10th Maths - Quarterly Exam 2025 - Answer Key | Sivagangai District | English Medium

10th Maths Quarterly Exam 2025: Complete Question Paper with Solutions

Solutions

Part - I

1. The range of the relation R = {(x, x²) | x is a prime number less than 13} is

• a) {2, 3, 5, 7}
• b) {2, 3, 5, 7, 11}
• c) {4, 9, 25, 49, 121}
• d) {1, 4, 9, 25, 49, 121}

Solution:

The prime numbers less than 13 are 2, 3, 5, 7, 11. These are the values of x (the domain).

The relation is defined by R = {(x, x²)}. The range is the set of all second elements, which are x².

For x = 2, x² = 4
For x = 3, x² = 9
For x = 5, x² = 25
For x = 7, x² = 49
For x = 11, x² = 121

The range is {4, 9, 25, 49, 121}.

Correct Answer: c) {4, 9, 25, 49, 121}

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

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

Solution:

fog(x) means f(g(x)).

First, substitute g(x) into f(x):

$f(g(x)) = f(\frac{1}{3x})$

Now, apply the function f(x) = 2x² to the input $\frac{1}{3x}$:

$f(\frac{1}{3x}) = 2 \left(\frac{1}{3x}\right)^2 = 2 \left(\frac{1}{9x^2}\right) = \frac{2}{9x^2}$

Correct Answer: c) $\frac{2}{9x^2}$

3. Which one of the following represents a constant function

• a) f(x) = x
• b) f(x) = c
• c) f(x) = mx + c
• d) $f(x) = \frac{1}{x}$

Solution:

A constant function is a function whose output value is the same for every input value. It is represented as f(x) = c, where c is a constant.

Correct Answer: b) f(x) = c

4. Using Euclid's division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are

• a) 0, 1, 8
• b) 1, 4, 8
• c) 0, 1, 3
• d) 1, 3, 5

Solution:

Let 'n' be any positive integer. By Euclid's division lemma, n can be written in the form 3q, 3q+1, or 3q+2.

Case 1: n = 3q
$n^3 = (3q)^3 = 27q^3 = 9(3q^3)$. The remainder is 0.

Case 2: n = 3q+1
$n^3 = (3q+1)^3 = (3q)^3 + 3(3q)^2(1) + 3(3q)(1)^2 + 1^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1$. The remainder is 1.

Case 3: n = 3q+2
$n^3 = (3q+2)^3 = (3q)^3 + 3(3q)^2(2) + 3(3q)(2)^2 + 2^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8$. The remainder is 8.

The possible remainders are 0, 1, and 8.

Correct Answer: a) 0, 1, 8

5. If 6 times of 6th term of an A.P. is equal to 7 times the 7th term, then the 13th term of the A.P. is

• a) 0
• b) 6
• c) 7
• d) 13

Solution:

Given: $6 \times t_6 = 7 \times t_7$

We know the nth term of an A.P. is $t_n = a + (n-1)d$.

$6(a + (6-1)d) = 7(a + (7-1)d)$
$6(a + 5d) = 7(a + 6d)$
$6a + 30d = 7a + 42d$
$30d - 42d = 7a - 6a$
$-12d = a$

We need to find the 13th term, $t_{13}$:

$t_{13} = a + (13-1)d = a + 12d$

Substitute $a = -12d$:

$t_{13} = (-12d) + 12d = 0$

Correct Answer: a) 0

6. The series 3 + 33 + 333 + ... (n terms) is

• a) a G.P.
• b) an A.P.
• c) both A.P and G.P
• d) neither A.P. nor G.P.

Solution:

Check for A.P.: The common difference must be constant. $d_1 = 33 - 3 = 30$. $d_2 = 333 - 33 = 300$. Since $d_1 \neq d_2$, it is not an A.P.

Check for G.P.: The common ratio must be constant. $r_1 = 33 / 3 = 11$. $r_2 = 333 / 33 \approx 10.09$. Since $r_1 \neq r_2$, it is not a G.P.

Correct Answer: d) neither A.P. nor G.P.

7. $y^2 + \frac{1}{y^2}$ is not equal to

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

Solution:

Let's check 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) $(y + \frac{1}{y})^2 = y^2 + 2(y)(\frac{1}{y}) + (\frac{1}{y})^2 = y^2 + 2 + \frac{1}{y^2}$. This is not equal.

c) $(y - \frac{1}{y})^2 + 2 = (y^2 - 2(y)(\frac{1}{y}) + (\frac{1}{y})^2) + 2 = (y^2 - 2 + \frac{1}{y^2}) + 2 = y^2 + \frac{1}{y^2}$. This is equal.

d) $(y + \frac{1}{y})^2 - 2 = (y^2 + 2(y)(\frac{1}{y}) + (\frac{1}{y})^2) - 2 = (y^2 + 2 + \frac{1}{y^2}) - 2 = y^2 + \frac{1}{y^2}$. This is equal.

The expression not equal to $y^2 + \frac{1}{y^2}$ is option b.

Correct Answer: b) $(y + \frac{1}{y})^2$

8. The solution of $(2x - 1)^2 = 9$ is equal to

• a) -1
• b) 2
• c) -1, 2
• d) none of these

Solution:

Given: $(2x - 1)^2 = 9$

Take the square root of both sides:

$2x - 1 = \pm\sqrt{9}$
$2x - 1 = \pm 3$

Case 1: $2x - 1 = 3$
$2x = 4$
$x = 2$

Case 2: $2x - 1 = -3$
$2x = -2$
$x = -1$

The solutions are -1 and 2.

Correct Answer: c) -1, 2

9. If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5cm, then AB is

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

Solution:

Given an isosceles triangle ∆ABC with ∠C = 90°.

Since it is an isosceles right-angled triangle, the sides adjacent to the right angle are equal. So, AC = BC = 5 cm.

AB is the hypotenuse. 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.

Correct Answer: d) $5\sqrt{2}$ cm

10. If in ∆ABC, DE || 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:

Since DE || BC, by the Basic Proportionality Theorem (Thales' Theorem):

$\frac{AD}{AB} = \frac{AE}{AC}$

Substitute the given values:

$\frac{2.1}{3.6} = \frac{AE}{2.4}$

$AE = \frac{2.1 \times 2.4}{3.6} = \frac{5.04}{3.6} = 1.4$ cm.

Correct Answer: a) 1.4 cm

11. The slope of the line joining (12, 3), (4, a) is 1/8. The value of 'a' is

• a) 1
• b) 4
• c) -5
• d) 2

Solution:

The formula for the slope (m) is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Given points $(x_1, y_1) = (12, 3)$ and $(x_2, y_2) = (4, a)$, and slope $m = 1/8$.

$\frac{1}{8} = \frac{a - 3}{4 - 12}$
$\frac{1}{8} = \frac{a - 3}{-8}$

Cross-multiply:

$1 \times (-8) = 8 \times (a - 3)$
$-8 = 8a - 24$
$8a = 24 - 8$
$8a = 16$
$a = 2$

Correct Answer: d) 2

12. Consider four straight lines:
i) $l_1: 3y = 4x + 5$
ii) $l_2: 4y = 3x - 1$
iii) $l_3: 4y + 3x = 7$
iv) $l_4: 4x + 3y = 2$
Which of the following statement is true?

• a) $l_1$ and $l_2$ are perpendicular
• b) $l_1$ and $l_4$ are parallel
• c) $l_2$ and $l_4$ are perpendicular
• d) $l_2$ and $l_3$ are parallel

Solution:

First, find the slope of each line by rewriting it in the slope-intercept form $y = mx + c$.

i) $l_1: 3y = 4x + 5 \implies y = \frac{4}{3}x + \frac{5}{3}$. So, the slope $m_1 = \frac{4}{3}$.

ii) $l_2: 4y = 3x - 1 \implies y = \frac{3}{4}x - \frac{1}{4}$. So, the slope $m_2 = \frac{3}{4}$.

iii) $l_3: 4y + 3x = 7 \implies 4y = -3x + 7 \implies y = -\frac{3}{4}x + \frac{7}{4}$. So, the slope $m_3 = -\frac{3}{4}$.

iv) $l_4: 4x + 3y = 2 \implies 3y = -4x + 2 \implies y = -\frac{4}{3}x + \frac{2}{3}$. So, the slope $m_4 = -\frac{4}{3}$.

Now, let's check each option:

a) $l_1$ and $l_2$ are perpendicular:
For perpendicular lines, $m_1 \times m_2 = -1$. Here, $m_1 \times m_2 = (\frac{4}{3}) \times (\frac{3}{4}) = 1$. Since $1 \neq -1$, this is FALSE.

b) $l_1$ and $l_4$ are parallel:
For parallel lines, $m_1 = m_4$. Here, $m_1 = \frac{4}{3}$ and $m_4 = -\frac{4}{3}$. Since the slopes are not equal, this is FALSE.

c) $l_2$ and $l_4$ are perpendicular:
For perpendicular lines, $m_2 \times m_4 = -1$. Here, $m_2 \times m_4 = (\frac{3}{4}) \times (-\frac{4}{3}) = -1$. This statement is TRUE.

d) $l_2$ and $l_3$ are parallel:
For parallel lines, $m_2 = m_3$. Here, $m_2 = \frac{3}{4}$ and $m_3 = -\frac{3}{4}$. Since the slopes are not equal, this is FALSE.

Correct Answer: c) $l_2$ and $l_4$ are perpendicular

13. The slope of the line which is parallel to line joining the points (0, 0) and (-8, 8) is

• a) -1
• b) 1
• c) 1/3
• d) -8

Solution:

First, find the slope of the line joining (0, 0) and (-8, 8).

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{-8 - 0} = \frac{8}{-8} = -1$.

Parallel lines have the same slope. Therefore, the slope of the required line is also -1.

Correct Answer: a) -1

14. $tan\theta cosec^2\theta - tan\theta$ is equal to

• a) secθ
• b) cot²θ
• c) sinθ
• d) cotθ

Solution:

Start with the given expression: $tan\theta cosec^2\theta - tan\theta$

Factor out $tan\theta$:

$tan\theta (cosec^2\theta - 1)$

Using the trigonometric identity $1 + cot^2\theta = cosec^2\theta$, we get $cot^2\theta = cosec^2\theta - 1$.

Substitute this into the expression:

$tan\theta (cot^2\theta)$

We know that $cot\theta = \frac{1}{tan\theta}$. So, $cot^2\theta = \frac{1}{tan^2\theta}$.

$tan\theta \times \frac{1}{tan^2\theta} = \frac{1}{tan\theta} = cot\theta$

Correct Answer: d) cotθ

Part - II

15. If A x B = {(3, 2), (3,4), (5,2), (5,4)}, then find A and B.

Solution:

The Cartesian product A x B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

Set A is the set of all first elements in the ordered pairs:

A = {3, 5}

Set B is the set of all second elements in the ordered pairs:

B = {2, 4}

16. A relation R is given by the set {(x, y) | y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Find its domain and range.

Solution:

The domain is the set of all possible input values (x).

Domain = {0, 1, 2, 3, 4, 5}

The range is the set of all possible output values (y), which we find by applying the rule y = x + 3 to each element in the domain.

• If x = 0, y = 0 + 3 = 3
• If x = 1, y = 1 + 3 = 4
• If x = 2, y = 2 + 3 = 5
• If x = 3, y = 3 + 3 = 6
• If x = 4, y = 4 + 3 = 7
• If x = 5, y = 5 + 3 = 8

Range = {3, 4, 5, 6, 7, 8}

17. In the given factor tree, find the numbers m and n.

Solution:

To solve this, we work from the bottom of the factor tree upwards, multiplying the factors at each step.

1. Find the value of the lowest box: The branches coming from the lowest box are 5 and 2.
   Value of the lowest box = $5 \times 2 = 10$.
2. Find the value of n: The node 'n' is the product of its branches, which are 5 and the lowest box (which we found to be 10).
   n = $5 \times 10 = 50$.
3. Find the value of the box below m: This box is the product of its branches, which are 3 and 'n' (which we found to be 50).
   Value of the box below m = $3 \times 50 = 150$.
4. Find the value of m: The top node 'm' is the product of its branches, which are 2 and the box below it (which we found to be 150).
   m = $2 \times 150 = 300$.

Therefore, the final values are m = 300 and n = 50.

18. If 3 + k, 18 - k, 5k + 1 are in A.P. then find k.

Solution:

If three terms are in an Arithmetic Progression (A.P.), the difference between consecutive terms is constant. An easier property is that twice the middle term equals the sum of the first and third terms.

$2 \times (18 - k) = (3 + k) + (5k + 1)$

$36 - 2k = 6k + 4$

$36 - 4 = 6k + 2k$

$32 = 8k$

$k = \frac{32}{8} = 4$

So, k = 4.

19. If $1+2+3+...+k = 325$, then find $1^3 + 2^3+3^3 + ... + k^3$.

Solution:

We are given the sum of the first k natural numbers:

$\sum_{i=1}^{k} i = 1+2+...+k = \frac{k(k+1)}{2} = 325$

We need to find the sum of the cubes of the first k natural numbers:

$\sum_{i=1}^{k} i^3 = 1^3+2^3+...+k^3 = \left(\frac{k(k+1)}{2}\right)^2$

Substitute the given value:

$1^3+2^3+...+k^3 = (325)^2 = 105625$

The sum is 105,625.

20. Find the breadth of a rectangle whose area is given as $\frac{(x-4)(x+3)}{3x-12}$ (sq. units) and its length is $\frac{x-3}{3}$ (units)

Solution:

We know that the area of a rectangle is given by the formula:

Area = Length × Breadth

Therefore, Breadth = $\frac{\text{Area}}{\text{Length}}$

First, let's simplify the given expression for the area:

Area = $\frac{(x-4)(x+3)}{3x-12} = \frac{(x-4)(x+3)}{3(x-4)}$

Canceling the term $(x-4)$, we get:

Area = $\frac{x+3}{3}$

Now, we can find the breadth:

Breadth = $\frac{\text{Area}}{\text{Length}} = \frac{\left(\frac{x+3}{3}\right)}{\left(\frac{x-3}{3}\right)}$

Breadth = $\frac{x+3}{3} \times \frac{3}{x-3}$

Breadth = $\frac{x+3}{x-3}$ units.

21. Determine the quadratic equation whose sum and product of roots are -9 and 20.

Solution:

The general form of a quadratic equation with a given sum and product of roots is:

$x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$

Given:

Sum of roots = -9

Product of roots = 20

Substituting these values into the formula:

$x^2 - (-9)x + 20 = 0$

$x^2 + 9x + 20 = 0$

This is the required quadratic equation.

22. Solve by formula method : $x^2 + 2x - 2 = 0$

Solution:

The given quadratic equation is $x^2 + 2x - 2 = 0$.

Comparing this with the standard form $ax^2 + bx + c = 0$, we have:

$a = 1, b = 2, c = -2$

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values of a, b, and c:

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-2)}}{2(1)}$

$x = \frac{-2 \pm \sqrt{4 + 8}}{2}$

$x = \frac{-2 \pm \sqrt{12}}{2}$

To simplify $\sqrt{12}$, we write it as $\sqrt{4 \times 3} = 2\sqrt{3}$.

$x = \frac{-2 \pm 2\sqrt{3}}{2}$

Factor out 2 from the numerator: $x = \frac{2(-1 \pm \sqrt{3})}{2}$

Cancel the common factor 2:

$x = -1 \pm \sqrt{3}$

The solutions are $x = -1 + \sqrt{3}$ and $x = -1 - \sqrt{3}$.

23. In ∆ABC, AD is the bisector of ∠A. If BD = 4cm, DC = 3cm and AB = 6cm, find AC.

Solution:

By the Angle Bisector Theorem, the internal bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides.

Therefore, for the bisector AD of ∠A, we have:

$\frac{AB}{AC} = \frac{BD}{DC}$

Given values: AB = 6 cm, BD = 4 cm, DC = 3 cm.

Substituting these values into the theorem:

$\frac{6}{AC} = \frac{4}{3}$

Cross-multiplying to solve for AC:

$6 \times 3 = 4 \times AC$

$18 = 4 \times AC$

$AC = \frac{18}{4} = \frac{9}{2} = 4.5$

The length of AC is 4.5 cm.

24. Show that the given points are collinear (-3, -4), (7,2) and (12, 5)

Solution:

Points are collinear if the area of the triangle formed by them is zero.

Let the points be A(-3, -4), B(7, 2), and C(12, 5).

The formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is:

Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substituting the coordinates:

Area = $\frac{1}{2} |(-3)(2 - 5) + (7)(5 - (-4)) + (12)(-4 - 2)|$

Area = $\frac{1}{2} |(-3)(-3) + (7)(9) + (12)(-6)|$

Area = $\frac{1}{2} |9 + 63 - 72|$

Area = $\frac{1}{2} |72 - 72|$

Area = $\frac{1}{2} |0| = 0$

Since the area of the triangle is 0 sq. units, the given points are collinear.

25. 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 cat's position, A(-6, -4), and the milk's position, B(5, 11).

Let $(x_1, y_1) = (-6, -4)$ and $(x_2, y_2) = (5, 11)$.

Using the two-point form of the equation of a line:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

$\frac{y - (-4)}{11 - (-4)} = \frac{x - (-6)}{5 - (-6)}$

$\frac{y + 4}{15} = \frac{x + 6}{11}$

Cross-multiplying:

$11(y + 4) = 15(x + 6)$

$11y + 44 = 15x + 90$

Rearranging the terms to the standard form $Ax + By + C = 0$:

$15x - 11y + 90 - 44 = 0$

$15x - 11y + 46 = 0$

This is the required equation of the path.

26. Check whether the given lines are parallel or perpendicular: $\frac{2x}{3} + \frac{y}{2} + \frac{1}{10} = 0$ and $\frac{x}{3} + \frac{y}{4} + \frac{1}{7} = 0$.

Solution:

To check if lines are parallel or perpendicular, we find their slopes ($m$).

Line 1: $\frac{2x}{3} + \frac{y}{2} + \frac{1}{10} = 0$

To find the slope, we isolate y:

$\frac{y}{2} = -\frac{2x}{3} - \frac{1}{10}$

$y = 2 \left(-\frac{2x}{3} - \frac{1}{10}\right) = -\frac{4x}{3} - \frac{2}{10} = -\frac{4x}{3} - \frac{1}{5}$

The slope of the first line is $m_1 = -\frac{4}{3}$.

Line 2: $\frac{x}{3} + \frac{y}{4} + \frac{1}{7} = 0$

Isolate y:

$\frac{y}{4} = -\frac{x}{3} - \frac{1}{7}$

$y = 4 \left(-\frac{x}{3} - \frac{1}{7}\right) = -\frac{4x}{3} - \frac{4}{7}$

The slope of the second line is $m_2 = -\frac{4}{3}$.

Conclusion:

Since the slopes are equal ($m_1 = m_2 = -\frac{4}{3}$), the given lines are parallel.

27. Prove that $tan^2\theta - sin^2\theta = tan^2\theta sin^2\theta$.

Solution:

We start with the Left Hand Side (LHS) and simplify it.

LHS = $tan^2\theta - sin^2\theta$

Rewrite $tan^2\theta$ as $\frac{sin^2\theta}{cos^2\theta}$:

LHS = $\frac{sin^2\theta}{cos^2\theta} - sin^2\theta$

Factor out $sin^2\theta$ from both terms:

LHS = $sin^2\theta \left(\frac{1}{cos^2\theta} - 1\right)$

Find a common denominator inside the parenthesis:

LHS = $sin^2\theta \left(\frac{1 - cos^2\theta}{cos^2\theta}\right)$

Using the Pythagorean identity $sin^2\theta + cos^2\theta = 1$, we know that $1 - cos^2\theta = sin^2\theta$.

LHS = $sin^2\theta \left(\frac{sin^2\theta}{cos^2\theta}\right)$

We know that $\frac{sin^2\theta}{cos^2\theta} = tan^2\theta$.

LHS = $sin^2\theta \cdot tan^2\theta = tan^2\theta sin^2\theta$

LHS = RHS. Hence, the identity is proved.

28. The following table represents a function from A = {5, 6, 8, 10} to B = {19, 15, 9, 11} where f(x) = 2x - 1. Find the values of a and b.
x: 5, 6, 8, 10
f(x): a, 11, b, 19

Solution:

The function is given by the rule $f(x) = 2x - 1$.

To find the value of 'a':

From the table, 'a' is the value of f(x) when x = 5.

$a = f(5)$

$a = 2(5) - 1 = 10 - 1 = 9$

To find the value of 'b':

From the table, 'b' is the value of f(x) when x = 8.

$b = f(8)$

$b = 2(8) - 1 = 16 - 1 = 15$

Therefore, the values are a = 9 and b = 15.

29. Let A = {1, 2, 3, 4} and B = {2, 5, 8, 11, 14} be two sets. Let f:A→B be a function given by f(x) = 3x - 1. Represent this function (i) by arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) in a graphical form.

Solution:

First, let's find the image for each element in the domain A = {1, 2, 3, 4} using the function f(x) = 3x - 1.

• f(1) = 3(1) - 1 = 2
• f(2) = 3(2) - 1 = 5
• f(3) = 3(3) - 1 = 8
• f(4) = 3(4) - 1 = 11

(i) Arrow Diagram:

Draw two ovals. The first oval (Set A) contains {1, 2, 3, 4}. The second oval (Set B) contains {2, 5, 8, 11, 14}. Draw arrows from each element in A to its corresponding image in B.

1 → 2
2 → 5
3 → 8
4 → 11

(ii) Table Form:

x	f(x)
1	2
2	5
3	8
4	11

(iii) Set of Ordered Pairs:

The function can be represented as a set of ordered pairs (x, f(x)).

f = {(1, 2), (2, 5), (3, 8), (4, 11)}

(iv) Graphical Form:

Plot the ordered pairs {(1, 2), (2, 5), (3, 8), (4, 11)} as points on a Cartesian coordinate plane.

30. Find x if gff(x) = fgg(x) given f(x) = 3x + 1 and g(x) = x + 3.

Solution:

First, we find the expression for the Left Hand Side (LHS), gff(x).

1. Find ff(x):
$ff(x) = f(f(x)) = f(3x+1) = 3(3x+1) + 1 = 9x + 3 + 1 = 9x + 4$.

2. Find gff(x):
$gff(x) = g(ff(x)) = g(9x+4) = (9x+4) + 3 = 9x + 7$.

Next, we find the expression for the Right Hand Side (RHS), fgg(x).

1. Find gg(x):
$gg(x) = g(g(x)) = g(x+3) = (x+3) + 3 = x + 6$.

2. Find fgg(x):
$fgg(x) = f(gg(x)) = f(x+6) = 3(x+6) + 1 = 3x + 18 + 1 = 3x + 19$.

Now, we set LHS = RHS and solve for x:

$9x + 7 = 3x + 19$

$9x - 3x = 19 - 7$

$6x = 12$

$x = \frac{12}{6} = 2$

The value of x is 2.

31. Let A = {-1, 1}, B = {0, 2}. If the function f:A→B defined by f(x) = ax + b is an onto function, then find the values of a and b.

Solution:

Since f is an onto function, the range of f must be equal to the codomain B = {0, 2}. This means that the two elements in the domain A, {-1, 1}, must map to the two distinct elements in B, {0, 2}. There are two possible cases.

Case 1: f(-1) = 0 and f(1) = 2

Substitute these into f(x) = ax + b:

$f(-1) \implies a(-1) + b = 0 \implies -a + b = 0$ (Equation 1)

$f(1) \implies a(1) + b = 2 \implies a + b = 2$ (Equation 2)

From Equation 1, we get $b = a$. Substitute this into Equation 2:

$a + (a) = 2 \implies 2a = 2 \implies a = 1$.

Since $b=a$, we have $b = 1$.

So, one possible solution is a = 1, b = 1.

Case 2: f(-1) = 2 and f(1) = 0

Substitute these into f(x) = ax + b:

$f(-1) \implies a(-1) + b = 2 \implies -a + b = 2$ (Equation 3)

$f(1) \implies a(1) + b = 0 \implies a + b = 0$ (Equation 4)

From Equation 4, we get $b = -a$. Substitute this into Equation 3:

$-a + (-a) = 2 \implies -2a = 2 \implies a = -1$.

Since $b=-a$, we have $b = -(-1) = 1$.

So, another possible solution is a = -1, b = 1.

The possible values are (a=1, b=1) or (a=-1, b=1).

32. The sum of first n, 2n and 3n terms of an A.P. are S₁, S₂ and S₃ respectively. Prove that S₃ = 3(S₂ - S₁).

Solution:

Let the A.P. have first term 'a' and common difference 'd'.

The sum of the first k terms of an A.P. is given by $S_k = \frac{k}{2}[2a + (k-1)d]$.

$S_1 = \frac{n}{2}[2a + (n-1)d]$

$S_2 = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$

$S_3 = \frac{3n}{2}[2a + (3n-1)d]$

Now, let's evaluate the Right Hand Side (RHS) of the equation we need to prove: $3(S_2 - S_1)$.

$S_2 - S_1 = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$

Factor out $\frac{n}{2}$:

$S_2 - S_1 = \frac{n}{2} \left[ 2(2a + (2n-1)d) - (2a + (n-1)d) \right]$

$S_2 - S_1 = \frac{n}{2} [ (4a + 4nd - 2d) - (2a + nd - d) ]$

$S_2 - S_1 = \frac{n}{2} [ 4a + 4nd - 2d - 2a - nd + d ]$

$S_2 - S_1 = \frac{n}{2} [ (4a-2a) + (4nd-nd) + (-2d+d) ]$

$S_2 - S_1 = \frac{n}{2} [ 2a + 3nd - d ] = \frac{n}{2} [ 2a + (3n-1)d ]$

Now, multiply this result by 3:

$3(S_2 - S_1) = 3 \times \frac{n}{2} [ 2a + (3n-1)d ] = \frac{3n}{2} [ 2a + (3n-1)d ]$

This expression is exactly the formula for $S_3$.

Therefore, $S_3 = 3(S_2 - S_1)$. Hence proved.

33. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these colour papers?

Solution:

The side lengths of the 15 square papers are 10 cm, 11 cm, ..., 24 cm.

The area of a square is side². To find the total area, we need to calculate the sum of the areas of all squares:

Total Area = $10^2 + 11^2 + 12^2 + ... + 24^2$

We can find this sum using the formula for the sum of the first n squares: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.

The required sum can be calculated as: $(\sum_{k=1}^{24} k^2) - (\sum_{k=1}^{9} k^2)$

1. Calculate $\sum_{k=1}^{24} k^2$ (Sum of squares from 1 to 24):

$\frac{24(24+1)(2(24)+1)}{6} = \frac{24(25)(49)}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900$

2. Calculate $\sum_{k=1}^{9} k^2$ (Sum of squares from 1 to 9):

$\frac{9(9+1)(2(9)+1)}{6} = \frac{9(10)(19)}{6} = \frac{1710}{6} = 285$

3. Find the total area:

Total Area = $4900 - 285 = 4615$

Rekha can decorate 4615 cm² of area.

34. Solve the following system of linear equations in three variables.
3x - 2y + z = 2
2x + 3y - z = 5
x + y + z = 6

Solution:

Let the equations be:

(1) $3x - 2y + z = 2$

(2) $2x + 3y - z = 5$

(3) $x + y + z = 6$

Step 1: Eliminate 'z' using equations (1) and (2).

Add equation (1) and (2):

$(3x - 2y + z) + (2x + 3y - z) = 2 + 5$

$5x + y = 7$ (Equation 4)

Step 2: Eliminate 'z' using equations (2) and (3).

Add equation (2) and (3):

$(2x + 3y - z) + (x + y + z) = 5 + 6$

$3x + 4y = 11$ (Equation 5)

Step 3: Solve the new system of two equations (4) and (5).

From equation (4), we can write $y = 7 - 5x$. Substitute this into equation (5):

$3x + 4(7 - 5x) = 11$

$3x + 28 - 20x = 11$

$-17x = 11 - 28$

$-17x = -17 \implies x = 1$

Step 4: Find 'y'.

Substitute $x=1$ into equation (4):

$5(1) + y = 7 \implies y = 7 - 5 = 2$

Step 5: Find 'z'.

Substitute $x=1$ and $y=2$ into equation (3):

$1 + 2 + z = 6 \implies 3 + z = 6 \implies z = 3$

The solution is x = 1, y = 2, z = 3.

35. Find the square root of $x^4 - 12x^3 + 42x^2 - 36x + 9$ by division method.

Solution:

We use the long division method to find the square root.

             x²  - 6x  + 3
           _____________________
    x²    | x⁴ - 12x³ + 42x² - 36x + 9
          | x⁴
          |_____________________
2x²-6x    |    -12x³ + 42x²
          |    -12x³ + 36x²
          |_____________________
2x²-12x+3 |           6x² - 36x + 9
          |           6x² - 36x + 9
          |_____________________
          |                     0
        

Step-by-step Explanation:

1. The square root of the first term, $x^4$, is $x^2$. This is the first term of our answer.
2. Subtract $(x^2)^2 = x^4$ and bring down the next two terms: $-12x^3 + 42x^2$.
3. Double the current answer ($x^2$) to get the new divisor's first part: $2x^2$.
4. Divide the first term of the new dividend ($-12x^3$) by the first term of the new divisor ($2x^2$) to get $-6x$. This is the second term of our answer.
5. The new divisor is $2x^2 - 6x$. Multiply it by $-6x$: $(-6x)(2x^2 - 6x) = -12x^3 + 36x^2$.
6. Subtract this from the current dividend. $( -12x^3 + 42x^2) - (-12x^3 + 36x^2) = 6x^2$. Bring down the next two terms: $-36x + 9$.
7. Double the current answer ($x^2 - 6x$) to get the new divisor's first part: $2x^2 - 12x$.
8. Divide the first term of the new dividend ($6x^2$) by the first term of the new divisor ($2x^2$) to get $3$. This is the third term of our answer.
9. The new divisor is $2x^2 - 12x + 3$. Multiply it by $3$: $3(2x^2 - 12x + 3) = 6x^2 - 36x + 9$.
10. Subtracting this gives a remainder of 0.

The square root is $|x^2 - 6x + 3|$.

36. The hypotenuse of a right angled triangle is 25cm and its perimeter 56 cm. Find the length of the smallest side.

Solution:

Let the sides of the right-angled triangle be a, b, and c, where c is the hypotenuse.

Given:

Hypotenuse, c = 25 cm

Perimeter, a + b + c = 56 cm

Substitute the value of c into the perimeter equation:

$a + b + 25 = 56 \implies a + b = 31$ (Equation 1)

By Pythagoras' theorem, for a right-angled triangle:

$a^2 + b^2 = c^2$

$a^2 + b^2 = 25^2 = 625$ (Equation 2)

We know the algebraic identity $(a+b)^2 = a^2 + b^2 + 2ab$.

Substitute the values from equations (1) and (2):

$(31)^2 = 625 + 2ab$

$961 = 625 + 2ab$

$2ab = 961 - 625 = 336$

$ab = 168$

Now we have a system of two equations: $a+b=31$ and $ab=168$. We need to find two numbers that add up to 31 and multiply to 168. We can solve this by inspection or by forming a quadratic equation $t^2 - 31t + 168 = 0$. Factoring 168, we find $7 \times 24 = 168$ and $7 + 24 = 31$.

So, the two sides are a = 7 cm and b = 24 cm.

The sides of the triangle are 7 cm, 24 cm, and 25 cm.

The length of the smallest side is 7 cm.

37. State and prove the Thales theorem.

Solution:

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.

38. Find the value of k, if the area of a quadrilateral is 28sq. units, whose vertices are taken in order (-4, -2), (-3, k), (3, -2) and (2, 3).

Solution:

Let the vertices be A(-4, -2), B(-3, k), C(3, -2), D(2, 3).

The formula for the area of a quadrilateral is:

Area = $\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$

Given Area = 28 sq. units.

$28 = \frac{1}{2} |((-4)(k) + (-3)(-2) + (3)(3) + (2)(-2)) - ((-2)(-3) + (k)(3) + (-2)(2) + (3)(-4))|$

Multiply both sides by 2:

$56 = |(-4k + 6 + 9 - 4) - (6 + 3k - 4 - 12)|$

$56 = |(-4k + 11) - (3k - 10)|$

$56 = |-4k + 11 - 3k + 10|$

$56 = |-7k + 21|$

This gives two possible equations:

Case 1: $-7k + 21 = 56$

$-7k = 56 - 21$

$-7k = 35 \implies k = -5$

Case 2: $-7k + 21 = -56$

$-7k = -56 - 21$

$-7k = -77 \implies k = 11$

The possible values for k are -5 and 11.

39. A (-3, 0), B (10, -2) and C (11, 2) are the vertices of ∆ABC. Find the equation of the median through B.

Solution:

The median through vertex B is the line segment that connects vertex B to the midpoint of the opposite side, AC.

Step 1: Find the midpoint of the side AC.

Let D be the midpoint of AC. The coordinates of A are (-3, 0) and the corrected coordinates of C are (11, 2).

Using the midpoint formula:

Midpoint D = $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{-3+11}{2}, \frac{0+2}{2})$

Midpoint D = $(\frac{8}{2}, \frac{2}{2}) = (4, 1)$

Step 2: Find the equation of the line passing through B(10, -2) and the midpoint D(4, 1).

We use the two-point form for the equation of a line:

$\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

Let $(x_1, y_1) = (10, -2)$ and $(x_2, y_2) = (4, 1)$.

$\frac{y - (-2)}{1 - (-2)} = \frac{x - 10}{4 - 10}$

$\frac{y + 2}{3} = \frac{x - 10}{-6}$

Now, cross-multiply:

$-6(y + 2) = 3(x - 10)$

We can divide both sides by 3 to simplify:

$-2(y + 2) = x - 10$

$-2y - 4 = x - 10$

Rearrange the terms into the standard form $Ax + By + C = 0$:

$x + 2y - 10 + 4 = 0$

$x + 2y - 6 = 0$

This is the required equation of the median through B.

40. Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x - 2y - 4 = 0 to the point of intersection of 7x - 3y = -12 and 2y = x+3.

Solution:

Step 1: Find the first point of intersection (P1).

(1) $3x + y + 2 = 0$

(2) $x - 2y - 4 = 0$

From (1), $y = -3x - 2$. Substitute this into (2):

$x - 2(-3x - 2) - 4 = 0$

$x + 6x + 4 - 4 = 0 \implies 7x = 0 \implies x = 0$.

Substitute $x=0$ back into $y = -3x - 2 \implies y = -3(0) - 2 = -2$.

So, the first point is P1 = (0, -2).

Step 2: Find the second point of intersection (P2).

(3) $7x - 3y = -12$

(4) $2y = x + 3 \implies x = 2y - 3$

Substitute x from (4) into (3):

$7(2y - 3) - 3y = -12$

$14y - 21 - 3y = -12$

$11y = 9 \implies y = \frac{9}{11}$.

Substitute y back into $x = 2y - 3 \implies x = 2(\frac{9}{11}) - 3 = \frac{18}{11} - \frac{33}{11} = -\frac{15}{11}$.

So, the second point is P2 = ($-\frac{15}{11}, \frac{9}{11}$).

Step 3: Find the equation of the line through P1(0, -2) and P2($-\frac{15}{11}, \frac{9}{11}$).

First, find the slope (m):

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{9}{11} - (-2)}{-\frac{15}{11} - 0} = \frac{\frac{9+22}{11}}{-\frac{15}{11}} = \frac{\frac{31}{11}}{-\frac{15}{11}} = -\frac{31}{15}$.

Using the point-slope form $y - y_1 = m(x - x_1)$ with P1(0, -2):

$y - (-2) = -\frac{31}{15}(x - 0)$

$y + 2 = -\frac{31}{15}x$

Multiply by 15: $15(y + 2) = -31x \implies 15y + 30 = -31x$.

The required equation is $31x + 15y + 30 = 0$.

41. Prove the following identities : $\frac{sin^3A + cos^3A}{sinA + cosA} + \frac{sin^3A - cos^3A}{sinA - cosA} = 2$

Solution:

We will use the sum and difference of cubes formulas:

• $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
• $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

Let's work with the Left Hand Side (LHS).

First term:

$\frac{sin^3A + cos^3A}{sinA + cosA} = \frac{(sinA + cosA)(sin^2A - sinAcosA + cos^2A)}{sinA + cosA}$

= $sin^2A - sinAcosA + cos^2A$

Second term:

$\frac{sin^3A - cos^3A}{sinA - cosA} = \frac{(sinA - cosA)(sin^2A + sinAcosA + cos^2A)}{sinA - cosA}$

= $sin^2A + sinAcosA + cos^2A$

Now, add the simplified terms:

LHS = $(sin^2A - sinAcosA + cos^2A) + (sin^2A + sinAcosA + cos^2A)$

The terms $-sinAcosA$ and $+sinAcosA$ cancel each other out.

LHS = $sin^2A + cos^2A + sin^2A + cos^2A$

Using the Pythagorean identity $sin^2A + cos^2A = 1$:

LHS = $1 + 1 = 2$

LHS = RHS. Hence, the identity is proved.

42. Find the sum of all natural numbers between 300 and 500 which are divisible by 11.

Solution:

We need to find the sum of an Arithmetic Progression (A.P.) where the terms are multiples of 11 between 300 and 500.

Step 1: Find the first term (a) of the A.P.

Divide 300 by 11: $300 \div 11 = 27$ with a remainder of 3. The first multiple of 11 after 300 is $300 + (11 - 3) = 308$. So, $a = 308$.

Step 2: Find the last term (l) of the A.P.

Divide 500 by 11: $500 \div 11 = 45$ with a remainder of 5. The last multiple of 11 before 500 is $500 - 5 = 495$. So, $l = 495$.

Step 3: Find the number of terms (n).

The common difference is $d = 11$. Using the formula $l = a + (n-1)d$:

$495 = 308 + (n-1)11$

$495 - 308 = (n-1)11$

$187 = (n-1)11$

$n-1 = \frac{187}{11} = 17$

$n = 17 + 1 = 18$

There are 18 terms in this A.P.

Step 4: Find the sum of the A.P. (Sₙ).

Using the formula $S_n = \frac{n}{2}(a+l)$:

$S_{18} = \frac{18}{2}(308 + 495)$

$S_{18} = 9(803)$

$S_{18} = 7227$

The sum of all natural numbers between 300 and 500 divisible by 11 is 7227.

Part - IV

43. a) Construct a triangle similar to a given triangle ABC with its sides equal to 6/5 of the corresponding sides of the triangle ABC (scale factor 6/5 > 1).

Solution:

Steps of Construction:

1. Draw any triangle ABC.
2. Draw a ray BX starting from B, making an acute angle with the base BC, on the side opposite to vertex A.
3. Since the scale factor is 6/5, locate 6 points (the greater of 6 and 5) $B_1, B_2, B_3, B_4, B_5, B_6$ on BX so that $BB_1 = B_1B_2 = ... = B_5B_6$.
4. Join $B_5$ (the smaller of 6 and 5) to C.
5. Draw a line through $B_6$ parallel to $B_5C$, to intersect the extended line segment BC at C'.
6. Draw a line through C' parallel to CA to intersect the extended line segment BA at A'.
7. The triangle A'BC' is the required similar triangle whose sides are 6/5 of the corresponding sides of ∆ABC.

43. b) (OR) Construct a triangle PQR such that QR = 5 cm, ∠P = 30° and the altitude from P to QR is of length 4.2 cm.

Solution:

Steps of Construction:

1. Draw a line segment QR = 5 cm.
2. At Q, draw a line QX making an angle of 30° with QR (i.e., ∠RQX = 30°).
3. Draw a line QY perpendicular to QX at Q.
4. Draw the perpendicular bisector of the line segment QR. Let it intersect QR at M and the line QY at O.
5. With O as the center and OQ (or OR) as the radius, draw a circle. The major arc of this circle contains the vertex P, where the angle subtended is 30°.
6. Draw a line segment MN of length 4.2 cm perpendicular to QR at M.
7. Draw a line parallel to QR passing through the point N. This line will intersect the circle at two points, P and P'.
8. Join PQ and PR.
9. ∆PQR is the required triangle.

44. a) Draw a graph of the linear function y = 1/2x. Identify the constant of variation and verify it with the graph. Also, i) find y when x = 9, ii) find x when y = 7.5.

Solution:

The function is $y = \frac{1}{2}x$. This is in the form y = kx, which represents a direct variation.

Constant of Variation (k):

The constant of variation is $k = \frac{1}{2}$ or 0.5.

Table of Values:

x	-4	-2	0	2	4	9
y = (1/2)x	-2	-1	0	1	2	4.5

Graphing:

Plot the points (-4,-2), (-2,-1), (0,0), (2,1), (4,2) on a graph sheet and draw a straight line passing through them and the origin.

Verification from Graph:

From the graph, we can see that for every 2 units increase in x, y increases by 1 unit. This verifies that the slope (constant of variation) is 1/2.

i) Find y when x = 9:

From the graph, locate x = 9 on the x-axis, move vertically up to the line, and then horizontally to the y-axis. You will find y = 4.5.

Calculation: $y = \frac{1}{2}(9) = 4.5$.

ii) Find x when y = 7.5:

From the graph, locate y = 7.5 on the y-axis, move horizontally to the line, and then vertically down to the x-axis. You will find x = 15.

Calculation: $7.5 = \frac{1}{2}x \implies x = 7.5 \times 2 = 15$.

44. b) (OR) A company initially started with 40 workers to complete the work by 150 days. Later, it decided to fasten up the work by increasing the number of workers as shown below.
Number of workers (x): 40, 50, 60, 75
Number of days (y): 150, 120, 100, 80
i) Graph the above data and identify the type of variation.
ii) From the graph, find the number of days required to complete the work if the company decides to opt for 120 workers.
iii) If the work has to be completed by 200 days, how many workers are required?

Solution:

i) Type of Variation:

Let's check the product of x and y for each pair:

• $40 \times 150 = 6000$
• $50 \times 120 = 6000$
• $60 \times 100 = 6000$
• $75 \times 80 = 6000$

Since the product $xy = 6000$ is a constant, this is an inverse variation. The equation of variation is $xy = 6000$.

Graph:

Plot the points (40, 150), (50, 120), (60, 100), and (75, 80) on a graph. Connect the points with a smooth curve. The resulting graph is a rectangular hyperbola.

ii) Days for 120 workers:

On the graph, locate x = 120 on the x-axis. Move vertically to the curve and then horizontally to the y-axis to find the corresponding number of days.

Calculation: $120 \times y = 6000 \implies y = \frac{6000}{120} = 50$.

It will take 50 days for 120 workers.

iii) Workers for 200 days:

On the graph, locate y = 200 on the y-axis. Move horizontally to the curve and then vertically down to the x-axis to find the corresponding number of workers.

Calculation: $x \times 200 = 6000 \implies x = \frac{6000}{200} = 30$.

30 workers are required to complete the work in 200 days.