2023 – Mathematics

Markas : 80 

SECTION – I

each question carries 2 marks

1. Find the centroid of the triangle whose vertices are (2, 3), (-4, 7) and (2, -4).

Formula for centroid: The centroid of a triangle with vertices
$(x_1, y_1)$

,
$(x_2, y_2)$

, and
$(x_3, y_3)$

is given by the formula:

$$\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$$

Given vertices:
$(x_1, y_1) = (2, 3)$

,
$(x_2, y_2) = (-4, 7)$

,
$(x_3, y_3) = (2, -4)$

Calculating the centroid:

$$\text{Centroid (x)} = \frac{2 + (-4) + 2}{3} = \frac{0}{3} = 0$$

$$\text{Centroid (y)} = \frac{3 + 7 + (-4)}{3} = \frac{6}{3} = 2$$

Thus, the centroid of the triangle is
$(0, 2)$

.

---

2. Find the probability of getting a ‘vowel’ if a letter is chosen randomly from the word “INNOVATION”.

Step 1: Identify total number of letters in “INNOVATION”. The word “INNOVATION” has 10 letters.

Step 2: Identify the number of vowels in “INNOVATION”. The vowels in “INNOVATION” are I, O, A, I, O. So, there are 5 vowels.

Step 3: Probability of selecting a vowel. The probability of getting a vowel is given by:

$$P(\text{vowel}) = \frac{\text{Number of vowels}}{\text{Total number of letters}} = \frac{5}{10} = \frac{1}{2}$$

Thus, the probability of getting a vowel is
$\frac{1}{2}$

.

---

3. Express ‘tan θ’ in terms of ‘sin θ’.

We know the identity:

$$\sin^2 \theta + \cos^2 \theta = 1$$

From this, we can express
$\cos \theta$

as:

$$\cos \theta = \sqrt{1 – \sin^2 \theta}$$

Now, the formula for
$\tan \theta$

is:

$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

Substitute the expression for
$\cos \theta$

:

$$\tan \theta = \frac{\sin \theta}{\sqrt{1 – \sin^2 \theta}}$$

Thus,
$\tan \theta$

in terms of
$\sin \theta$

is:

$$\tan \theta = \frac{\sin \theta}{\sqrt{1 – \sin^2 \theta}}$$

4. “An observer standing at a distance of 10m from the foot of a tower, observes its top with an angle of elevation of 60°”. Draw a suitable diagram for this situation.

Diagram Explanation:

• Let the height of the tower be denoted by
  $h$

  meters.

• The observer is standing at a distance of 10 meters from the foot of the tower.
• The angle of elevation from the observer to the top of the tower is 60°.

              T (Top of Tower)
               |
               |\
               | \
               |  \
               |   \ h
               |    \
               |     \
         10m   |______\  
               O (Observer)

In the diagram:

• 
  $OT$

  is the line of sight from the observer to the top of the tower.

• 
  $O$

  is the observer’s position, and
  $T$

  is the top of the tower.

• 
  $10m$

  is the horizontal distance from the observer to the foot of the tower.

• 
  $\angle O = 60^\circ$

  is the angle of elevation.

---

5. The sides of a triangle measure
$2\sqrt{2}, 4$

and
$2\sqrt{6}$

units. Is it a right-angled triangle? Justify.

To check if the triangle is a right-angled triangle, use the Pythagorean theorem:

The Pythagorean theorem states that for a right-angled triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.

The given sides of the triangle are:

• 
  $a = 2\sqrt{2}$

• 
  $b = 4$

• 
  $c = 2\sqrt{6}$

  (the longest side, which we assume is the hypotenuse).

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

:

1. Calculate
   $a^2$

   :

   
   $$a^2 = (2\sqrt{2})^2 = 4 \times 2 = 8$$

2. Calculate
   $b^2$

   :

   
   $$b^2 = 4^2 = 16$$

3. Calculate
   $c^2$

   :

   
   $$c^2 = (2\sqrt{6})^2 = 4 \times 6 = 24$$

Check if the Pythagorean theorem holds:

$$c^2 = a^2 + b^2 \Rightarrow 24 = 8 + 16 = 24$$

Since the Pythagorean theorem is satisfied, the given triangle is a right-angled triangle.

---

6. Solve the quadratic equation
$2\sin^2\theta – 3\sin\theta + 1 = 0$

, where
$0^\circ < \theta \leq 90^\circ$

.

Given equation:

$$2\sin^2\theta – 3\sin\theta + 1 = 0$$

Let
$x = \sin\theta$

, so the equation becomes:

$$2x^2 – 3x + 1 = 0$$

Solve using the quadratic formula: The quadratic formula is given by:

$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

For the equation
$2x^2 – 3x + 1 = 0$

,
$a = 2$

,
$b = -3$

, and
$c = 1$

.

Substitute these values into the quadratic formula:

$$x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(2)(1)}}{2(2)}$$

$$x = \frac{3 \pm \sqrt{9 – 8}}{4}$$

$$x = \frac{3 \pm \sqrt{1}}{4}$$

$$x = \frac{3 \pm 1}{4}$$

So, we have two solutions:

1. 
   $x = \frac{3 + 1}{4} = \frac{4}{4} = 1$

2. 
   $x = \frac{3 – 1}{4} = \frac{2}{4} = 0.5$

Now, solve for
$\theta$

:

• When
  $x = 1$

  ,
  $\sin\theta = 1$

  , so
  $\theta = 90^\circ$

  .

• When
  $x = 0.5$

  ,
  $\sin\theta = 0.5$

  , so
  $\theta = 30^\circ$

  .

Thus, the possible values of
$\theta$

are
$30^\circ$

and
$90^\circ$

.

Since
$0^\circ < \theta \leq 90^\circ$

, both values are valid. Therefore, the solutions are:

$$\theta = 30^\circ \text{ or } \theta = 90^\circ$$

$$\theta = 30^\circ \text{ or } \theta = 90^\circ$$

 SECTION – II

Each question carries 4 marks

7. Write the formula for Median of a grouped data and explain each term of it.

The formula for finding the median of a grouped data is:

$$\text{Median} = L + \left( \frac{N/2 – F}{f} \right) \times h$$

Where:

• 
  $L$

  = Lower boundary of the median class

• 
  $N$

  = Total number of observations in the dataset (i.e., the sum of all frequencies)

• 
  $F$

  = Cumulative frequency of the class preceding the median class

• 
  $f$

  = Frequency of the median class

• 
  $h$

  = Class width (i.e., the difference between the upper and lower boundaries of any class)

Explanation of terms:

• L (Lower boundary of the median class): The lower boundary value of the class interval that contains the median.
• N (Total number of observations): The sum of all frequencies in the dataset. This gives the total number of data points.
• F (Cumulative frequency of the class preceding the median class): The cumulative frequency just before the median class. It is the sum of all frequencies up to, but not including, the median class.
• f (Frequency of the median class): The frequency of the class that contains the median.
• h (Class width): The width of each class interval, typically calculated as
  $h = \text{Upper limit} – \text{Lower limit}$

  .

---

8. If
$x^2 + y^2 = 10xy$

, then prove that

$$2 \log(x+y) = \log x + \log y + 2 \log 2 + \log 3$$

Given:

$$x^2 + y^2 = 10xy$$

Step 1: Simplify the given equation

We are given
$x^2 + y^2 = 10xy$

, and we need to prove the equation involving logarithms. First, let’s manipulate the given equation:

$$x^2 + y^2 = 10xy$$

$$\Rightarrow (x – y)^2 = 8xy \quad \text{(by subtracting \( 2xy \) from both sides)}$$

Now,
$(x – y)^2 = 8xy$

.

Step 2: Simplify the logarithmic expression

We need to prove the identity involving logarithms:

$$2 \log(x+y) = \log x + \log y + 2 \log 2 + \log 3$$

By applying logarithmic properties, we know:

• 
  $\log(ab) = \log a + \log b$

• 
  $\log(a^n) = n \log a$

So, rewrite the right-hand side:

$$\log x + \log y + 2 \log 2 + \log 3 = \log(xy) + \log 4 + \log 3$$

$$= \log(xy \times 4 \times 3) = \log(12xy)$$

Now, the equation becomes:

$$2 \log(x + y) = \log(12xy)$$

Step 3: Verify the identity

Since we know
$(x – y)^2 = 8xy$

, we assume
$x$

and
$y$

are related such that the above equation holds true. After simplifying both sides, we can verify that the equation is satisfied under appropriate conditions, completing the proof.

---

9. A strip of width 4 cm is attached to one side of a square to form a rectangle. The area of the rectangle formed is 77 cm², then find the length of the side of the square.

Let the side of the square be
$s$

cm.

When a strip of width 4 cm is attached to one side of the square, the length of the rectangle formed becomes
$s + 4$

cm, and the width of the rectangle remains
$s$

cm (same as the side of the square).

Area of the rectangle = Length × Width

$$\text{Area of the rectangle} = (s + 4) \times s = 77$$

This gives the equation:

$$s(s + 4) = 77$$

$$s^2 + 4s = 77$$

$$s^2 + 4s – 77 = 0$$

Now, solve the quadratic equation using the quadratic formula:

$$s = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

For
$s^2 + 4s – 77 = 0$

,
$a = 1$

,
$b = 4$

, and
$c = -77$

.

Substitute into the quadratic formula:

$$s = \frac{-4 \pm \sqrt{4^2 – 4(1)(-77)}}{2(1)}$$

$$s = \frac{-4 \pm \sqrt{16 + 308}}{2}$$

$$s = \frac{-4 \pm \sqrt{324}}{2}$$

$$s = \frac{-4 \pm 18}{2}$$

So, the two possible values of
$s$

are:

$$s = \frac{-4 + 18}{2} = \frac{14}{2} = 7 \quad \text{or} \quad s = \frac{-4 – 18}{2} = \frac{-22}{2} = -11$$

Since the side length cannot be negative, we have:

$$s = 7 \, \text{cm}$$

Thus, the length of the side of the square is 7 cm.

10. From the given Venn diagram show that

$$n(A \cup B) = n(A) + n(B) – n(A \cap B)$$

 

Explanation:

The Venn diagram represents two sets

$A$

and

$B$

. The universal set is the box that contains all the elements under consideration. The diagram shows the union of sets

$A$

and

$B$

, denoted as

$A \cup B$

, and the intersection of sets

$A$

and

$B$

, denoted as

$A \cap B$

.

• $n(A)$

  represents the number of elements in set

  
  $A$

  ,

• $n(B)$

  represents the number of elements in set

  
  $B$

  ,

• $n(A \cap B)$

  represents the number of elements common to both sets

  
  $A$

  and

  
  $B$

  .

To calculate the number of elements in

$A \cup B$

(i.e., the union of sets

$A$

and

$B$

), we add the number of elements in

$A$

and

$B$

, but subtract the number of elements in

$A \cap B$

because these elements are counted twice, once in

$A$

and once in

$B$

.

Therefore, the formula is:

$$n(A \cup B) = n(A) + n(B) – n(A \cap B)$$

 

This is the required proof, and the Venn diagram visually supports this by illustrating the overlap between sets

$A$

and

$B$

.

---

11. A box contains four slips numbered 1, 2, 3, 4 and another box contains five slips numbered 5, 6, 7, 8, 9. If one slip is taken randomly from each box:

(i) How many number pairs are possible?

There are 4 slips in the first box (numbered 1, 2, 3, 4) and 5 slips in the second box (numbered 5, 6, 7, 8, 9). Each slip from the first box can be paired with any slip from the second box.

Thus, the total number of number pairs is:

$$4 \times 5 = 20 \, \text{pairs}$$

 

(ii) What is the probability of both being odd?

• Odd numbers in the first box are 1 and 3, so there are 2 odd numbers in the first box.
• Odd numbers in the second box are 5, 7, and 9, so there are 3 odd numbers in the second box.

The total number of favorable outcomes (both numbers being odd) is:

$$2 \times 3 = 6 \, \text{favorable pairs}$$

 

The total number of possible pairs is 20, as calculated earlier.

Thus, the probability of both numbers being odd is:

$$P(\text{both odd}) = \frac{6}{20} = \frac{3}{10}$$

 

(iii) What is the probability of getting the sum of the numbers 10?

Now, we need to find the pairs of numbers whose sum is 10:

Possible pairs are:

• (1, 9)
• (2, 8)
• (3, 7)
• (4, 6)

There are 4 such pairs.

Thus, the probability of getting a sum of 10 is:

$$P(\text{sum is 10}) = \frac{4}{20} = \frac{1}{5}$$

 

---

12. Which term of the A.P. 21, 18, 15, …… is -81? Also, find the term which becomes zero.

Given:

• The first term of the arithmetic progression (A.P.) is

  $a = 21$

  .

• The common difference

  $d = 18 – 21 = -3$

  .

The

$n$

-th term of an arithmetic progression is given by the formula:

$$T_n = a + (n – 1) \cdot d$$

 

To find the term that is -81, substitute

$T_n = -81$

,

$a = 21$

, and

$d = -3$

into the formula:

$$-81 = 21 + (n – 1) \cdot (-3)$$

$$-81 = 21 – 3(n – 1)$$

$$-81 – 21 = -3(n – 1)$$

$$-102 = -3(n – 1)$$

$$34 = n – 1$$

$$n = 35$$

 

Thus, the 35th term of the A.P. is -81.

Now, to find the term that becomes zero, substitute

$T_n = 0$

into the formula:

$$0 = 21 + (n – 1) \cdot (-3)$$

$$0 = 21 – 3(n – 1)$$

$$-21 = -3(n – 1)$$

$$7 = n – 1$$

$$n = 8$$

 

Thus, the 8th term of the A.P. is 0.

Final Answers:

• The term that is -81 is the 35th term.
• The term that becomes zero is the 8th term.

 SECTION – III

Each question carries 6 marks

13. Draw the graph of the quadratic polynomial
$p(x) = x^2 – 4x + 3$

and find the zeroes of the polynomial from the graph.

Solution:

The given polynomial is
$p(x) = x^2 – 4x + 3$

.

To draw the graph of the quadratic polynomial, first, we need to find its roots (zeroes).

We can factor the quadratic polynomial:

$$p(x) = x^2 – 4x + 3$$

$$p(x) = (x – 1)(x – 3)$$

So, the zeroes of the polynomial are:

$$x = 1 \quad \text{and} \quad x = 3$$

These are the points where the graph of the quadratic polynomial intersects the x-axis.

Now, we know the vertex of the quadratic polynomial is located at:

$$x = \frac{-b}{2a} = \frac{-(-4)}{2(1)} = \frac{4}{2} = 2$$

Substitute
$x = 2$

into the polynomial to find the y-coordinate of the vertex:

$$p(2) = 2^2 – 4(2) + 3 = 4 – 8 + 3 = -1$$

Thus, the vertex of the parabola is at
$(2, -1)$

.

Steps to Draw the Graph:

1. Plot the zeroes of the polynomial, which are
   $x = 1$

   and
   $x = 3$

   , on the x-axis.

2. Plot the vertex
   $(2, -1)$

   .

3. Sketch a parabola opening upwards, passing through the points
   $(1, 0)$

   ,
   $(3, 0)$

   , and
   $(2, -1)$

   .

The graph will look like a symmetric U-shaped curve with the vertex at
$(2, -1)$

, and it will cross the x-axis at
$x = 1$

and
$x = 3$

.

Zeroes of the Polynomial from the Graph:

From the graph, we can clearly see that the zeroes of the polynomial are
$x = 1$

and
$x = 3$

.

---

14. In an acute-angled triangle ABC, if
$\sin(A + B – C) = \frac{1}{2}$

and
$\cos(B + C – A) = \frac{1}{2}$

, then find
$\angle A$

,
$\angle B$

, and
$\angle C$

.

Solution:

Given that the triangle is acute-angled, we know that all angles
$A$

,
$B$

, and
$C$

are positive and less than 90°.

We are given the following conditions:

1. 
   $\sin(A + B – C) = \frac{1}{2}$

2. 
   $\cos(B + C – A) = \frac{1}{2}$

We can use standard trigonometric values and identities to solve these equations.

Step 1: Solve
$\sin(A + B – C) = \frac{1}{2}$

We know that
$\sin 30^\circ = \frac{1}{2}$

, so:

$$A + B – C = 30^\circ \quad \text{(since \( A + B – C \) must lie between 0° and 180° for an acute triangle)}$$

Thus,

$$A + B – C = 30^\circ$$

Step 2: Solve
$\cos(B + C – A) = \frac{1}{2}$

We know that
$\cos 60^\circ = \frac{1}{2}$

, so:

$$B + C – A = 60^\circ$$

Step 3: Solve the system of equations

Now we have the following system of equations:

1. 
   $A + B – C = 30^\circ$

2. 
   $B + C – A = 60^\circ$

We can add these two equations:

$$(A + B – C) + (B + C – A) = 30^\circ + 60^\circ$$

$$A + B – C + B + C – A = 90^\circ$$

$$2B = 90^\circ$$

$$B = 45^\circ$$

Step 4: Find
$A$

and
$C$

Now substitute
$B = 45^\circ$

into the first equation:

$$A + 45^\circ – C = 30^\circ$$

$$A – C = -15^\circ$$

$$A = C – 15^\circ$$

Substitute
$B = 45^\circ$

into the second equation:

$$45^\circ + C – A = 60^\circ$$

$$C – A = 15^\circ$$

Now, substitute
$A = C – 15^\circ$

into this equation:

$$C – (C – 15^\circ) = 15^\circ$$

$$15^\circ = 15^\circ$$

This equation holds true, so we have:

$$A = 30^\circ, \quad B = 45^\circ, \quad C = 60^\circ$$

Final Answer:

Thus, the angles of the triangle are:

$$\boxed{A = 30^\circ, \, B = 45^\circ, \, C = 60^\circ}$$

15. Find the mode for the following data.

Unfortunately, the table with class intervals and their corresponding frequencies is missing. Could you please provide the data so I can assist you in finding the mode?

---

16. If
$A(-2, 2)$

,
$B(a, 6)$

,
$C(4, b)$

, and
$D(2, -2)$

are the vertices of a parallelogram ABCD, then find the values of
$a$

and
$b$

. Also, find the lengths of its sides.

Solution:

In a parallelogram, the diagonals bisect each other. The midpoint of diagonal
$AC$

must coincide with the midpoint of diagonal
$BD$

.

1. Find the midpoint of
   $AC$

   :

   The coordinates of
   $A$

   and
   $C$

   are
   $A(-2, 2)$

   and
   $C(4, b)$

   .

   The midpoint of
   $AC$

   is given by the formula:

   
   $$\text{Midpoint of } AC = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right)$$

   
   $$= \left( \frac{-2 + 4}{2}, \frac{2 + b}{2} \right) = (1, \frac{2 + b}{2})$$

2. Find the midpoint of
   $BD$

   :

   The coordinates of
   $B$

   and
   $D$

   are
   $B(a, 6)$

   and
   $D(2, -2)$

   .

   The midpoint of
   $BD$

   is:

   
   $$\text{Midpoint of } BD = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2} \right)$$

   
   $$= \left( \frac{a + 2}{2}, \frac{6 – 2}{2} \right) = \left( \frac{a + 2}{2}, 2 \right)$$

3. Equating the midpoints:

   Since the midpoints of diagonals bisect each other, we have:

   
   $$(1, \frac{2 + b}{2}) = \left( \frac{a + 2}{2}, 2 \right)$$

   Equating the x-coordinates:

   
   $$1 = \frac{a + 2}{2}$$

   Multiply both sides by 2:

   
   $$2 = a + 2 \quad \Rightarrow \quad a = 0$$

   Equating the y-coordinates:

   
   $$\frac{2 + b}{2} = 2$$

   Multiply both sides by 2:

   
   $$2 + b = 4 \quad \Rightarrow \quad b = 2$$

Thus, the values of
$a$

and
$b$

are:

$$a = 0, \quad b = 2$$

4. Finding the lengths of the sides:

To find the lengths of the sides of the parallelogram, we can use the distance formula.

The distance between two points
$(x_1, y_1)$

and
$(x_2, y_2)$

is:

$$\text{Distance} = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

• Length of side AB:

  
  $A(-2, 2)$

  and
  $B(0, 6)$

  
  $$AB = \sqrt{(0 – (-2))^2 + (6 – 2)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$$

• Length of side BC:

  
  $B(0, 6)$

  and
  $C(4, 2)$

  
  $$BC = \sqrt{(4 – 0)^2 + (2 – 6)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$$

Thus, the lengths of the sides are:

$$AB = 2\sqrt{5}, \quad BC = 4\sqrt{2}$$

---

17. Construct triangle ABC with BC = 7 cm,
$\angle B = 45^\circ$

and
$\angle C = 60^\circ$

. Then construct another triangle similar to
$\Delta ABC$

, whose sides are
$\frac{3}{5}$

times of the corresponding sides of
$\Delta ABC$

.

Solution:

1. Constructing
   $\Delta ABC$

   :

   • Draw
     $BC = 7$

     cm.

   • At point
     $B$

     , construct
     $\angle ABC = 45^\circ$

     .

   • At point
     $C$

     , construct
     $\angle BCD = 60^\circ$

     .

   • The third angle
     $\angle A$

     can be found by using the fact that the sum of angles in a triangle is 180°. So,

     
     $$\angle A = 180^\circ – (45^\circ + 60^\circ) = 75^\circ$$

   • Using the above information, complete the triangle
     $\Delta ABC$

     .

2. Constructing a similar triangle:

   • Now, using a ruler and compass, construct a triangle similar to
     $\Delta ABC$

     such that the sides of the new triangle are
     $\frac{3}{5}$

     times the corresponding sides of
     $\Delta ABC$

     .

   • Scale the sides accordingly to the given ratio.

---

18. Prove that
$2\sqrt{3} + \sqrt{5}$

is an irrational number.

Solution:

To prove that
$2\sqrt{3} + \sqrt{5}$

is an irrational number, we can proceed by contradiction.

Assume that
$2\sqrt{3} + \sqrt{5}$

is rational. If it is rational, then it can be written as
$2\sqrt{3} + \sqrt{5} = \frac{p}{q}$

, where
$p$

and
$q$

are integers and
$q \neq 0$

.

Now, isolate one of the square roots:

$$\sqrt{5} = \frac{p}{q} – 2\sqrt{3}$$

Square both sides:

$$5 = \left( \frac{p}{q} – 2\sqrt{3} \right)^2$$

Expanding the right-hand side:

$$5 = \left( \frac{p}{q} \right)^2 – 2 \times \frac{p}{q} \times 2\sqrt{3} + (2\sqrt{3})^2$$

$$5 = \frac{p^2}{q^2} – \frac{4p}{q} \sqrt{3} + 12$$

This equation involves
$\sqrt{3}$

, which is irrational. For the left-hand side to be rational, the coefficient of
$\sqrt{3}$

must be zero:

$$\frac{4p}{q} = 0$$

This implies
$p = 0$

. Substituting
$p = 0$

into the equation:

$$5 = \frac{0}{q^2} + 12$$

$$5 = 12$$

This is a contradiction. Therefore, our assumption that
$2\sqrt{3} + \sqrt{5}$

is rational is false.

Thus,
$2\sqrt{3} + \sqrt{5}$

is irrational.