparing the equation x²+ xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 1, b = -1
Let m₁ and m₂ be the slopes of the lines represented by x² + xy – y² = 0
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-1}{-1}\) and m₁m₂ = \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{-1}\) = -1 ..(1)
Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y)(x + m₂y) = 0
∴ x² + (m₁ + m₂) + m₁m₂y² = 0
∴ x² + 1xy + (-1)y² = 0 …[By (1)]
∴ x² + xy – y² = 0

Question 5.
Find k if
(i) The sum of the slopes of the lines given by 3x² + kxy – y² = 0 is zero.
Solution:
Comparing the equation 3x² + kxy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 3, 2h = k, b = -1
Let m₁ and m₂ be the slopes of the lines represented by 3x² + kxy – y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-k}{-1}\) = k
Now, m₁ + m₂ = 0 … (Given)
∴ k = 0.

(ii) The sum of slopes of the lines given by 2x² + kxy – 3y² = 0 is equal to their product.
Question is modified.
The sum of slopes of the lines given by x² + kxy – 3y² = 0 is equal to their product.
Solution:
Comparing the equation x² + kxy – 3y² = 0, with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = k, b = -3
Let m₁ and m₂ be the slopes of the lines represented by x² + kxy – 3y² = 0.
∴ m₁ + m₂ = \(-\frac{2 h}{b}=\frac{-k}{-3}=\frac{k}{3}\)
and m₁m₂ = \(\frac{a}{b}=\frac{1}{-3}=\frac{-1}{3}\)
Now, m₁ + m₂ = m₁m₂ … (Given)
∴ \(\frac{k}{3}=\frac{-1}{3}\)
∴ k = -1.

(iii) The slope of one of the lines given by 3x² – 4xy + ky² = 0 is 1.
Solution:
The auxiliary equation of the lines given by 3x² – 4xy + ky² = 0 is km² – 4m + 3 = 0.
Given, slope of one of the lines is 1.
∴ m = 1 is the root of the auxiliary equation km² – 4m + 3 = 0.
∴ k(1)² – 4(1) + 3 = 0
∴ k – 4 + 3 = 0
∴ k = 1.

(iv) One of the lines given by 3x² – kxy + 5y² = 0 is perpendicular to the 5x + 3y = 0.
Solution:
The auxiliary equation of the lines represented by 3x² – kxy + 5y² = 0 is 5m² – km + 3 = 0.
Now, one line is perpendicular to the line 5x + 3y = 0, whose slope is \(-\frac{5}{3}\).
∴ slope of that line = m = \(\frac{3}{5}\)
∴ m = \(\frac{3}{5}\) is the root of the auxiliary equation 5
5m² – km + 3 = 0.
∴ 5\(\left(\frac{3}{5}\right)^{2}\) – k\(\left(\frac{3}{5}\right)\) + 3 = 0
∴ \(\frac{9}{5}-\frac{3 k}{5}\) + 3 = 0
∴ 9 – 3k + 15 = 0
∴ 3k = 24
∴ k = 8.

(v) The slope of one of the lines given by 3x² + 4xy + ky² = 0 is three times the other.
Solution:
3x² + 4xy + ky² = 0
∴ divide by x²

∴ y = mx
∴ \(\frac{\mathrm{y}}{\mathrm{x}}\) = m
put \(\frac{\mathrm{y}}{\mathrm{x}}\) = m in equation (1)
Comparing the equation km² + 4m + 3 = 0 with ax² + 2hxy+ by² = 0, we get,
a = k, 2h = 4, b = 3
m₁ = 3m₂ ..(given condition)
m₁ + m₂ = \(\frac{-2 h}{k}=-\frac{4}{k}\)
m₁m₂ = \(\frac{a}{b}=\frac{3}{k}\)
m₁ + m₂ = \(-\frac{4}{\mathrm{k}}\)
4m₂ = \(-\frac{4}{\mathrm{k}}\) …(m₁ = 3m₂)
m₂ = \(-\frac{1}{\mathrm{k}}\)
m₁m₂ = \(\frac{3}{k}\)
\(3 \mathrm{~m}_{2}^{2}=\frac{3}{\mathrm{k}}\) …(m₁ = 3m₂)
\(3\left(-\frac{1}{\mathrm{k}}\right)^{2}=\frac{3}{\mathrm{k}}\) …(m₂ = \(-\frac{1}{k}\))
\(\frac{1}{k^{2}}=\frac{1}{k}\)
k² = k
k = 1 or k = 0

(vi) The slopes of lines given by kx² + 5xy + y² = 0 differ by 1.
Solution:
Comparing the equation kx² + 5xy +y² = 0 with ax² + 2hxy + by²
a = k, 2h = 5 i.e. h = \(\frac{5}{2}\)
m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{5}{1}\) = -5
and m₁m₂ = \(\frac{a}{b}=\frac{k}{1}\) = k
the slope of the line differ by (m₁ – m₂) = 1 …(1)
∴ (m₁ – m₂)² = (m₁ + m₂)² – 4m₁m₂
(m₁ – m₂)² = (-5)² – 4(k)
(m₁ – m₂)² = 25 – 4k
1 = 25 – 4k ..[By (1)]
4k = 24
k = 6

(vii) One of the lines given by 6x² + kxy + y² = 0 is 2x + y = 0.
Solution:
The auxiliary equation of the lines represented by 6x² + kxy + y² = 0 is
m² + km + 6 = 0.
Since one of the line is 2x + y = 0 whose slope is m = -2.
∴ m = -2 is the root of the auxiliary equation m² + km + 6 = 0.
∴ (-2)² + k(-2) + 6 = 0
∴ 4 – 2k + 6 = 0
∴ 2k = 10 ∴ k = 5

Question 6.
Find the joint equation of the pair of lines which bisect angle between the lines given by x² + 3xy + 2y² = 0
Solution:
x² + 3xy + 2y² = 0
∴ x² + 2xy + xy + 2y² = 0
∴ x(x + 2y) + y(x + 2y) = 0
∴ (x + 2y)(x + y) = 0
∴ separate equations of the lines represented by x² + 3xy + 2y² = 0 are x + 2y = 0 and x + y = 0.
Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,

the distance of P (x, y) from the line x + 2y = 0
= the distance of P(x, y) from the line x + y = 0

∴ 2(x + 2y)² = 5(x + y)²
∴ 2(x² + 4xy + 4y²) = 5(x² + 2xy + y²)
∴ 2x² + 8xy + 8y² = 5x² + 10xy + 5y²
∴ 3x² + 2xy – 3y² = 0.
This is the required joint equation of the lines which bisect the angles between the lines represented by x² + 3xy + 2y² = 0.

Question 7.
Find the joint equation of the pair of lies through the origin and making equilateral triangle with the line x = 3.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the line x = 3.
∴ OA and OB make an angle of 30° and 150° with the positive direction of X-axis
∴ slope of OA = tan 30° = 1/\(\sqrt {3}\)
∴ equation of the line OA is y = \(\frac{1}{\sqrt{3}}\)x
∴ \(\sqrt {3}\)y = x ∴ x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= – tan 30°= -1/\(\sqrt {3}\)
∴ equation of the line OB is y = \(\frac{-1}{\sqrt{3}}\)x
∴ \(\sqrt {3}\)y = -x ∴ x + \(\sqrt {3}\)y = 0
∴ required combined equation of the lines is
(x – \(\sqrt {3}\)y) (x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

Question 8.
Show that the lines x² – 4xy + y² = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle.
Solution:
We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60° with x + y = 10 whose slope is -1.
Let OA (or OB) has slope m.
∴ its equation is y = mx … (1)
Also, tan 60° = \(\left|\frac{m-(-1)}{1+m(-1)}\right|\)
∴ \(\sqrt {3}\) = \(\left|\frac{m+1}{1-m}\right|\)
Squaring both sides, we get,
3 = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
∴ 3(1 – 2m + m²) = m² + 2m + 1
∴ 3 – 6m + 3m² = m² + 2m + 1
∴ 2m² – 8m + 2 = 0
∴ m² – 4m + 1 = 0
∴ \(\left(\frac{y}{x}\right)^{2}\) – 4\(\left(\frac{y}{x}\right)\) + 1 = 0 …[By (1)]
∴ y² – 4xy + x² = 0
∴ x² – 4xy + y\left(\frac{y}{x}\right) = 0 is the joint equation of the two lines through the origin each making an angle of 60° with x + y = 10
∴ x² – 4xy + y² = 0 and x + y = 10 form a triangle OAB which is equilateral.
Let seg OM ⊥^r line AB whose question is x + y = 10

Question 9.
If the slope of one of the lines represented by ax² + 2hxy + by² = 0 is three times the other then prove that 3h² = 4ab.
Solution:
Let m₁ and m₂ be the slopes of the lines represented by ax² + 2hxy + by² = 0.
∴ m₁ + m₂ = \(-\frac{2 h}{b}\) and m₁m₂ = \(\frac{a}{b}\)
We are given that m₂ = 3m₁
∴ m₁ + 3m₁ = \(-\frac{2 h}{b}\) 4m₁ = \(-\frac{2 h}{b}\)
∴ m₁ = \(-\frac{h}{2 b}\) …(1)
Also, m₁(3m₁) = \(\frac{a}{b}\) ∴ 3m₁² = \(\frac{a}{b}\)
∴ 3\(\left(-\frac{h}{2 b}\right)^{2}\) = \(\frac{a}{b}\) ….[By (1)]
∴ \(\frac{3 h^{2}}{4 b^{2}}=\frac{a}{b}\)
∴ 3h² = 4ab, as b ≠0.

Question 10.
Find the combined equation of the bisectors of the angles between the lines represented by 5x² + 6xy – y² = 0.
Solution:
Comparing the equation 5x² + 6xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = 6, b = -1
Let m₁ and m₂ be the slopes of the lines represented by 5x² + 6xy – y² = 0.

The separate equations of the lines are
y = m₁x and y = m₂x, where m1 ≠ m2
i.e. m₁x – y = 0 and m₁x – y = 0.
Let P (x, y) be any point on one of the bisector of the angles between the lines.
∴ the distance of P from the line m₁x – y = 0 is equal to the distance of P from the line m₂x – y = 0.

∴ (m₂² + 1)(m₁x – y)² = (m₁² + 1)(m₂x – y)²
∴ (m₂² + 1)(m₁²x² – 2m₁xy + y²) = (m₁² + 1)(m₂²x² – 2m₂xy + y²)
∴ m₁²m₂²x² – 2m₁m₁²y²xy + m₂²y² + m₁²x² – 2m₁²xy + y²
= m₁²m₂²x² – 2m₁²m₂xy + m₁²y² + m₂²x² – 2m₂xy + y²
∴ (m₁² – m₂²)x² + 2m₁m₂(m₁ – m₂)xy – 2(m₁ – m₂)xy – (m₁² – m₂²)y² = 0
Dividing throughout by m₁ – m₂ (≠0), we get,
(m₁ + m₂)x² + 2m₁m₂xy – 2xy – (m₁ + m₂)y² = 0
∴ 6x² – 10xy – 2xy – 6y² = 0 …[By (1)]
∴ 6x² – 12xy – 6y² = 0
∴ x² – 2xy – y² = 0
This is the joint equation of the bisectors of the angles between the lines represented by 5x² + 6xy – y² = 0.

Question 11.
Find a, if the sum of the slopes of the lines represented by ax² + 8xy + 5y² = 0 is twice their product.
Solution :
Comparing the equation ax² + 8xy + 5y² = 0 with ax² + 2hxy + by² = 0,
we get, a = a, 2h = 8, b = 5
Let m₁ and m₂ be the slopes of the lines represented by ax² + 8xy + 5y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{8}{5}\)
and m₁m₂ = \(\frac{a}{b}=\frac{a}{5}\)
Now, (m₁ + m₂) = 2(m₁m₂)
\(-\frac{8}{5}\) = \(2\left(\frac{a}{5}\right)\)
a = -4

Question 12.
If the line 4x – 5y = 0 coincides with one of the lines given by ax² + 2hxy + by² = 0, then show that 25a + 40h +16b = 0.
Solution :
The auxiliary equation of the lines represented by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0
Given that 4x – 5y = 0 is one of the lines represented by ax² + 2hxy + by² = 0.
The slope of the line 4x – 5y = 0 is \(\frac{-4}{-5}=\frac{4}{5}\)
∴ m = \(\frac{4}{5}\) is a root of the auxiliary equation bm² + 2hm + a = 0.
∴ b\(\left(\frac{4}{5}\right)^{2}\) + 2h\(\left(\frac{4}{5}\right)\) + a = 0
∴ \(\frac{16 b}{25}+\frac{8 h}{5}\) + a = 0
∴ 16b + 40h + 25a = 0 i.e.
∴ 25a + 40h + 16b = 0

Question 13.
Show that the following equations represent a pair of lines. Find the acute angle between them :
(i) 9x² – 6xy + y² + 18x – 6y + 8 = 0
Solution:
Comparing this equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 9, h = -3, b = 1, g = 9, f = -3 and c = 8.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
9 & -3 & 9 \\
-3 & 1 & -3 \\
9 & -3 & 8
\end{array}\right|\)
= 9(8 – 9) + 3(-24 + 27) + 9(9 – 9)
= 9(-1) + 3(3) + 9(0)
= -9 + 9 + 0 = 0
and h² – ab = (-3)² – 9(1) = 9 – 9 = 0
∴ the given equation represents a pair of lines.
Let θ be the acute angle between the lines.

∴ tan θ = tan0°
∴ θ = 0°.

(ii) 2x² + xy – y² + x + 4y – 3 = 0
Solution:
Comparing this equation with
ax² + 2hxy + by² + 2gx + 2fy+ c = 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2 and c = -3

= -2 + 1 + 1
= -2 + 2= 0
∴ the given equation represents a pair of lines.
Let θ be the acute angle between the lines.

∴ tan θ = tan 3
∴ θ = tan^-1(3)

(iii) (x – 3)² + (x – 3)(y – 4) – 2(y – 4)² = 0.
Solution :
Put x – 3 = X and y – 4 = Y in the given equation, we get,
X² + XY – 2Y² = 0
Comparing this equation with ax² + 2hxy + by² = 0, we get,
a = 1, h = \(\frac{1}{2}\), b = -2
This is the homogeneous equation of second degreeand h² – ab = \(\left(\frac{1}{2}\right)^{2}\) – 1(-2)
= \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
Hence, it represents a pair of lines passing through the new origin (3, 4).
Let θ be the acute angle between the lines.

∴ tanθ = 3 ∴ θ = tan^-1(3)

Question 14.
Find the combined equation of pair of lines through the origin each of which makes angle of 60° with the Y-axis.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.
Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.
∴ slope of OA = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= tan 30° = \(-\frac{1}{\sqrt{3}}\)
∴ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
∴ required combined equation is
(x – \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

Question 15.
If lines representedby ax² + 2hxy + by² = 0 make angles of equal measures with the co-ordinate
axes then show that a = ± b.
OR
Show that, one of the lines represented by ax² + 2hxy + by² = 0 will make an angle of the same measure with the X-axis as the other makes with the Y-axis, if a = ± b.
Solution:

Let OA and OB be the two lines through the origin represented by ax² + 2hxy + by² = 0.
Since these lines make angles of equal measure with the coordinate axes, they make angles ∝ and \(\frac{\pi}{2}\) – ∝ with the positive direction of X-axis or ∝ and \(\frac{\pi}{2}\) + ∝ with thepositive direction of X-axis.
∴ slope of the line OA = m₁ = tan ∝
and slope of the line OB = m₂
= tan(\(\frac{\pi}{2}\) – ∝) or tan(\(\frac{\pi}{2}\) + ∝)
i.e. m₂ = cot ∝ or m₂ = -cot ∝
∴ m₁m₂ – tan ∝ x cot ∝ = 1
OR m₁m₂ = tan ∝ (-cot ∝) = -1
i.e. m₁m₂ = ± 1
But m₁m₂ = \(\frac{a}{b}\)
∴ \(\frac{a}{b}\)= ±1 ∴ a = ±b
This is the required condition.

Question 16.
Show that the combined equation of a pair of lines through the origin and each making an angle of ∝ with the line x + y = 0 is x² + 2(sec 2∝) xy + y² = 0.
Solution:
Let OA and OB be the required lines.
Let OA (or OB) has slope m.
∴ its equation is y = mx … (1)
It makes an angle ∝ with x + y = 0 whose slope is -1. m +1
∴ tan ∝ = \(\left|\frac{m+1}{1+m(-1)}\right|\)
Squaring both sides, we get,
tan²∝ = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
∴ tan²∝(1 – 2m + m²) = m² + 2m + 1
∴ tan²∝ – 2m tan²∝ + m²tan²∝ = m² + 2m + 1
∴ (tan²∝ – 1)m² – 2(1 + tan²∝)m + (tan²∝ – 1) = 0

∴ y² + 2xysec2∝ + x² = 0
∴ x² + 2(sec2∝)xy + y² = 0 is the required equation.

Question 17.
Show that the line 3x + 4y+ 5 = 0 and the lines (3x + 4y)² – 3(4x – 3y)² =0 form an equilateral triangle.
Solution:
The slope of the line 3x + 4y + 5 = 0 is \(\frac{-3}{4}\)
Let m be the slope of one of the line making an angle of 60° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m₁ is 60°.

On squaring both sides, we get,
3 = \(\frac{(4 m+3)^{2}}{(4-3 m)^{2}}\)
∴ 3 (4 – 3m)² = (4m + 3)²
∴ 3(16 – 24m + 9m²) = 16m² + 24m + 9
∴ 48 – 72m + 27m² = 16m² + 24m + 9
∴ 11m² – 96m + 39 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
∴ the combined equation of the two lines is
11\(\left(\frac{y}{x}\right)^{2}\) – 96\(\left(\frac{y}{x}\right)\) + 39 = 0
∴ \(\frac{11 y^{2}}{x^{2}}-\frac{96 y}{x}\) + 39 = 0
∴ 11y² – 96xy + 39x² = 0
∴ 39x² – 96xy + 11y² = 0.
∴ 39x² – 96xy + 11y² = 0 is the joint equation of the two lines through the origin each making an angle of 60° with the line 3x + 4y + 5 = 0.
The equation 39x² – 96xy + 11y² = 0 can be written as :
-39x² + 96xy – 11y² = 0
i.e., (9x² – 48x²) + (24xy + 72xy) + (16y² – 27y²) = 0
i.e. (9x² + 24xy + 16y²) – (48x² – 72xy + 27y²) = 0
i.e. (9x² + 24xy + 16y²) – 3(16x² – 24xy + 9y²) = 0
i.e. (3x + 4y)² – 3(4x – 3y)² = 0
Hence, the line 3x + 4y + 5 = 0 and the lines
(3x + 4y)² – 3(4x – 3y)² form the sides of an equilateral triangle.

Question 18.
Show that lines x² – 4xy + y² = 0 and x + y = \(\sqrt {6}\) form an equilateral triangle. Find its area and perimeter.
Solution:
x² – 4xy + y² = 0 and x + y = \(\sqrt {6}\) form a triangle OAB which is equilateral.
Let OM be the perpendicular from the origin O to AB whose equation is x + y = \(\sqrt {6}\)

In right angled triangle OAM,
sin 60° = \(\frac{\mathrm{OM}}{\mathrm{OA}}\) ∴ \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{\mathrm{OA}}\)
∴ OA = 2
∴ length of the each side of the equilateral triangle OAB = 2 units.
∴ perimeter of ∆ OAB = 3 × length of each side
= 3 × 2 = 6 units.

Question 19.
If the slope of one of the lines given by ax² + 2hxy + by² = 0 is square of the other then show that a²b + ab² + 8h³ = 6abh.
Solution:
Let m be the slope of one of the lines given by ax² + 2hxy + by² = 0.
Then the other line has slope m²

Multiplying by b³, we get,
-8h³ = ab² + a²b – 6abh
∴ a²b + ab² + 8h³ = 6abh
This is the required condition.

Question 20.
Prove that the product of lengths of perpendiculars drawn from P (x₁, y₁) to the lines repersented by ax² + 2hxy + by² = 0 is \(\left|\frac{a x_{1}^{2}+2 h x_{1} y_{1}+b y_{1}^{2}}{\sqrt{(a-b)^{2}+4 h^{2}}}\right|\)
Solution:
Let m₁ and m₂ be the slopes of the lines represented by ax² + 2hxy + by² = 0.
∴ m¹ + m² = \(-\frac{2 h}{b}\) and m¹m² = \(\frac{a}{b}\) …(1)
The separate equations of the lines represented by
ax² + 2hxy + by² = 0 are
y = m₁x and y = m₂x
i.e. m₁x – y = 0 and m₂x – y = 0
Length of perpendicular from P(x₁, ₁) on

Question 21.
Show that the difference between the slopes of lines given by (tan²θ + cos²θ )x² – 2xytanθ + (sin²θ )y² = 0 is two.
Solution:
Comparing the equation (tan²θ + cos²θ)x² – 2xy tan θ + (sin²θ) y² = 0 with ax² + 2hxy + by² = 0, we get,
a = tan²θ + cos²θ, 2h = -2 tan θ and b = sin2θ
Let m₁ and m₂ be the slopes of the lines represented by the given equation.

Question 22.
Find the condition that the equation ay² + bxy + ex + dy = 0 may represent a pair of lines.
Solution:
Comparing the equation
ay² + bxy + ex + dy = 0 with
Ax² + 2Hxy + By² + 2Gx + 2Fy + C = 0, we get,
A = 0, H = \(\frac{b}{2}\), B = a,G = \(\frac{e}{2}\), F = \(\frac{d}{2}\), C = 0
The given equation represents a pair of lines,

i.e. if bed – ae² = 0
i.e. if e(bd – ae) = 0
i.e. e = 0 or bd – ae = 0
i.e. e = 0 or bd = ae
This is the required condition.

Question 23.
If the lines given by ax² + 2hxy + by² = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + 3b) = 4h².
Solution:
Since the lines ax² + 2hxy + by² = 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines ax² + 2hxy + by² = 0 is 60°.

∴ 3(a + b)² = 4(h² – ab)
∴ 3(a² + 2ab + b²) = 4h² – 4ab
∴ 3a² + 6ab + 3b² + 4ab = 4h²
∴ 3a² + 10ab + 3b² = 4h²
∴ 3a² + 9ab + ab + 3b² = 4h²
∴ 3a(a + 3b) + b(a + 3b) = 4h²
∴ (3a + b)(a + 3b) = 4h²
This is the required condition.

Question 24.
If line x + 2 = 0 coincides with one of the lines represented by the equation x² + 2xy + 4y + k = 0 then show that k = -4.
Solution:
One of the lines represented by
x² + 2xy + 4y + k = 0 … (1)
is x + 2 = 0.
Let the other line represented by (1) be ax + by + c = 0.
∴ their combined equation is (x + 2)(ax + by + c) = 0
∴ ax² + bxy + cx + 2ax + 2by + 2c = 0
∴ ax² + bxy + (2a + c)x + 2by + 2c — 0 … (2)
As the equations (1) and (2) are the combined equations of the same two lines, they are identical.
∴ by comparing their corresponding coefficients, we get,

∴ 1 = \(\frac{-4}{k}\)
∴ k = -4.

Question 25.
Prove that the combined equation of the pair of lines passing through the origin and perpendicular to the lines represented by ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0
Solution:
Let m₁ and m₂ be the slopes of the lines represented by ax² + 2hxy + by² = 0.

Now, required lines are perpendicular to these lines.
∴ their slopes are and \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(-\frac{1}{m_{1}}\)x and y = \(-\frac{1}{m_{2}}\)x
i.e. m₁y= -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y)(x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x²\(\frac{-2 h}{b}\)x + \(\frac{a}{b}\)y² = 0
∴ bx² – 2hxy + ay² = 0.

Question 26.
If equation ax² – y² + 2y + c = 1 represents a pair of perpendicular lines then find a and c.
Solution:
The given equation represents a pair of lines perpendicular to each other.
∴ coefficient of x² + coefficient of y² = 0
∴ a – 1 = 0 ∴ a = 1
With this value of a, the given equation is
x² – y² + 2y + c – 1 = 0
Comparing this equation with
Ax² + 2Hxy + By² + 2Gx + 2Fy + C = 0, we get,
A = 1, H = 0, B = -1, G = 0, F = 1, C = c – 1
Since the given equation represents a pair of lines,
D = \(\left|\begin{array}{ccc}
A & H & G \\
H & B & F \\
G & F & C
\end{array}\right|\) = 0
∴ \(\left|\begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 1 \\
0 & 1 & c-1
\end{array}\right|\) = 0
∴ 1(-c + 1 – 1) – 0 + 0 = 0
∴ -c = 0
∴ c = 0.
Hence, a = 1, c = 0.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

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

Question 1.
Find the co-factors of the elements of the following matrices
(i) \(\left[\begin{array}{cc}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Here, a₁₁ = -11, M₁₁ = 4
∴ A₁₁ = (-1)¹⁺¹(4) = 4
a₁₂ = 2, M₁₂ = -3
∴ A₁₂ = (-1)¹⁺²(- 3) = 3
a₂₁ = – 3, M₂₁ = -2
∴ A₂₁ = (- 1)²⁺¹(2) = -2
a₂₂ = 4, M₂₂ = -1
∴ A₂₂ = (-1)²⁺²(-1) = -1.

(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
The co-factor of a_ij is given by A_ij = (-1)^i+jM_ij

Question 2.
Find the matrix of co-factors for the following matrices
(i) \(\left[\begin{array}{rr}
1 & 3 \\
4 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rr}
1 & 3 \\
4 & -1
\end{array}\right]\)
Here, a₁₁ = 1, M₁₁ = -1
∴ A₁₁ = (-1)¹⁺¹(-1) = -1
a₁₂ = 3, M₁₂ = 4
∴ A₁₂ = (-1)¹⁺²(4) = -4
a₂₁ = 4, M₂₁ = 3
∴ A₂₁ = (-1)²⁺¹(3) = -3
a₂₂ = -1, M₂₂ = 1
∴ A₂₂ = (-1)²⁺¹(1) = 1
∴ the co-factor matrix = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\)
= \(\left(\begin{array}{rr}
-1 & -4 \\
-3 & 1
\end{array}\right)\)

(ii) \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)



A₁₁ = -14, A₁₂ = -10, A₁₃ = -6,
A₂₁ = 6, A₂₂ = -5, A₂₃ = -3,
A₃₁ = -2, A₃₂ = -7, A₃₃ = 1.
∴ the co-factor matrix
= \(\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]\) = \(\left[\begin{array}{rrr}
-14 & -10 & -6 \\
6 & -5 & -3 \\
-2 & -7 & 1
\end{array}\right]\)

Question 3.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
Here, a₁₁ = 2, M₁₁= 5
∴ A₁₁ = (-1)¹⁺¹(5) = 5
a₁₂ = -3, M₁₂ = 3
∴ A₁₂ = (-1)¹⁺²(3) = -3
a₂₁ = 3, M₂₁ = -3
∴ A A₂₁ = (-1)²⁺¹(-3) = 3
a₂₂ = 5, M₂₂ = 2
∴ A₂₂ = (-1)²⁺¹ = 2
∴ the co-factor matrix = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\)
= \(\left[\begin{array}{rr}
5 & -3 \\
3 & 2
\end{array}\right]\)
∴ adj A = \(\left(\begin{array}{rr}
5 & 3 \\
-3 & 2
\end{array}\right)\)

(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:


A₁₁ = -3, A₁₂ = -12, A₁₃ = 6,
A₂₁ = -1, A₂₂ = 3, A₂₃ = 2,
A₃₁ = -11, A₃₂ = -9, A₃₃ = 1
∴ the co-factor matrix = \(\left[\begin{array}{lll}
\mathrm{A}_{11} & \mathrm{~A}_{12} & \mathrm{~A}_{15} \\
\mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \\
\mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{33}
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-3 & -12 & 6 \\
-1 & 3 & 2 \\
-11 & -9 & 1
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{rrr}
-3 & -1 & -11 \\
-12 & 3 & -9 \\
6 & 2 & 1
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\), verify that A (adj A) = (adj A) A = | A | ∙ I
Solution:
A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\)




From (1), (2) and (3), we get,
A(adj A) = (adj A)A = |A|∙I.
Note: This relation is valid for any non-singular matrix A.

Question 5.
Find the inverse of the following matrices by the adjoint method
(i) \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right|\) = -2 + 15 = 13 ≠ 0
∴ A^-1 exists.
First we have to find the co-factor matrix
= [A_ij]_2×2, where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = 2
A₁₂ = (-1)¹⁺²M₁₂ = -(-3) = 3
A₂₁ = (-1)²⁺¹M₂₁ = -5
A₂₂ = (-1)²⁺²M₂₂ = -1
Hence, the co-factor matrix

(ii) \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
|A| = \(\) = 6 + 8 = 14 ≠ 0
∴ A^-1 exist
First we have to find the co-factor matrix
= [A_ij] _2×2 where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = 3
A₁₂ = (-1)¹⁺²M = -4
A₂₁ = (-2)²⁺¹M₂₁ = (-2) = 2
A₂₂ = (-1)²⁺²M₂₂ = 2
Hence the co-factor matrix
= \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\) = \(\left[\begin{array}{cc}
3 & -4 \\
2 & 2
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) (adj A) = \(\frac{1}{14}\left(\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right)\)

(iii) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)



∴ A^-1 = \(\frac{1}{3}\left[\begin{array}{rrr}
3 & 0 & 0 \\
-3 & 1 & 0 \\
9 & 2 & -3
\end{array}\right]\)

(iv) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
= 1(10 – 0) – 0 + 0
= 1(10) – 0 + 0
= 10 ≠ 0
∴ A^-1 exists.
First we have to find the co-factor matrix


∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) (adj A)
= \(\frac{1}{10}\left(\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right)\)
∴ A^-1 = \(\frac{1}{10}\left(\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right)\)

Question 6.
Find the inverse of the following matrices
(i) \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)

(ii) \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]\)

∴ A^-1 = \(\left(\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right)\)

(iii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)

(iv) \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Ex 2.1

Question 1.
Apply the given elementary transformation on each of the following matrices.
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\), R₁ ↔ R₂
Solution:
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\)
By R₁ ↔ R₂, we get,
A ~ \(\left[\begin{array}{rr}
-1 & 3 \\
1 & 0
\end{array}\right]\)

Question 2.
B = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 5 & 4
\end{array}\right]\), R₁ → R₁ → R₂
Solution:
B = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 5 & 4
\end{array}\right]\),
R₁ → R₁ → R₂ gives,
B ~ \(\left[\begin{array}{rrr}
-1 & -6 & -1 \\
2 & 5 & 4
\end{array}\right]\)

Question 3.
A = \(\left[\begin{array}{ll}
5 & 4 \\
1 & 3
\end{array}\right]\), C₁ ↔ C₂; B = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 5
\end{array}\right]\), R₁ ↔ R₂. What do you observe?
Solution:
A = \(\left[\begin{array}{ll}
5 & 4 \\
1 & 3
\end{array}\right]\)
By C₁ ↔ C₂, we get,
A ~ \(\left[\begin{array}{ll}
4 & 5 \\
3 & 1
\end{array}\right]\) …(1)
B = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 5
\end{array}\right]\)
By R₁ ↔ R₂, we get,
B ~ \(\left[\begin{array}{ll}
4 & 5 \\
3 & 1
\end{array}\right]\) …(2)
From (1) and (2), we observe that the new matrices are equal.

Question 4.
A = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & 1 & 3
\end{array}\right]\), 2C₂
B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 4 & 5
\end{array}\right]\), -3R₁
Find the addition of the two new matrices.
Solution:
A = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & 1 & 3
\end{array}\right]\)
By 2C₂, we get,
A ~ \(\left[\begin{array}{rrr}
1 & 4 & -1 \\
0 & 2 & 3
\end{array}\right]\)
B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 4 & 5
\end{array}\right]\)
By -3R₁, we get,
B ~ \(\left[\begin{array}{rrr}
-3 & 0 & -6 \\
2 & 4 & 5
\end{array}\right]\)
Now, addition of the two new matrices

Question 5.
A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\), 3R₃ and then C₃ + 2C₂.
Solution:
A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\)
By 3R₃, we get
A ~ \(\left[\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
9 & 9 & 3
\end{array}\right]\)
By C₃ + 2C₂, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 3+2(-1) \\
2 & 1 & 0+2(1) \\
9 & 9 & 3+2(9)
\end{array}\right)\)
∴ A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
9 & 9 & 21
\end{array}\right)\)

Question 6.
A = \(\left(\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right)\), C₃ + 2C₂ and then 3R₃. What do you conclude from Ex. 5 and Ex. 6 ?
Solution:
A = \(\left(\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right)\)
By C₃ + 2C₂, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 3+2(-1) \\
2 & 1 & 0+2(1) \\
3 & 3 & 1+2(3)
\end{array}\right)\)
∴ A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
3 & 3 & 7
\end{array}\right)\)
By 3R₃, we get
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
9 & 9 & 21
\end{array}\right)\)
We conclude from Ex. 5 and Ex. 6 that the matrix remains same by interchanging the order of the elementary transformations. Hence, the transformations are commutative.

Question 7.
Use suitable transformation on \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) into an upper triangular matrix.
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
By R₂ – 3R₁, we get,
A ~ \(\left[\begin{array}{rr}
1 & 2 \\
0 & -2
\end{array}\right]\)
This is an upper triangular matrix.

Question 8.
Convert \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\) into an identity matrix by suitable row transformations.
Solution:
Let A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
By R₂ – 2R₁, we get,
A ~ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 5
\end{array}\right]\)
By \(\left(\frac{1}{5}\right)\)R₂, we get,
A ~ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right]\)
By R₁ + R₂, we get,
A ~ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
This is an identity matrix.

Question 9.
Transform \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\) into an upper triangular matrix by suitable row transformations.
Solution:
Let A = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\)
By R₂ – 2R₁ and R₃ – 3R₁, we get
A ~ \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 5 & -2
\end{array}\right]\)
By R₃ – \(\left(\frac{5}{3}\right)\)R₂, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 0 & -\frac{1}{3}
\end{array}\right)\)
This is an upper triangular matrix.

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

Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Miscellaneous Exercise 1

Question 1.
Select and write the correct answer from the given alternatives in each of the following questions:
i) If p ∧ q is false and p ∨ q is true, the ________ is not true.
(A) p ∨ q
(B) p ↔ q
(C) ~p ∨ ~q
(D) q ∨ ~p
Solution:
(b) p ↔ q.

(ii) (p ∧ q) → r is logically equivalent to ________.
(A) p → (q → r)
(B) (p ∧ q) → ~r
(C) (~p ∨ ~q) → ~r
(D) (p ∨ q) → r
Solution:
(a) p → (q → r) [Hint: Use truth table.]

(iii) Inverse of statement pattern (p ∨ q) → (p ∧ q) is ________.
(A) (p ∧ q) → (p ∨ q)
(B) ~(p ∨ q) → (p ∧ q)
(C) (~p ∧ ~q) → (~p ∨ ~q)
(D) (~p ∨ ~q) → (~p ∧ ~q)
Solution:
(c) (~p ∧ ~q) → (~p ∨ ~ q)

(iv) If p ∧ q is F, p → q is F then the truth values of p and q are ________.
(A) T, T
(B) T, F
(C) F, T
(D) F, F
Solution:
(b) T, F

(v) The negation of inverse of ~p → q is ________.
(A) q ∧ p
(B) ~p ∧ ~q
(C) p ∧ q
(D) ~q → ~p
Solution:
(a) q ∧ p

(vi) The negation of p ∧ (q → r) is ________.
(A) ~p ∧ (~q → ~r)
(B) p ∨ (~q ∨ r)
(C) ~p ∧ (~q → ~r)
(D) ~p ∨ (~q ∧ ~r)
Solution:
(d) ~p ∨ (q ∧ ~r)

(vii) If A = {1, 2, 3, 4, 5} then which of the following is not true?
(A) Ǝ x ∈ A such that x + 3 = 8
(B) Ǝ x ∈ A such that x + 2 < 9
(C) Ɐ x ∈ A, x + 6 ≥ 9
(D) Ǝ x ∈ A such that x + 6 < 10
Solution:
(c) Ǝ x ∈ A, x + 6 ≥ 9.

Question 2.
Which of the following sentences are statements in logic? Justify. Write down the truth
value of the statements :
(i) 4! = 24.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(ii) π is an irrational number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(iii) India is a country and Himalayas is a river.
Solution:
It is a statement which is false, hence its truth value is ‘F’. ….[T ∧ F ≡ F]

(iv) Please get me a glass of water.
Solution:
It is an imperative sentence, hence it is not a statement.

(v) cos²θ – sin²θ = cos2θ for all θ ∈ R.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(vi) If x is a whole number the x + 6 = 0.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

Question 3.
Write the truth values of the following statements :
(i) \(\sqrt {5}\) is an irrational but \(3\sqrt {5}\) is a complex number.
Solution:
Let p : \(\sqrt {5}\) is an irrational.
q : \(3\sqrt {5}\) is a complex number.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(ii) Ɐ n ∈ N, n² + n is even number while n² – n is an odd number.
Solution:
Let p : Ɐ n ∈ N, n² + n is an even number.
q : Ɐ n ∈ N, n² – n is an odd number.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F].

(iii) Ǝ n ∈ N such that n + 5 > 10.
Solution:
Ǝ n ∈ N, such that n + 5 > 10 is a true statement, hence its truth value is T.
(All n ≥ 6, where n ∈ N, satisfy n + 5 > 10).

(iv) The square of any even number is odd or the cube of any odd number is odd.
Solution:
Let p : The square of any even number is odd.
q : The cube of any odd number is odd.
Then the symbolic form of the given statement is p ∨ q.
The truth values of p and q are F and T respectively.
∴ the truth value of p ∨ q is T. … [F ∨ T ≡ T].

(v) In ∆ ABC if all sides are equal then its all angles are equal.
Solution:
Let p : ABC is a triangle and all its sides are equal.
q : Its all angles are equal.
Then the symbolic form of the given statement is p → q
If the truth value of p is T, then the truth value of q is T.
∴ the truth value of p → q is T. … [T → T ≡ T].

(vi) Ɐ n ∈ N, n + 6 > 8.
Solution:
Ɐ n ∈ N, 11 + 6 > 8 is a false statement, hence its truth value is F.
{n = 1 ∈ N, n = 2 ∈ N do not satisfy n + 6 > 8).

Question 4.
If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}, determine the truth value of each of the following statement :
(i) Ǝ x ∈ A such that x + 8 = 15.
Solution:
True

(ii) Ɐ x ∈ A, x + 5 < 12.
Solution:
False

(iii) Ǝ x ∈ A, such that x + 7 ≥ 11.
Solution:
True

(iv) Ɐ x ∈ A, 3x ≤ 25.
Solution:
False

Question 5.
Write the negations of the following :
(i) Ɐ n ∈ A, n + 7 > 6.
Solution:
The negation of the given statements are :
Ǝ n ∈ A, such that n + 7 ≤ 6.
OR Ǝ n ∈ A, such that n + 7 ≯ 6.

(ii) Ǝ x ∈ A, such that x + 9 ≤ 15.
Solution:
Ɐ x ∈ A, x + 9 > 15.

(iii) Some triangles are equilateral triangle.
Solution:
All triangles are not equilateral triangles.

Question 6.
Construct the truth table for each of the following :
(i) p → (q → p)
Solution:

(ii) (~p ∨ ~q) ↔ [~(p ∧ q)]
Solution:

(iii) ~(~p ∧ ~q) ∨ q
Solution:

(iv) [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)]
Solution:

(v) [(~p ∨ q) ∧ (q → r)] → (p → r)
Solution:

Question 7.
Determine whether the following statement patterns are tautologies contradictions or contingencies :
(i) [(p → q) ∧ ~q)] → ~p
Solution:

All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q)] → ~p is a tautology.

(ii) [(p ∨ q) ∧ ~p] ∧ ~q
Solution:

All the entries in the last column of the above truth table are F.
∴ [(p ∨ q) ∧ ~p] ∧ ~q is a contradiction.

(iii) (p → q) ∧ (p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

(iv) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:

All the entries in the last column of the above truth table are T.
∴ [p → (q → r)] ↔ [(p ∧ q) → r] is a tautology.

(v) [(p ∧ (p → q)] → q
Solution:

All the entries in the last column of the above truth table are T.
∴ [(p ∧ (p → q)] → q is a tautology.

(vi) (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q) is a tautology.

(vii) [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r
Solution:

The entries in the last column are neither T nor all F.
∴ [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r is a contingency.

(viii) (p → q) ∨ (q → p)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p → q) ∨ (q → p) is a tautology.

Question 8.
Determine the truth values ofp and q in the following cases :
(i) (p ∨ q) is T and (p ∧ q) is T
Solution:

Since p ∨ q and p ∧ q both are T, from the table the truth values of both p and q are T.

(ii) (p ∨ q) is T and (p ∨ q) → q is F
Solution:

Since the truth values of (p ∨ q) is T and (p ∨ q) → q is F, from the table, the truth values of p and q are T and F respectively.

(iii) (p ∧ q) is F and (p ∧ q) → q is T
Solution:

Since the truth values of (p ∧ q) is F and (p ∧ q) → q is T, from the table, the truth values of p and q are either T and F respectively or F and T respectively or both F.

Question 9.
Using truth tables prove the following logical equivalences :
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:

The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q).

(ii) (p ∧ q) → r ≡ p → (q → r)
Solution:

The entries in the columns 5 and 7 are identical.
∴ (p ∧ q) → r ≡ p → (q → r).

Question 10.
Using rules in logic, prove the following :
(i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p)
Solution:
By the rules of negation of biconditional,
~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
∴ ~ [(p ∧ ~ q) ∨ (q ∧ ~p)] ≡ p ↔ q
∴ ~(p ∧ ~q) ∧ ~(q ∧ ~p) ≡ p ↔ q … (Negation of disjunction)
≡ p ↔ q ≡ ~(p ∧ ~ q) ∧ ~ (q ∧ ~p).

(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
(p ∨ q) ∧ ~ p
≡ (p ∧ ~p) ∨ (q ∧ ~p) … (Distributive Law)
≡ F ∨ (q ∧ ~p) … (Complement Law)
≡ q ∧ ~ p … (Identity Law)
≡ ~p ∧ q …(Commutative Law)
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
~ (p ∨ q) ∨ (~p ∧ q)
≡ (~p ∧ ~q) ∨ (~p ∧ q) … (Negation of disjunction)
≡ ~p ∧ (~q ∨ q) … (Distributive Law)
≡ ~ p ∧ T … (Complement Law)
≡ ~ p … (Identity Law)
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Question 11.
Using the rules in logic, write the negations of the following :
(i) (p ∨ q) ∧ (q ∨ ~r)
Solution:
The negation of (p ∨ q) ∧ (q ∨ ~ r) is
~ [(p ∨ q) ∧ (q ∨ ~r)]
≡ ~ (p ∨ q) ∨ ~ (q ∨ ~r) … (Negation of conjunction)
≡ (~p ∧ ~q) ∨ [~q ∧ ~(~r)] … (Negation of disjunction)
≡ {~ p ∧ ~q) ∨ (~q ∧ r) … (Negation of negation)
≡ (~q ∧ ~p) ∨ (~q ∧ r) … (Commutative law)
≡ (~ q) ∧ (~ p ∨ r) … (Distributive Law)

(ii) p ∧ (q ∨ r)
Solution:
The negation of p ∧ (q ∨ r) is
~ [p ∧ (q ∨ r)]
≡ ~ p ∨ ~(q ∨ r) … (Negation of conjunction)
≡ ~p ∨ (~q ∧ ~r) … (Negation of disjunction)

(iii) (p → q) ∧ r
Solution:
The negation of (p → q) ∧ r is
~ [(p → q) ∧ r]
≡ ~ (p → q) ∨ (~ r) … (Negation of conjunction)
≡ (p ∧ ~q) ∨ (~ r) … (Negation of implication)

(iv) (~p ∧ q) ∨ (p ∧ ~q)
Solution:
The negation of (~ p ∧ q) ∨ (p ∧ ~ q) is
~ [(~p ∧ q) ∨ (p ∧ ~q)]
≡ ~(~p ∧ q) ∧ ~ (p ∧ ~q) … (Negation of disjunction)
≡ [~(~p) ∨ ~q] ∧ [~p ∨ ~(q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~ p ∨ q) … (Negation of negation)

Question 12.
Express the following circuits in the symbolic form. Prepare the switching table :
(i)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p : the switch S₁‘ is closed or the switch S₁ is open
~ q: the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form of the given circuit is :
(p ∧ q) ∨ (~p) ∨ (p ∧ ~q).

(ii)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed.
Then the symbolic form of the given statement is : (p ∨ q) ∧ (p ∨ r).

Question 13.
Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p: the switch S₁‘ is closed or the switch S₁ is open
~ q: the switch S₂‘ is closed or the switch S₂ is open.
Then the given circuit in symbolic form is :
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~q)
Using the laws of logic, we have,
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~ q)
= (p ∧ ~q) ∨ [(~p ∧ q) ∨ (~p ∧ ~q) …(By Complement Law)
= (p ∧ ~q) ∨ [~p ∧ (q ∨ ~q)} (By Distributive Law)
= (p ∧ ~q) ∨ (~p ∧ T) …(By Complement Law)
= (p ∧ ~q) ∨ ~ p …(By Identity Law)
= (p ∨ ~p) ∧ (~q ∨ ~p) …(By Distributive Law)
= ~q ∨ ~p …(By Identity Law)
= ~p ∨ ~p …(By Commutative Law)
Hence, the simplified circuit for the given circuit is :

(ii)

Solution:
(ii) Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
s : the switch S₄ is closed
t : the switch S₅ is closed
~ p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open
~ r : the switch S₃‘ is closed or the switch S₃ is open
~ s : the switch S₄‘ is closed or the switch S₄ is open
~ t : the switch S₅‘ is closed or the switch S₅ is open.
Then the given circuit in symbolic form is
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~t] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)]
Using the laws of logic, we have,
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~ t] ∧ [(p A q) ∨ (r ∧ s ∧ t)]
= [(p∧ q) ∨ ~(r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] … (By De Morgan’s Law)
= (p ∧ q) ∨ [ ~(r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] … (By Distributive Law)
= (p ∧ q) ∨ F … (By Complement Law)
= p ∧ q … (By Identity Law)
Hence, the alternative simplified circuit is :

Question 14.
Check whether the following switching circuits are logically equivalent – Justify.
(A)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
(A) The symbolic form of the given switching circuits are
p ∧ (q ∨ r) and (p ∧ q) ∨ (p ∧ r) respectively.
By Distributive Law, p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Hence, the given switching circuits are logically equivalent.

(B)

Solution:
The symbolic form of the given switching circuits are
(p ∨ q) ∧ (p ∨ r) and p ∨ (q ∧ r)
By Distributive Law,
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
Hence, the given switching circuits are logically equivalent.

Question 15.
Give alternative arrangement of the switching following circuit, has minimum switches.

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~p : the switch S₁‘ is closed, or the switch S₁ is open
~q : the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form Of the given circuit is :
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
≡ (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) y (p ∧ ~q ∧ r) …(By Commutative Law)
≡ (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Complement Law)
≡ F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [(~p ∨ p) ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) … (By Distributive Law)
≡ [T ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) = (q ∧ r) ∨ (p ∧ ~q ∧ r) …(By Complement Law)
≡ (q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [q ∨ (p ∧ ~q)] ∧ r … (By Distributive Law)
≡ [q ∨ p) ∧ ((q ∨ ~q)] ∧ r … (By Distributive Law)
≡ [(q ∨ p) ∧ T] ∧ r …(By Complement Law)
≡ (q ∨ p) ∧ r … (By Identity Law)
≡ (p ∨ q) ∧ r …(By Commutative Law)
∴ the alternative arrangement of the new circuit with minimum switches is :

Question 16.
Simplify the following so that the new circuit circuit.

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form of the given switching circuit is :
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
Using the laws of logic, we have,
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
≡ (~p ∨ q ∨ p ∨ ~q) ∨ (p ∨ q)
≡ [(~p ∨ p) ∨ (q ∨ ~q)] ∨ (p ∨ q) … (By Commutative Law)
≡ (T ∨ T) ∨ (p ∨ q) … (By Complement Law)
≡ T ∨ (p ∨ q) … (By Identity Law)
≡ T … (By Identity Law)
∴ the current always flows whether the switches are open or closed. So, it is not necessary to use any switch in the circuit.
∴ the simplified form of given circuit is :

Question 17.
Represent the following switching circuit in symbolic form and construct its switching table. Write your conclusion from the switching table.

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~ q : the switch S₂‘ is closed or the switch S₂ is open
~ r : the switch S₃‘ is closed or the switch S₃ is open.
Then, the symbolic form of the given switching circuit is : [p ∨ (~ q) ∨ (~ r)] ∧ [p ∨ (q ∧ r)]

From the table, the’ final column’ and the column of p are identical. Hence, the given circuit is equivalent to the simple circuit with only one switch S₁.
the simplified form of the given circuit is :

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.5

Question 1.
Express the following circuits in the symbolic form of logic and write the input-output table.
(i)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~p : the switch S₁‘ is closed or the switch S₁is open
~q : the switch S₂‘ is closed or the switch S₂ is open
~r : the switch S₃‘ is closed or the switch S₃ is open
l : the lamp L is on
(i) The symbolic form of the given circuit is : p ∨ (q ∧ r) = l
l is generally dropped and it can be expressed as : p ∨ (q ∧ r).

(ii)

Solution:
The symbolic form of the given circuit is : (~ p ∧ q) ∨ (p ∧ ~ q).

(iii)

Solution:
The symbolic form of the given circuit is : [p ∧ (~q ∨ r)] ∨ (~q ∧ ~ r).

(iv)

Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ q ∧ (r ∨ ~p).

(v)

Solution:
The symbolic form of the given circuit is : [p ∨ (~p ∧ ~q)] ∨ (p ∧ q).

(vi)

Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ (q ∨ r) ∧ (r ∨ p)

Question 2.
Construct the switching circuit of the following :
(i) (~p∧ q) ∨ (p∧ ~r)
Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open
~ r : the switch S₃‘ is closed or the switch S₃ is open.
Then the switching circuits corresponding to the given statement patterns are :

(ii) (p∧ q) ∨ [~p ∧ (~q ∨ p ∨ r)]
Solution:

(iii) [(p ∧ r) ∨ (~q ∧ ~r)] ∧ (~p ∧ ~r)
Solution:

(iv) (p ∧ ~q ∧ r) ∨ [p ∧ (~q ∨ ~r)]
Solution:

(v) p ∨ (~p ) ∨ (~q) ∨ (p ∧ q)
Solution:

(vi) (p ∧ q) ∨ (~p) ∨ (p ∧ ~q)
Solution:

Question 3.
Give an alternative equivalent simple circuits for the following circuits :
(i)

Solution:
(i) Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p : the switch S₁‘ is closed or the switch Si is open Then the symbolic form of the given circuit is :
p ∧ (~p ∨ q).
Using the laws of logic, we have,
p ∧ (~p ∨ q)
= (p ∧ ~ p) ∨ (p ∧ q) …(By Distributive Law)
= F ∨ (p ∧ q) … (By Complement Law)
= p ∧ q… (By Identity Law)
Hence, the alternative equivalent simple circuit is :

(ii)

Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~q : the switch S₂‘ is closed or the switch S₂ is open
~r : the switch S₃‘ is closed or the switch S₃ is open.
Then the symbolic form of the given circuit is :
[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p).
Using the laws of logic, we have
[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p)
≡ [p ∧ (q ∨ r)] ∨ [ ~(r ∨ q) ∧ p] …. (By De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] … (By Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~(q ∨ r)) … (By Distributive Law)
≡ p ∧ T … (By Complement Law)
≡ p … (By Identity Law)
Hence, the alternative equivalent simple circuit is :

Question 4.
Write the symbolic form of the following switching circuits construct its switching table and interpret it.
i)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form of the given circuit is :
(p ∨ ~q) ∨ (~p ∧ q)

Since the final column contains all’ 1′, the lamp will always glow irrespective of the status of switches.

ii)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~p : the switch S₁ is closed or the switch S₁ is open.
~q : the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form of the given circuit is : p ∨ (~p ∧ ~q) ∨ (p ∧ q)

Since the final column contains ‘0’ when p is 0 and q is ‘1’, otherwise it contains ‘1′.
Hence, the lamp will not glow when S₁ is OFF and S₂ is ON, otherwise the lamp will glow.

iii)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~q : the switch S₂‘ is closed or the switch S₂ is open
~r: the switch S₃‘ is closed or the switch S₃ is open.
Then the symbolic form of the given circuit is : [p ∨ (~q) ∨ r)] ∧ [p ∨ (q ∧ r)]

From the switching table, the ‘final column’ and the column of p are identical. Hence, the lamp will glow which S₁ is ‘ON’.

Question 5.
Obtain the simple logical expression of the following. Draw the corresponding switching circuit.
(i) p ∨ (q ∧ ~ q)
Solution:
Using the laws of logic, we have, p ∨ (q ∧ ~q)
≡ p ∨ F … (By Complement Law)
≡ p … (By Identity Law)
Hence, the simple logical expression of the given expression is p.
Let p : the switch S₁ is closed
Then the corresponding switching circuit is :

(ii) (~p ∧ q) ∨ (~p ∧ ~q) ∨ (p ∧ ~q)]
Solution:
Using the laws of logic, we have,
(~p ∧ q) ∨ (~p ∨ ~q) ∨ (p ∧ ~q)
≡ [~p ∧ (q ∨ ~q)] ∨ (p ∧ ~ q)… (By Distributive Law)
≡ (~p ∧ T) ∨ (p ∧ ~q) … (By Complement Law)
≡ ~p ∨ (p ∧ ~q) … (By Identity Law)
≡ (~p ∨ p) ∧ (~p ∧~q) … (By Distributive Law)
≡ T ∧ (~p ∧ ~q) … (By Complement Law)
≡ ~p ∨ ~q … (By Identity Law)
Hence, the simple logical expression of the given expression is ~ p ∨ ~q.
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open,
Then the corresponding switching circuit is :

(iii) [p (∨ (~q) ∨ ~r)] ∧ (p ∨ (q ∧ r)
Solution:
Using the laws of logic, we have,
[p ∨ (~ (q) ∨ (~r)] ∧ [p ∨ (q ∧ r)]
= [p ∨ { ~(q ∧ r)}] ∧ [p ∨ (q ∧ r)] … (By De Morgan’s Law)
= p ∨ [~(q ∧ r) ∧ (q ∧ r) ] … (By Distributive Law)
= p ∨ F … (By Complement Law)
= p … (By Identity Law)
Hence, the simple logical expression of the given expression is p.
Let p : the switch S₁ is closed
Then the corresponding switching circuit is :

(iv) (p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) ∨ (p ∧ q ∧ r)
Question is Modified
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r)∨ (p ∧ q ∧ r)
Solution:
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r)
= (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Commutative Law)
= (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Complement Law)
= F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)
= (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)
= (~ p ∨ p) ∧ (q ∧ r) … (By Distributive Law)
= T ∧ (q ∧ r) … (By Complement Law)
= q ∧ r … (By Identity Law)
Hence, the simple logical expression of the given expression is q ∧ r.
Let q : the switch S₂ is closed
r : the switch S₃ is closed.
Then the corresponding switching circuit is :

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.4

Question 1.
Using rules of negation write the negations of the following with justification.
(i) ~q → p
Solution:
The negation of ~q → p is
~(~q → p) ≡ ~ q ∧ ~p…. (Negation of implication)

(ii) p ∧ ~q
Solution:
The negation of p ∧ ~q is
~(p ∧ ~q) ≡ ~p ∨ ~(~q) … (Negation of conjunction)
≡ ~ p ∨ q … (Negation of negation)

(iii) p ∨ ~q
Solution:
The negation of p ∨ ~ p is
~ (p ∨ ~(q) ≡ ~p ∧ ~(~(q) … (Negation of disjunction)
≡ ~ p ∧ q … (Negation of negation)

(iv) (p ∨ ~q) ∧ r
Solution:
The negation of (p ∨ ~ q) ∧ r is
~[(p ∨ ~q) ∧ r] ≡ ~(p ∨ ~q) ∨ ~r … (Negation of conjunction)
≡ [ ~p ∧ ~(~q)] ∨ ~ r… (Negation of disjunction)
≡ (~ p ∧ q) ∧ ~ r … (Negation of negation)

(v) p → (p ∨ ~q)
Solution:
The negation of p → (p ∨ ~q) is
~ [p → (p ∨ ~q)] ≡ p ∧ ~ (p ∧ ~p) … (Negation of implication)
≡ p ∧ [ ~ p ∧ ~ (~(q)] … (Negation of disjunction)
≡ p ∧ (~ p ∧ q) (Negation of negation)

(vi) ~(p ∧ q) ∨ (p ∨ ~q)
Solution:
The negation of ~(p ∧ q) ∨ (p ∨ ~q) is
~[~(p ∧ q) ∨ (p ∨ ~q)] ≡ ~[~(p ∧ q)] ∧ ~(p ∨ ~q) … (Negation of disjunction)
≡ ~[~(p ∧ q)] ∧ [ p ∧ ~(~q)] … (Negation of disjunction)
≡ (p ∧ q) ∧ (~ p ∧ q) … (Negation of negation)

(vii) (p ∨ ~q) → (p ∧ ~q)
Solution:
The negation of (p ∨ ~q) → (p ∧ ~q) is
~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [ ~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~q) ∧ (~p ∨ q) … (Negation of negation)

(viii) (~ p ∨ ~q) ∨ (p ∧ ~q)
Solution:
The negation of (~ p ∨ ~q) ∨ (p ∧ ~ q) is
~ [(~p ∨ ~q) ∨ (p ∧ ~ q)]
≡ ~(~p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of disjunction)
≡ [~(~p) ∧ ~(~q)] ∧ [~p ∨ ~(~q)] … (Negation of disjunction and conjunction)
≡ (p ∧ q) ∧ (~p ∨ q) … (Negation of negation)

Question 2.
Rewrite the following statements without using if .. then.
(i) If a man is a judge then he is honest.
Solution:
Since p → ≡ ~p ∨ q, the given statements can be written as :
A man is not a judge or he is honest.

(ii) It 2 is a rational number then \(\sqrt {2}\) is irrational number.
Solution:
2 is not a rational number or \(\sqrt {2}\) is irrational number.

(iii) It f(2) = 0 then f(x) is divisible by (x – 2).
Solution:
f(2) ≠ 0 or f(x) is divisible by (x – 2).

Question 3.
Without using truth table prove that :
(i) p ↔ q ≡ (p∧ q) ∨ (~ p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p ↔ q) ∧ (q ↔ p) … (Biconditional Law)
≡ (~p ∨ q) ∧ (~q ∨ p) … (Conditional Law)
≡ [~p ∧ (~q ∨ p)] ∨ [q ∧ (~q ∨ p)] … (Distributive Law)
≡ [(~p ∧ ~q) ∨ (~p ∧ p)] ∨ [(q ∧ ~q) ∨ (q ∧ p)] … (Distributive Law)
≡ [(~p ∧ ~q) ∨ F] ∨ [F ∨ (q ∧ p)] … (ComplementLaw)
≡ (~ p ∧ ~ q) ∨ (q ∧ p) … (Identity Law)
≡ (~ p ∧ ~ q) ∨ (p ∧ q) … (Commutative Law)
≡ (p ∧ q) ∨ (~p ∧ ~q) … (Commutative Law)
≡ RHS.

(ii) (p ∨ q) ∧ (p ∨ ~q) ≡ p
Solution:
LHS = (p ∨ q) ∧ (p ∨ ~q)
≡ p ∨ (q ∧ ~q) … (Distributive Law)
≡ p ∨ F … (Complement Law)
≡ p … (Identity Law)
≡ RHS.

(iii) (p ∧ q) ∨ (~ p ∧ q) ∨ (p ∧ ~q) ≡ p ∨ q
Solution:
LHS = (p ∧ q) v (~p ∧ q) ∨ (p ∧ ~q)
≡ [(p ∨ ~p) ∧ q] ∨ (p ∧ ~q) … (Distributive Law)
≡ (T ∧ q) ∨ (p ∧ ~q) … (Complement Law)
≡ q ∨ (p ∧ ~q) … (Identity Law)
≡ (q ∨ p) ∧ (q ∨ ~q) … (Distributive Law)
≡ (q ∨ p) ∧ T .. (Complement Law)
≡ q ∨ p … (Identity Law)
≡ p ∨ q … (Commutative Law)
≡ RHS.

(iv) ~[(p ∨ ~q) → (p ∧ ~q)] ≡ (p ∨ ~q) ∧ (~p ∨ q)
Solution:
LHS = ~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~p ∨ q)… (Negation of negation)
≡ RHS.