Kaliyuga Ekalavya

NCERT Class X Chapter 2: Polynomials Example 1(ii) Question: Look at the graphs in Fig. 2.9 given below. Each is the graph of \( y = p(x) \), where \( p(x) \) is a polynomial. For the given graph, find the number of zeroes of \( p(x) \). Given: The graph of the polynomial \( y = p(x) \) is given. To Find: The number of zeroes of the polynomial \( p(x) \). Formula: Number of zeroes of a polynomial = Number of points where the graph of the polynomial intersects the x-axis. Solution: Step 1: Observe the given graph of the polynomial. $$ y = p(x) $$ Step 2: Identify the points where the graph intersects the x-axis. Step 3: The graph cuts the x-axis at two distinct points. Step 4: Each point of intersection with the x-axis represents one zero of the polynomial. Result: The number of zeroes of the polynomial \( p(x) \) is 2 . Next question s...

NCERT Class X Chapter 2: Polynomials Example 1(i) Question: Look at the graph in Fig. 2.9 given below. It is the graph of \( y = p(x) \), where \( p(x) \) is a polynomial. Find the number of zeroes of \( p(x) \). Given: The graph of the polynomial function \( y = p(x) \) is given. To Find: The number of zeroes of the polynomial \( p(x) \). Formula: The number of zeroes of a polynomial equals the number of points where its graph intersects the x-axis. Solution: Step 1: Observe the given graph of the polynomial \( y = p(x) \). Step 2: Check how many times the graph cuts or touches the x-axis. Step 3: The graph intersects the x-axis at exactly one point. Step 4: Each point of intersection with the x-axis represents one zero of the polynomial. Result: The polynomial \( p(x) \) has exactly one zero . Next question solution: NCERT Class X Chapter 2: Polynomials Example 1(ii) ...

NCERT Class X Chapter 7: Coordinate Geometry Example 1 Question: Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed. Given: The points are \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \). To Find: Whether the points form a triangle and, if so, the type of triangle formed. Formula: Distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \): $$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Solution: Step 1: Let the points be \( A(3, 2) \), \( B(-2, -3) \), and \( C(2, 3) \). Step 2: Find the length of \( AB \) using the distance formula. $$ AB = \sqrt{( -2 - 3 )^2 + ( -3 - 2 )^2} = \sqrt{ ( -5 )^2 + ( -5 )^2 } = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} $$ Step 3: Find the length of \( BC \). $$ BC = \sqrt{ (2 - ( -2 ))^2 + (3 - ( -3 ))^2 } = \sqrt{ (4)^2 + (6)^2 } = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} $$ Step 4: Find the length of \( CA \). $$ ...

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 1 Question: Check graphically whether the pair of equations \( x + 3y = 6 \) and \( 2x - 3y = 12 \) are consistent. Given: The pair of linear equations: 1. \( x + 3y = 6 \) 2. \( 2x - 3y = 12 \) To Find: To check graphically whether the given pair of equations is consistent (i.e., if they have a solution or not). Formula: To solve graphically, plot both equations on the Cartesian plane. If the lines intersect at a point, the system is consistent and has a unique solution. If the lines are parallel, the system is inconsistent (no solution). If the lines coincide, there are infinitely many solutions. Solution: Step 1: Write the equations in standard form. $$ \begin{align*} \text{Equation 1:} & \quad x + 3y = 6 \\ \text{Equation 2:} & \quad 2x - 3y = 12 \end{align*} $$ Step 2: Find at least two solutions for each equation. For Equation 1: \( x + 3y = 6 \) If \( ...

NCERT Class X Chapter 12: Surface Areas And Volumes Example 1 Question: Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Use \( \pi = \frac{22}{7} \)) Given: Total height of top, \( H = 5 \) cm Diameter of top \( = 3.5 \) cm Radius of top, \( r = \frac{3.5}{2} = 1.75 \) cm \( \pi = \frac{22}{7} \) To Find: The total surface area of the top to be coloured. Formula: Curved surface area of a cone: \( \pi r l \) Curved surface area of a hemisphere: \( 2 \pi r^2 \) Slant height of cone: \( l = \sqrt{r^2 + h^2} \) Solution: Step 1: Find the height of the cone part. The hemisphere sits on top of the cone, so the height of the cone is: $$ h = H - r = 5 - 1.75 = 3.25 \text{ cm} $$ Step 2: ...

NCERT Class X Chapter 9: Some Applications of Trigonometry Example 1 Question: A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower. Given: Distance from the point to the foot of the tower = 15 m Angle of elevation of the top of the tower = 60° To Find: Height of the tower Formula: For a right-angled triangle, $$ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$ Solution: Step 1: Let the height of the tower be \( h \) metres. The distance from the point to the foot of the tower is 15 m. We form a right-angled triangle where: Opposite side = height of tower = \( h \) Adjacent side = distance from point to foot = 15 m Angle of elevation = 60° Step 2: Write the trigonometric ratio for tan 60°. $$ \tan 60^\circ = \frac{h}{15} $$ Step 3: ...

NCERT Class X Chapter 1: Real Numbers Example 1 Question: Consider the numbers \(4^n\), where \(n\) is a natural number. Check whether there is any value of \(n\) for which \(4^n\) ends with the digit zero. Given: The numbers considered are of the form \(4^n\), where \(n\) is a natural number (\(n \in \mathbb{N}\)). To Find: Determine whether there exists any natural number \(n\) such that \(4^n\) ends with the digit zero. Formula: To end with the digit zero, a number must be divisible by 10. \(10 = 2 \times 5\) Solution: Step 1: State the condition for a number to end in zero. A number ends with the digit zero if and only if it is divisible by 10. That is, it must have both 2 and 5 as its prime factors. $$ 10 = 2 \times 5 $$ Step 2: Express \(4^n\) in terms of its prime factors. We know that \(4 = 2^2\), so: $$ 4^n = (2^2)^n $$ Step 3: Apply t...

NCERT Class X Chapter 11: Area Related To Circles Example 1 Question: Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14). Given: Radius, \( r = 4 \) cm Angle, \( \theta = 30^\circ \) \( \pi = 3.14 \) To Find: 1. Area of the sector (minor sector) 2. Area of the corresponding major sector Formula: Area of a sector:   $$ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 $$ Area of major sector:   $$ \text{Area of major sector} = \pi r^2 - \text{Area of minor sector} $$ Solution: Step 1: Write the formula for the area of the sector. $$ \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 $$ Step 2: Substitute the given values: \( \theta = 30^\circ \), \( r = 4 \) cm, \( \pi = 3.14 \). $$ \text{Area of sector} = \frac{30^\circ}{360^\circ} \times 3.14 \times (4)^2 $$ Step 3: Simplify the fraction and calculate \(...

NCERT Class X Chapter 14: Probability Example 1 Question: Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail. Given: A coin is tossed once. To Find: The probability of getting a head and the probability of getting a tail. Formula: Probability = Outcome we want  Total Outcomes Solution: If a coin is tossed, (like in cricket) the result we will either head (H) or tail (T). So, total possible outcomes when a coin is tossed once are {H, T}. The outcome we want is called favorable outcomes Number of favorable outcomes for getting a head = 1 (H).  Because if a coin is tossed 1 time it can be head only once. Probability of getting a head =    Outcome we want  Total Outcomes Probability of getting a head =    Number of favorable outcomes for getting a head Total Outcomes Probability of getting a head = 1 2 Numb...

NCERT Class X Chapter 5: Arithmetic Progression Example 1 Question: For the AP: 3/2, 1/2, -1/2, -3/2,.... write the first term a and common difference d Given: The Arithmetic Progression (AP) is: 3 2 , 1 2 , -1 2 , -3 2 , ... To Find: The first term (a) and the common difference (d) of the given AP. Formula: In an AP, the first term is denoted by 'a' and the common difference is denoted by 'd'. The common difference is calculated by subtracting any term from its succeeding term: d = a n+1 - a n Solution: The first term of the AP is a = 3 2 To find the common difference, we subtract the first term from the second term: d = a 2 - a 1   ⇒ d = 1 2 - 3 2   ⇒ d = 1 - 3 2   ⇒ d = -2 2   ⇒ d = -1 Result: First term (a) = 3 2 Common difference (d) = -1 Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exa...

NCERT Class X Chapter 4: Quadratic Equation Example 1 (ii) Question: Represent the following situations mathematically: A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day. Given: Cost of production of each toy = 55 - (number of toys produced) Total cost of production on a particular day = 750 rupees To Find: The number of toys produced on that day. Formula: Total cost of production = (Number of toys) × (Cost of production of each toy) Solution: Let,the number of toys produced in a day = x Then,the cost of production of each toy = 55 - x Total cost of production = (Number of toys) × (Cost of each toy) Total cost of production = x(55...

NCERT Class X Chapter 4: Quadratic Equation Example 1 (i) Question: Represent the following situations mathematically: John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. Given: Total marbles John and Jivanti initially have = 45 Both lost 5 marbles each. Product of the number of marbles they now have = 124 To Find: Represent the following situations mathematically Number of marbles each had to start with. Solution: Let, John's initial marbles = 'x', John's initial marbles + Jivanti's initial marbles = 45 Therefore Jivanti's initial marbles = 45 - x After losing 5 marbles each: John's marbles = x - 5 Jivanti's marbles = (45 - x) - 5 = 40 - x John's marbles × Jivanti's marbles = 124 ⇒ (x - 5)(40 - x) = 124 ⇒ 40x - x 2 - 200 + 5x = 124 Gr...