Inter 1st Year

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Products of Vectors Solutions Exercise 5(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Products of Vectors Solutions Exercise 5(b)

I.

Question 1.
If \(|\overline{\mathbf{p}}|=2,|\overline{\mathbf{q}}|=3\) and \((\bar{p}, \bar{q})=\frac{\pi}{6}\), then find \(|\bar{p} \times \bar{q}|^{2}\).
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\), then find \(|\overline{\mathbf{a}} \times \overline{\mathbf{b}}|\).
Solution:

Question 3.
If \(\bar{a}=2 \bar{i}-3 \bar{j}+\overline{\mathbf{k}}\) and \(\bar{b}=\bar{i}+4 \bar{j}-2 \bar{k}\), then find \((\overline{\mathbf{a}}+\overline{\mathbf{b}}) \times(\overline{\mathbf{a}}-\overline{\mathbf{b}})\).
Solution:

Question 4.
If \(4 \bar{i}+\frac{2 p}{3} \bar{j}+p \bar{k}\) is parallel to the vector \(\overline{\mathbf{i}}+2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}\), find p.
Solution:

Question 5.
Compute \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}}+\overline{\mathbf{c}})+\overline{\mathbf{b}} \times(\overline{\mathbf{c}}+\overline{\mathbf{a}})+\overline{\mathbf{c}} \times(\overline{\mathbf{a}}+\overline{\mathbf{b}})\)
Solution:

Question 6.
If \(\overline{\mathbf{p}}=\mathbf{x} \overline{\mathbf{i}}+\mathbf{y} \overline{\mathbf{j}}+\mathbf{z} \overline{\mathbf{k}}\), find the value of \(|\overline{\boldsymbol{p}} \times \overline{\mathbf{k}}|^{2}\)
Solution:

Question 7.
Compute \(2 \bar{j} \times(3 \bar{i}-4 \bar{k})+(\bar{i}+2 \hat{j}) \times \bar{k}\).
Solution:

Question 8.
Find a unit vector perpendicular to both \(\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(2 \bar{i}+\bar{j}+3 \bar{k}\).
Solution:

Question 9.
If θ is the angle between the vectors \(\overline{\mathbf{i}}+\overline{\mathbf{j}}\) and \(\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then find sin ?.
Solution:

Question 10.
Find the area of the parallelogram having \(\bar{a}=2 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=-\overline{\mathbf{i}}+\overline{\mathbf{k}}\) as adjacent sides.
Solution:
Vector area of the parallelogram having \(\bar{a}=2 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=-\overline{\mathbf{i}}+\overline{\mathbf{k}}\) as adjacent sides.

Question 11.
Find the area of the parallelogram, whose diagonals are \(3 \overline{\mathbf{i}}+\overline{\mathbf{j}}-2 \overline{\mathbf{k}}\) and \(\bar{i}-3 \bar{j}+4 \bar{k}\).
Solution:

Question 12.
Find the area of the triangle having \(3 \bar{i}+4 \bar{j}\) and \(-5 \bar{i}+7 \bar{j}\) as two of its sides.
Solution:

Question 13.
Find unit vector perpendicular to the plane determined by the vectors \(\bar{a}=4 \bar{i}+3 \bar{j}-\bar{k}\) and \(\overline{\mathbf{b}}=2 \tilde{i}-6 \overline{\mathbf{j}}-3 \overline{\mathbf{k}}\).
Solution:

Question 14.
Find the area of the triangle whose vertices are A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2).
Solution:
Suppose \(\bar{i}, \bar{j}, \bar{k}\) are unit vectors along the co-ordinate axes.
Position vectors of A, B, C are

II.

Question 1.
If \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}=\overline{\mathbf{0}}\), then prove that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{b}} \times \overline{\mathbf{c}}=\overline{\mathbf{c}} \times \overline{\mathbf{a}}\).
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \bar{i}+\bar{j}-\bar{k}, \quad \bar{b}=-\bar{i}+2 \bar{j}-4 \bar{k}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}+\mathbf{j}+\overline{\mathbf{k}}\), then find \((\bar{a} \times \bar{b}) \cdot(\bar{b} \times \bar{c})\).
Solution:

Question 3.
Find the vector area and the area of the parallelogram having \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+2 \overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\mathbf{2} \overline{\mathbf{k}}\) as adjacent sides.
Solution:

Question 4.
If \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{b}} \times \overline{\mathbf{c}} \neq \overline{\mathbf{0}}\), show that, \(\overline{\mathbf{a}}+\overline{\mathbf{c}}=\mathbf{p} \overline{\mathbf{b}}\), where p is some scalar.
Solution:

Question 5.
Let \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) be vectors, satisfying \(|\overline{\mathbf{a}}|=|\overline{\mathbf{b}}|=5\) and \((\bar{a}, \bar{b})=45^{\circ}\). Find the area of the triangle having \(\overline{\mathbf{a}}-\mathbf{2} \overline{\mathbf{b}}\) and \(3 \bar{a}+2 \bar{b}\) as two of its sides.
Solution:

Question 6.
Find the vector having magnitude ?6 units and perpendicular to both \(\mathbf{2} \overline{\mathbf{i}}-\overline{\mathbf{k}}\) and \(\mathbf{3} \overline{\mathbf{i}}-\overline{\mathbf{j}}-\overline{\mathbf{k}}\).
Solution:

Question 7.
Find a unit vector perpendicular to the plane determined by the points P(1, -1, 2), Q(2, 0, -1) and R(0, 2, 1).
Solution:

Question 8.
If \(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\overline{\mathbf{a}} \cdot \overline{\mathbf{c}}\) and \(\overline{\mathbf{a}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{c}}, \overline{\mathbf{a}} \neq 0\), then show that \(\overline{\mathbf{b}}=\overline{\mathbf{c}}\).
Solution:

Question 9.
Find a vector of magnitude 3 and perpendicular to both the vector \(\overline{\mathbf{b}}=2 \bar{i}-2 \bar{j}+\bar{k}\) and \(\bar{c}=2 \bar{i}+2 \bar{j}+3 \bar{k}\).
Solution:

Question 10.
If \(|\overline{\mathbf{a}}|\) = 13, \(|\overline{\mathbf{b}}|\) = 5 and \(\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}=\mathbf{6 0}\), then find \(|\overline{\mathbf{a}} \times \overline{\mathbf{b}}|\).
Solution:

Question 11.
Find a unit vector perpendicular to the plane passing through the points (1, 2, 3), (2, -1, 1) and (1, 2, -4).
Solution:
Let ‘O’ be the origin and let A, B, C be the given points.

III.

Question 1.
If \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) represent the vertices A, B and C respectively of ∆ABC, then prove that \(|(\overline{\mathbf{a}} \times \overline{\mathbf{b}})+(\overline{\mathbf{b}} \times \overline{\mathbf{c}})+(\overline{\mathbf{c}} \times \overline{\mathbf{a}})|\) is twice the area of ∆ABC.
Solution:

Question 2.
If \(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+4 \overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then compute \(\overline{\mathbf{a}} \times(\overline{\mathbf{b}} \times \overline{\mathbf{c}})\) and verify that it is perpendicular to \(\bar{a}\).
Solution:

Question 3.
If \(\overline{\mathbf{a}}=7 \overline{\mathbf{i}}-2 \overline{\mathbf{j}}+3 \overline{\mathbf{k}}, \overline{\mathbf{b}}=2 \overline{\mathbf{i}}+8 \overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) then compute \(\overline{\mathbf{a}} \times \mathbf{b}, \overline{\mathbf{a}} \times \overline{\mathbf{c}}\) and \(\bar{a} \times(\bar{b}+\bar{c})\). Verify whether cross product is distributive over vector addition.
Solution:

Question 4.
If \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{c}}=\overline{\mathbf{j}}-\overline{\mathbf{k}}\), then find vector \(\overline{\mathbf{b}}\) such that \(\overline{\mathbf{a}} \times \overline{\mathbf{b}}=\overline{\mathbf{c}}\) and \(\bar{a} \cdot \bar{b}=3\).
Solution:

Question 5.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are three vectors of equal magnitudes and each of them is inclined at an angle of 60° to the others. If \(|\bar{a}+\bar{b}+\bar{c}|=\sqrt{6}\), then find \(|\overline{\mathbf{a}}|\).
Solution:

Question 6.
For any two vectors \(\bar{a}\) and \(\bar{b}\), show that \(\left(1+|\bar{a}|^{2}\right)\left(1+|\bar{b}|^{2}\right)\) = \(|\mathbf{1}-\overline{\mathbf{a}} \cdot \overline{\mathbf{b}}|^{2}+|\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{a}} \times \overline{\mathbf{b}}|^{2}\)
Solution:

Question 7.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are unit vectors such that \(\overline{\mathbf{a}}\) is perpendicular to the plane of \(\overline{\mathbf{b}}, \overline{\mathbf{c}}\) and the angle between \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) is \(\frac{\pi}{3}\), then find \(|\bar{a}+\bar{b}+\bar{c}|\).
Solution:
Given that \(|\bar{a}|=|\bar{b}|=|\bar{c}|=1\)

Question 8.
\(\overline{\mathbf{a}}=3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}, \overline{\mathbf{b}}=-\overline{\mathbf{i}}+3 \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\), \(\bar{c}=4 \bar{i}+5 \bar{j}-2 \bar{k}\) and \(\bar{d}=\bar{i}+3 \bar{j}+5 \bar{k}\) then compute the following.
(i) \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \times(\bar{c} \times \bar{d})\) and
(ii) \((\overline{\mathbf{a}} \times \overline{\mathbf{b}}) \cdot \overline{\mathbf{c}}-(\overline{\mathbf{a}} \times \overline{\mathbf{d}}) \cdot \overline{\mathbf{b}}\)
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Products of Vectors Solutions Exercise 5(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Products of Vectors Solutions Exercise 5(a)

I.

Question 1.
Find the angle between the vectors \(\bar{i}+2 \bar{j}+3 \bar{k}\) and \(3 \bar{i}-\bar{j}+2 \bar{k}\).
Solution:
Let \(\overline{\mathrm{a}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+3 \overline{\mathrm{k}}\) and \(\overline{\mathrm{b}}=3 \overline{\mathrm{i}}-\overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) and ‘θ’ be the angle between them (i.e.,) \((\bar{a}, \bar{b})\) = θ

Question 2.
If the vectors \(\mathbf{2} \overline{\mathbf{i}}+\lambda \overline{\mathbf{j}}-\overline{\mathbf{k}}\) and \(4 \bar{i}-2 \bar{j}+2 \bar{k}\) are perpendicular to each other, then find λ.
Solution:

Question 3.
For what values of λ, the vectors \(\overline{\mathbf{i}}-\lambda \overline{\mathbf{j}}+2 \overline{\mathbf{k}}\) and \(8 \overline{\mathbf{i}}+6 \overline{\mathbf{j}}-\overline{\mathbf{k}}\) are at right angles?
Solution:

Question 4.
\(\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-\overline{\mathbf{j}}+\overline{\mathbf{k}}, \overline{\mathbf{b}}=\overline{\mathbf{i}}-3 \overline{\mathbf{j}}-5 \overline{\mathbf{k}}\). Find the vector C such that \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) form the sides of a triangle.
Solution:

Question 5.
Find the angle between the planes \(\bar{r} \cdot(2 \bar{i}-\bar{j}+2 \bar{k})=3\) and \(\overline{\mathrm{r}} \cdot(3 \overline{\mathrm{i}}+6 \bar{j}+\bar{k})=4\).
Solution:

Question 6.
Let \(\overline{\mathbf{e}}_{1}\) and \(\overline{\mathbf{e}}_{2}\) be unit vectors makingangle θ. If \(\frac{1}{2}\left|\bar{e}_{1}-\bar{e}_{2}\right|=\sin \lambda \theta\), then find λ.
Solution:

Question 7.
Let \(\overline{\mathbf{a}}=\overline{\mathbf{i}}+\overline{\mathbf{j}}+\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}=\mathbf{2} \overline{\mathbf{i}}+3 \overline{\mathbf{j}}+\overline{\mathbf{k}}\). Find
(i) The projection vector of \(\overline{\mathbf{b}}\) on \(\overline{\mathbf{a}}\) and its magnitude.
(ii) The vector components of \(\overline{\mathbf{b}}\) in the direction of a and perpendicular to \(\overline{\mathbf{a}}\).
Solution:

Question 8.
Find the equation of the plane through the point (3, -2, 1) and perpendicular to the vector (4, 7, -4).
Solution:

Question 9.
If \(\overline{\mathbf{a}}=2 \bar{i}+2 \bar{j}-3 \bar{k}\); \(\overline{\mathbf{b}}=3 \overline{\mathbf{i}}-\overline{\mathbf{j}}+2 \overline{\mathbf{k}}\), then find the angle between \(2 \overline{\mathbf{a}}+\overline{\mathbf{b}}\) and \(\bar{a}+2 \bar{b}\).
Solution:

II.

Question 1.
Find unit vector parallel to the XOY- plane and perpendicular to the vector \(4 \bar{i}-3 \bar{j}+\bar{k}\).
Solution:
Any vector parallel to XOY-plane will be of the form \(p \bar{i}+q \bar{j}\)
∴ The vector parallel to the XOY-plane and perpendicular to the vector \(4 \bar{i}-3 \bar{j}+\bar{k}\) is \(3 \bar{i}+4 \bar{j}\)
Its magnitude = \(|3 \bar{i}+\overline{4 j}|=\sqrt{9+16}=5\)
∴ Unit vector parallel to the XOY-plane and perpendicular to the vector \(4 \bar{i}-3 \bar{j}+\bar{k}\) is \(\pm \frac{(3 \overline{\mathrm{i}}+4 \overline{\mathrm{j}})}{5}\)

Question 2.
If \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathrm{c}}=0,|\overline{\mathbf{a}}|=3,|\overline{\mathbf{b}}|=5\) and \(|\bar{c}|=7\), then find the angle between \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:

Question 3.
If \(|\overline{\mathbf{a}}|\) = 2, \(|\overline{\mathbf{b}}|\) = 3 and \(|\overline{\mathbf{c}}|\) = 4 and each of \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) is perpendicular to the sum of the other two vectors, then find the magnitude of \(\overline{\mathbf{a}}+\overline{\mathbf{b}}+\overline{\mathbf{c}}\).
Solution:

Question 4.
Find the equation of the plane passing through the point \(\overline{\mathbf{a}}=\mathbf{2} \overline{\mathbf{i}}+3 \bar{j}-\overline{\mathbf{k}}\) and perpendicular to the vector \(3 \bar{i}-2 \bar{j}-2 \bar{k}\) and the distance of this plane from the origin.
Solution:
Equation of the plane passing through the point \(\overline{\mathbf{a}}=\mathbf{2} \overline{\mathbf{i}}+3 \bar{j}-\overline{\mathbf{k}}\) and perpendicular to the vector \(\bar{n}=3 \bar{i}-2 \bar{j}-2 \bar{k}\) is

Question 5.
\(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\) are the position vectors of four coplanar points such that \((\mathbf{a}-\overline{\mathbf{d}}) \cdot(\bar{b}-\bar{c})=(\bar{b}-\bar{d}) \cdot(\bar{c}-\bar{a})=0\). Show that the point \(\bar{d}\) represents the orthocentre of the triangle with \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) as its vertices.
Solution:
Position vectors of A, B, C, D are \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) respectively.


⇒ BD is perpendicular to AC
∴ BD is another altitude of ∆ABC.
Altitudes AD and BD intersect at D.
∴ D is the orthocentre of ∆ABC.

III.

Question 1.
Show that the points (5, -1, 1), (7, -4, 7), (1, -6, 10) and (-1, -3, 4) are the vertices of a rhombus.
Solution:
Let A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4) are the given points.

∵ AB = BC = CD = DA = 7 units
AC ≠ BD
∴ A, B, C, D points are the vertices of a rhombus.

Question 2.
Let \(\bar{a}=4 \bar{i}+5 \bar{j}-\bar{k}, \quad \bar{b}=\bar{i}-4 \bar{j}+5 \bar{k}\) and \(\overline{\mathbf{c}}=\mathbf{3} \overline{\mathbf{i}}+\overline{\mathbf{j}}-\overline{\mathbf{k}}\). Find the vector which is perpendicular to both \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) and whose magnitude is twenty one times the magnitude of \(\overline{\mathbf{c}}\).
Solution:

Question 3.
G is the centroid of ΔABC and a, b, c are the lengths of the sides BC, CA and AB respectively prove that a² + b² + c² = 3 (OA² + OB² + OC²) – 9(OG)² where O is any point.
Solution:
Given that \(\overline{\mathrm{BC}}=\overline{\mathrm{a}}, \overline{\mathrm{CA}}=\overline{\mathrm{b}}, \overline{\mathrm{AB}}=\overline{\mathrm{c}}\)
Let ‘O’ be the origin and let \(\overline{\mathrm{OA}}=\overline{\mathrm{p}}, \overline{\mathrm{OB}}=\overline{\mathrm{q}} \text { and } \overline{\mathrm{OC}}=\overline{\mathrm{r}}\)
Then P.V. of the centroid of ΔABC is

From (1) and (2)
∴ a² + b² + c² = 3(OA² + OB² + OC²) – 9(OG)².

Question 4.
A line makes angles θ₁, θ₂, θ₃, and θ₄ with the diagonals of a cube. Show that cos²θ₁ + cos²θ₂ + cos²θ₃ + cos²θ₄ = \(\frac{4}{3}\).
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Addition of Vectors Solutions Exercise 4(b)

I.

Question 1.
Find the vector equation of the line passing through the point \(2 \bar{i}+3 \bar{j}+\bar{k}\) and parallel to the vector \(4 \bar{i}-2 \bar{j}+3 \bar{k}\).
Solution:

Question 2.
OABC is a parallelogram. If \(\overline{O A}=\bar{a}\) and \(\overline{O C}=\bar{c}\), find the vector equation of the side BC.
Solution:

Question 3.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are the position vectors of the vertices A, B and C respectively of ∆ABC, theind the vector equation of the median through the vertex A.
Solution:

Question 4.
Find the vertor equation of the line joining the points \(2 \bar{i}+\bar{j}+3 \bar{k}\) and \(-4 \bar{i}+3 \bar{j}-\bar{k}\).
Solution:

Question 5.
Find the vector equation of the plane passing through the points \(\overline{\mathbf{i}}-2 \overline{\mathbf{j}}+5 \overline{\mathbf{k}},-5 \overline{\mathbf{j}}-\overline{\mathbf{k}} \text { and }-3 \overline{\mathbf{i}}+5 \overline{\mathbf{j}}\).
Solution:

Question 6.
Find the vector equation of the plane through the points (0, 0, 0), (0, 5, 0) and (2, 0, 1).
Solution:

II.

Question 1.
If \(\overline{\mathbf{a}}, \overline{\mathbf{b}}, \overline{\mathbf{c}}\) are noncoplanar find the point of intersection of the line passing through the points \(2 \bar{a}+3 \bar{b}-\bar{c}\), \(3 \bar{a}+4 \bar{b}-2 \bar{c}\) with the line joining the points \(\bar{a}-2 \bar{b}+3 \bar{c}, \bar{a}-6 \bar{b}+6 \bar{c}\).
Solution:

Question 2.
ABCD is a trapezium in which AB and CD are parallel. Prove by vector methods, that the mid points of the sides AB, CD and the intersection of the diagonals are collinear.
Solution:




⇒ M, P, N are collinear
Hence the midpoints of parallel sides of a trapezium and the point of intersection of the diagonals are collinear.

Question 3.
In a quadrilateral ABCD, if the midpoints of one pair of opposite sides and the point of intersection of the diagonals are collinear, using vector methods, prove that the quadrilateral ABCD is a trapezium.
Solution:

III.

Question 1.
Find the vector equation of the plane which passes through the points \(2 \bar{i}+4 \bar{j}+2 \bar{k}, 2 \bar{i}+3 \bar{j}+5 \bar{k}\) and parallel to the vector \(3 \overline{\mathbf{i}}-2 \overline{\mathbf{j}}+\overline{\mathbf{k}}\). Also find the point where this plane meets the line joining the points \(2 \overline{\mathbf{i}}+\overline{\mathbf{j}}+3 \overline{\mathbf{k}}\) and \(4 \bar{i}-2 \bar{j}+3 \bar{k}\).
Solution:

Question 2.
Find the vector equation of the plane passing through points \(4 \overline{\mathbf{i}}-3 \overline{\mathbf{j}}-\overline{\mathbf{k}}\), \(3 \overline{\mathbf{i}}+7 \overline{\mathbf{j}}-10 \overline{\mathbf{k}}\) and \(2 \bar{i}+5 \bar{j}-7 \bar{k}\), and show that the point \(\overline{\mathbf{i}}+2 \bar{j}-3 \overline{\mathbf{k}}\) lies in the plane.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(i) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(i)

Solve the following systems of homogeneous equations.

Question 1.
2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0
Solution:
The Coefficient matrix is \(\left[\begin{array}{ccc}
2 & 3 & -1 \\
1 & -1 & -2 \\
3 & 1 & 3
\end{array}\right]\)
det of \(\left[\begin{array}{ccc}
2 & 3 & -1 \\
1 & -1 & -2 \\
3 & 1 & 3
\end{array}\right]=\left[\begin{array}{ccc}
2 & 3 & -1 \\
1 & -1 & -2 \\
3 & 1 & 3
\end{array}\right]\)
= 2(-3 + 2) – 3(3 + 6) – 1(1 + 3)
= -2 – 27 – 4
= -33 ≠ 0, ρ(A) = 3
Hence the system has the trivial solution x = y = z = 0 only.

Question 2.
3x + y – 2z = 0, x + y + z = 0, x – 2y + z = 0
Hint: If the determinant of the coefficient matrix ≠ 0 then the system has a trivial solution (i.e.) ρ(A) = 3.
Solution:
The coefficient matrix is \(\left[\begin{array}{ccc}
3 & 1 & -2 \\
1 & 1 & 1 \\
1 & -2 & 1
\end{array}\right]\)
\(\left|\begin{array}{ccc}
3 & 1 & -2 \\
1 & 1 & 1 \\
1 & -2 & 1
\end{array}\right|\)
= 3(1 + 2) – 1(1 – 1) – 2(-2 – 1)
= 9 + 6
= 15 ≠ 0, ρ(A) = 3
Hence the system has the trivial solutions x = y = z = 0 only.

Question 3.
x + y – 2z = 0, 2x + y – 3z = 0, 5x + 4y – 9z = 0
Solution:
The coefficient matrix is \(\left[\begin{array}{rrr}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\) = A (say)
|A| = \(\left|\begin{array}{rrr}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right|\)
= 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5)
= 3 + 3 – 6
= 0
∴ Rank of A = 2 as the submatrix \(\left[\begin{array}{ll}
1 & 1 \\
2 & 1
\end{array}\right]\) is non-singular, ρ(A) < 3
Hence the system has a non-trivial solution.
A = \(\left[\begin{array}{lll}
1 & 1 & -2 \\
2 & 1 & -3 \\
5 & 4 & -9
\end{array}\right]\)
R₂ → R₂ – 2R₁, R₃ → R₃ – 3R₁
A ~ \(\left[\begin{array}{ccc}
1 & 1 & -2 \\
0 & -1 & 1 \\
0 & -1 & -1
\end{array}\right]\)
The system of equation is equivalent to the given system of equations are x + y – 2z = 0, -y + z = 0
Let z = k
⇒ y = k, x = k
∴ x = y = z = k for real number k.

Question 4.
x + y – z = 0, x – 2y + z = 0, 3x + 6y – 5z = 0
Solution:
Coefficient matrix A = \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
1 & -2 & 1 \\
3 & 6 & -5
\end{array}\right]\)
R₂ → R₂ – R₁, R₃ → R₃ – 3R₁
A ~ \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
0 & -3 & 2 \\
0 & 3 & -2
\end{array}\right]\)
⇒ det A = 0 as R₂, R₃ are identical.
and rank (A) = 2 as the submatrix \(\left[\begin{array}{cc}
1 & 1 \\
0 & -3
\end{array}\right]\) is non-singular.
Hence the system has a non-trivial solution, ∵ ρ(A) < 3
The system of equations equivalent to the given system of equations are
x + y – z = 0
3y – 2z = 0
Let z = k
⇒ y = \(\frac{2 k}{3}\)
x = \(\frac{k}{3}\)
∴ x = \(\frac{k}{3}\), y = \(\frac{2 k}{3}\), z = k for any real number of k.

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(h) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(h)

Solve the following systems of equations.
(i) by using Cramer’s rule and matrix inversion method, when the coefficient matrix is non-singular.
(ii) by using the Gauss-Jordan method. Also, determine whether the system has a unique solution or an infinite number of solutions, or no solution, and find the solutions if exist.

Question 1.
5x – 6y + 4z = 15
7x + 4y – 3z = 19
2x + y + 6z = 46
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
5 & -6 & 4 \\
7 & 4 & -3 \\
2 & 1 & 6
\end{array}\right|\)
= 5(24 + 3) + 6(42 + 6) + 4(7 – 8)
= 135 + 288 – 4
= 419
Δ₁ = \(\left|\begin{array}{ccc}
15 & -6 & 4 \\
19 & 4 & -3 \\
46 & 1 & 6
\end{array}\right|\)
= 15(24 + 3) + 6(114 + 138) + 4(19 – 184)
= 405 + 1512 – 660
= 1917 – 660
= 1257
Δ₂ = \(\left|\begin{array}{ccc}
5 & 15 & 4 \\
7 & 19 & -3 \\
2 & 46 & 6
\end{array}\right|\)
= 5(114 + 138) – 15(42 + 6) + 4(322 – 38)
= 1260 – 720 + 1136
= 1676
Δ₃ = \(\left|\begin{array}{ccc}
5 & -6 & 15 \\
7 & 4 & 19 \\
2 & 1 & 46
\end{array}\right|\)
= 5(184 – 19) + 6(322 – 38) + 15(7 – 8)
= 825 + 1704 – 15
= 2529 – 15
= 2514

Solution is x = 3, y = 4, z = 6.

(ii) Matrix inversion method:



Solution is x = 3, y = 4, z = 6

(iii) Gauss-Jordan method:

∴ Unique solution exists.
Solution is x = 3, y = 4, z = 6.

Question 2.
x + y + z = 1
2x + 2y + 3z = 6
x + 4y + 9z = 3
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(18 – 3) + 1(8 – 2)
= 6 – 15 + 6
= -3
Δ₁ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
6 & 2 & 3 \\
3 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(54 – 9) + 1(24 – 6)
= 6 – 45 + 18
= -21
Δ₂ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 6 & 3 \\
1 & 3 & 9
\end{array}\right|\)
= 1(54 – 9) – 1(18 – 3) + 1(6 – 6)
= 45 – 15
= 30
Δ₃ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 6 \\
1 & 4 & 3
\end{array}\right|\)
= 1(6 – 24) – 1(6 – 6) + 1(8 – 2)
= -18 – 0 + 6
= -12

Solution is x = 7, y = -10, z = 4

(ii) Matrix inversion method:



∴ Solution is x = 7, y = -10, z = 4

(iii) Gauss-Jordan method:
Augmented matrix is A = \(\left[\begin{array}{llll}
1 & 1 & 1 & 1 \\
2 & 2 & 3 & 6 \\
1 & 4 & 9 & 3
\end{array}\right]\)
R₂ → R₂ – 2R₁, R₃ → R₃ – R₁

Unique solution exists.
∴ Solution is x = 7, y = -10, z = 4

Question 3.
x – y + 3z = 5
4x + 2y – z = 0
-x + 3y + z = 5
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
1 & -1 & 3 \\
4 & 2 & -1 \\
-1 & 3 & 1
\end{array}\right|\)
= 1(2 + 3) + 1(4 – 1) + 3(12 + 2)
= 5 + 3 + 42
= 50
Δ₁ = \(\left|\begin{array}{ccc}
5 & -1 & 3 \\
0 & 2 & -1 \\
5 & 3 & 1
\end{array}\right|\)
= 5(2 + 3) + 1(0 + 5) + 3(0 – 10)
= 25 + 5 – 30
= 0
Δ₂ = \(\left|\begin{array}{ccc}
1 & 5 & 3 \\
4 & 0 & -1 \\
-1 & 5 & 1
\end{array}\right|\)
= 1(0 + 5) – 5(4 – 1) + 3(20 – 0)
= 5 – 15 + 60
= 50
Δ₃ = \(\left|\begin{array}{ccc}
1 & -1 & 5 \\
4 & 2 & 0 \\
-1 & 3 & 5
\end{array}\right|\)
= 1(10 – 0) + 1(20 – 0) + 5(12 + 2)
= 10 + 20 + 70
= 100

∴ Solution is x = 0, y = 1, z = 2.

(ii) Matrix inversion method:



Solution is x = 0, y = 1, z = 2

(iii) Gauss Jordan method:

Unique solution exists.
∴ Solution is x = 0, y = 1, z = 2

Question 4.
2x + 6y + 11 = 0
6x + 20y – 6z + 3 = 0
6y – 18z + 1 = 0
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
2 & 6 & 0 \\
6 & 20 & -6 \\
0 & 6 & -18
\end{array}\right|\)
= 2(-360 + 36) – 6(-108 – 0)
= -648 + 648
= 0
∴ Cramer’s rule and matrix inversion method cannot be used.
∵ Δ = 0

(ii) Gauss Jordan method:

ρ(A) = 2, ρ(AB) = 3
ρ(A) ≠ ρ(AB)
∴ The given system of equations does not have a solution.

Question 5.
2x – y + 3z = 9
x + y + z = 6
x – y + z = 2
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
1 & 1 & 1 \\
1 & -1 & 1
\end{array}\right|\)
= 2(1 + 1) + 1(1 – 1) + 3(-1 – 1)
= 4 + 0 – 6
= -2
Δ₁ = \(\left|\begin{array}{ccc}
9 & -1 & 3 \\
6 & 1 & 1 \\
2 & -1 & 1
\end{array}\right|\)
= 9(1 + 1) + 1(6 – 2) + 3(-6 – 2)
= 18 + 4 – 24
= -2
Δ₂ = \(\left|\begin{array}{lll}
2 & 9 & 3 \\
1 & 6 & 1 \\
1 & 2 & 1
\end{array}\right|\)
= 2(6 – 2) – 9(1 – 1) + 3(2 – 6)
= 8 – 0 – 12
= -4
Δ₃ = \(\left|\begin{array}{ccc}
2 & -1 & 9 \\
1 & 1 & 6 \\
1 & -1 & 2
\end{array}\right|\)
= 2(2 + 6) + 1(2 – 6) + 9(-1 – 1)
= 16 – 4 – 18
= -6

Solution is x = 1, y = 2, z = 3.

(ii) Matrix inversion method:


Solution is x = 1, y = 2, z = 3.

(iii) Gauss-Jordan method:

∴ The given equations have a unique solution.
Solution is x = 1, y = 2, z = 3

Question 6.
2x – y + 8z = 13
3x + 4y + 5z = 18
5x – 2y + 7z = 20
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
2 & -1 & 8 \\
3 & 4 & 5 \\
5 & -2 & 7
\end{array}\right|\)
= 2(28 + 10) + 1(21 – 25) + 8(-6 – 20)
= 76 – 4 – 208
= -136
Δ₁ = \(\left|\begin{array}{ccc}
13 & -1 & 8 \\
18 & 4 & 5 \\
20 & -2 & 7
\end{array}\right|\)
= 13(28 + 10) + 1(126 – 100) + 8(-36 – 80)
= 494 + 26 – 928
= -408
Δ₂ = \(\left|\begin{array}{lll}
2 & 13 & 8 \\
3 & 18 & 5 \\
5 & 20 & 7
\end{array}\right|\)
= 2(126 – 100) – 13(21 – 25) + 8(60 – 90)
= 52 + 52 – 240
= -136
Δ₃ = \(\left|\begin{array}{ccc}
2 & -1 & 13 \\
3 & 4 & 18 \\
5 & -2 & 20
\end{array}\right|\)
= 2(80 + 36) + 1(60 – 90) + 13(-6 – 20)
= 232 – 30 – 338
= -136

∴ Solution is x = 3, y = 1, z = 1

(ii) Matrix inversion method:



∴ Solution is x = 3, y = 1, z = 1

(iii) Gauss Jordan method:

∴ The given equations have a unique solution and Solution is x = 3, y = 1, z = 1.

Question 7.
2x – y + 3z = 8
-x + 2y + z = 4
3x + y – 4z = 0
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
-1 & 2 & 1 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-8 – 1) + 1(4 – 3) + 3(-1 – 6)
= -18 + 1 – 21
= -38
Δ₁ = \(\left|\begin{array}{ccc}
8 & -1 & 3 \\
4 & 2 & 1 \\
0 & 1 & -4
\end{array}\right|\)
= 8(-8 – 1) + 1(-16 – 0) + 3(4 – 0)
= -72 – 16 + 12
= -76
Δ₂ = \(\left|\begin{array}{ccc}
2 & 8 & 3 \\
-1 & 4 & 1 \\
3 & 0 & -4
\end{array}\right|\)
= 2(-16 – 0) – 8(4 – 3) + 3(-0 – 12)
= -32 – 8 – 36
= -76
Δ₃ = \(\left|\begin{array}{ccc}
2 & -1 & 8 \\
-1 & 2 & 4 \\
3 & 1 & 0
\end{array}\right|\)
= 2(0 – 4) + 1(0 – 12) + 8(-1 – 6)
= -8 – 12 – 56
= -76

∴ Solution is x = 2, y = 2, z = 2.

(ii) Matrix inversion method:


Solution is x = 2, y = 2, z = 2

(iii) Gauss Jordan method:


∴ The given equations have a unique solution and solution is x = 2, y = 2, z = 2.

Question 8.
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0
Solution:
(i) Cramer’s rule:
Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 5 & 7 \\
2 & 1 & -1
\end{array}\right|\)
= 1(-5 – 7) – 1(-2 – 14) + 1(2 – 10)
= -12 + 16 – 8
= -4
Δ₁ = \(\left|\begin{array}{ccc}
9 & 1 & 1 \\
52 & 5 & 7 \\
0 & 1 & -1
\end{array}\right|\)
= 9(-5 – 7) – 1(-52 – 0) + 1(52 – 0)
= -108 + 52 + 52
= -4
Δ₂ = \(\left|\begin{array}{ccc}
1 & 9 & 1 \\
2 & 52 & 7 \\
2 & 0 & -1
\end{array}\right|\)
= 1(-52 – 0) – 9(-2 – 14) + 1(0 – 104)
= -52 + 144 – 104
= -12
Δ₃ = \(\left|\begin{array}{ccc}
1 & 1 & 9 \\
2 & 5 & 52 \\
2 & 1 & 0
\end{array}\right|\)
= 1(0 – 52) – 1(0 – 104) + 9(2 – 10)
= -52 + 104 – 72
= -20

(ii) Matrix inversion method:


Solution is x = 1, y = 3, z = 5

(iii) Gauss Jordan method:

∴ The given equations have a unique solution and solution is x = 1, y = 3, z = 5.

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(g) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(g)

Examine whether the following systems of equations are consistent or inconsistent and if consistent find the complete solutions.

Question 1.
x + y + z = 4
2x + 5y – 2z = 3
x + 7y – 7z = 5
Solution:

ρ(A) = 2, ρ(AB) = 3
ρ(A) ≠ ρ(AB)
∴ The given system of equations are in consistent.

Question 2.
x + y + z = 6
x – y + z = 2
2x – y + 3z = 9
Solution:

Question 3.
x + y + z = 1
2x + y + z = 2
x + 2y + 2z = 1
Solution:

ρ(A) = 2 = ρ(AB) < 3
The given system of equations are consistent and have infinitely many solutions.
The solutions are given by [(x, y, z) 1x = 1, y + z = 0].

Question 4.
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0
Solution:

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent have a unique solution.
∴ Solution is given by x = 1, y = 3, z = 5.

Question 5.
x + y + z = 6
x + 2y + 3z = 10
x + 2y + 4z = 1
Solution:
Augmented matrix A = \(\left[\begin{array}{cccc}
1 & 1 & 1 & 6 \\
1 & 2 & 3 & 10 \\
1 & 2 & 4 & 1
\end{array}\right]\)
By R₂ → R₂ – R₁, R₃ → R₃ – R₂, we obtain

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent.
They have a unique solution.
∴ Solution is given by x = -7, y = 22, z = -9.

Question 6.
x – 3y – 8z = -10
3x + y – 4z = 0
2x + 5y + 6z = 13
Solution:
The Augmented matrix

ρ(A) = ρ(AB) = 2 < 3
∴ The given system of equations are consistent have infinitely many solutions.
x + y = 2 and y + 2z = 3
Taking z = k, y = 3 – 2z = 3-2k
x = 2 – y
= 2 – (3 – 2k)
= 2 – 3 + 2k
= 2k – 1
∴ The solutions are given by x = -1 + 2k, y = 3 – 2k, z = k where ‘k’ is any scalar.

Question 7.
2x + 3y + z = 9
x + 2y + 3z = 6
3x + y + 2z = 8
Solution:

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent have a unique solution.
∴ Solution is given by x = \(\frac{35}{18}\), y = \(\frac{29}{18}\), z = \(\frac{5}{18}\)

Question 8.
x + y + 4z = 6
3x + 2y – 2z = 9
5x + y + 2z = 13
Solution:

∴ ρ(A) = ρ(AB) = 3
∴ The given system of equations are consistent have a unique solution.
∴ Solution is given by x = 2, y = 2, z = \(\frac{1}{2}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(f) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(f)

I. Find the rank of each of the following matrices.

Question 1.
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\) = 0 – 0 = 0
and |1| = 1 ≠ 0
∴ ρ(A) = 1

Question 2.
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1 ≠ 0
∴ ρ(A) = 2

Question 3.
\(\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right|\) = 0 – 0 = 0
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 4.
\(\left[\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right|\) = 0 – 1 = -1 ≠ 0
∴ ρ(A) = 2

Question 5.
\(\left[\begin{array}{ccc}
1 & 0 & -4 \\
2 & -1 & 3
\end{array}\right]\)
Solution:
\(\left|\begin{array}{cc}
1 & -4 \\
2 & 3
\end{array}\right|\) = 3 + 8 = -11 ≠ 0
∴ ρ(A) = 2

Question 6.
\(\left[\begin{array}{lll}
1 & 2 & 6 \\
2 & 4 & 3
\end{array}\right]\)
Solution:
\(\left|\begin{array}{ll}
2 & 6 \\
4 & 3
\end{array}\right|\) = 6 – 24 = -18 ≠ 0
∴ ρ(A) = 2

II.

Question 1.
\(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
= 1(1 – 0) – 0(0 – 0) + 0(0 – 0)
= 1 – 0 + 0
= 1 ≠ 0
∴ ρ(A) = 3

Question 2.
\(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right|\)
= 1(6 – 0) – 2(8 + 1) + 0(0 + 3)
= 6 – 18
= -12 ≠ 0
∴ ρ(A) = 3

Question 3.
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right|\)
= 1(6 – 4) – 2(4 – 3) + 0(8 – 9)
= 2 – 2 + 0
= 0
∴ ρ(A) ≠ 3, ρ(A) < 3
Take \(\left|\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right|\) = 3 – 4 = -1 ≠ 0
∴ ρ(A) = 2

Question 4.
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\), det A = 0, ρ(A) ≠ 3
All 2 × 2 sub-matrix det. is zero
∴ ρ(A) ≠ 2
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 5.
\(\left[\begin{array}{cccc}
1 & 2 & 0 & -1 \\
3 & 4 & 1 & 2 \\
-2 & 3 & 2 & 5
\end{array}\right]\)
Solution:
Take sub-matrix B = \(\left|\begin{array}{ccc}
1 & 2 & 0 \\
3 & 4 & 1 \\
-2 & 3 & 2
\end{array}\right|\)
= 1(8 – 3) – 2(6 + 2)
= 5 – 16
= -11 ≠ 0
Rank of the given matrix is 3.

Question 6.
\(\left[\begin{array}{cccc}
0 & 1 & 1 & -2 \\
4 & 0 & 2 & 5 \\
2 & 1 & 3 & 1
\end{array}\right]\)
Solution:
Take sub matrix A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
4 & 0 & 2 \\
2 & 1 & 3
\end{array}\right]\)
= -1(12 – 4) + 1(4 – 0)
= -8 + 4
= -4 ≠ 0
∴ ρ(A) = 3

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(c)

I.

Question 1.
Find the domains of the following real-valued functions.
(i) f(x) = \(\frac{1}{\left(x^{2}-1\right)(x+3)}\)
Solution:
f(x) = \(\frac{1}{\left(x^{2}-1\right)(x+3)}\) ∈ R
⇔ (x² – 1) (x + 3) ≠ 0
⇔ (x + 1) (x – 1) (x + 3) ≠ 0
⇔ x ≠ -1, 1, -3
∴ Domain of f is R – {-1, 1, -3}

(ii) f(x) = \(\frac{2 x^{2}-5 x+7}{(x-1)(x-2)(x-3)}\)
⇔ (x – 1) (x – 2) (x – 3) ≠ 0
⇔ x ≠ 1, x ≠ 2, x ≠ 3
∴ Domain of f is R – {1, 2, 3}

(iii) f(x) = \(\frac{1}{\log (2-x)}\)
Solution:
f(x) = \(\frac{1}{\log (2-x)}\)
⇔ log (2 – x) ≠ 0 and 2 – x > 0
⇔ (2 – x) ≠ 1 and 2 > x
⇔ x ≠ 1 and x < 2
x ∈ (-∞, 1) ∪ (1, 2) (or) x ∈ (-∞, 2) – {1}
∴ Domain of f is {(-∞, 2) – {1}}

(iv) f(x) = |x – 3|
Solution:
f(x) = |x – 3| ∈ R
⇔ x ∈ R
∴ The domain of f is R

(v) f(x) = \(\sqrt{4 x-x^{2}}\)
Solution:
f(x) = \(\sqrt{4 x-x^{2}}\) ∈ R
⇔ 4x – x² ≥ 0
⇔ x(4 – x) ≥ 0
⇔ x ∈ [0, 4]
∴ Domain of f is [0, 4]

(vi) f(x) = \(\frac{1}{\sqrt{1-x^{2}}}\)
Solution:
f(x) = \(\frac{1}{\sqrt{1-x^{2}}}\) ∈ R
⇔ 1 – x² > 0
⇔ (1 + x) (1 – x) > 0
⇔ x ∈ (-1, 1)
∴ Domain of f is {x/x ∈ (-1, 1)}

(vii) f(x) = \(\frac{3^{x}}{x+1}\)
Solution:
f(x) = \(\frac{3^{x}}{x+1}\) ∈ R
⇔ 3^x ∈ R, ∀ x ∈ R and x + 1 ≠ 0
⇔ x ≠ -1
∴ Domain of f is R – {-1}

(viii) f(x) = \(\sqrt{x^{2}-25}\)
Solution:
f(x) = \(\sqrt{x^{2}-25}\) ∈ R
⇔ x² – 25 ≥ 0
⇔ (x + 5) (x – 5) ≥ 0
⇔ x ∈ (-∞, -5] ∪ [5, ∞)
⇔ x ∈ R – (-5, 5)
∴ Domain of f is R – (- 5, 5)

(ix) f(x) = \(\sqrt{x-[x]}\)
Solution:
f(x) = \(\sqrt{x-[x]}\) ∈ R
⇔ x – [x] ≥ 0
⇔ x ≥ [x]
⇔ x ∈ R
∴ Domain of f is R.

(x) f(x) = \(\sqrt{[x]-x}\)
Solution:
f(x) = \(\sqrt{[x]-x}\) ∈ R
⇔ [x] – x ≥ 0
⇔ [x] ≥ x
⇔ x ≤ [x]
⇔ x ∈ Z
∴ The domain of f is z (Where z denotes a set of integers)

Question 2.
Find the ranges of the following real-valued functions.
(i) log|4 – x²|
Solution:
Let y = f(x) = log|4 – x²|
f(x) ∈ R
⇔ 4 – x² ≠ 0
⇔ x ≠ ±2
∵ y = log|4 – x²|
⇒ |4 – x²| = e^y
∵ e^y > 0 ∀ y ∈ R
∴ The range of f is R.

(ii) \(\sqrt{[x]-x}\)
Solution:
Let y = f(x) = \(\sqrt{[x]-x}\)
f(x) ∈ R
⇔ [x] – x ≥ 0
⇔ x ≤ [x]
⇔ x ∈ z
∴ Domain of f is z. Then range of f is {0}

(iii) \(\frac{\sin \pi[x]}{1+[x]^{2}}\)
Solution:
Let f(x) = \(\frac{\sin \pi[x]}{1+[x]^{2}}\) ∈ R
⇔ x ∈ R
∴ The domain of f is R
For x ∈ R, [x] is an integer,
sin π[x] = 0, ∀ x ∈ R [∵ sin nπ = 0, ∀ n ∈ z]
∴ Range of f is {0}

(iv) \(\frac{x^{2}-4}{x-2}\)
Solution:
Let y = f(x) = \(\frac{x^{2}-4}{x-2}\) ∈ R
⇔ y = \(\frac{(x+2)(x-2)}{x-2}\)
⇔ x ≠ 2
∴ The domain of f is R – {2}
Then y = x + 2, [∵ x ≠ 2 ⇒ y ≠ 4]
Then its range R – {4}

(v) \(\sqrt{9+x^{2}}\)
Solution:
Let y = f(x) = \(\sqrt{9+x^{2}}\) ∈ R
The domain of f is R
When x = 0, f(0) = √9 = 3
For all values of x ∈ R – {0}, f(x) > 3
∴ The range of f is [3, ∞)

Question 3.
If f and g are real-valued functions defined by f(x) = 2x – 1 and g(x) = x² then find
(i) (3f – 2g)(x)
(ii) (fg) (x)
(iii) \(\left(\frac{\sqrt{f}}{g}\right)(x)\)
(iv) (f + g + 2) (x)
Solution:
(i) (3f – 2g)(x)
f(x) = 2x – 1, g(x) = x²
(3f – 2g) (x) = 3f(x) – 2g(x)
= 3(2x – 1) – 2x²
= -2x² + 6x – 3

(ii) (fg) (x)
= f(x) . g(x)
= (2x – 1) (x²)
= 2x³ – x²

(iii) \(\left(\frac{\sqrt{f}}{g}\right)(x)\)
\(\frac{\sqrt{f(x)}}{g(x)}=\frac{\sqrt{2 x-1}}{x^{2}}\)

(iv) (f + g + 2) (x)
= f(x) + g(x) + 2
= (2x – 1) + x² + 2
= x² + 2x + 1
= (x + 1)²

Question 4.
If f = {(1, 2), (2, -3), (3, -1)} then find
(i) 2f
(ii) 2 + f
(iii) f²
(iv) √f
Solution:
Given f = {(1, 2), (2, -3), (3, -1)}
(i) 2f = {(1, 2 × 2), (2, 2(-3), (3, 2(-1))}
= {(1, 4), (2, -6), (3, -2)}

(ii) 2 + f = {(1, 2 + 2), (2, 2 + (-3)), (3, 2 + (-1)}
= {(1, 4), (2, -1), (3, 1)}

(iii) f² = {(1, 22), (2, (-3)2), (3, (-1)2)}
= {(1, 4), (2, 9), (3, 1)}

(iv) √f = {(1, √2)}
∵ √-3 and √-1 are not real

II.

Question 1.
Find the domains of the following real-valued functions.
(i) f(x) = \(\sqrt{x^{2}-3 x+2}\)
Solution:
f(x) = \(\sqrt{x^{2}-3 x+2}\) ∈ R
⇔ x² – 3x + 2 ≥ 0
⇔ (x- 1) (x – 2) ≥ 0
⇔ x ∈ (-∞, 1 ] ∪ [2, ∞]
∴ The domain of f is R – (1, 2)

(ii) f(x) = log(x² – 4x + 3)
Solution:
f(x) = log(x² – 4x + 3) ∈ R
⇔ x² – 4x + 3 > 0
⇔ (x – 1) (x – 3) > 0
⇔ x ∈ (-∞, 1) ∪ (3, ∞)
∴ Domain of f is R – [1, 3]

(iii) f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\)
Solution:
f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\) ∈ R
⇔ 2 + x ≥ 0, 2 – x ≥ 0, x ≠ 0
⇔ x ≥ -2, x ≤ 2, x ≠ 0
⇔ -2 ≤ x ≤ 2, x ≠ 0
⇔ x ∈ [-2, 2] – {0}
Domain of f is [-2, 2] – {0}

(iv) f(x) = \(\frac{1}{\sqrt[3]{(x-2)} \log _{(4-x)} 10}\)
Solution:
f(x) = \(\frac{1}{\sqrt[3]{(x-2)} \log _{(4-x)} 10}\) ∈ R
⇔ 4 – x > 0, 4 – x ≠ 1 and x – 2 ≠ 0
⇔ x < 4, x ≠ 3, x ≠ 2
∴ Domain of f is (-∞, 4) – {2, 3}

(v) f(x) = \(\sqrt{\frac{4-x^{2}}{[x]+2}}\)
Solution:
f(x) = \(\sqrt{\frac{4-x^{2}}{[x]+2}}\) ∈ R
Case (i) 4 – x² ≥ 0 and [x] + 2 > 0 (or) Case (ii) 4 – x² ≤ 0 and [x] + 2 < 0
Case (i): 4 – x² ≥ 0 and [x] + 2 > 0
⇔ (2 – x) (2 + x) ≥ 0 and [x] > -2
⇔ x ∈ [-2, 2] and x ∈ [-1, ∞]
⇔ x ∈ [-1, 2] ……..(1)
Case (ii): 4 – x² ≤ 0 and [x] + 2 < 0
⇔ (2 + x) (2 – x) ≤ 0 and [x] < – 2
⇔ x ∈ (-∞, -2] ∪ [2, ∞] and x ∈ (-∞, -2)
⇔ x ∈ (-∞, -2) ……(2)
from (1) and (2),
Domain of f is (-∞, -2) ∪ [-1, 2]

(vi) f(x) = \(\sqrt{\log _{0.3}\left(x-x^{2}\right)}\)
Solution:
f(x) = \(\sqrt{\log _{0.3}\left(x-x^{2}\right)}\) ∈ R
Then log_0.3(x – x²) ≥ 0
⇒ x – x² ≤ (0.3)⁰
⇒ x – x² ≤ 1
⇒ -x² + x – 1 ≤ 0
⇒ x² – x + 1 ≥ 0
This is true for all x ∈ R ……..(1)
and x – x² ≥ 0
⇒ x² – x ≤ 0
⇒ x(x – 1) ≤ 0
⇒ x ∈ (0, 1) …….(2)
From (1) and (2)
Domain of f is R ^ (0, 1) = (0, 1)
∴ The domain of f is (0, 1)

(vii) f(x) = \(\frac{1}{x+|x|}\)
Solution:
f(x) = \(\frac{1}{x+|x|}\) ∈ R
⇔ x + |x| ≠ 0
⇔ x ∈ (0, ∞)
∵ |x| = x, if x ≥ 0
|x| = -x, if x < 0
∴ The domain of f is (0, ∞)

Question 2.
Prove that the real valued function f(x) = \(\frac{x}{e^{x}-1}+\frac{x}{2}+1\) is an even function on R \ {0}.
Solution:
f(x) ∈ R, e^x – 1 ≠ 0
⇒ e^x ≠ 1
⇒ x ≠ 0

⇒ f(x) is an even function on R – {0}

Question 3.
Find the domain and range of the following functions.
(i) f(x) = \(\frac{\tan \pi[x]}{1+\sin \pi[x]+\left[x^{2}\right]}\)
Solution:
f(x) = \(\frac{\tan \pi[x]}{1+\sin \pi[x]+\left[x^{2}\right]}\) ∈ R
⇔ x ∈ R, since [x] is an integer tan π[x] and sin π[x] each is zero for ∀ x ∈ R and f(x) ∈ R
Domain of f is R
Its range = {0}

(ii) f(x) = \(\frac{x}{2-3 x}\)
Solution:

(iii) f(x) = |x| + |1 + x|
Solution:
f(x) = |x| + |1 + x| ∈ R
⇔ x ∈ R
∴ Domain of f is R
∵ |x| = x, if x ≥ 0
= -x, if x < 0
|1 + x| = 1 + x, if x ≥ -1
= -(1 + x) if x < -1
For x = 0, f(0) = |0| + |1 + 0| = 1
x = 1, f(1) = |1| + |1 + 1| = 1 + 2 = 3
x = 2, f(2) = |2| + |1 + 2| = 2 + 3 = 5
x = -2, f(-2) = |-2| + |1 + (-2)| = 2 + 1 = 3
x = -1, f(-1) = |-1| + |1 +(-1)| = 1 + 0 = 1
∴ The range of f is [1, ∞]

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(b)

I.

Question 1.
If f(x) = e^x and g(x) = log_ex, then show that f o g = g o f and find f^-1 and g^-1.
Solution:
Given f(x) = e^x and g(x) = log_ex
Now (f o g) (x) = f(g(x))
= f(log_ex) [∵ g(x) = \(\log _{e} x\)]
= \(e^{\left(\log _{e} x\right)}\)
= x
∴ (fog) (x) = x ………(1)
and (g o f) (x) = g(f(x))
= g(e^x) [∵ f(x) = e^x]
= log_e (e^x) [∵ g(x) = log_ex]
= x log_e (e)
= x(1)
= x
∴ (g o f) (x) = x …….(2)
From (1) and (2)
f o g = g o f
Given f(x) = e^x
Let y = f(x) = e^x ⇒ x = f^-1(y)
and y = e^x ⇒ x = log_e (y)
∴ f^-1(y) = log_e (y) ⇒ f^-1(x) = log_e (x)
Let y = g(x) = log_e (x)
∵ y = g(x) ⇒ x = g^-1(y)
∵ y = log_e (x) ⇒ x = e^y
∴ g^-1(y) = e^y ⇒ g^-1(x) = e^x
∴ f^-1(x) = log_e (x) and g^-1(x) = e^x

Question 2.
If f(y) = \(\frac{y}{\sqrt{1-y^{2}}}\), g(y) = \(\frac{y}{\sqrt{1+y^{2}}}\) then show that (fog) (y) = y
Solution:
f(y) = \(\frac{y}{\sqrt{1-y^{2}}}\) and g(y) = \(\frac{y}{\sqrt{1+y^{2}}}\)
Now, (fog) (y) = f(g(y))

∴ (fog) (y) = y

Question 3.
If f : R → R, g : R → R are defined by f(x) = 2x² + 3 and g(x) = 3x – 2, then find
(i) (fog)(x)
(ii) (gof) (x)
(iii) (fof) (0)
(iv) go(fof) (3)
Solution:
f : R → R, g : R → R and f(x) = 2x² + 3; g(x) = 3x – 2
(i) (f o g) (x) = f(g(x))
= f(3x – 2) [∵ g(x) = 3x – 2]
= 2(3x- 2)² + 3 [∵ f(x) = 2x² + 3]
= 2(9x² – 12x + 4) + 3
= 18x² – 24x + 8 + 3
= 18x² – 24x + 11

(ii) (gof) (x) = g(f(x))
= g(2x² + 3) [∵ f(x) = 2x² + 3]
= 3(2x² + 3) – 2 [∵ g(x) = 3x – 2]
= 6x² + 9 – 2
= 6x² + 7

(iii) (fof) (0) = f(f(0))
= f(2(0) + 3) [∵ f(x) = 2x² + 3]
= f(3)
= 2(3)² + 3
= 18 + 3
= 21

(iv) g o (f o f) (3)
= g o (f (f(3)))
= g o (f (2(3)² + 3)) [∵ f(x) = 2x² + 3]
= g o (f(21))
= g(f(21))
= g(2(21)² + 3)
= g(885)
= 3(885) – 2 [∵ g(x) = 3x – 2]
= 2653

Question 4.
If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x² + 1, then find
(i) (f o f) (x² + 1)
(ii) f o g (2)
(iii) g o f (2a – 3)
Solution:
f : R → R, g : R → R and f(x) = 3x – 1 ; g(x) = x² + 1
(i) (f o f) (x² + 1)
= f(f(x² + 1))
= f[3(x² + 1) – 1] [∵ f(x) = 3x – 1]
= f(3x² + 2)
= 3(3x² + 2) – 1
= 9x² + 5

(ii) (f o g) (2)
= f(g(2))
= f(2² + 1) [∵ g(x) = x² + 1]
= f(5)
= 3(5) – 1
= 14 [∵ f(x) = 3x – 1]

(iii) (g o f) (2a – 3)
= g(f(2a – 3))
= g[3(2a – 3) – 1] [∵ f(x) = 3x – 1]
= g(6a – 10)
= (6a – 10)² + 1 [∵ g(x) = x² + 1]
= 36a² – 120a + 100 + 1
= 36a² – 120a + 101

Question 5.
If f(x) = \(\frac{1}{x}\), g(x) = √x for all x ∈ (0, ∞) then find (g o f) (x).
Solution:
f(x) = \(\frac{1}{x}\), g(x) = √x, ∀ x ∈ (0, ∞)
(g o f) (x) = g(f(x))
= g(\(\frac{1}{x}\)) [∵ f(x) = \(\frac{1}{x}\)]
= \(\sqrt{\frac{1}{x}}\)
= \(\frac{1}{\sqrt{x}}\) [∵ g(x) = √x]
∴ (gof) (x) = \(\frac{1}{\sqrt{x}}\)

Question 6.
f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) for all x ∈ R, find (g o f) (x).
Solution:
f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) ∀ x ∈ R
(g o f) (x) = g(f(x))
= g(2x – 1) [∵ f(x) = 2x – 1]
= \(\frac{(2 x-1)+1}{2}\)
= x [∵ g(x) = \(\frac{x+1}{2}\)]
∴ (g o f) (x) = x

Question 7.
If f(x) = 2, g(x) = x², h(x) = 2x for all x ∈ R, then find (f o (g o h)) (x).
Solution:
f(x) = 2, g(x) = x², h(x) = 2x, ∀ x ∈ R
[f o (g o h) (x)]
= [f o g (h(x))]
= f o g (2x) [∵ h(x) = 2x]
= f[g(2x)]
= f((2x)²) [∵ g(x) = x²]
= f(4x²) = 2 [∵ f(x) = 2]
∴ [f o (g o h) (x)] = 2

Question 8.
Find the inverse of the following functions.
(i) a, b ∈ R, f : R → R defined by f(x) = ax + b, (a ≠ 0).
Solution:
a, b ∈ R, f : R → R and f(x) = ax + b, a ≠ 0
Let y = f(x) = ax + b
⇒ y = f(x)
⇒ x = f^-1(y) ……..(i)
and y = ax + b
⇒ x = \(\frac{y-b}{a}\) ……..(ii)
From (i) and (ii)
f^-1(y) = \(\frac{y-b}{a}\)
⇒ f^-1(x) = \(\frac{x-b}{a}\)

(ii) f : R → (0, ∞) defined by f(x) = 5^x
Solution:
f : R → (0, ∞) and f(x) = 5^x
Let y = f (x) = 5^x
y = f(x) ⇒ x = f^-1(y) ……(i)
and y = 5^x ⇒ log₅ (y) = x ……..(ii)
From (i) and (ii)
f^-1(y) = log₅(y) ⇒ f^-1(x) = log5 (x)

(iii) f : (0, ∞) → R defined by f(x) = log₂ (x).
Solution:
f : (0, ∞) → R and f(x) = log₂ (x)
Let y = f(x) = log₂ (x)
∵ y = f(x) ⇒ x = f^-1(y) ……..(i)
and y = log₂(x) ⇒ x = 2y
From (i) and (ii)
f^-1(y) = 2y ⇒ f^-1(x) = 2x

Question 9.
If f(x) = 1 + x + x² + …… for |x| < 1 then show that f^-1(x) = \(\frac{x-1}{x}\)
Solution:
f(x) = 1 + x + x² + ……..

Question 10.
If f : [1, ∞) ⇒ [1, ∞) defined by f(x) = \(2^{x(x-1)}\) then find f^-1(x).
Solution:

II.

Question 1.
If f(x) = \(\frac{x-1}{x+1}\), x ≠ ±1, then verify (f o f^-1) (x) = x.
Solution:
Given f(x) = \(\frac{x-1}{x+1}\), x ≠ ±1
Let y = f(x) = \(\frac{x-1}{x+1}\)
∵ y = f(x) ⇒ x = f^-1(y) ……(i)
and y = \(\frac{x-1}{x+1}\)

Question 2.
If A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f : A → B, g : B → C are defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}, then show that f and g are bijective functions and (gof)^-1 = f^-1 o g^-1.
Solution:
A = {1, 2, 3}, B = {α, β, γ},
f : A → B and f = {(1, α), (2, γ), (3, β)}
⇒ f(1) = α, f(2) = γ, f(3) = β
∵ Distinct elements of A have distinct f – images in B, f: A → B is an injective function.
Range of f = {α, γ, β} = B(co-domain)
∴ f : A → B is a surjective function.
Hence f : A → B is a bijective function.
B = {α, β, γ}, C = {p, q, r}, g : B → C and g : {(α, q), (β, r), (γ, p)}
⇒ g(α) = q, g(β) = r, g(γ) = p
∴ Distinct elements of B have distinct g – images in C, g : B → C is an injective function.
Range of g = {q, r, p} = C, (co-domain)
∴ g : B → C is a surjective function
Hence g : B → C is a bijective function
Now f = {(1, α), (2, γ), (3, β)}
g = {(α, q), (β, r), (γ, p)}
g o f = {(1, q), (2, p), (3, r)}
∴ (g o f)^-1 = {(q, 1), (r, 3), (p, 2)} ………(1)
g^-1 = {(q, α), (r, β), (p, γ)}
f^-1 = {(α, 1), (γ, 2),(β, 3)}
Now f^-1 o g^-1 = {(q, 1), (r, 3), (p, 2)} …….(2)
From eq’s (1) and (2)
(gof)^-1 = f^-1 o g^-1

Question 3.
If f : R → R, g : R → R defined by f(x) = 3x – 2, g(x) = x² + 1, then find
(i) (g o f^-1) (2)
(ii) (g o f)(x – 1)
Solution:
f : R → R, g : R → R and f(x) = 3x – 2
f is a bijective function ⇒ its inverse exists
Let y = f(x) = 3x – 2
∵ y = f(x) ⇒ x = f^-1(y) …….(i)
and y = 3x – 2
⇒ x = \(\frac{y+2}{3}\) ……..(ii)
From (i) and (ii)
f^-1(y) = \(\frac{y+2}{3}\)
⇒ f^-1(x) = \(\frac{x+2}{3}\)
Now (gof^-1) (2)
= g(f^-1(2))

∴ (g o f^-1) (2) = \(\frac{25}{9}\)

(ii) (g o f) (x -1)
= g(f(x – 1))
= g(3(x – 1) – 2) [∵ f(x) = 3x – 2]
= g(3x – 5)
= (3x – 5)² + 1 [∵ g(x) = x² + 1]
= 9x² – 30x + 26
∴ (g o f) (x – 1) = 9x² – 30x + 26

Question 4.
Let f = {(1, a), (2, c), (4, d), (3, b)} and g^-1 = {(2, a), (4, b), (1, c), (3, d)} then show that (gof)^-1 = f^-1 o g^-1
Solution:
f = {(1, a), (2, c), (4, d), (3, b)}
∴ f^-1 = {(a, 1), (c, 2), (d, 4), (b, 3)}
g^-1 = {(2, a), (4, b), (1, c), (3, d)}
∴ g = {(a, 2), (b, 4), (c, 1), (d, 3)}
(g o f) = {(1, 2), (2, 1), (4, 3), (3, 4)}
∴ (gof)^-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ……….(1)
f^-1 o g^-1 = {(2, 1), (4, 3), (1, 2), (3, 4)} ……..(2)
From eq’s (1) and (2), we observe (gof)^-1 = f^-1 o g^-1

Question 5.
Let f : R → R, g : R → R be defined by f(x) = 2x – 3, g(x) = x³ + 5 then find (f o g)^-1 (x).
Solution:
f : R → R, g : R → R and f(x) = 2x – 3 and g(x) = x³ + 5
Now (fog) (x) = f(g(x))
= f(x³ + 5) [∵ g(x) = x² + 5]
= 2(x³ + 5) – 3 [∵ f(x) = 2x – 3]
= 2x³ + 7
∴ (f o g) (x) = 2x³ + 7
Let y = (f o g) (x) = 2x³ + 7
∵ y = (fog)(x)
⇒ x = (fog)^-1 (y) …….(1)
and y = 2x³ + 7
⇒ x³ = \(\frac{y-7}{2}\)
⇒ x = \(\left(\frac{y-7}{2}\right)^{\frac{1}{3}}\) …..(2)
From eq’s (1) and (2),
(f o g)^-1 (y) = \(\left(\frac{y-7}{2}\right)^{\frac{1}{3}}\)
∴ (f o g)^-1 (x) = \(\left(\frac{x-7}{2}\right)^{\frac{1}{3}}\)

Question 6.
Let f(x) = x², g(x) = 2^x. Then solve the equation (f o g) (x) = (g o f) (x)
Solution:
Given f(x) = x² and g(x) = 2^x
Now (f o g) (x) = f(g(x))
= f(2^x) [∵ g(x) = 2^x]
= (2^x)²
= 2^2x [∵ f(x) = x²]
∴ (f o g) (x) = 2^2x ……(1)
and (g o f) (x) = g(f(x))
= g(x²) [∵ f(x) = x²]
= \((2)^{x^{2}}\) [∵ g(x) = 2^x]
∴ (g o f) (x) = \((2)^{x^{2}}\)
∵ (f o g) (x) = (g o f) (x)
⇒ 2^2x = \((2)^{x^{2}}\)
⇒ 2^x = x²
⇒ x² – 2x = 0
⇒ x(x – 2) = 0
⇒ x = 0, x = 2
∴ x = 0, 2

Question 7.
If f(x) = \(\frac{x+1}{x-1}\), (x ≠ ±1) then find (fofof) (x) and (fofofof) (x).
Solution:
f(x) = \(\frac{x+1}{x-1}\), (x ≠ ±1)
(i) (fofof) (x) = (fof) [f(x)]
= (fof) \(\left(\frac{x+1}{x-1}\right)\) [∵ f(x) = \(\left(\frac{x+1}{x-1}\right)\)]

(ii) (fofofof) (x) = f[(f o f o f) (x)]
= f [f(x)] {from (1)}

In the above problem if a number of f is even its answer is x and if a number of f is odd its answer is f(x).

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(a)

I.

Question 1.
If the function f is defined by

then find the values of
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
(v) f(-5)
Solution:
(i) f(3)
For x > 1, f(x) = x + 2
∴ f(3) = 3 + 2 = 5

(ii) f(0)
For -1 ≤ x ≤ 1, f(x) = 2
∴ f(0) = 2

(iii) f(-1.5)
For -3 < x < -1, f(x) = x – 1
∴ f(-1.5) = -1.5 – 1 = -2.5

(iv) f(2) + f(-2) For x > 1, f(x) = x + 2
∴ f(2) = 2 + 2 = 4
For -3 < x < -1, f(x) = x – 1
∴ f(-2)= -2 – 1 = -3
f(2) + f(-2) = 4 + (-3) = 1

(v) f(-5) is not defined, since domain of x is {X/X ∈ (-3, ∞)}

Question 2.
If f: R{0}R is defined by f(x) = \(x^{3}-\frac{1}{x^{3}}\); then show that f(x) + \(f\left(\frac{1}{x}\right)\) = 0.
Solution:
Given f(x) = \(x^{3}-\frac{1}{x^{3}}\) ……(i)
Now \(f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^{3}-\frac{1}{\left(\frac{1}{x}\right)^{3}}=\frac{1}{x^{3}}-x^{3}\) ……(2)
Add (1) and (2)
\(f(x)+f\left(\frac{1}{x}\right)=\left(x^{3}-\frac{1}{x^{3}}\right)+\left(\frac{1}{x^{3}}-x^{3}\right)\) = 0
∴ f(x) + \(f\left(\frac{1}{x}\right)\) = 0

Question 3.
If f : R → R is defined by f(x) = \(\frac{1-x^{2}}{1+x^{2}}\), then show that f(tan θ) = cos 2θ.
Solution:
Given f(x) = \(\frac{1-x^{2}}{1+x^{2}}\)

∴ f(tan θ) = cos 2θ

Question 4.
If f : R\{±1} → R is defined by f(x) = \(\log \left|\frac{1+x}{1-x}\right|\), then show that \(f\left(\frac{2 x}{1+x^{2}}\right)\) = 2f(x)
Solution:
f : R\{±1} → R and f(x) = \(\log \left|\frac{1+x}{1-x}\right|\)

∴ \(f\left(\frac{2 x}{1+x^{2}}\right)\) = 2f(x)

Question 5.
If A = {-2, -1, 0, 1, 2} and f : A → B is a surjection defined by f(x) = x² + x + 1, then find B.
Solution:
A = {-2, -1, 0, 1, 2} and f : A → B, f(x) = x² + x + 1
f : A → B is a surjection
f(-2) = (-2)² + (-2) + 1
= 4 – 2 + 1
= 3
f(-1) = (-1)² + (-1) + 1
= 1 – 1 + 1
= 1
f(0) = 0² + 0 + 1
= 0 + 0 + 1
= 1
f(1) = 1² + 1 + 1
= 1 + 1 + 1
= 3
f(2) = 2² + 2 + 1
= 4 + 2 + 1
= 7
∴ B = f(A) = {3, 1, 7}

Question 6.
If A = {1, 2, 3, 4} and f : A → R is a function defined by f(x) = \(\frac{x^{2}-x+1}{x+1}\), then find the range of f.
Solution:
A= {1, 2, 3, 4}

∴ Range of f = f(A) = \(\left\{\frac{1}{2}, 1, \frac{7}{4}, \frac{13}{5}\right\}\)

Question 7.
If f(x + y) = f(xy) ∀ x, y ∈ R then prove that f is a constant function.
Solution:
Given f(x + y) = f(x y), x, y ∈ R
take x = y = 0
⇒ f(0) = f(0) ………(1)
Let x = 1, y = 0
⇒ f(1) = f(0) ……..(2)
Let x = 1, y = 1
⇒ f(2) = f(1) ………(3)
from (1), (2), (3)
f(0) = f(1) = f(2)
⇒ f(0) = f(2)
Similarly f(3) = f(0)
f(4) = f(0)
and so on
f(n) = f(0)
∴ f is a constant function

II.

Question 1.
If A = {x | -1 ≤ x ≤ 1}, f(x) = x², g(x) = x³, which of the following are surjections?
(i) f : A → A
(ii) g : A → A
Solution:
(i) ∵ A = {x | -1 ≤ x ≤ 1} and f(x) = x²
This implies f(x) is a function from A to A
(i.e.,) f : A → A
Now let y ∈ A
If f(x) = y then x² = y
x = √y
So, if y = -1 then x = √-1 ∉ A
∴ f : A → A is not a surjection.

(ii) ∵ A = {x | -1 ≤ x ≤ 1} and g(x) = x³
⇒ g : A → A
Let y ∈ A. Then g(x) = y
⇒ x³ = y
⇒ x = \((y)^{1 / 3}\) ∈ A
So if y = -1 then x = -1 ∈ A
y = 0, then x = 0 ∈ A
y = 1, then x = 1 ∈ A
∴ g : A → A is a surjections.

Question 2.
Which of the following are injections or surjections or bijections? Justify your answers.
(i) f : R → R defined by f(x) = \(\frac{2 x+1}{3}\)
Solution:
f(x) = \(\frac{2 x+1}{3}\)
Let x₁, x₂ ∈ R
∵ f(x₁) = f(x₂)
⇒ \(\frac{2 x_{1}+1}{3}=\frac{2 x_{2}+1}{3}\)
⇒ 2x₁ + 1 = 2x₂ + 1
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂
∵ f(x₁) = f(x₂) ⇒ x₁ = x₂, ∀ x₁, x₂ ∈ R
So f(x) = \(\frac{2 x+1}{3}\), f : R → R is an injection
If y ∈ R (co-domain) then y = \(\frac{2 x+1}{3}\)
⇒ x = \(\frac{3 y-1}{2}\)
Then f(x) = \(\frac{2 x+1}{3}=\frac{2\left(\frac{3 y-1}{2}\right)+1}{3}=y\)
∴ f is a surjection
∴ f : R → R defined by f(x) = \(\frac{2 x+1}{3}\) is a bijection

(ii) f : R → (0, ∞) defined by f(x) = 2x
Solution:
Let x₁, x₂ ∈ R
∵ f(x₁) = f(x₂)
⇒ \(2^{x_{1}}=2^{x_{2}}\)
⇒ x₁ = x₂
∴ f(x₁) = f(x₂) ⇒ x₁ = x₂ ∀ x₁, x₂ ∈ R
∴ f(x) = 2x, f : R → (0, ∞) is injection
If y ∈ (0, ∞) and y = 2x ⇒ x = log₂ (y)
Then f(x) = 2x
= \(2^{\log _{2}(y)}\)
= y
∴ f is a surjection
Hence f is a bijection.

(iii) f : (0, ∞) → R defined by f(x) = log_ex
Solution:
Let x₁, x₂ e (0, ∞)
f(x₁) = f(x₂)
⇒ \(\log _{e}\left(x_{1}\right)=\log _{e}\left(x_{2}\right)\)
⇒ x₁ = x₂
∵ f(x₁) = f(x₂)
⇒ x₁ = x₂ ∀ x₁, x₂ ∈ (0, ∞)
∴ f(x) is injection.
Let y ∈ R.
y = log_ex ⇒ x = e^y
Then f(x) = log_ex
= log_e(e^y)
= y . log_ee
= y(1)
= y
∴ f is a surjection.
∴ f is a bijection.

(iv) f : [0, ∞) → [0, ∞) defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ [0, ∞) (i.e.,) domain of f.
Now f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = x₂
∵ x₁, x₂ ≥ 0
∴ f(x) = x², f : {0, ∞) → {0, ∞) is injection
Let y ∈ (0, ∞), co-domain of f
Let y = x² ⇒ x = √y, ∵ y ≥ 0
Then f(x) = x²
= \((\sqrt{y})^{2}\)
= y
∴ f is surjection.
Hence f is a bijection.

(v) f : R → [0, ∞) defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ R.
f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = ±x₂, ∵ x₁, x₂ ∈ R
Hence f is not injection
Let y ∈ [0, ∞)
y = x²
⇒ x = ±√y, where y ∈ [0, ∞)
Then f(x) = x²
= \((\sqrt{y})^{2}\)
= y
∴ f is surjection
Hence f is not a bijection

(vi) f : R → R defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ R, (domain of f)
∴ f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = ±x₂, ∵ x₁, x₂ ∈ R
∴ f(x) is not injection
For elements that belong to (-∞, 0) codomain of f has no pre-image in f.
∴ f is not a surjection
Hence f is neither injection nor surjection.

Question 3.
Is g = {(1, 1) (2, 3) (3, 5) (4, 7)} is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}. If this is given by the formula g(x) = ax + b, then find a and b.
Solution:
A = {1, 2, 3, 4}; B = {1, 3, 5, 7}
g : {(1, 1), (2, 3), (3, 5), (4, 7)}
∵ g(1) = 1, g(2) = 3, g(3) = 5, g(4) = 7
So for each element a ∈ A, there exists a unique b ∈ B such (a, b) ∈ g
∴ g : A → B is a function
Given g(x) = ax + b, ∀ x ∈ A
g(1) = (a) + b = 1
⇒ a + b = 1 ……..(1)
g(2) = 2a + b = 3
⇒ 2a + b = 3 …….(2)
Solve (1) and (2)
a = 2, b = -1

Question 4.
If the function f : R → R defined by f(x) = \(\frac{3^{x}+3^{-x}}{2}\), then show that f(x + y) + f(x – y) = 2f(x) f(y).
Solution:
f : R → R and f(x) = \(\frac{3^{x}+3^{-x}}{2}\)

∴ f(x + y) + f(x – y) = 2 f(x).f(y)

Question 5.
If the function f : R → R defined by f(x) = \(\frac{4^{x}}{4^{x}+2}\), then show that f(1 – x) = 1 – f(x) and hence reduce the value of \(f\left(\frac{1}{4}\right)+2 f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)\)
Solution:


∴ f(1 – x) = 1 – f(x)

Question 6.
If the function f : {-1, 1} → {0, 2), defined by f(x) = ax + b is a surjection, then find a and b.
Solution:
f : {-1, 1} → {0, 2} and f(x) = ax + b is a surjection
Given f(-1) = 0 and f(1) = 2 (or) f(-1) = 2, f(1) = 0
Case (i):
f(-1) = 0 and f(1) = 2
a(-1) + b = 0 ⇒ -a + b = 0 ……..(1)
a(1) + b = 2 ⇒ a + b = 2 ……(2)
Solve eq’s (1) and (2), we get a = 1, b = 1
Case (ii):
f(-1) = 2 and f(1) = 0
a(-1) + b = 2 ⇒ -a + b = 2 ……(3)
a(1) + b = 0 ⇒ a + b = 0 ……….(4)
Solve eq’s (3) and (4), we get a = -1, b = 1
Hence a = ±1 and b = 1

Question 7.
If f(x) = cos (log x), then show that \(f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]=0\)
Solution:
Given f(x) = cos(log x)
\(f\left(\frac{1}{x}\right)=\cos \left(\log \left(\frac{1}{x}\right)\right)\)
= cos(log 1 – log x)
= cos(-log x)
= cos (log x) (∵ log 1 = 0)
Similarly
\(f\left(\frac{1}{y}\right)\) = cos (log y)
\(f\left(\frac{x}{y}\right)=\cos \log \left(\frac{x}{y}\right)\)
= cos (log x – log y)
and f(x y) = cos log (x y) = cos (log x + log y)
\(f\left(\frac{x}{y}\right)\) + f(x y) = cos (log x – log y) + cos (log x + log y)
= 2 cos (log x) cos (log y)
[∵ cos (A – B) + cos (A + B) = 2 cos A . cos B]
LHS = \(f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]\)
= cos (log x) cos (log y) – \(\frac{1}{2}\) [2 cos (log x) cos (log y)]
= 0

Andhra Pradesh BIEAP AP Inter 1st Year Commerce Study Material 5th Lesson Partnership Textbook Questions and Answers.

AP Inter 1st Year Commerce Study Material 5th Lesson Partnership

Essay Answer Questions

Question 1.
Define Partnership. Discuss its merits and limitations. [Mar. 2019; May 17 – T.S. M Mar. 15 – A.P. & T.S.]
Answer:
Partnership is an association of two or more persons who pool their financial and managerial resources and agree to carry on a business with profit motive. The persons who are entering into partnership individually called ‘Partners’ and collectively known as ‘Firm’.

Partnership – Definition :
Partnership is “the relation between two or more persons who have agreed to share the profits of a business carried on by all or any one of them acting for all” – Section 4 of the Indian Partnership Act, 1932.

Merits:
1) Easy to form :
A partnership can be formed easily without many legal formalities. Since it is not compulsory to get the firm registered, a simple agreement, either in oral, writing or implied is sufficient to create a partnership firm.

2) Availability of larger resources :
A partnership firm consists of more than one person, it may be to pool more resources as compared to sole proprietorship.

3) Better decisions :
In partnership firm each partner has a right to take part in the management of the business. All important decisions are taken in consultation with and with the consent of all partners. Thus, collective wisdom prevails and there is less scope for reckless and hasty decisions.

4) Flexibility:
The partnership firm is a flexible organisation. Changes in the business can be adopted easily. At any time the partners can change the size or nature of business or area of its operation after taking the necessary consent of all the partners.

5) Sharing of risks:
The losses of the firm are shared by all the partners equally or as per the agreed ratio.

6) Protection of interest:
In partnership form of business organisation, the rights of each partner and his/her interests are fully protected. If a partner is dissatisfied with any decision, he can ask for dissolution of the firm or can withdraw from the partnership.

7) Secrecy :
Business secrets of the firm are known to the partners only. It is not required to disclose any information to the outsiders. It is also not mandatory to publish the annual accounts of the Partnership firm.

Limitations:
1) Unlimited liability :
The partners liability is unlimited. This is the most important drawback of partnership. The partners are personally liable for the debts and obligations of the firm. In other words, their personal property can also be utilised for payment of firm’s liabilities.

2) Limited capital:
Since the total number of partners cannot exceed 20, the capacity to raise funds remains limited as compared to a joint stock company where there is no limit on the number of share holders.

3) Non-transferability of share :
In partnership firm, the partners cannot transfer his share of interest to other without consent of remaining partners.

4) Possibility of conflicts:
Differences and disputes among the partners are common. These conflicts harm to the firm. Difference of opinion may give rise to quarrels and lead to dissolution of the firm.

Question 2.
Is registration of Partnership compulsory under the Partnership Act, 1932? Explain the procedure required for registration of a firm.
Answer:
Partnership is an association of two or more persons who pool their financial and managerial resources and agree to carry on a business, and share its profits or losses. The persons who form a partnership are individually known as ‘Partners’ and collectively a firm or partnership firm.

The Indian Partnership Act, 1932 does not make it compulsory for a firm to be registered; but there are certain disabilities which attach to an unregistered firm. These disabilities make its virtually compulsory for a firm to be registered. Registration can take place at any time.

The procedure for registration of a firm is as follows:

1. The firm will have to apply to the Registrar of Firms of the state concerned in the prescribed form.
2. The firm will have to apply to the Registrar of firms of the state concerned in the prescribed form. For this, a form containing the following particulars, accompanied by a fee of ₹3/- has to be sent to the Registrar of Firms.
   a) The name of the firm
   b) Location of the firm
   c) Names of other places where the firm carries on business
   d) The name in full and addresses of the partners
   e) The date on which various partners joined the firm.
   f) The duration of the firm
3. The duly filled in form must be signed by all the partners. The filled in form along with prescribed registration fee must be deposited in the office of the Registrar of Firms.
4. The Registrar will scrutinise the application, and if he is satisfied that all formalities relating to registration have been duly complied with, he will put the name of the firm in his register and issue the Certificate of Registration.

Question 3.
Discuss different types of Partners. [Mar. 2019; May 17 – A.P. Mar. 17 – T.S.]
Answer:
A Partnership firm can have different types of partners with different roles and liabilities. Some of them may take part in the management while other may contribute capital.

1) Active Partners or Working Partners:
The partners who actively participate in the day-to-day operations of the business are known as active partners or working partners.

2) Sleeping Partners :
The partners, who simply provide capital and do not participate in the management activities of the firm are called sleeping partners.

3) Nominal Partners :
Nominal partners allow the firm to use their name as partner. They neither invest any capital nor participate in the day-to-day operations. They are not entitled to share the profits of the firm. However they are liable to third parties for all the acts of the firm.

4) Partners in Profits :
A person who shares the profits of the business without being liable for the losses is known as partner in profits. This is applicable only to the minors who are admitted to the benefits of the firm and their liability is limited to their capital contribution.

5) Limited Partners :
The partners whose liability is limitied in a firm are called limited partners. They are also known as special partners.

6) General Partners :
The partners having unlimited liability are called general – partners. According to Indian Partnership Act, 1932 the liability of the partner is unlimited. So they are general partners (excpet minor partner).

7) Partner by Estoppel:
A person, who behaves in the public is such a way as to give an impression that he/she is a partner of the firm, is called partner by Estoppel. Such partners are not entitled to share the profits of the firm, but are, fully liable if somebody suffers because of his/her false representation.

8) Partner by Holding out:
Sometimes, the firm may use the name of a person in its activities, creating a sense in the public that he is also a partner. If that person accepts the same, he becomes automatically responsible to the third parties. Such person is known as “Partner by Holding out”.

4. What is Partnership Deed? And also explain its contents. [Mar. 2018 – T.S.]
Answer:
Partnership is an association of two or more persons who pool their financial and managerial resources and agree to carry on a business, and share its profits or losses. Partnership was established among partners through an agreement. Such agreement may be in the form of oral or written. If partnership agreement is in registration it is known as Partnership Deed.

Partnership Deed is a document containing the terms and conditions of a partnership. It is an agreement in writing signed by the partners duly stamped and registered. The Partnership deed defines certain rights, duties and obligations of partners and governs relations among them in the conduct of business affairs of the firm.

The Partnership deed must not contain any term which is contrary to the provisions of the Partnership Act. Each partner should have a copy of the deed.

The following points are generally included in the deed.

Partnership Deed – Contents

1. Name of the firm
2. Nature of the business
3. Names and addresses of partners
4. Location of business
5. Duration of partnership, if decided
6. Amount of capital to be contributed by each partner
7. Profit and loss sharing ratio
8. Duties, powers and obligations of partners
9. Salaries and withdrawals of the partners
10. Preparation of accounts and their auditing
11. Procedure for dissolution of the firm
12. Procedure for settlement of disputes

Short Answer Questions

Question 1.
Define Partnership and state its important features. [Mar. 17 – A.P.]
Answer:
Partnership is an agreement between two or more persons to carry on business with profit motive, carried on by all or any one of them acting for all.

Partnership – Definition :
“Partnership is the relation existing between persons competent to make contract, who agree to carry on a lawful business in common with a view to private gain.” – L.H. Haney

“The relation between persons who have agreed to share the profits of a business carried on by all of them acting for all.” – Indian Partnership Act, 1932, Section 4

Partnership Firm – Features:
The following are the important features of partnership organisation.

1. Formation
2. Unlimited liability
3. Existence of lawful business
4. Principal agent relationship
5. Voluntary registration

1) Formation :
The partnership form of business organisation is governed by the provision of Indian Partnership Act, 1932. It comes into existence through a legal agreement where in the terms and conditions governing the relationship among partners. Partnership formation is very easy.

2) Unlimited Liability :
The liability of partner is unlimited, joint and several. Personal assets may be used for repaying debts in cases the business assets are insufficient. All the partners are responsible for the debts and they contribute in proportion to their share in business and as such are Habile to that extent.

3) Existence of lawful business:
The business to be carried on by a partnership must always be lawful. Any agreement to indugle in sumuggling, black marketing, etc. cannot be called partnership business in the eyes of law.

4) Principal agent relationship :
Each partner is an agent of the firm. He can act on behalf of the firm. He is responsible for his own acts and also the acts on behalf of the other partners. There must be an agency relationship between the partners.

5) Voluntary registration :
The registration of a partnership firm is not compulsory. But an unregistered firm suffers from some limitations which make it virtually compulsory to be registered.

Question 2.
Discuss the registration procedure of partnership.
Answer:
The Indian Partnership Act, 1932 does not make it compulsory for a firm to be registered; but there are certain disabilities which attach to an unregistered firm. These disabilities make it virtually compulsory for a firm to be registered. Registration can take place at any time. The procedure for registration of a firm is as follows:

1. The firm will have to apply to the Rigistrar of Firms of the state concerned in the prescribed form.
2. The firm will have to apply to the Registrar of Firms of the state concerned in the prescribed form. For this, a form containing the following particulars, accompanied by a fee of Rs. 3/- has to be sent to the Registrar of Firms.
   a) The name of the firm
   b) Location of the firm
   c) Names of other places where the firm carries on business
   d) The name in full and addresses of the partners
   e) The date on which various partners joined the firm
   f) The duration of the firm
3. The duly filled in form must be signed by all the partners. The filled in form along with prescribed registration fee must be deposited in the office of the Registrar of Firms.
4. The Registrar will scrutinise the application, and if he is satisfied that all formalities relating to registration have been duly complied with, he will put the name of the firm in his register and issue the Certificate of Registration.

Question 3.
Briefly explain the rights of partners.
Answer:
The rights and duties of the partners of a firm are usually governed by the partnership agreement among the partners. In case the Partnership Deed does not specify them, then the partners will have rights and duties as laid down in the Indian Partnership Act, 1932.

Rights of Partners :

1. Every partner has a right to take part in the management of the business.
2. Right to be consulted and expressed his opinion on any matter related to the firm.
3. Partner has a right to inspect the books of accounts or to copy them.
4. Right to have an equal share in the profits or losses of the firm, unless and otherwise agreed by the partners.
5. Right to receive interest on loan and advances made by partner to the firm.
6. Right to the partnership property unless and otherwise mentioned in the partnership deed.
7. Every partner has power or authority, in an emergency, to do any such acts, for the purpose of protecting the firm from losses.
8. Right to prevent the introduction of a new partner without consent of other partners.
9. Right to act an agent of the partnership firm in the ordinary course of business.
10. Right to be indemnified for the expenses incurred and losses sustained by partner to the firm.

Question 4.
Briefly explain the duties of partners.
Answer:
Generally, the Partnership Deed contains rights and duties of the partners. If deed is not prepared, the provisions of the Partnership Act will apply. Also when deed is in silent on any point, the relevant provisions of the act will apply.

Duties of Partners

1. Every partner has to attend diligently to his duties in the conduct of the business.
2. Should act in a just and faithful manner towards other partner and partners.
3. Should bound to share the losses of the firm equally unless and otherwise agreed upon by all partners.
4. No partner shall carry on any business competing with the firm. If he does so, he has to render to the firm any profits arising out of such business.
5. Must maintain true and correct accounts relating to the firm’s business.
6. No partner should make secret profits by way of commission or otherwise from the firm’s business.
7. Every partner is bound to keep and render true and proper accounts of the firm to his copartners.
8. No partner is allowed to assign or transfer his rights and interest in the firm to an ‘ outsider without the consent of other partners.

Question 5.
Explain the ways of dissolution of a Partnership Firm.
Answer:
The partnership is established through an agreement among partners. The partnership firm is established through partnership. A distinction should be made between the “Dissolution of partnership” and “Dissolution of firm”.

Dissolution of Partnership:
Dissolution of partnership implies the termination of the original partnership agreement or change in contractual relationship among partners. A partnership is dissolved by the insolvency, retirement, incapacity, death, expulsion, etc. of a partner or on the expiry/ completion of the term of partnership.

A partnership can be dissolved without dissolving the firm.
In dissolution of partnership, the business of the firm does not come to an end. The remaining partners continue the business by entering into a new agreement. On the Other hand, dissolution of firm implies dissolution between all the partners. Thus, business of the partnership firm comes to an end. The remaining partners continue the business by entering into a new agreement.

Dissolution of Firm:
Dissolution of firm implies dissolution between all the partners. The business of the partnership firm comes to an end. Its assets are realised and the creditors are paid off. Thus, dissolution of firm always involves dissolution of partnership but the dissolution of partnership does not necessarily mean dissolution of the firm.

Dissolution of the firm takes place under certain circumstances.
1) Dissolution by agreement:
A partnership firm may be dissolved with the mutual consent of all the partners or in accordance with the terms of the agreement.

2) Dissolution by notice :
In case partnership-at-will, a firm may be dissolved, if any partner gives a notice in writing to other partners indicating his intention to dissolve the firm.

3) Contingent dissolution :
A firm may be dissolved on the expiry of the firm, completion of the venture, death of a partner, adjudication of a partner as insolvent.

4) Compulsory dissolution:
A firm stands automatically dissolved if all partners or all but one partner are declared insolvent, or business becomes unlawful.

5) Dissolution through Court:
Court may order the dissolution of a firm, when any partner becomes members unsound, permanently incapable of performing his duties, guilty of misconduct, wilfully and persistently commits breach of the partnership agreement, unauthorised transfers the whole of his interest or share in the firm to a third person.

Very Short Answer Questions

Question 1.
Partnership Firm
Answer:
Partnership is an association of two or more persons who pool their financial and managerial resources and agree to carry on a business, and share its profits or losses. Partnership was established among partners through an agreement. Such agreement may be in the form of oral or written. If partnership agreement is in registration, it is known as partnership deed.

Question 2.
Partnership Deed [May 17 -A.P.]
Answer:
Partnership Deed is a document containing the terms and conditions of a partnership. It is an agreement in writing signed by the partners duly stamped and registered. The partnership deed defines certain rights, duties and obligations of partners and gov- erj^s relations among them in the conduct of business affairs of the firm.

Question 3.
Active Partner [Mar. 2018, -A.P. & T.S.]
Answer:
The partners who actively participate in the day-to-day operations of the business are knovyn as active partners or working partners.

Question 4.
Sleeping Partner
Answer:
The partners, who simply provide capital and do not participate in the management activities of the firm are called sleeping partners.

Question 5.
Partner by Estoppel
Answer:
A person who behaves in the public in such a way as to give an impression that he/she is a partner of the firm is called partner by estoppel. Such partners are not entitled to share the profits of the firm, but are fully liable if somebody suffers because of his/her false representation.

Question 6.
Prartner by Holding out
Answer:
Sometimes, the firm may use the name of a person in its activities, creating a sense in the public that he is also a partner. If that person accepts the same, he becomes automatically responsible to the third parties. Such person is known as “Partner by Holding Out”.

Andhra Pradesh BIEAP AP Inter 1st Year Civics Study Material 7th Lesson Justice Textbook Questions and Answers.

AP Inter 1st Year Civics Study Material 7th Lesson Justice

Long Answer Questions

Question 1.
Define Justice and describe various types of Justice.
Answer:
Introduction:
Justice is a dynamic concept in contemporary society. It has received the attention of several political philosophers, social reformers, economic thinkers, and psychological experts. They have considered the basic instinct of individuals belonging to the various sections residing in several parts of the world. Besides, almost all states, irrespective of their political and economic doctrines, have been striving to achieve justice and establish a society based on justice.

Meaning :
The word “Justice” is derived from the Latin word “Jus” which means “to bind”

Definitions :
We may advance some of the definitions of Justice in the following lines.
1. Plato:
“Justice is giving to every man his due. It is a combination of reason, courage, appetite and will in terms of the state”.

2. Aristotle:
“Justice is no other than each and every individual in society discharging his moral duties.”

3. Caphalous :
“Justice means speaking the truth and paying one’s debts.”

4. Polymarchus :
“Justice means to help friends and harm enemies.”

5. Barker :
“Justice means a combination and coordination of political values.”

Types of Justice
There are different types of Justice. They relate to Natural, Social, Political, and Legal spheres. Let us analyse these types of Justice.

1. Natural Justice :
Natural Justice is based on the notion that every person in the world possesses some rights for availing the natural resources. Natural resources provide support to the life of each and every creature on earth. As the human beings are the only rational creatures, it is their responsibility to see that natural resources have to be judiciously exploited. Human beings must keep in mind the requirements of future generations in this regard.

2. Social Justice :
Social Justice envisages a balance between rights of individuals and social control. It facilitates the fulfillment of the legitimate expectations of the individuals under the existing laws. It ensures several benefits and extends protection to the individuals against the interference or encroachment from others in society. It is consistent with the unity and the integrity of the nation. It fulfills the needs of the society.

Social Justice enforces the principle of equality before law. It also ensures eradication of social evils like poverty, unemployment, starvation, disease, etc. It also extends protection to the downtrodden and weaker sections of society. Ultimately it provides those conditions essential for the all round development of individuals.

3. Political Justice :
Political Justice symbolises politicl equality. It implies provision of political rights to all the adult citizens in a state. It facilitates free and fair participation of the citizens in the governance of the country. It is manifested to the full extent in times of elections. It allows the citizens for their active participation in day-to-day administration. It is based on the premise that everyone is counted as one and none for more than one. It may be noted that political justice prevails in the State when the following conditions are prevalent

1. Rule of law
2. Independent Judiciary
3. Popular elections to the representative bodies
4. Political parties
5. Freedom of press and assembly
6. Democratic rule etc.

4. Economic Justice :
Economic Justice refers to the absence of economic discrimination between individuals on irrational and unnatural grounds. It stands for the equal treatment of individuals irrespective of differences in the income, money, wealth, property etc. In its positive aspect, it implies payment of adequate emoluments to the workers strongly abhorring disparities in the distribution of wealth and incomes. It does not allow exploitation of the weaker sections. It sees that nobody is deprived of the basic necessities of life. It hints out that everyone must be provided with adequate food, clothing, shelter and other minimum needs. It conceives just economic order in the society. It supports the principle “from each according to his ability, to each according to his needs.”

5. Legal Justice:
Legal Justice is manifested in the laws of the state. It is supplemented by customs of the society. It is embodied in the Constitution and legislative enactments in a state. It determines the legal contours of Justice. Legal Justice basically has two implications. Firstly, it implies that there is just application of the laws in society on the basis of rule of law. There will be no discrimination between individuals in the applications of laws. Secondly, laws are made in consonance with the principles of natural justice.

Question 2.
What is meant by Justice? How is it evolved?
Answer:
Meaning :
The word “Justice is derived from a Latin word “JUS” which means ” to bind”.

It refers to the formulation and implementation of rules and regulations endorsed by the constitution and the judicial organisations.

Development of Justice :
In ancient India, Justice, being associated with dharma as enunciated in Hindu scriptures, was considered to be the duty of the King. The King used to maintain a just social order based on dharma. It was the primary duty of the King to maintain justice by punishing the wrong doers and rewarding the virtuous persons.

Justice normally means giving each person his due. However, its understanding differs from person to a person. Justice is viewed from the human aspect of every individual. Immanuel Kant, a German philosopher, stated that human beings possess dignity. When all persons ‘are endowed with dignity, they will be entitled to adequate opportunities for developing their talents and for pursuing their goals. Thus, Justice demands that each individual should be given equal consideration.

In Medieval age St. Augustine derived the concept of Justice from Plato. He emphasized on the proper relations between individuals for the harmonious working of society. Thomas Aquinas was considered as the first political philosopher who separated Justice from religion. By 16^th century, the concept of justice got completely secularized. The social contractualist like Hobbes identified Justice with the orders of the sovereign.

His successors John Locke, Rousseau, Emmanuel Kant, and others regarded Justice as a synthesis of liberty and equality. The advocates of Natural law developed the idea of individual justice. The Socialists conceived justice from economic point of view. While the conventionalists explained the concept of justice from individual perspective, the modernists viewed it from social perspective.

There is no single precise definition to the concept of justice. It was defined and discussed by various writers in different ways basing on the place, time context, culture, etc. It is considered as the sum total of the principles and beliefs advanced for the survival of the society. These principles and beliefs in turn led to the making of rights, freedoms and laws.

Question 3.
Write an essay on Social Justice.
Answer:
Social Justice is generally equated with the notion of equality. Equality is an indisputable and inherent element of social justice. The term ‘social justice’ has wider meaning. ‘Social justice’ connotes fairness, mutual obligation and responsibility in a society. It firmly believes that everyone is responsible to the others. Everyone must be provided adequate opportunities. Social Justice, in brief, aims at achieving a just society by eliminating injustice. It prevails when people have the belief of sharing the things in the society. They must be entitled to equitable treatment, human rights and fair allocation of common resources.

In this context modem political scientists like John Rawls and David Miller gave two prominent statements.

John Rawls advanced the theory of social justice commonly known as “Justice or Fariness”. To him, social justice implies equal access to the liberties, rights and opportunities as well as taking care of the interests of the deprived and disadvantaged sections of the society. He maintained that what is just or unjust in the human activities is determined on the basis of utility of such activities. He stated that social justice enables human beings equal access to civil liberties and human rights to lead a happy and healthy life. He emphasised that disadvantaged groups in society will be taken care of through the extension of social justice.

John Rawls concept of social justice is built around the idea of a social contract whereby all people sign a covenant for following and obeying certain rules for the betterment of the society as a whole. These rules or principles specify the basic rights and obligations involving the main political and social institutions. They regulate the allocation of benefits arising from social co-operation.

David Miller pointed out that social justice is concerned with the distribution of good (advantages) and bad (disadvantages) in society. He further analysed more specifically how these things are distributed in the society. According to him, social justice is concerned with the allocation of resources among people by social and political institutions. People, through social justice, receive many benefits in the fields of education, employment, wealth, health, welfare, transport etc.

Short Answer Questions

Question 1.
Explain the major concepts of Justice.
Answer:
Meaning :
The word “Justice” is derived from a Latin word “JUS” which means “to bind”.

Definition :
“Justice means speaking the truth and paying one’s debts”. – Caphalous

Major Concepts of Justice :
There are two major concepts of Justice. They are i) Numerial concept ii) Geometrical concept. They may be explained as follows :

1. Numerical concept:
Numerical concept of justice regards that everyone has equal share. The ancient Greek city states adopted this concept in public matters. The rulers of these city states filled up various offices with as many persons as maximum possible to demonstrate equality. They have not considered special knowledge, qualifications etc., for holding public offices. Jeremy Bentham, a famous British political philosopher, advocated this concept in modem times. He stated thus : “Everyone is to count for one, nobody for more than one.” Many modem liberal democratic states have been functioning on the basis of this concept.

2. Geometrical concept:
Geometrical concept is based on the notion of proportionate equality. It advocates equal share to equals and unequal share to unequals. It means that the distribution of power and patronage in public offices should be allocated in proportion to the worth or contribution of the individuals. Plato and Aristotle favoured this concept. Aristotle stated this concept in the following words: “If flutes are to be distributed, they should be distributed only among those who have the capacity of flute playing.” Efforts were made for allocating of benefits and responsibilities on equal basis keeping in view the worth of the recipients.

Question 2.
How is justice evolved?
Answer:
In ancient India, Justice, being associated with dharma as enunciated in Hindu scriptures, was considered to be the duty of the King. The King used to maintain a just social order based on dharma. It was the primary duty of the King to maintain justice by punishing the wrong doers and rewarding the virtuous persons.

Justice normally means giving each person his due. However, its understanding differs from person to a person. Justice is viewed from the human aspect of every individual. Immanuel Kant, a German philosopher, stated that human beings possess dignity. When all persons ‘are endowed with dignity, they will be entitled to adequate opportunities for developing their talents and for pursuing their goals. Thus, Justice demands that each individual should be given equal consideration.

In Medieval age St. Augustine derived the concept of Justice from Plato. He emphasized on the proper relations between individuals for the harmonious working of society. Thomas Aquinas was considered as the first political philosopher who separated Justice from religion. By 16^th century, the concept of justice got completely secularized. The social contractualist like Hobbes identified Justice with the orders of the sovereign. His successors John Locke, Rousseau, Emmanuel Kant and others regarded Justice as a synthesis of liberty and equality.

The advocates of Natural law developed the idea of individual justice. The Socialists conceived justice from economic point of view. While the conventionalists explained the concept of justice from individual perspective, the modernists viewed it from social perspective.

There is no single precise definition to the concept of justice. It was defined and discussed by various writers in different ways basing on the place, time context, culture etc. It is considered as the sum total of the principles and beliefs advanced for the survival of the society. These principles and beliefs in turn led to the making of rights, freedoms and laws.

Question 3.
Describe any three types of Justice.
Answer:
1. Natural Justice :
Natural Justice is based on the notion that every person in the world possesses some rights for availing the natural resources. Natural resources provide support to the life of each and every creature on earth. As the human beings are the only rational creatures, it is their responsibility to see that natural resources have to be judiciously exploited. Human beings must keep in mind the requirements of future generations in this regard.

2. Social Justice :
Social Justice envisages a balance between rights of individuals and social control. It facilitates the fulfillment of the legitimate expectations of the individuals under the existing laws. It ensures several benefits and extends protection to the individuals against the interference or encroachment from others in society. It is consistent with the unity and the integrity of the nation. It fulfills the needs of the society.

Social Justice enforces the principle of equality before law. It also ensures eradication of social evils like poverty, unemployment, starvation, disease etc. It also extends protection to the downtrodden and weaker sections of society. Ultimately it provides those conditions essential for the all round development of individuals.

3. Political Justice:
Political Justice symbolises political equality. It implies provision of political rights to all the adult citizens in a state. It facilitates free and fair participation of the citizens in the governance of the country. It is manifested to the full extent in times of elections. It allows the citizens for their active participation in day-to-day administration. It is based on the premise that everyone is counted as one and none for more than one. It may be noted that political justice prevails in the State when the following conditions are prevalent

1. Rule of law
2. Independent Judiciary
3. Popular elections to the representative bodies
4. Political parties
5. Freedom of press and assembly
6. Democratic rule etc.

Question 4.
Point out any three sources of Justice.
Answer:
Meaning :
The word “Justice” is derived from a Latin word “JUS” which means “to bind”.

Definition :
“Justice means speaking the truth and paying one’s debts” – Caphalous Sources of Justice:

Earnest Barker gives four sources of Justice. They are mentioned as below.

1. Nature
2. Ethics
3. Religion
4. Economic elements

1. Nature:
The Greek stoics perceived nature to be a source of Justice. Their perception of nature was a combination of moral philosophy and religious beliefs. For them nature, God and reason were inseparable entities. They pointed out that men who lived according to nature shared similar views of reason and God. They viewed that nature embodies three things. They are

1. Man should be free, 2. Man should be treated equally, 3. Man should be associated with his fellow beings by the common element of reason. These three things in turn have remained as a basis for liberty, equality and fraternity in society in course of time.

2. Ethics :
Idealist thinkers like Plato, Emanuel Kant, Thomas Hilly Green, Earnest Barker and other propounded that justice originated from ethical practices. They pointed out that values accepted by the society over a period of time have intum become the impersonal source of positive Justice. The state enforced this positive justice in course of time.

3. Religion :
Religion is regarded as another source of Justice. This source has been in force since medieval age. The church authorities held the notion that it was God who propounded the notions of justice, right and wrong. God, through church, initiated the concept of justice as the rule of the theory of might. Thomas Acqinas a philosopher turned saint believed that the Church is the manifestation of religion. According to him, life based on laws is the best one. The king must lead the people in right directions. He must exercise his authority in compliance to the church authority.

4. Economic elements :
Economic elements are also treated as a source of justice. These elements attained significance with the advent of industrial revolution which led to glaring economic disparities between different sections of society. Industrial revolution, inspite of its tremendous achievements, led to the growth of miseries, poverty and immorality in society. It forced the people to have a strong zeal of enterprise. Adam Smith, David Ricardo, Thomas Robert Malthus and other classical economists analysed justice in terms of economic factors.

Later, revolutionary thinkers like Karl Marx and Frederich Engles strongly advocated the role of economic elements as a basis to the justice. These thinkers began to prove the deficiencies in capitalist society. They argued that justice prevails only when economic equality is achieved through a classless society. But their

Question 5.
How is social Justice pursued?
Answer:
Social justice remains a mirage in a society having glaring disparities between different sections. Justice can’t be understood in absolute terms. Justice along with equality is a strong desire of every one in modem society. A society dominated by unjust relations between different sections can not achieve progress. In such a society the disadvantaged and deprived sections develop frustration in their day to day life. This leads to mutual conflicts between the majority poor and a few affluent persons. Hence a just society which ensures basic minimum facilities to all to lead happy and secure life is a must. In such a society adequate opportunities will be provided to various sections for realizing their goals.

Though many agree with the veiw that the State should lend a helping hand to the disadvantaged sections of the society to attain some degree of parity with others, there remains a disagreement over the methods pursued for achieving the goal. Extensive debate has taken place in the contemporary society. Such a debate revolved on the topic of inviting open competition through Free State organisation or private enterprises. But the fact lies in between the two. Both state and private involvement are necessary for achieving social justice in a state.

Very Short Answer Questions

Question 1.
Define Justice.
Answer:
“Justice is giving to every man his due. It is a combination of reason, courage, appetite and will in terms of the state” -Plato

Question 2.
What is Distributive Justice? [A.P. 19, 18]
Answer:
Distributive justice implies the distribution of goods and wealth of citizens by the state on merit basis. Aristotle stated that Justice is a sort of proportion. He regarded it as the most powerful instrument against revolutions. But modern writers like John Rawls denied Aristotle’s view. He pointed out that inequalities are inherent in the society. He remarked that inequalities must be balanced by some restrictive arrangements in the political system.

Question 3.
What is Corrective Justice?
Answer:
Corrective justice comprises restoring each person the lost rights due to the infringement of his rights by others. Aristotle viewed this justice as essentially negative which is concerned with voluntary commercial transactions like hire, sale and furnishing of property. In brief, corrective justice embodies moral excellence of individuals.

Question 4.
How are economic elements considered as a source of Justice?
Answer:
Economic elements are considered to be one of the important sources of Justice. These elements attained significance with the advent of industrial revolution which led to the vast economic disparities between different sections of the people.

Question 5.
What do you mean by Political Justice?
Answer:
Political Justice symbolises political equality. It implies provision of political rights to all the adult citizens in a state. It facilitates free and fair participation of the citizens in the governance of the country. It is manifested to the full extent in times of elections. It allows the citizens for their active participation in day-to-day administration. It is based on the premise that everyone is counted as one and none for more than one. It may be noted that political justice prevails in the State when the following conditions are prevalent 1. Rule of law 2. Independent Judiciary 3. Popular elections to the representative bodies 4. Political parties 5. Freedom of press and assembly 6. Democratic rule etc.

Question 6.
What is meant by Social Justice? [A.P. & T.S. Mar. 15]
Answer:
Social Justice envisages a balance between rights of individuals and social control. It facilitates the fulfillment of the legitimate expectations of the individuals under the existing laws. It ensures several benefits and extends protection to the individuals against the interference or encroachment from others in society. It is consistent with the unity and the integrity of the nation. It fulfills the needs of the society.

Social Justice enforces the principle of equality before law. It also ensures eradication of social evils like poverty, unemployment, starvation, disease etc. It also extends protection to the downtrodden and weaker sections of society. Ultimately it provides those conditions essential for the all round development of individuals.

Question 7.
What are the implications of Legal Justice?
Answer:
Legal Justice has two implications

• It implies that there is just application of the laws in the society on the basis of rule of law.
• Laws are made in accordance with the principle of Natural Justice.

Question 8.
What are the views of John Rawls on Social Justice? [A.P. Mar 18]
Answer:
John Rawls Admitted that:

• Social Justice implies equal access to the liberties, rights and opportunities to the deprived sections of the society.
• Social Justice is built around the idea of a social contract committed by the people for obeying certain rules.

Question 9.
Point out the views of David Miller on Social Justice?
Answer:
David Miller pointed out that social justice is concerned with the distribution of good (advantages) and bad (disadvantages) in society. He further analysed more specifically how these things are distributed in society. According to him, social justice is concerned with the allocation of resources among people by social and political institutions. People, through social justice, receive many benefits in the fields of education, employment, wealth, health, welfare, transport etc. However social justice has some negative repercussions. These relate to the interference of government in the private life and prescription of compulsory military service to individuals.

Miller’s theory applied to both public and private spheres. His theory regards social justice as a social virtue that speaks of what a person possesses and what he will owe to others in society.

Question 10.
In what respect is religion considered as a source of Justice?
Answer:
Religion is regarded as another source of Justice. This source has been in force since medieval age. The church authorities held the notion that it was God who propounded the notions of justice, right and wrong. God, through church, initiated the concept of justice as the rule of the theory of might. Thomas Acqinas a philosopher turned saint believed that the Church is the manifestation of religion. According to him, life based on laws is the best one. The king must lead the people in right directions. He must exercise his authority in compliance to the church authority.