10th Maths - Quarterly Exam 2024 - English Medium Original Question Paper | Theni District
Part - I
Choose the correct answer: (14 x 1 = 14)
1. A = {a,b,p}, B = {2,3}, C = {p,q,r,s} then n[(A U C) x B] is
Solution:
Given, A = {a, b, p}, B = {2, 3}, C = {p, q, r, s}.First, find A U C:
A U C = {a, b, p} U {p, q, r, s} = {a, b, p, q, r, s}.
Next, find the number of elements in A U C:
n(A U C) = 6.
The number of elements in B is n(B) = 2.
Now, we need to find n[(A U C) x B].
Using the formula n(X x Y) = n(X) * n(Y):
n[(A U C) x B] = n(A U C) * n(B) = 6 * 2 = 12.
c) 12
2. If \( f(x) = 2x^2 \) and \( g(x) = \frac{1}{3x} \), then fog is
Solution:
Given, \( f(x) = 2x^2 \) and \( g(x) = \frac{1}{3x} \).The composition function fog is defined as f(g(x)).
Substitute g(x) into f(x):
fog = f(g(x)) = \( f(\frac{1}{3x}) \).
Now, replace x in f(x) with \( \frac{1}{3x} \):
\( f(\frac{1}{3x}) = 2 \left( \frac{1}{3x} \right)^2 = 2 \left( \frac{1}{9x^2} \right) = \frac{2}{9x^2} \).
c) \( \frac{2}{9x^2} \)
3. A function f: R→R defined by f(x) = ax² + bx + c, (a ≠ 0) is called a
Solution:
A function defined by a polynomial of degree 2, of the form f(x) = ax² + bx + c where a ≠ 0, is called a quadratic function.d) quadratic function
4. \( 7^{4K} \equiv \underline{\hspace{1cm}} \pmod{100} \)
Solution:
We need to find the remainder when \( 7^{4K} \) is divided by 100.Let's find the pattern for powers of 7 mod 100.
\( 7^1 \equiv 7 \pmod{100} \)
\( 7^2 = 49 \equiv 49 \pmod{100} \)
\( 7^3 = 49 \times 7 = 343 \equiv 43 \pmod{100} \)
\( 7^4 = 43 \times 7 = 301 \equiv 1 \pmod{100} \)
Now, we can write \( 7^{4K} \) as \( (7^4)^K \).
\( (7^4)^K \equiv (1)^K \pmod{100} \)
\( (1)^K = 1 \).
Therefore, \( 7^{4K} \equiv 1 \pmod{100} \).
a) 1
5. The sum of first n natural numbers are also called ________
Solution:
The sum of the first n natural numbers gives the sequence: n=1: 1 n=2: 1+2=3 n=3: 1+2+3=6 n=4: 1+2+3+4=10 The numbers 1, 3, 6, 10, ... can be represented by dots arranged in an equilateral triangle. These are known as Triangular numbers.c) Triangular numbers
6. The value of (1³ + 2³ + 3³ + ... + 15³) – (1 + 2 + 3 + ... + 15) is
Solution:
We use the formulas for the sum of the first n natural numbers and the sum of the cubes of the first n natural numbers.Sum of cubes: \( \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \)
Sum of numbers: \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
Here, n = 15.
1³ + 2³ + ... + 15³ = \( \left( \frac{15(15+1)}{2} \right)^2 = \left( \frac{15 \times 16}{2} \right)^2 = (15 \times 8)^2 = 120^2 = 14400 \).
1 + 2 + ... + 15 = \( \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120 \).
The required value is 14400 - 120 = 14280.
c) 14280
7. \( \frac{3y-3}{y} + \frac{7y-7}{3y^2} \) is
Solution:
Let's simplify the given expression: $$ \frac{3y-3}{y} + \frac{7y-7}{3y^2} = \frac{3(y-1)}{y} + \frac{7(y-1)}{3y^2} $$ The least common multiple (LCM) of the denominators y and 3y² is 3y². $$ = \frac{3(y-1) \cdot 3y}{3y^2} + \frac{7(y-1)}{3y^2} $$ $$ = \frac{9y(y-1) + 7(y-1)}{3y^2} $$ Factor out the common term (y-1) in the numerator: $$ = \frac{(9y+7)(y-1)}{3y^2} = \frac{9y^2 - 9y + 7y - 7}{3y^2} = \frac{9y^2 - 2y - 7}{3y^2} $$ None of the given options match this result. There might be a typo in the question or options. Assuming the question intended a different operation or had different terms. However, based on the question as written, the correct simplification is \( \frac{9y^2 - 2y - 7}{3y^2} \).Note: None of the options are correct based on the provided question.
8. Graph of a linear equation is a ________
Solution:
A linear equation is an equation of the form y = mx + c. Its graph is always a straight line.a) straight line
9. The square root of \( \frac{256 x^8y^4z^{10}}{25 x^6y^6z^6} \) is equal to
Solution:
First, simplify the expression inside the square root: $$ \frac{256 x^8y^4z^{10}}{25 x^6y^6z^6} = \frac{256}{25} x^{8-6} y^{4-6} z^{10-6} = \frac{256}{25} x^2 y^{-2} z^4 = \frac{256 x^2 z^4}{25 y^2} $$ Now, take the square root of the simplified expression: $$ \sqrt{\frac{256 x^2 z^4}{25 y^2}} = \frac{\sqrt{256} \sqrt{x^2} \sqrt{z^4}}{\sqrt{25} \sqrt{y^2}} $$ Assuming the variables represent positive quantities, this becomes: $$ = \frac{16 x z^2}{5 y} $$ The absolute value signs are usually omitted in this context.d) \( \frac{16 xz^2}{5 y} \)
10. If ΔABC is an isosceles triangle with ∠C = 90° and AC = 5cm, then AB is
Solution:
In an isosceles triangle with ∠C = 90°, the sides adjacent to the right angle are equal. So, AC = BC = 5 cm.By the Pythagorean theorem, AB² = AC² + BC².
AB² = 5² + 5² = 25 + 25 = 50.
AB = √50 = √(25 × 2) = 5√2 cm.
d) 5√2 cm
11. In a ΔABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. The length of the side AC is
Solution:
According to the Angle Bisector Theorem, the bisector of an angle in a triangle divides the opposite side in the same ratio as the other two sides.Therefore, \( \frac{AB}{AC} = \frac{BD}{DC} \).
Substituting the given values: \( \frac{8}{AC} = \frac{6}{3} \).
\( \frac{8}{AC} = 2 \).
AC = \( \frac{8}{2} \) = 4 cm.
b) 4 cm
12. The area of triangle formed by the points (-5,0), (0,-5) and (5,0) is
Solution:
The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by the formula:Area = \( \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \).
Area = \( \frac{1}{2} |-5(-5-0) + 0(0-0) + 5(0-(-5))| \).
Area = \( \frac{1}{2} |-5(-5) + 0 + 5(5)| \).
Area = \( \frac{1}{2} |25 + 0 + 25| = \frac{1}{2} |50| = 25 \) sq.units.
b) 25 sq.units
13. The slope of the line joining (12,3), (4,a) is \( \frac{1}{8} \). The value of 'a' is
Solution:
The formula for the slope (m) of a line joining two points (x₁, y₁) and (x₂, y₂) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).Given slope m = \( \frac{1}{8} \).
\( \frac{1}{8} = \frac{a - 3}{4 - 12} \).
\( \frac{1}{8} = \frac{a - 3}{-8} \).
Multiply both sides by -8: \( \frac{-8}{8} = a - 3 \).
-1 = a - 3.
a = 3 - 1 = 2.
d) 2
14. tanθ cosec²θ – tanθ is equal to
Solution:
Factor out tanθ from the expression:tanθ cosec²θ – tanθ = tanθ (cosec²θ - 1).
We know the trigonometric identity: 1 + cot²θ = cosec²θ.
Rearranging this, we get cot²θ = cosec²θ - 1.
Substitute this back into the expression:
tanθ (cot²θ) = tanθ * \( (\frac{1}{tanθ})^2 \) = tanθ * \( \frac{1}{tan^2θ} \) = \( \frac{1}{tanθ} \).
Since \( \frac{1}{tanθ} = cotθ \), the expression simplifies to cotθ.
d) cotθ
Part - II
Answer any 10 questions. (Q.No. 28 is compulsory) (10 x 2 = 20)
15. A relation R is given by the set {(x,y) / y = x + 3, x ∈ {0,1,2,3,4,5}}. Determine its domain and range.
Solution:
The relation R is defined by y = x + 3, with the set of possible x values being {0, 1, 2, 3, 4, 5}.Domain: The domain is the set of all possible input values (x values). Domain = {0, 1, 2, 3, 4, 5}.
Range: The range is the set of all possible output values (y values). We find the y value for each x 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
16. Given the function f : x → x² – 5x + 6, evaluate
i) f(-1)
ii) f(2a)
Solution:
The function is f(x) = x² – 5x + 6.i) f(-1):
Substitute x = -1 into the function:
f(-1) = (-1)² - 5(-1) + 6 = 1 + 5 + 6 = 12.
ii) f(2a):
Substitute x = 2a into the function:
f(2a) = (2a)² - 5(2a) + 6 = 4a² - 10a + 6.
17. Find k if fof(k) = 5 where f(k) = 2k - 1
Solution:
Given f(k) = 2k - 1.First, find the expression for fof(k), which is f(f(k)).
f(f(k)) = f(2k - 1)
Substitute (2k-1) for k in the function definition:
f(2k - 1) = 2(2k - 1) - 1 = 4k - 2 - 1 = 4k - 3.
We are given that fof(k) = 5.
So, 4k - 3 = 5.
4k = 5 + 3 = 8.
k = 8 / 4 = 2.
18. Find the HCF of 252525 and 363636
Solution:
We can factorize the numbers:252525 = 25 × 10101
363636 = 36 × 10101
The highest common factor (HCF) is the largest number that divides both. From the factorization, it's clear that 10101 is a common factor.
Let's verify the HCF of 25 and 36. 25 = 5²
36 = 2² × 3²
The HCF of 25 and 36 is 1.
Therefore, the HCF of 252525 and 363636 is 1 × 10101 = 10101.
19. What is the time 15 hours before 11 p.m?
Solution:
First, convert 11 p.m. to 24-hour format: 11 p.m. is 23:00 hours.We need to find the time 15 hours before 23:00.
Time = 23 - 15 = 8 hours.
In 12-hour format, 8:00 is 8 a.m.
So, the time 15 hours before 11 p.m. is 8 a.m.
20. Find the sum \( 3 + 1 + \frac{1}{3} + ... \infty \)
Solution:
This is an infinite geometric series.First term, a = 3.
Common ratio, r = \( \frac{1}{3} \).
Since |r| = |1/3| < 1, the sum to infinity exists.
The formula for the sum to infinity of a G.P. is \( S_\infty = \frac{a}{1-r} \).
\( S_\infty = \frac{3}{1 - \frac{1}{3}} = \frac{3}{\frac{2}{3}} = 3 \times \frac{3}{2} = \frac{9}{2} \).
21. Subtract \( \frac{1}{x^2 + 2} \) from \( \frac{2x^3 + x^2 + 3}{(x^2 + 2)^2} \)
Solution:
The problem is: \( \frac{2x^3 + x^2 + 3}{(x^2 + 2)^2} - \frac{1}{x^2 + 2} \)To subtract the fractions, we need a common denominator, which is \( (x^2 + 2)^2 \).
We rewrite the second term with the common denominator: $$ \frac{1}{x^2 + 2} = \frac{1 \cdot (x^2 + 2)}{(x^2 + 2)(x^2 + 2)} = \frac{x^2 + 2}{(x^2 + 2)^2} $$ Now, perform the subtraction: $$ = \frac{2x^3 + x^2 + 3}{(x^2 + 2)^2} - \frac{x^2 + 2}{(x^2 + 2)^2} $$ $$ = \frac{(2x^3 + x^2 + 3) - (x^2 + 2)}{(x^2 + 2)^2} $$ $$ = \frac{2x^3 + x^2 + 3 - x^2 - 2}{(x^2 + 2)^2} $$ Simplify the numerator: $$ = \frac{2x^3 + 1}{(x^2 + 2)^2} $$
22. Solve x² + 2x - 2 = 0 by formula method.
Solution:
The given quadratic equation is x² + 2x - 2 = 0.We compare this to the standard form ax² + bx + c = 0.
Here, a = 1, b = 2, c = -2.
The quadratic formula is: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Substitute the values of a, b, and c: $$ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-2)}}{2(1)} $$ $$ x = \frac{-2 \pm \sqrt{4 + 8}}{2} $$ $$ x = \frac{-2 \pm \sqrt{12}}{2} $$ Simplify the square root: \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \)
$$ x = \frac{-2 \pm 2\sqrt{3}}{2} $$ Factor out 2 from the numerator: $$ x = \frac{2(-1 \pm \sqrt{3})}{2} $$ Cancel the common factor: $$ x = -1 \pm \sqrt{3} $$ The solutions are \( x = -1 + \sqrt{3} \) and \( x = -1 - \sqrt{3} \).
23. If ΔABC is similar to ΔDEF such that BC = 3 cm, EF = 4 cm and area of ΔABC = 54 cm². Find the area of ΔDEF.
Solution:
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.$$ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2 $$ Substitute 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(ΔDEF): $$ \text{Area}(\triangle DEF) = \frac{54 \times 16}{9} $$ $$ \text{Area}(\triangle DEF) = 6 \times 16 $$ $$ \text{Area}(\triangle DEF) = 96 \text{ cm}^2 $$
24. In ΔABC, D and E are points on the sides AB and AC respectively such that DE||BC. If \( \frac{AD}{DB} = \frac{3}{4} \) and AC = 15 cm, find AE.
Solution:
Given DE || BC, by Basic Proportionality Theorem (Thales' Theorem), we have: $$ \frac{AD}{DB} = \frac{AE}{EC} $$ Using a corollary of the theorem, we also have: $$ \frac{AD}{AB} = \frac{AE}{AC} $$ We are given \( \frac{AD}{DB} = \frac{3}{4} \). Let AD = 3k and DB = 4k.Then, AB = AD + DB = 3k + 4k = 7k.
The ratio \( \frac{AD}{AB} = \frac{3k}{7k} = \frac{3}{7} \).
Now, substitute this into the corollary equation: $$ \frac{3}{7} = \frac{AE}{15} $$ Solve for AE: $$ AE = \frac{3 \times 15}{7} $$ $$ AE = \frac{45}{7} \text{ cm} $$
25. Show that the points (-2,5), (6,-1) and (2,2) are collinear.
Solution:
To show that three points are collinear, we can prove that the slope between the first two points is equal to the slope between the last two points.Let the points be A(-2, 5), B(6, -1), and C(2, 2).
The formula for slope is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
Slope of AB: $$ m_{AB} = \frac{-1 - 5}{6 - (-2)} = \frac{-6}{6 + 2} = \frac{-6}{8} = -\frac{3}{4} $$ Slope of BC: $$ m_{BC} = \frac{2 - (-1)}{2 - 6} = \frac{2 + 1}{-4} = \frac{3}{-4} = -\frac{3}{4} $$ Since the slope of AB is equal to the slope of BC, and they share a common point B, the points A, B, and C lie on the same straight line and are therefore collinear.
26. Find the slope and y-intercept of \( \sqrt{3}x + (1 - \sqrt{3})y = 3 \)
Solution:
To find the slope and y-intercept, we need to rearrange the equation into the slope-intercept form, y = mx + c.Given equation: \( \sqrt{3}x + (1 - \sqrt{3})y = 3 \)
Isolate the y-term: $$ (1 - \sqrt{3})y = -\sqrt{3}x + 3 $$ Divide by \( (1 - \sqrt{3}) \): $$ y = \frac{-\sqrt{3}}{1 - \sqrt{3}}x + \frac{3}{1 - \sqrt{3}} $$ Now, we rationalize the denominators for the slope (m) and y-intercept (c).
Slope (m): $$ m = \frac{-\sqrt{3}}{1 - \sqrt{3}} = \frac{-\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{-\sqrt{3} - 3}{1 - 3} = \frac{-(\sqrt{3} + 3)}{-2} = \frac{\sqrt{3} + 3}{2} $$ Y-intercept (c): $$ c = \frac{3}{1 - \sqrt{3}} = \frac{3(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{3(1 + \sqrt{3})}{1 - 3} = \frac{3(1 + \sqrt{3})}{-2} = -\frac{3(1 + \sqrt{3})}{2} $$ Thus, Slope = \( \frac{3 + \sqrt{3}}{2} \) and y-intercept = \( -\frac{3(1 + \sqrt{3})}{2} \).
27. Prove that secθ - cosθ = tanθ sinθ
Solution:
We will start with the Left Hand Side (LHS) and show that it simplifies to the Right Hand Side (RHS).LHS = secθ - cosθ
Convert secθ to its equivalent in terms of cosθ: $$ \text{LHS} = \frac{1}{\cos\theta} - \cos\theta $$ Find a common denominator: $$ \text{LHS} = \frac{1}{\cos\theta} - \frac{\cos^2\theta}{\cos\theta} $$ $$ \text{LHS} = \frac{1 - \cos^2\theta}{\cos\theta} $$ Using the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \), we know that \( 1 - \cos^2\theta = \sin^2\theta \).
$$ \text{LHS} = \frac{\sin^2\theta}{\cos\theta} $$ We can split the numerator: $$ \text{LHS} = \left(\frac{\sin\theta}{\cos\theta}\right) \cdot \sin\theta $$ Since \( \frac{\sin\theta}{\cos\theta} = \tan\theta \), we get: $$ \text{LHS} = \tan\theta \sin\theta $$ LHS = RHS. Hence, the identity is proved.
28. Find the excluded values of the following expression: \( \frac{7P+2}{8P^2 + 13P + 5} \) (Compulsory)
Solution:
Excluded values are the values of the variable that make the denominator of a rational expression equal to zero.Set the denominator to zero:
\( 8P^2 + 13P + 5 = 0 \)
We can factorize this quadratic equation. We need two numbers that multiply to (8 × 5 = 40) and add up to 13. These numbers are 8 and 5.
\( 8P^2 + 8P + 5P + 5 = 0 \)
\( 8P(P + 1) + 5(P + 1) = 0 \)
\( (8P + 5)(P + 1) = 0 \)
This gives two possible solutions:
8P + 5 = 0 => P = -5/8
P + 1 = 0 => P = -1
The excluded values are -1 and -5/8.
Part - III
Answer any 10 questions. (Q.No. 42 is compulsory) (10 x 5 = 50)
29. Let A = {x ∈ W / x < 2}, B = {x ∈ N / 1 < x ≤ 4} and C = {3,5}, verify that A x (B ∩ C) = (A x B) ∩ (A x C)
Solution:
First, let's write the sets in roster form:A = {x is a whole number less than 2} = {0, 1}
B = {x is a natural number, 1 < x ≤ 4} = {2, 3, 4}
C = {3, 5}
L.H.S = A x (B ∩ C)
First, find B ∩ C:
B ∩ C = {2, 3, 4} ∩ {3, 5} = {3}
Now, find A x (B ∩ C):
A x (B ∩ C) = {0, 1} x {3} = {(0, 3), (1, 3)} --- (1)
R.H.S = (A x B) ∩ (A x C)
First, find A x B:
A x B = {0, 1} x {2, 3, 4} = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
Next, find A x C:
A x C = {0, 1} x {3, 5} = {(0, 3), (0, 5), (1, 3), (1, 5)}
Now, find the intersection of these two sets:
(A x B) ∩ (A x C) = {(0, 3), (1, 3)} --- (2)
From (1) and (2), L.H.S = R.H.S. Hence verified.
30. 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:
Given A = {1, 2, 3, 4}, B = {2, 5, 8, 11, 14}, and f(x) = 3x - 1.First, we find the images for each element in set A:
- f(1) = 3(1) - 1 = 2
- f(2) = 3(2) - 1 = 6 - 1 = 5
- f(3) = 3(3) - 1 = 9 - 1 = 8
- f(4) = 3(4) - 1 = 12 - 1 = 11
An arrow diagram shows the mapping from elements of set A to elements of set B.
ii) Table Form:
The function can be represented in a table with inputs (x) and their corresponding outputs (f(x)).
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| f(x) | 2 | 5 | 8 | 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:
The function is represented by plotting the ordered pairs as points on a Cartesian plane.
31. A function f: [-5, 9] → R is defined as follows:
\( f(x) = \begin{cases} 6x+1 & \text{if } -5 \le x < 2 \\ 5x^2-1 & \text{if } 2 \le x < 6 \\ 3x-4 & \text{if } 6 \le x \le 9 \end{cases} \)
Find:
i) f(-3) + f(2)
ii) f(7) - f(1)
iii) 2f(4) + f(8)
iv) \( \frac{2f(-2) - f(6)}{f(4) + f(-2)} \)
Solution:
i) f(-3) + f(2)For x = -3 (since -5 ≤ -3 < 2), use f(x) = 6x + 1. So, f(-3) = 6(-3) + 1 = -18 + 1 = -17.
For x = 2 (since 2 ≤ 2 < 6), use f(x) = 5x² - 1. So, f(2) = 5(2)² - 1 = 5(4) - 1 = 20 - 1 = 19.
f(-3) + f(2) = -17 + 19 = 2.
ii) f(7) - f(1)
For x = 7 (since 6 ≤ 7 ≤ 9), use f(x) = 3x - 4. So, f(7) = 3(7) - 4 = 21 - 4 = 17.
For x = 1 (since -5 ≤ 1 < 2), use f(x) = 6x + 1. So, f(1) = 6(1) + 1 = 7.
f(7) - f(1) = 17 - 7 = 10.
iii) 2f(4) + f(8)
For x = 4 (since 2 ≤ 4 < 6), use f(x) = 5x² - 1. So, f(4) = 5(4)² - 1 = 5(16) - 1 = 80 - 1 = 79.
For x = 8 (since 6 ≤ 8 ≤ 9), use f(x) = 3x - 4. So, f(8) = 3(8) - 4 = 24 - 4 = 20.
2f(4) + f(8) = 2(79) + 20 = 158 + 20 = 178.
iv) \( \frac{2f(-2) - f(6)}{f(4) + f(-2)} \)
For x = -2 (since -5 ≤ -2 < 2), use f(x) = 6x + 1. So, f(-2) = 6(-2) + 1 = -12 + 1 = -11.
For x = 6 (since 6 ≤ 6 ≤ 9), use f(x) = 3x - 4. So, f(6) = 3(6) - 4 = 18 - 4 = 14.
f(4) was calculated above as 79.
Numerator = 2f(-2) - f(6) = 2(-11) - 14 = -22 - 14 = -36.
Denominator = f(4) + f(-2) = 79 + (-11) = 68.
The value is \( \frac{-36}{68} = \frac{-9}{17} \).
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 Arithmetic Progression have the first term 'a' and common difference 'd'.The formula for the sum of the first k terms of an A.P. is \( S_k = \frac{k}{2}[2a + (k-1)d] \).
According to the question:
\( S_1 = \) sum of n terms = \( \frac{n}{2}[2a + (n-1)d] \)
\( S_2 = \) sum of 2n terms = \( \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d] \)
\( S_3 = \) sum of 3n terms = \( \frac{3n}{2}[2a + (3n-1)d] \)
Now, let's evaluate the Right Hand Side (RHS) of the equation we need to prove:
RHS = 3(S₂ - S₁)
$$ \text{RHS} = 3 \left( n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d] \right) $$ Factor out \( \frac{n}{2} \) from the terms inside the bracket: $$ \text{RHS} = 3 \cdot \frac{n}{2} \left( 2[2a + (2n-1)d] - [2a + (n-1)d] \right) $$ $$ \text{RHS} = \frac{3n}{2} \left( (4a + 4nd - 2d) - (2a + nd - d) \right) $$ $$ \text{RHS} = \frac{3n}{2} \left( 4a + 4nd - 2d - 2a - nd + d \right) $$ Combine like terms inside the bracket: $$ \text{RHS} = \frac{3n}{2} \left( (4a - 2a) + (4nd - nd) + (-2d + d) \right) $$ $$ \text{RHS} = \frac{3n}{2} \left( 2a + 3nd - d \right) $$ $$ \text{RHS} = \frac{3n}{2} [2a + (3n-1)d] $$ This expression is exactly the formula for S₃.
Thus, RHS = S₃. Hence, S₃ = 3(S₂ - S₁) is proved.
33. In a G.P the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is \( \frac{57}{2} \). Find the three terms.
Solution:
Let the three consecutive terms in the G.P. be \( \frac{a}{r}, a, ar \).Product of the terms: $$ (\frac{a}{r}) \cdot a \cdot (ar) = 27 $$ $$ a^3 = 27 $$ $$ a = 3 $$ Sum of the product of two terms at a time: $$ (\frac{a}{r})(a) + (a)(ar) + (ar)(\frac{a}{r}) = \frac{57}{2} $$ $$ \frac{a^2}{r} + a^2r + a^2 = \frac{57}{2} $$ Substitute a = 3: $$ \frac{3^2}{r} + 3^2r + 3^2 = \frac{57}{2} $$ $$ \frac{9}{r} + 9r + 9 = \frac{57}{2} $$ $$ 9(\frac{1}{r} + r + 1) = \frac{57}{2} $$ $$ \frac{1}{r} + r + 1 = \frac{57}{2 \times 9} = \frac{19}{6} $$ $$ \frac{1}{r} + r = \frac{19}{6} - 1 = \frac{19 - 6}{6} = \frac{13}{6} $$ $$ \frac{1+r^2}{r} = \frac{13}{6} $$ $$ 6(1+r^2) = 13r $$ $$ 6r^2 - 13r + 6 = 0 $$ Factor the quadratic equation: $$ 6r^2 - 9r - 4r + 6 = 0 $$ $$ 3r(2r - 3) - 2(2r - 3) = 0 $$ $$ (3r - 2)(2r - 3) = 0 $$ So, \( r = \frac{2}{3} \) or \( r = \frac{3}{2} \).
Case 1: a = 3, r = 3/2. The terms are \( \frac{3}{3/2}, 3, 3(\frac{3}{2}) \), which are 2, 3, 9/2.
Case 2: a = 3, r = 2/3. The terms are \( \frac{3}{2/3}, 3, 3(\frac{2}{3}) \), which are 9/2, 3, 2.
The three terms are 2, 3, and 9/2.
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).
Adding (1) and (2): $$ (3x - 2y + z) + (2x + 3y - z) = 2 + 5 $$ $$ 5x + y = 7 \quad \text{--- (4)} $$ Step 2: Eliminate 'z' using equations (2) and (3).
Adding (2) and (3): $$ (2x + 3y - z) + (x + y + z) = 5 + 6 $$ $$ 3x + 4y = 11 \quad \text{--- (5)} $$ Step 3: Solve the new system of equations (4) and (5).
From equation (4), we get \( y = 7 - 5x \).
Substitute this into equation (5): $$ 3x + 4(7 - 5x) = 11 $$ $$ 3x + 28 - 20x = 11 $$ $$ -17x = 11 - 28 $$ $$ -17x = -17 $$ $$ x = 1 $$ Step 4: Substitute x = 1 back into the expression for y.
y = 7 - 5(1) = 2.
Step 5: Substitute x = 1 and y = 2 into equation (3) to find z.
1 + 2 + z = 6
3 + z = 6
z = 3.
The solution is x = 1, y = 2, z = 3.
35. If 9x⁴ + 12x³ + 28x² + 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.
- The square root of the first term, 9x⁴, is 3x². This is the first term of the root.
- Subtract (3x²)² = 9x⁴. Bring down the next two terms: 12x³ + 28x².
- Double the current root (3x²) to get 6x². Divide the first term of the new dividend (12x³) by 6x² to get +2x. This is the second term of the root.
- The new divisor is 6x² + 2x. Multiply this by 2x: (6x² + 2x)(2x) = 12x³ + 4x².
- Subtract this: (12x³ + 28x²) - (12x³ + 4x²) = 24x². Bring down the remaining terms: ax + b.
- Double the current root (3x² + 2x) to get 6x² + 4x. Divide the first term of the new dividend (24x²) by 6x² to get +4. This is the third term of the root.
- The new divisor is 6x² + 4x + 4. Multiply this by 4: (6x² + 4x + 4)(4) = 24x² + 16x + 16.
b - 16 = 0 => b = 16
The values are a = 16 and b = 16.
36. If α, β are the roots of 7x² + ax + 2 = 0 and if β - α = \(-\frac{13}{7}\), find the value of a.
Solution:
For the quadratic equation 7x² + ax + 2 = 0:Sum of roots: α + β = \(-\frac{\text{coefficient of x}}{\text{coefficient of } x^2} = -\frac{a}{7}\)
Product of roots: αβ = \(\frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{2}{7}\)
We are given: β - α = \(-\frac{13}{7}\)
We use the identity: \((\alpha + \beta)^2 = (\beta - \alpha)^2 + 4\alpha\beta\)
Substitute the values we have: $$ \left(-\frac{a}{7}\right)^2 = \left(-\frac{13}{7}\right)^2 + 4\left(\frac{2}{7}\right) $$ $$ \frac{a^2}{49} = \frac{169}{49} + \frac{8}{7} $$ To add the fractions on the right, find a common denominator (49): $$ \frac{a^2}{49} = \frac{169}{49} + \frac{8 \times 7}{7 \times 7} $$ $$ \frac{a^2}{49} = \frac{169 + 56}{49} $$ $$ \frac{a^2}{49} = \frac{225}{49} $$ $$ a^2 = 225 $$ $$ a = \pm\sqrt{225} $$ $$ a = \pm 15 $$
37. State and prove Thales theorem.
Solution:
Statement (Thales Theorem or Basic Proportionality Theorem):If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Given: In a triangle ΔABC, a line DE is drawn parallel to BC, intersecting sides AB and AC at D and E respectively.
To Prove: \( \frac{AD}{DB} = \frac{AE}{EC} \)
Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.
Proof:
We know the area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \).
$$ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (1)} $$ Similarly, $$ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (2)} $$ Now, ΔBDE and ΔDEC are on the same base DE and between the same parallel lines DE and BC.
Therefore, Area(ΔBDE) = Area(ΔDEC).
From equation (1) and (2), if the denominators Area(ΔBDE) and Area(ΔDEC) are equal, then the ratios must be equal. $$ \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle DEC)} $$ Therefore, from (1) and (2): $$ \frac{AD}{DB} = \frac{AE}{EC} $$ Hence, the theorem is proved.
38. Find the area of the quadrilateral formed by the points (8,6), (5,11), (-5,12) and (-4,3)
Solution:
Let the vertices be A(8,6), B(5,11), C(-5,12), and D(-4,3). The points are in counter-clockwise order.The formula for the area of a quadrilateral with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃), (x₄,y₄) is: $$ \text{Area} = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4)| $$ Let's use the matrix form for easier calculation: $$ \text{Area} = \frac{1}{2} \left| \begin{matrix} 8 & 5 & -5 & -4 & 8 \\ 6 & 11 & 12 & 3 & 6 \end{matrix} \right| $$ $$ \text{Area} = \frac{1}{2} |[(8)(11) + (5)(12) + (-5)(3) + (-4)(6)] - [(5)(6) + (-5)(11) + (-4)(12) + (8)(3)]| $$ $$ \text{Area} = \frac{1}{2} |[88 + 60 - 15 - 24] - [30 - 55 - 48 + 24]| $$ $$ \text{Area} = \frac{1}{2} |[148 - 39] - [54 - 103]| $$ $$ \text{Area} = \frac{1}{2} |[109] - [-49]| $$ $$ \text{Area} = \frac{1}{2} |109 + 49| $$ $$ \text{Area} = \frac{1}{2} |158| = 79 $$ The area of the quadrilateral is 79 sq. units.
39. You are downloading a song. The percent y (in decimal form) of mega bytes remaining to get downloaded in x seconds is given by y = -0.1x + 1
i) Find the total MB of the song
ii) After how many seconds will 75% of the songs gets downloaded?
iii) After how many seconds the song will be downloaded completely?
Solution:
The given equation is y = -0.1x + 1, where y is the percentage of the song remaining (as a decimal) and x is time in seconds.i) Find the total MB of the song
The given equation relates the percentage of the file remaining to time. It does not contain any information about the file size in Mega Bytes (MB). Therefore, the total MB of the song cannot be determined from the given information.
ii) After how many seconds will 75% of the song get downloaded?
If 75% of the song is downloaded, then 100% - 75% = 25% of the song is remaining.
In decimal form, this is y = 0.25.
Substitute y = 0.25 into the equation: $$ 0.25 = -0.1x + 1 $$ $$ 0.1x = 1 - 0.25 $$ $$ 0.1x = 0.75 $$ $$ x = \frac{0.75}{0.1} = 7.5 $$ So, 75% of the song will be downloaded after 7.5 seconds.
iii) After how many seconds the song will be downloaded completely?
When the song is downloaded completely, 0% of the song is remaining.
In decimal form, this is y = 0.
Substitute y = 0 into the equation: $$ 0 = -0.1x + 1 $$ $$ 0.1x = 1 $$ $$ x = \frac{1}{0.1} = 10 $$ The song will be downloaded completely after 10 seconds.
40. Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6,-4)
Solution:
The perpendicular bisector passes through the midpoint of the line segment AB and is perpendicular to it.Step 1: Find the midpoint of AB.
Midpoint M = \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
M = \( \left( \frac{-4+6}{2}, \frac{2+(-4)}{2} \right) = \left( \frac{2}{2}, \frac{-2}{2} \right) = (1, -1) \).
Step 2: Find the slope of the line AB.
Slope \( m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 2}{6 - (-4)} = \frac{-6}{10} = -\frac{3}{5} \).
Step 3: Find the slope of the perpendicular bisector.
The slope of the perpendicular line (\( m_{\perp} \)) is the negative reciprocal of \( m_{AB} \).
\( m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3} \).
Step 4: Find the equation of the perpendicular bisector.
The line passes through the midpoint M(1, -1) and has a slope of 5/3. Using the point-slope form \( y - y_1 = m(x - x_1) \):
$$ y - (-1) = \frac{5}{3}(x - 1) $$ $$ y + 1 = \frac{5}{3}(x - 1) $$ Multiply the entire equation by 3 to eliminate the fraction: $$ 3(y + 1) = 5(x - 1) $$ $$ 3y + 3 = 5x - 5 $$ Rearrange to the standard form: $$ 5x - 3y - 8 = 0 $$ This is the required equation of the perpendicular bisector.
41. If cotθ + tanθ = x and secθ – cosθ = y, then prove that \( (x^2y)^{2/3} - (xy^2)^{2/3} = 1 \)
Solution:
First, simplify the expressions for x and y in terms of sinθ and cosθ.Simplify x: $$ x = \cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} $$ $$ x = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} $$ Using the identity \( \sin^2\theta + \cos^2\theta = 1 \): $$ x = \frac{1}{\sin\theta\cos\theta} $$ Simplify y: $$ y = \sec\theta - \cos\theta = \frac{1}{\cos\theta} - \cos\theta $$ $$ y = \frac{1 - \cos^2\theta}{\cos\theta} $$ Using the identity \( 1 - \cos^2\theta = \sin^2\theta \): $$ y = \frac{\sin^2\theta}{\cos\theta} $$ Now, let's evaluate the terms inside the brackets of the expression we need to prove.
Evaluate \( x^2y \): $$ x^2y = \left(\frac{1}{\sin\theta\cos\theta}\right)^2 \left(\frac{\sin^2\theta}{\cos\theta}\right) = \left(\frac{1}{\sin^2\theta\cos^2\theta}\right) \left(\frac{\sin^2\theta}{\cos\theta}\right) = \frac{1}{\cos^3\theta} = \sec^3\theta $$ Evaluate \( xy^2 \): $$ xy^2 = \left(\frac{1}{\sin\theta\cos\theta}\right) \left(\frac{\sin^2\theta}{\cos\theta}\right)^2 = \left(\frac{1}{\sin\theta\cos\theta}\right) \left(\frac{\sin^4\theta}{\cos^2\theta}\right) = \frac{\sin^3\theta}{\cos^3\theta} = \tan^3\theta $$ Now substitute these results back into the Left Hand Side (LHS) of the identity:
LHS = \( (x^2y)^{2/3} - (xy^2)^{2/3} \)
$$ \text{LHS} = (\sec^3\theta)^{2/3} - (\tan^3\theta)^{2/3} $$ $$ \text{LHS} = \sec^{(3 \times 2/3)}\theta - \tan^{(3 \times 2/3)}\theta $$ $$ \text{LHS} = \sec^2\theta - \tan^2\theta $$ Using the Pythagorean identity \( 1 + \tan^2\theta = \sec^2\theta \), we can rearrange it to \( \sec^2\theta - \tan^2\theta = 1 \).
LHS = 1.
Since LHS = 1 and RHS = 1, the identity is proved.
42. Find the sum of 10³ + 11³ + 12³ + ... + 20³ (Compulsory)
Solution:
We can find the sum by subtracting the sum of the first 9 cubes from the sum of the first 20 cubes.Sum = (1³ + 2³ + ... + 20³) - (1³ + 2³ + ... + 9³)
We use the formula for the sum of the cubes of the first n natural numbers: \( \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \).
For n = 20:
1³ + ... + 20³ = \( \left( \frac{20(20+1)}{2} \right)^2 = \left( \frac{20 \times 21}{2} \right)^2 = (10 \times 21)^2 = 210^2 = 44100 \).
For n = 9:
1³ + ... + 9³ = \( \left( \frac{9(9+1)}{2} \right)^2 = \left( \frac{9 \times 10}{2} \right)^2 = (9 \times 5)^2 = 45^2 = 2025 \).
Required sum = 44100 - 2025 = 42075.
Part - IV
Answer all the questions. (2 x 8 = 16)
43. a) Construct a triangle similar to a given triangle PQR with its sides equal to \( \frac{7}{3} \) of the corresponding sides of the triangle PQR (Scale factor \( \frac{7}{3} > 1 \)).
Steps of Construction:
- Draw any triangle PQR.
- Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
- Since the scale factor is 7/3, locate 7 points (the greater of 7 and 3) Q₁, Q₂, ..., Q₇ on QX such that QQ₁ = Q₁Q₂ = ... = Q₆Q₇.
- Join Q₃ (the 3rd point, corresponding to the denominator) to R.
- Draw a line through Q₇ (the 7th point, corresponding to the numerator) parallel to Q₃R to intersect the extended line segment QR at R'.
- Draw a line through R' parallel to PR to intersect the extended line segment QP at P'.
- The resulting triangle P'QR' is the required triangle similar to ΔPQR, with sides 7/3 times the corresponding sides of ΔPQR.
43. b) (OR) Construct a triangle ΔPQR such that QR = 5 cm, ∠P = 30° and the altitude from P to QR is of length 4.2 cm.
Steps of Construction:
- Draw a line segment QR = 5 cm.
- At Q, draw a line QX such that ∠RQX = 30°.
- Draw a line QY perpendicular to QX at Q.
- Draw the perpendicular bisector of QR. Let it intersect QY at O and QR at M.
- With O as the center and OQ as the radius, draw a circle. This circle will pass through Q and R. All angles subtended by the arc QR in the major segment will be 30°.
- On the perpendicular bisector from M, mark a point L such that ML = 4.2 cm (the altitude).
- Draw a line through L parallel to QR. This line will intersect the circle at two points, P and P'.
- Join PQ and PR (or P'Q and P'R).
- ΔPQR is the required triangle.
44. a) A bus is travelling at a uniform speed of 50 km/hr. Draw the distance-time graph and hence find
i) The constant of variation
ii) How far will it travel in 90 minutes?
iii) The time required to cover a distance of 300 km from the graph.
Solution:
Let x be the time in hours and y be the distance in km. Since speed is uniform, y is directly proportional to x. The relationship is y = 50x.Graph:
We create a table of values:
| Time x (hours) | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Distance y (km) | 50 | 100 | 150 | 200 |
i) The constant of variation:
The equation is y = kx, where k is the constant of variation. Here, y = 50x. So, the constant of variation k = 50.
ii) How far will it travel in 90 minutes?
90 minutes = 1.5 hours. From the graph, find the y-value corresponding to x = 1.5. Draw a vertical line from x=1.5 to the graph line, then a horizontal line to the y-axis. It will meet at y = 75.
Calculation: y = 50 * 1.5 = 75 km.
iii) The time required to cover a distance of 300 km:
From the graph, find the x-value corresponding to y = 300. Draw a horizontal line from y=300 to the graph line, then a vertical line to the x-axis. It will meet at x = 6.
Calculation: 300 = 50x => x = 300/50 = 6 hours.
44. b) (OR) 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:
The equation is xy = 24, which represents indirect variation. y = 24/x.Graph:
We create a table of values for x > 0, y > 0:
| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 |
|---|---|---|---|---|---|---|---|
| y | 24 | 12 | 8 | 6 | 4 | 3 | 2 |
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.
So, when x = 3, y = 8.
ii) Find x when y = 6:
On the graph, locate y = 6 on the y-axis. Draw a horizontal line to 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.
So, when y = 6, x = 4.