10th Maths - Quarterly Exam 2025 - English Medium Original Question Paper | Tirunelveli District

10th Maths Quarterly Exam Model Paper with Solutions (2025-26) - Tirunelveli District

Part-I (14 x 1 = 14 Marks)

Answer all the questions.

1) Let $n(A) = m$ and $n(B) = n$, then the total number of non-empty relations that can be defined from A to B is

• a) $m^n$
• b) $n^m$
• c) $2^{mn} - 1$
• d) $2^{mn}$

Solution:

The total number of elements in the Cartesian product $A \times B$ is $n(A \times B) = n(A) \times n(B) = mn$.
The total number of relations from A to B is the number of subsets of $A \times B$, which is $2^{n(A \times B)} = 2^{mn}$.
This includes the empty relation. To find the number of non-empty relations, we subtract the empty relation from the total number of relations.
Number of non-empty relations = $2^{mn} - 1$.

c) $2^{mn} - 1$

2) If $f(x) = 2x^2$ and $g(x) = \frac{1}{3x}$, then $f \circ g$ is

• a) $\frac{3}{2x^2}$
• b) $\frac{2}{3x^2}$
• c) $\frac{2}{9x^2}$
• d) $\frac{1}{6x^2}$

Solution:

$f \circ g = f(g(x))$
Substitute $g(x)$ into $f(x)$:
$f(g(x)) = f\left(\frac{1}{3x}\right) = 2\left(\frac{1}{3x}\right)^2 = 2\left(\frac{1}{9x^2}\right) = \frac{2}{9x^2}$

c) $\frac{2}{9x^2}$

3) If $g = \{(1,1), (2,3), (3,5), (4,7)\}$ is a function given by $g(x) = \alpha x + \beta$ then the values of $\alpha$ and $\beta$ are

• a) $(-1,2)$
• b) $(2,-1)$
• c) $(-1,-2)$
• d) $(1,2)$

Solution:

From the set, we know $g(1)=1$ and $g(2)=3$. Using $g(x) = \alpha x + \beta$:
For (1,1): $1 = \alpha(1) + \beta \implies \alpha + \beta = 1$ ---(1)
For (2,3): $3 = \alpha(2) + \beta \implies 2\alpha + \beta = 3$ ---(2)
Subtracting equation (1) from (2):
$(2\alpha + \beta) - (\alpha + \beta) = 3 - 1 \implies \alpha = 2$
Substitute $\alpha = 2$ into equation (1):
$2 + \beta = 1 \implies \beta = 1 - 2 = -1$
So, the values are $(\alpha, \beta) = (2, -1)$.

b) $(2,-1)$

4) If the HCF of 65 and 117 is expressible in the form of $65m - 117$, then the value of m is

• a) 4
• b) 2
• c) 1
• d) 3

Solution:

First, find the HCF of 65 and 117 using Euclid's division algorithm.
$117 = 1 \times 65 + 52$
$65 = 1 \times 52 + 13$
$52 = 4 \times 13 + 0$
The HCF is 13.
Now, set the given expression equal to the HCF:
$65m - 117 = 13$
$65m = 117 + 13$
$65m = 130$
$m = \frac{130}{65} = 2$

b) 2

5) The $n^{th}$ term of an AP is $2n-1$. The sum of the first 'n' terms of that AP is

• a) $n^2$
• b) $2n-1$
• c) $n+2$
• d) $n(n+2)$

Solution:

The $n^{th}$ term is $t_n = 2n-1$. First term, $a = t_1 = 2(1) - 1 = 1$.
The terms are 1, 3, 5, ... which is an AP of the first n odd numbers.
The sum of the first n terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.
Here, $a=1$ and the common difference $d = 3 - 1 = 2$.
$S_n = \frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}[2n] = n^2$.
Alternatively, the sum of the first n odd natural numbers is $n^2$.

a) $n^2$

6) $y^2 + \frac{1}{y^2}$ is not equal to

• a) $\frac{y^4+1}{y^2}$
• b) $\left(y+\frac{1}{y}\right)^2$
• c) $\left(y-\frac{1}{y}\right)^2$
• d) $\left(y+\frac{1}{y}\right)^2 - 2$

Solution:

Let's evaluate each option:
a) $\frac{y^4+1}{y^2} = \frac{y^4}{y^2} + \frac{1}{y^2} = y^2 + \frac{1}{y^2}$. This is equal.
b) $\left(y+\frac{1}{y}\right)^2 = y^2 + 2(y)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2 = y^2 + 2 + \frac{1}{y^2}$. This is not equal.
c) $\left(y-\frac{1}{y}\right)^2 = y^2 - 2(y)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2 = y^2 - 2 + \frac{1}{y^2}$. This is not equal.
d) $\left(y+\frac{1}{y}\right)^2 - 2 = \left(y^2 + 2 + \frac{1}{y^2}\right) - 2 = y^2 + \frac{1}{y^2}$. This is equal.
Typically, in such MCQs, the most direct incorrect identity is the answer. Option (b) is the classic algebraic error $(a+b)^2 \neq a^2 + b^2$.

b) $\left(y+\frac{1}{y}\right)^2$

7) Graph of a linear equation is a

• a) straight line
• b) circle
• c) parabola
• d) hyperbola

Solution:

A linear equation, such as $ax + by + c = 0$, always represents a straight line when plotted on a Cartesian plane.

a) straight line

8) If $\triangle ABC$ is an isosceles triangle with $\angle C = 90^\circ$ and $AC = 5$ cm, then AB is

• a) 2.5 cm
• b) 5 cm
• c) 10 cm
• d) $5\sqrt{2}$ cm

Solution:

Since $\triangle ABC$ is an isosceles right-angled triangle at C, the sides adjacent to the right angle are equal. So, $AC = BC = 5$ cm.
By Pythagoras theorem, $AB^2 = AC^2 + BC^2$.
$AB^2 = 5^2 + 5^2 = 25 + 25 = 50$.
$AB = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$ cm.

d) $5\sqrt{2}$ cm

9) If in $\triangle ABC$, $DE \parallel BC$, $AB = 3.6$ cm, $AC = 2.4$ cm and $AD = 2.1$ cm, then the length of AE is

• a) 1.4 cm
• b) 1.8 cm
• c) 1.2 cm
• d) 1.05 cm

Solution:

By Basic Proportionality Theorem (Thales's Theorem), since $DE \parallel BC$, we have:
$\frac{AD}{AB} = \frac{AE}{AC}$
$\frac{2.1}{3.6} = \frac{AE}{2.4}$
$AE = \frac{2.1 \times 2.4}{3.6} = \frac{5.04}{3.6} = \frac{50.4}{36} = 1.4$ cm.

a) 1.4 cm

10) The point of intersection of $3x - y = 4$ and $x + y = 8$ is

• a) (5,3)
• b) (2,4)
• c) (3,5)
• d) (4,4)

Solution:

We have two equations:
(1) $3x - y = 4$
(2) $x + y = 8$
Adding equation (1) and (2):
$(3x - y) + (x + y) = 4 + 8$
$4x = 12 \implies x = 3$
Substitute $x = 3$ into equation (2):
$3 + y = 8 \implies y = 5$
The point of intersection is (3,5).

c) (3,5)

11) (2,1) is the point of intersection of two lines

• a) $x - y - 3 = 0; 3x - y - 7 = 0$
• b) $x + y = 3; 3x + y = 7$
• c) $3x + y = 3; x + y = 7$
• d) $x + 3y - 3 = 0; x - y - 7 = 0$

Solution:

Substitute the point (2,1) into the equations of each option to check if it satisfies both.
a) $(2) - (1) - 3 = -2 \ne 0$. Incorrect.
b) $(2) + (1) = 3$. Correct. And $3(2) + (1) = 6 + 1 = 7$. Correct. This is the answer.
c) $3(2) + (1) = 7 \ne 3$. Incorrect.
d) $(2) + 3(1) - 3 = 2 \ne 0$. Incorrect.

b) $x + y = 3; 3x + y = 7$

12) The area of a triangle formed by the points (2,-3), (3,2) and (-2,5) is

• a) 11
• b) 12
• c) 14
• d) 13

Solution:

Area of triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:
Area = $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
Area = $\frac{1}{2} |2(2-5) + 3(5-(-3)) + (-2)(-3-2)|$
Area = $\frac{1}{2} |2(-3) + 3(8) + (-2)(-5)|$
Area = $\frac{1}{2} |-6 + 24 + 10| = \frac{1}{2} |28| = 14$ sq. units.

c) 14

13) $\tan \theta \cdot \text{cosec}^2 \theta - \tan \theta$ is equal to

• a) $\sec \theta$
• b) $\cot^2 \theta$
• c) $\sin \theta$
• d) $\cot \theta$

Solution:

Factor out $\tan \theta$:
$\tan \theta (\text{cosec}^2 \theta - 1)$
Using the identity $1 + \cot^2 \theta = \text{cosec}^2 \theta$, we get $\text{cosec}^2 \theta - 1 = \cot^2 \theta$.
The expression becomes: $\tan \theta \cdot \cot^2 \theta$
Since $\cot \theta = \frac{1}{\tan \theta}$, we have:
$\tan \theta \cdot \left(\frac{1}{\tan \theta}\right)^2 = \tan \theta \cdot \frac{1}{\tan^2 \theta} = \frac{1}{\tan \theta} = \cot \theta$.

d) $\cot \theta$

14) If $\sin \theta + \cos \theta = a$ and $\sec \theta + \text{cosec} \theta = b$, then the value of $b(a^2 - 1)$ is equal to

• a) 2a
• b) 3a
• c) 0
• d) 2ab

Solution:

Given $\sin \theta + \cos \theta = a$.
Squaring both sides: $(\sin \theta + \cos \theta)^2 = a^2$
$\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = a^2$
$1 + 2\sin \theta \cos \theta = a^2 \implies 2\sin \theta \cos \theta = a^2 - 1$.
Now, consider $b = \sec \theta + \text{cosec} \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = \frac{a}{\sin \theta \cos \theta}$.
So, $b \sin \theta \cos \theta = a$.
We need to find the value of $b(a^2 - 1)$.
Substitute $a^2-1 = 2\sin \theta \cos \theta$:
$b(a^2 - 1) = b(2\sin \theta \cos \theta) = 2(b \sin \theta \cos \theta)$.
Since $b \sin \theta \cos \theta = a$, the expression becomes $2a$.

a) 2a

Part-II (10 x 2 = 20 Marks)

Answer any 10 questions. Question no. 28 is compulsory.

15) If $B \times A = \{(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)\}$ find A and B.

Solution:

In the Cartesian product $B \times A$, the set B consists of all the first elements of the ordered pairs, and the set A consists of all the second elements.
First elements (for set B): $\{-2, 0, 3\}$
Second elements (for set A): $\{3, 4\}$

So, $A = \{3, 4\}$ and $B = \{-2, 0, 3\}$.

16) A plane is flying at a speed of 500 km per hour. Express the distance 'd' travelled by the plane as a function of time t in hours.

Solution:

We know that Distance = Speed $\times$ Time.
Given: Speed = 500 km/hr.
Distance = d km.
Time = t hours.
So, the relationship is $d = 500 \times t$.
As a function of time t, we can write:

$d(t) = 500t$

17) 'a' and 'b' are two positive integers such that $a^b \times b^a = 800$. Find 'a' and 'b'.

Solution:

First, find the prime factorization of 800.
$800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^5 \times 5^2$.
We are given $a^b \times b^a = 800$.
Comparing $a^b \times b^a$ with $2^5 \times 5^2$, we can see a direct correspondence.
Let $a=2$ and $b=5$. Then $a^b \times b^a = 2^5 \times 5^2 = 32 \times 25 = 800$. This matches.
Let $a=5$ and $b=2$. Then $a^b \times b^a = 5^2 \times 2^5 = 25 \times 32 = 800$. This also matches.

The values are $a=2, b=5$ (or $a=5, b=2$).

18) Find the sum of 8 terms of the GP 1, -3, 9, -27,....

Solution:

This is a Geometric Progression (GP) with:
First term, $a = 1$.
Common ratio, $r = \frac{-3}{1} = -3$.
Number of terms, $n = 8$.
The formula for the sum of n terms of a GP is $S_n = \frac{a(r^n - 1)}{r-1}$.
$S_8 = \frac{1((-3)^8 - 1)}{-3 - 1} = \frac{6561 - 1}{-4} = \frac{6560}{-4} = -1640$.

The sum of the first 8 terms is -1640.

19) Find the L.C.M of $x^4 - 1$, $x^2 - 2x + 1$.

Solution:

Let the two polynomials be $p(x)$ and $q(x)$.
First polynomial: $p(x) = x^4 - 1$
We can factorize this using the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$.
$p(x) = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1)$
Factorizing $(x^2 - 1)$ further:
$p(x) = (x-1)(x+1)(x^2+1)$

Second polynomial: $q(x) = x^2 - 2x + 1$
This is a perfect square trinomial of the form $a^2 - 2ab + b^2 = (a-b)^2$.
$q(x) = (x-1)^2$

To find the L.C.M (Least Common Multiple), we take the product of the highest powers of all the factors present in the polynomials.
The factors are $(x-1)$, $(x+1)$, and $(x^2+1)$.
Highest power of $(x-1)$ is $(x-1)^2$.
Highest power of $(x+1)$ is $(x+1)$.
Highest power of $(x^2+1)$ is $(x^2+1)$.

L.C.M = $(x-1)^2(x+1)(x^2+1)$

20) Solve $2x^2 - 3x - 3 = 0$ by formula method.

Solution:

The given quadratic equation is $2x^2 - 3x - 3 = 0$.
We compare this with the standard form $ax^2 + bx + c = 0$.
Here, $a = 2$, $b = -3$, and $c = -3$.
The quadratic formula is given by:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Substituting the values of a, b, and c:
$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-3)}}{2(2)}$
$x = \frac{3 \pm \sqrt{9 - (-24)}}{4}$
$x = \frac{3 \pm \sqrt{9 + 24}}{4}$
$x = \frac{3 \pm \sqrt{33}}{4}$

The solutions are $x = \frac{3 + \sqrt{33}}{4}$ and $x = \frac{3 - \sqrt{33}}{4}$.

21) If $\alpha, \beta$ are the roots of the equation, $3x^2 + 7x - 2 = 0$, find the values of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.

Solution:

For the quadratic equation $3x^2 + 7x - 2 = 0$, we have $a=3, b=7, c=-2$.
Sum of the roots: $\alpha + \beta = -\frac{b}{a} = -\frac{7}{3}$.
Product of the roots: $\alpha\beta = \frac{c}{a} = -\frac{2}{3}$.

We need to find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
First, simplify the expression by taking a common denominator:
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$
We know the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
Substitute this into our expression:
$\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$
Now, substitute the values of the sum and product of the roots:
$= \frac{(-\frac{7}{3})^2 - 2(-\frac{2}{3})}{-\frac{2}{3}} = \frac{\frac{49}{9} + \frac{4}{3}}{-\frac{2}{3}}$
$= \frac{\frac{49+12}{9}}{-\frac{2}{3}} = \frac{\frac{61}{9}}{-\frac{2}{3}}$
$= \frac{61}{9} \times (-\frac{3}{2}) = \frac{61}{3} \times (-\frac{1}{2})$
$= -\frac{61}{6}$

The value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ is $-\frac{61}{6}$.

22) If $\triangle ABC$ is similar to $\triangle DEF$ such that $BC = 3$ cm, $EF = 4$ cm, and the area of $\triangle ABC = 54$ cm². Find the area of $\triangle DEF$.

Solution:

According to the theorem on the areas of similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given $\triangle ABC \sim \triangle DEF$.
Therefore, $\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2$.
Substituting the given values:
$\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2$
$\frac{54}{\text{Area}(\triangle DEF)} = \frac{9}{16}$
Now, solve for the area of $\triangle DEF$:
Area($\triangle DEF$) $= 54 \times \frac{16}{9}$
Area($\triangle DEF$) $= 6 \times 16$
Area($\triangle DEF$) $= 96$ cm²

The area of $\triangle DEF$ is 96 cm².

23) In $\triangle ABC$, D and E are points on the sides AB and AC respectively such that $DE \parallel BC$, if $\frac{AD}{DB} = \frac{3}{4}$ and $AC = 15$ cm, find AE.

Solution:

Given that in $\triangle ABC$, $DE \parallel BC$.
By the corollary of the Basic Proportionality Theorem (Thales's Theorem), we have:
$\frac{AD}{AB} = \frac{AE}{AC}$
We are given the ratio $\frac{AD}{DB} = \frac{3}{4}$. Let $AD = 3k$ and $DB = 4k$.
Then, the full length of side AB is $AB = AD + DB = 3k + 4k = 7k$.
Now, the ratio $\frac{AD}{AB} = \frac{3k}{7k} = \frac{3}{7}$.
Substitute this into the corollary formula:
$\frac{3}{7} = \frac{AE}{AC}$
We are given $AC = 15$ cm.
$\frac{3}{7} = \frac{AE}{15}$
$AE = \frac{3 \times 15}{7} = \frac{45}{7}$
$AE \approx 6.43$ cm.

The length of AE is $\frac{45}{7}$ cm or approximately 6.43 cm.

24) Find the slope of a line joining the points (14,10) and (14,−6).

Solution:

Let the points be $(x_1, y_1) = (14, 10)$ and $(x_2, y_2) = (14, -6)$.
The formula for the slope (m) of a line joining two points is:
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Substituting the coordinates:
$m = \frac{-6 - 10}{14 - 14} = \frac{-16}{0}$
Since the denominator is zero, the division is undefined. A line with an undefined slope is a vertical line.

The slope is undefined.

25) If the straight lines $12y = -(p + 3)x + 12$ and $12x - 7y = 16$ are perpendicular, then find 'p'.

Solution:

For two lines to be perpendicular, the product of their slopes must be -1 (i.e., $m_1 \times m_2 = -1$).

First line: $12y = -(p + 3)x + 12$
To find its slope ($m_1$), we rewrite it in the form $y = mx + c$.
$y = \frac{-(p + 3)}{12}x + \frac{12}{12}$
So, $m_1 = -\frac{p+3}{12}$.

Second line: $12x - 7y = 16$
To find its slope ($m_2$), we rewrite it in the form $y = mx + c$.
$-7y = -12x + 16$
$y = \frac{-12}{-7}x + \frac{16}{-7}$
$y = \frac{12}{7}x - \frac{16}{7}$
So, $m_2 = \frac{12}{7}$.

Now, apply the condition for perpendicular lines:
$m_1 \times m_2 = -1$
$\left(-\frac{p+3}{12}\right) \times \left(\frac{12}{7}\right) = -1$
$-\frac{p+3}{7} = -1$
$p+3 = 7$
$p = 7 - 3 = 4$

The value of 'p' is 4.

26) Prove that $\frac{\cos \theta}{1 + \sin \theta} = \sec \theta - \tan \theta$.

Solution:

We will start with the Left Hand Side (LHS) and show it is equal to the Right Hand Side (RHS).
LHS = $\frac{\cos \theta}{1 + \sin \theta}$
Multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 - \sin \theta)$.
LHS = $\frac{\cos \theta}{1 + \sin \theta} \times \frac{1 - \sin \theta}{1 - \sin \theta}$
LHS = $\frac{\cos \theta (1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)}$
In the denominator, we use the identity $(a+b)(a-b) = a^2 - b^2$:
LHS = $\frac{\cos \theta (1 - \sin \theta)}{1 - \sin^2 \theta}$
Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we know that $1 - \sin^2 \theta = \cos^2 \theta$.
LHS = $\frac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta}$
Cancel one $\cos \theta$ from the numerator and denominator:
LHS = $\frac{1 - \sin \theta}{\cos \theta}$
Split the fraction into two parts:
LHS = $\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}$
Using the identities $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
LHS = $\sec \theta - \tan \theta$

LHS = RHS. Hence Proved.

27) Prove that $\cot^2 \theta - \frac{1}{\sin^2 \theta} = -1$.

Solution:

We start with the Left Hand Side (LHS).
LHS = $\cot^2 \theta - \frac{1}{\sin^2 \theta}$
Using the reciprocal identity $\csc \theta = \frac{1}{\sin \theta}$, we can write $\csc^2 \theta = \frac{1}{\sin^2 \theta}$.
Substitute this into the expression:
LHS = $\cot^2 \theta - \csc^2 \theta$
Now, we use the Pythagorean identity involving $\cot \theta$ and $\csc \theta$:
$1 + \cot^2 \theta = \csc^2 \theta$
Rearranging this identity to match our expression:
$\cot^2 \theta - \csc^2 \theta = -1$
Therefore, LHS = -1.

LHS = -1 = RHS. Hence Proved.

28) A cat is located at the point $(-6,-4)$ in xy plane. A bottle of milk is kept at $(5,11)$. The cat wishes to consume the milk travelling through the shortest possible distance. Find the equation of the path it needs to take.

Solution:

The shortest path between two points is a straight line. We need to find the equation of the line passing through the points $(-6,-4)$ and $(5,11)$.
Let $(x_1, y_1) = (-6, -4)$ and $(x_2, y_2) = (5, 11)$.
Using the two-point form of a line: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$
$\frac{y - (-4)}{11 - (-4)} = \frac{x - (-6)}{5 - (-6)}$
$\frac{y + 4}{15} = \frac{x + 6}{11}$
$11(y + 4) = 15(x + 6)$
$11y + 44 = 15x + 90$
$15x - 11y + 90 - 44 = 0$

The equation of the path is $15x - 11y + 46 = 0$.

Part-III (10 x 5 = 50 Marks)

Answer any TEN questions. Question No. 42 is compulsory.

29) Let $A = \{x \in W \mid x < 2\}$, $B = \{x \in N \mid 1 < x \le 4\}$ and $C = \{3,5\}$. Verify that $A \times (B \cap C) = (A \times B) \cap (A \times C)$.

Solution:

First, list the elements of the sets:
$A = \{x \in W \mid x < 2\} = \{0, 1\}$ (W is Whole numbers)
$B = \{x \in N \mid 1 < x \le 4\} = \{2, 3, 4\}$ (N is Natural numbers)
$C = \{3, 5\}$

LHS: $A \times (B \cap C)$
$B \cap C = \{2, 3, 4\} \cap \{3, 5\} = \{3\}$
$A \times (B \cap C) = \{0, 1\} \times \{3\} = \{(0,3), (1,3)\}$ ---(1)

RHS: $(A \times B) \cap (A \times C)$
$A \times B = \{0, 1\} \times \{2, 3, 4\} = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\}$
$A \times C = \{0, 1\} \times \{3, 5\} = \{(0,3), (0,5), (1,3), (1,5)\}$
$(A \times B) \cap (A \times C) = \{(0,3), (1,3)\}$ ---(2)

From (1) and (2), we see that LHS = RHS.

Hence, $A \times (B \cap C) = (A \times B) \cap (A \times C)$ is verified.

30) If $f(x) = 2x + 3$, $g(x) = 1 - 2x$ and $h(x) = 3x$. Prove that $f \circ (g \circ h) = (f \circ g) \circ h$.

Solution:

Given functions are $f(x) = 2x + 3$, $g(x) = 1 - 2x$, and $h(x) = 3x$.

LHS: $f \circ (g \circ h)$
First, find $g \circ h = g(h(x))$:
$g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x$.
Now, find $f \circ (g \circ h) = f(g(h(x)))$:
$f(1 - 6x) = 2(1 - 6x) + 3 = 2 - 12x + 3 = 5 - 12x$. ---(1)

RHS: $(f \circ g) \circ h$
First, find $f \circ g = f(g(x))$:
$f(g(x)) = f(1 - 2x) = 2(1 - 2x) + 3 = 2 - 4x + 3 = 5 - 4x$.
Now, find $(f \circ g) \circ h = (f \circ g)(h(x))$:
$(f \circ g)(h(x)) = (f \circ g)(3x) = 5 - 4(3x) = 5 - 12x$. ---(2)

From (1) and (2), we can see that LHS = RHS.

Hence, $f \circ (g \circ h) = (f \circ g) \circ h$ is proved.

31) If the function f is defined by $f(x) = \begin{cases} x+2 & ; x > 1 \\ 2 & ; -1 \le x \le 1 \\ x-1 & ; -3 < x < -1 \end{cases}$ find the values of (i) $f(3)$ (ii) $f(0)$ (iii) $f(-1.5)$ (iv) $f(2) + f(-2)$.

Solution:

We use the definition of $f(x)$ based on the value of x.

(i) $f(3)$
Since $3 > 1$, we use the definition $f(x) = x + 2$.
$f(3) = 3 + 2 = 5$.

(ii) $f(0)$
Since $-1 \le 0 \le 1$, we use the definition $f(x) = 2$.
$f(0) = 2$.

(iii) $f(-1.5)$
Since $-3 < -1.5 < -1$, we use the definition $f(x) = x - 1$.
$f(-1.5) = -1.5 - 1 = -2.5$.

(iv) $f(2) + f(-2)$
First, find $f(2)$. Since $2 > 1$, $f(2) = 2 + 2 = 4$.
Next, find $f(-2)$. Since $-3 < -2 < -1$, $f(-2) = -2 - 1 = -3$.
$f(2) + f(-2) = 4 + (-3) = 1$.

(i) 5, (ii) 2, (iii) -2.5, (iv) 1

32) Find the sum to n terms of the series $3 + 33 + 333 + \dots$ to n terms.

Solution:

Let $S_n = 3 + 33 + 333 + \dots + n$ terms.
Take 3 common: $S_n = 3(1 + 11 + 111 + \dots + n$ terms).
Multiply and divide by 9:
$S_n = \frac{3}{9}(9 + 99 + 999 + \dots + n$ terms).
Rewrite the terms inside the bracket:
$S_n = \frac{1}{3}[(10 - 1) + (100 - 1) + (1000 - 1) + \dots + n$ terms].
Group the powers of 10 and the -1s separately:
$S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + n \text{ terms}) - (1 + 1 + 1 + \dots + n \text{ terms})]$.
The first part is a Geometric Progression (GP) with first term $a=10$, common ratio $r=10$.
The sum of this GP is $S_{GP} = \frac{a(r^n - 1)}{r-1} = \frac{10(10^n - 1)}{10-1} = \frac{10(10^n - 1)}{9}$.
The sum of the second part is simply $n$.
Substitute these back into the expression for $S_n$:
$S_n = \frac{1}{3}\left[\frac{10(10^n - 1)}{9} - n\right]$.
$S_n = \frac{1}{27}[10(10^n - 1) - 9n]$.

$S_n = \frac{1}{3}\left[\frac{10(10^n - 1)}{9} - n\right]$ or $\frac{10}{27}(10^n - 1) - \frac{n}{3}$

33) Find the sum of $15^2 + 16^2 + 17^2 + \dots + 28^2$.

Solution:

The required sum can be found by subtracting the sum of squares from 1 to 14 from the sum of squares from 1 to 28.
Sum = $(1^2 + 2^2 + \dots + 28^2) - (1^2 + 2^2 + \dots + 14^2)$.
We use the formula for the sum of the squares of the first n natural numbers: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.

For n = 28:
Sum up to 28 = $\frac{28(28+1)(2 \times 28+1)}{6} = \frac{28 \times 29 \times 57}{6} = 14 \times 29 \times 19 = 7714$.

For n = 14:
Sum up to 14 = $\frac{14(14+1)(2 \times 14+1)}{6} = \frac{14 \times 15 \times 29}{6} = 7 \times 5 \times 29 = 1015$.

Required Sum:
$7714 - 1015 = 6699$.

The sum is 6699.

34) Solve the following system of linear equations in three variables. $x+y+z=5$; $2x - y + z = 9$; $x - 2y + 3z = 16$

Solution:

Let the equations be:
(1) $x + y + z = 5$
(2) $2x - y + z = 9$
(3) $x - 2y + 3z = 16$

Step 1: Eliminate 'y' using (1) and (2).
Adding (1) and (2):
$(x+y+z) + (2x-y+z) = 5+9$
$3x + 2z = 14$ ---(4)

Step 2: Eliminate 'y' using (2) and (3).
Multiply equation (2) by 2:
$4x - 2y + 2z = 18$ ---(5)
Subtract (3) from (5):
$(4x - 2y + 2z) - (x - 2y + 3z) = 18 - 16$
$3x - z = 2$ ---(6)

Step 3: Solve equations (4) and (6).
(4) $3x + 2z = 14$
(6) $3x - z = 2$
Subtract (6) from (4):
$(3x + 2z) - (3x - z) = 14 - 2$
$3z = 12 \implies z = 4$.
Substitute $z=4$ into (6):
$3x - 4 = 2 \implies 3x = 6 \implies x = 2$.

Step 4: Find 'y'.
Substitute $x=2$ and $z=4$ into equation (1):
$2 + y + 4 = 5 \implies y + 6 = 5 \implies y = -1$.

The solution is $x=2, y=-1, z=4$.

35) If $9x^4 + 12x^3 + 28x^2 + ax + b$ is a perfect square, find the values of a and b.

Solution:

We use the long division method to find the square root.
Since the polynomial is a perfect square, the remainder must be zero.
This means the coefficients of the remainder terms must be zero.
$(a-16)x = 0 \implies a - 16 = 0 \implies a = 16$.
$b - 16 = 0 \implies b = 16$.

The values are $a=16$ and $b=16$.

36) If the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0$ are real and equal. Prove that either $a = 0$ (or) $a^3 + b^3 + c^3 = 3abc$.

Solution:

For the roots to be real and equal, the discriminant $(\Delta)$ must be zero. $\Delta = B^2 - 4AC = 0$.
Here, $A = c^2 - ab$, $B = -2(a^2 - bc)$, $C = b^2 - ac$.
$[-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0$
$4(a^2 - bc)^2 - 4(c^2b^2 - ac^3 - ab^3 + a^2bc) = 0$
Divide the entire equation by 4:
$(a^2 - bc)^2 - (c^2b^2 - ac^3 - ab^3 + a^2bc) = 0$
Expand the terms:
$(a^4 - 2a^2bc + b^2c^2) - c^2b^2 + ac^3 + ab^3 - a^2bc = 0$
$a^4 - 2a^2bc + b^2c^2 - c^2b^2 + ac^3 + ab^3 - a^2bc = 0$
Combine like terms:
$a^4 + ab^3 + ac^3 - 3a^2bc = 0$
Factor out 'a':
$a(a^3 + b^3 + c^3 - 3abc) = 0$
This implies that either the first factor is zero or the second factor is zero.
Case 1: $a = 0$
Case 2: $a^3 + b^3 + c^3 - 3abc = 0 \implies a^3 + b^3 + c^3 = 3abc$.

Hence, it is proved that either $a=0$ or $a^3+b^3+c^3=3abc$.

37) State and prove basic proportionality theorem.

Solution:

Statement: A straight line drawn parallel to a side of a triangle intersecting the other two sides, divides the sides in the same ratio.

Given: In ∆ABC, D is a point on AB and E is a point on AC.

To prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)

Construction: Draw a line DE || BC

No.	Statement	Reason
1.	∠ABC = ∠ADE = ∠1	Corresponding angles are equal because DE || BC
2.	∠ACB = ∠AED = ∠2	Corresponding angles are equal because DE || BC
3.	∠DAE = ∠BAC = ∠3	Both triangles have a common angle
	∆ABC ~ ∆ADE	By AAA similarity
	\(\frac{AB}{AD} = \frac{AC}{AE}\)	Corresponding sides are proportional
	\(\frac{AD+DB}{AD} = \frac{AE+EC}{AE}\)	Split AB and AC using the points D and E.
	\(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\)	On simplification
	\(\frac{DB}{AD} = \frac{EC}{AE}\)	Cancelling 1 on both sides
	\(\frac{AD}{DB} = \frac{AE}{EC}\)	Taking reciprocals. Hence proved.

Hence, the theorem is proved.

38) Find the area of the quadrilateral whose vertices are at $(-9,-2), (-8,-4), (2,2)$ and $(1,-3)$.

Solution:

First, we should arrange the vertices in counter-clockwise order to ensure a positive area. A rough sketch shows the order to be A$(-9,-2)$, B$(-8,-4)$, C$(1,-3)$, D$(2,2)$.
Let the vertices be $(x_1, y_1) = (-9, -2)$, $(x_2, y_2) = (-8, -4)$, $(x_3, y_3) = (1, -3)$, and $(x_4, y_4) = (2, 2)$.
We use the Shoelace formula for the area of a quadrilateral:
Area = $\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$

First part: $(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1)$
$= (-9)(-4) + (-8)(-3) + (1)(2) + (2)(-2)$
$= 36 + 24 + 2 - 4 = 58$.

Second part: $(y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)$
$= (-2)(-8) + (-4)(1) + (-3)(2) + (2)(-9)$
$= 16 - 4 - 6 - 18 = -12$.

Area = $\frac{1}{2} |58 - (-12)| = \frac{1}{2} |58 + 12| = \frac{1}{2} |70| = 35$.

The area of the quadrilateral is 35 sq. units.

39) A(-3,0), B(10,-2) and C(12,3) are the vertices of $\triangle ABC$. Find the equation of the altitude through A and B.

Solution:

1. Equation of the altitude through A (let it be AD):
The altitude from A is perpendicular to the side BC.
Slope of BC ($m_{BC}$) = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{12 - 10} = \frac{5}{2}$.
The slope of the altitude AD ($m_{AD}$) is the negative reciprocal of the slope of BC.
$m_{AD} = -\frac{1}{m_{BC}} = -\frac{2}{5}$.
The altitude passes through point A(-3,0). Using the point-slope form $y - y_1 = m(x - x_1)$:
$y - 0 = -\frac{2}{5}(x - (-3))$
$5y = -2(x + 3)$
$5y = -2x - 6 \implies 2x + 5y + 6 = 0$.

2. Equation of the altitude through B (let it be BE):
The altitude from B is perpendicular to the side AC.
Slope of AC ($m_{AC}$) = $\frac{3 - 0}{12 - (-3)} = \frac{3}{15} = \frac{1}{5}$.
The slope of the altitude BE ($m_{BE}$) = $-\frac{1}{m_{AC}} = -5$.
The altitude passes through point B(10,-2). Using the point-slope form:
$y - (-2) = -5(x - 10)$
$y + 2 = -5x + 50$
$5x + y + 2 - 50 = 0 \implies 5x + y - 48 = 0$.

Equation of altitude through A: $2x + 5y + 6 = 0$.
Equation of altitude through B: $5x + y - 48 = 0$.

40) Prove that $\frac{\cos^3 A - \sin^3 A}{\cos A - \sin A} - \frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = 2 \sin A \cos A$.

Solution:

We will simplify the Left Hand Side (LHS).
We use the algebraic identities for the sum and difference of cubes:
$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$

First Term:
$\frac{\cos^3 A - \sin^3 A}{\cos A - \sin A} = \frac{(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A)}{(\cos A - \sin A)}$
$= \cos^2 A + \cos A \sin A + \sin^2 A$
Since $\cos^2 A + \sin^2 A = 1$, this simplifies to $1 + \cos A \sin A$.

Second Term:
$\frac{\cos^3 A + \sin^3 A}{\cos A + \sin A} = \frac{(\cos A + \sin A)(\cos^2 A - \cos A \sin A + \sin^2 A)}{(\cos A + \sin A)}$
$= \cos^2 A - \cos A \sin A + \sin^2 A$
This simplifies to $1 - \cos A \sin A$.

Now, substitute these simplified forms back into the original expression:
LHS = $(1 + \cos A \sin A) - (1 - \cos A \sin A)$
LHS = $1 + \cos A \sin A - 1 + \cos A \sin A$
LHS = $2 \sin A \cos A$.

LHS = RHS. Hence Proved.

41) If $\frac{\cos \theta}{1 + \sin \theta} = \frac{1}{a}$ then prove that $\frac{a^2 - 1}{a^2 + 1} = \sin \theta$.
(Note: The question is solved based on the corrected version where the result is sinθ, which implies $a = \frac{1+\sin\theta}{\cos\theta}$)

Solution:

Given $\frac{\cos \theta}{1 + \sin \theta} = \frac{1}{a}$.
This implies $a = \frac{1 + \sin \theta}{\cos \theta}$.
We need to prove that $\frac{a^2 - 1}{a^2 + 1} = \sin \theta$.
Let's find expressions for $a^2-1$ and $a^2+1$.
$a^2 = \left(\frac{1 + \sin \theta}{\cos \theta}\right)^2 = \frac{(1 + \sin \theta)^2}{\cos^2 \theta}$.

Numerator: $a^2 - 1$
$a^2 - 1 = \frac{(1 + \sin \theta)^2}{\cos^2 \theta} - 1 = \frac{(1 + \sin \theta)^2 - \cos^2 \theta}{\cos^2 \theta}$
$= \frac{(1 + 2\sin\theta + \sin^2\theta) - \cos^2\theta}{\cos^2\theta}$
Substitute $\cos^2\theta = 1 - \sin^2\theta$:
$= \frac{1 + 2\sin\theta + \sin^2\theta - (1 - \sin^2\theta)}{\cos^2\theta} = \frac{2\sin\theta + 2\sin^2\theta}{\cos^2\theta} = \frac{2\sin\theta(1 + \sin\theta)}{\cos^2\theta}$.

Denominator: $a^2 + 1$
$a^2 + 1 = \frac{(1 + \sin \theta)^2}{\cos^2 \theta} + 1 = \frac{(1 + \sin \theta)^2 + \cos^2 \theta}{\cos^2 \theta}$
$= \frac{(1 + 2\sin\theta + \sin^2\theta) + \cos^2\theta}{\cos^2\theta}$
Since $\sin^2\theta + \cos^2\theta = 1$:
$= \frac{1 + 2\sin\theta + 1}{\cos^2\theta} = \frac{2 + 2\sin\theta}{\cos^2\theta} = \frac{2(1 + \sin\theta)}{\cos^2\theta}$.

Combine them:
$\frac{a^2 - 1}{a^2 + 1} = \frac{\frac{2\sin\theta(1 + \sin\theta)}{\cos^2\theta}}{\frac{2(1 + \sin\theta)}{\cos^2\theta}}$
$= \frac{2\sin\theta(1 + \sin\theta)}{\cos^2\theta} \times \frac{\cos^2\theta}{2(1 + \sin\theta)} = \sin\theta$.

Hence, $\frac{a^2 - 1}{a^2 + 1} = \sin \theta$ is proved.

42) Find the equation of a line passing through the point of intersection of the lines $4x + 7y - 3 = 0$ and $2x - 3y + 1 = 0$ that has equal intercepts on the axes.

Solution:

Step 1: Find the point of intersection.
(1) $4x + 7y - 3 = 0$
(2) $2x - 3y + 1 = 0$
Multiply equation (2) by 2: (3) $4x - 6y + 2 = 0$
Subtract equation (3) from (1):
$(4x + 7y - 3) - (4x - 6y + 2) = 0$
$13y - 5 = 0 \implies y = \frac{5}{13}$
Substitute $y = \frac{5}{13}$ into equation (2):
$2x - 3(\frac{5}{13}) + 1 = 0$
$2x - \frac{15}{13} + 1 = 0 \implies 2x = \frac{15}{13} - 1 = \frac{2}{13} \implies x = \frac{1}{13}$
The point of intersection is $P(\frac{1}{13}, \frac{5}{13})$.

Step 2: Find the equation of the required line.
A line with equal intercepts 'a' has the equation $\frac{x}{a} + \frac{y}{a} = 1$, which simplifies to $x + y = a$.
This line passes through the point $P(\frac{1}{13}, \frac{5}{13})$. Substitute these coordinates to find 'a'.
$\frac{1}{13} + \frac{5}{13} = a$
$a = \frac{6}{13}$
The equation of the line is $x + y = \frac{6}{13}$.
Multiplying by 13 to clear the fraction, we get $13x + 13y = 6$.

The required equation is $13x + 13y - 6 = 0$.

Part-IV (2 x 8 = 16 Marks)

Answer all the questions.

43) a) Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{2}{3}$ of the corresponding sides of the triangle PQR. (Scale factor $\frac{2}{3} < 1$). (OR)
b) Construct a $\triangle ABC$ such that $AB = 5.5$ cm, $\angle C = 25^\circ$ and the altitude from C to AB is 4 cm.

Solution:

a) Construction of a similar triangle (Scale factor $\frac{2}{3}$)

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b) Construction of $\triangle ABC$
Steps of Construction:

1. Draw a line segment $AB = 5.5$ cm.
2. At point A, draw a line AE such that $\angle BAE = 25^\circ$.
3. Draw a line AF perpendicular to AE.
4. Draw the perpendicular bisector of AB. Let it intersect AF at point O and AB at M.
5. With O as the center and OA as the radius, draw a circle. The major arc AKB of this circle contains the points that subtend an angle of $25^\circ$ at the circumference.
6. Draw a line GH parallel to AB at a distance of 4 cm. (This can be done by drawing a perpendicular from M, marking a point L at 4cm, and drawing a line parallel to AB through L).
7. This line GH intersects the circle at two points, C and C'.
8. Join AC and BC (or AC' and BC').
9. $\triangle ABC$ (or $\triangle ABC'$) is the required triangle.

44) a) A school announces that for a certain competition, the cash price will be distributed for all the participants equally as shown below:

No of participants (x)	2	4	6	8	10
Amount for each participant in Rs. (y)	180	90	60	45	36

(OR)
b) Draw the Graph of $xy = 24$, $x, y > 0$. Using the Graph find, (i) y when x = 3 and (ii) x when y = 6.

Solution:

a) Analysis of the Data Table
From the given table, let's find the product of x and y for each pair:
$2 \times 180 = 360$
$4 \times 90 = 360$
$6 \times 60 = 360$
$8 \times 45 = 360$
$10 \times 36 = 360$
Since the product $xy = 360$ is a constant, the number of participants (x) and the amount for each (y) are in inverse variation. The total prize money is Rs. 360.

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b) Graph of $xy = 24$
Step 1: Create a table of values for $xy = 24$ (or $y = \frac{24}{x}$).

x	1	2	3	4	6	8	12	24
y	24	12	8	6	4	3	2	1

Step 2: Plot the points and draw the curve.
Plot the points (1, 24), (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2), (24, 1) on a graph paper with appropriate scales on the x and y axes. Join the points with a smooth curve. The resulting curve is a rectangular hyperbola.

Step 3: Find the values from the graph.
(i) Find y when x = 3: On the graph, locate x = 3 on the x-axis. Draw a vertical line up to the curve. From that point on the curve, draw a horizontal line to the y-axis. The line meets the y-axis at 8.
By calculation: $y = \frac{24}{3} = 8$.

(ii) Find x when y = 6: On the graph, locate y = 6 on the y-axis. Draw a horizontal line to the right to meet the curve. From that point on the curve, draw a vertical line down to the x-axis. The line meets the x-axis at 4.
By calculation: $x = \frac{24}{6} = 4$.

From the graph: (i) When $x=3$, $y=8$. (ii) When $y=6$, $x=4$.

Original Question Paper