10th Maths Quarterly Exam 2024 Original Question Paper with Solution - Perambalur District
Part - I 14 x 1 = 14
Choose the most suitable answers from the given four alternatives.
1. \(f(x) = (x+1)^3 - (x-1)^3\) represents a function which is
Solution:
We have \(f(x) = (x+1)^3 - (x-1)^3\).
Using the formulas \( (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \) and \( (a-b)^3 = a^3-3a^2b+3ab^2-b^3 \):
\( f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) \)
\( f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 \)
\( f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1) \)
\( f(x) = 6x^2 + 2 \)
Since the highest power of x is 2, the function is quadratic.
2. If A = { x, y, z }, B = { 3, 5 }, C = { z, a, b, c }, then \( n[ (A \cup C) \times B ] \) is
Solution:
Given: A = {x, y, z}, B = {3, 5}, C = {z, a, b, c}
First, find \( A \cup C \):
\( A \cup C = \{x, y, z\} \cup \{z, a, b, c\} = \{x, y, z, a, b, c\} \)
The number of elements in \( A \cup C \) is \( n(A \cup C) = 6 \).
The number of elements in B is \( n(B) = 2 \).
Now, find \( n[ (A \cup C) \times B ] \):
\( n[ (A \cup C) \times B ] = n(A \cup C) \times n(B) = 6 \times 2 = 12 \)
3. \( 7^{4k} \equiv \) _____ (mod 100)
Solution:
We need to find the remainder when \(7^{4k}\) is divided by 100.
\( 7^1 = 7 \)
\( 7^2 = 49 \)
\( 7^3 = 343 \equiv 43 \) (mod 100)
\( 7^4 = 7^2 \times 7^2 = 49 \times 49 = 2401 \)
When 2401 is divided by 100, the remainder is 1. So, \( 7^4 \equiv 1 \) (mod 100).
Now, \( 7^{4k} = (7^4)^k \equiv 1^k \) (mod 100)
\( 7^{4k} \equiv 1 \) (mod 100)
4. The value of \( (1^3 + 2^3 + 3^3 + \dots + 15^3) - (1 + 2 + 3 + \dots + 15) \)
Solution:
We use the formulas for the sum of the first n natural numbers and the sum of their cubes:
\( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
\( \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \)
Here, n = 15.
First, calculate \( S = 1 + 2 + \dots + 15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120 \).
Next, calculate \( 1^3 + 2^3 + \dots + 15^3 = S^2 = 120^2 = 14400 \).
The required value is \( (1^3 + \dots + 15^3) - (1 + \dots + 15) = 14400 - 120 = 14280 \).
5. The value of a, b If \(4x^4 - 24x^3 + 76x^2 + ax + b\) is a perfect square.
Solution:
We use the long division method for finding the square root.
$$ \begin{array}{c|l} \multicolumn{2}{r}{2x^2 - 6x + 10} \\ \cline{2-2} 2x^2 & 4x^4 - 24x^3 + 76x^2 + ax + b \\ & -4x^4 \\ \cline{2-2} 4x^2 - 6x & -24x^3 + 76x^2 \\ & -(-24x^3 + 36x^2) \\ \cline{2-2} 4x^2 - 12x + 10 & 40x^2 + ax + b \\ & -(40x^2 - 120x + 100) \\ \cline{2-2} & (a+120)x + (b-100) \\ \end{array} $$
For the polynomial to be a perfect square, the remainder must be 0.
\( (a+120)x + (b-100) = 0 \)
This means \( a+120 = 0 \implies a = -120 \).
And \( b-100 = 0 \implies b = 100 \).
6. The solution of \( (2x - 1)^2 = 9 \) is equal to
Solution:
Given \( (2x - 1)^2 = 9 \).
Taking the square root of both sides:
\( 2x - 1 = \pm\sqrt{9} \)
\( 2x - 1 = \pm 3 \)
Case 1: \( 2x - 1 = 3 \implies 2x = 4 \implies x = 2 \)
Case 2: \( 2x - 1 = -3 \implies 2x = -2 \implies x = -1 \)
The solutions are -1 and 2.
7. If in \(\triangle ABC\), DE || BC, AB = 3.6cm, AC = 2.4 cm and AD = 2.1cm, then the length of AE is
Solution:
By the Basic Proportionality Theorem (Thales' Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally.
Given DE || BC, we have: \( \frac{AD}{AB} = \frac{AE}{AC} \)
Substituting 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
8. The point of intersection of \(3x - y = 4\) and \(x + y = 8\) is
Solution:
We have two linear equations:
(1) \( 3x - y = 4 \)
(2) \( x + y = 8 \)
Add equation (1) and (2) to eliminate y:
\( (3x - y) + (x + y) = 4 + 8 \)
\( 4x = 12 \implies x = 3 \)
Substitute x = 3 into equation (2):
\( 3 + y = 8 \implies y = 5 \)
The point of intersection is (3, 5).
9. The slope of the line which is perpendicular to a line joining the points (0,0) and (-8,8) is
Solution:
First, find the slope (\(m_1\)) of the line joining (0,0) and (-8,8).
\( m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{-8 - 0} = \frac{8}{-8} = -1 \)
Let \(m_2\) be the slope of the perpendicular line. The product of the slopes of two perpendicular lines is -1.
\( m_1 \times m_2 = -1 \)
\( -1 \times m_2 = -1 \implies m_2 = 1 \)
10. The value of \( (1 + \tan\theta + \sec\theta)(1 + \cot\theta - \csc\theta) \) is
Solution:
Convert to sin and cos:
\( = \left(1 + \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta}\right) \left(1 + \frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta}\right) \)
\( = \left(\frac{\cos\theta + \sin\theta + 1}{\cos\theta}\right) \left(\frac{\sin\theta + \cos\theta - 1}{\sin\theta}\right) \)
Let \( (\sin\theta + \cos\theta) = A \). The numerator is of the form \( (A+1)(A-1) = A^2 - 1 \).
\( = \frac{(\sin\theta + \cos\theta)^2 - 1^2}{\sin\theta \cos\theta} \)
\( = \frac{\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta - 1}{\sin\theta \cos\theta} \)
Since \( \sin^2\theta + \cos^2\theta = 1 \):
\( = \frac{1 + 2\sin\theta\cos\theta - 1}{\sin\theta \cos\theta} = \frac{2\sin\theta\cos\theta}{\sin\theta \cos\theta} = 2 \)
11. The value of \( \sin^2\theta + \frac{1}{1+\tan^2\theta} \) is
Solution:
We use the trigonometric identity \( 1 + \tan^2\theta = \sec^2\theta \).
Substitute this into the expression:
\( \sin^2\theta + \frac{1}{\sec^2\theta} \)
We know that \( \frac{1}{\sec\theta} = \cos\theta \), so \( \frac{1}{\sec^2\theta} = \cos^2\theta \).
The expression becomes:
\( \sin^2\theta + \cos^2\theta \)
Using the fundamental Pythagorean identity, \( \sin^2\theta + \cos^2\theta = 1 \).
12. If \( \sqrt{4m^2 - 24m + 36} = 0 \)
Solution:
The question is slightly ambiguous. It could be asking to simplify the expression \( \sqrt{4m^2 - 24m + 36} \) or solve the equation. The options are expressions, suggesting a simplification is intended.
Let's simplify the expression inside the square root:
\( 4m^2 - 24m + 36 \)
Factor out the common term 4:
\( 4(m^2 - 6m + 9) \)
The expression in the parenthesis is a perfect square: \( (m-3)^2 \).
So, the expression becomes \( 4(m-3)^2 \), which can be written as \( [2(m-3)]^2 \).
Now, take the square root:
\( \sqrt{[2(m-3)]^2} = |2(m-3)| \)
Assuming a non-negative result, this simplifies to \( 2(m-3) \).
Note: If we solve the equation \( \sqrt{4m^2 - 24m + 36} = 0 \), we get \( 4m^2 - 24m + 36 = 0 \), which gives \( (m-3)^2 = 0 \), so \( m=3 \). This doesn't match the format of the options.
13. The HCF of numbers of the form \(2^m\) and \(3^n\) is
Solution:
The Highest Common Factor (HCF) is the largest positive integer that divides each of the integers.
The prime factorization of the first number, \(2^m\), consists only of the prime factor 2.
The prime factorization of the second number, \(3^n\), consists only of the prime factor 3.
For two numbers to have a common factor other than 1, they must share at least one common prime factor. Since \(2^m\) and \(3^n\) do not share any common prime factors, their only common factor is 1.
14. What is the time 100 hours after 6 a.m.?
Solution:
There are 24 hours in a day. We need to find out how many full days and extra hours are in 100 hours.
We can do this using division:
\( 100 \div 24 \)
\( 100 = 4 \times 24 + 4 \)
This means 100 hours is equal to 4 full days and 4 extra hours.
The starting time is 6 a.m.
After 4 full days, the time will be 6 a.m. again.
We then add the remaining 4 hours to this time:
6 a.m. + 4 hours = 10 a.m.
Part - II 10 x 2 = 20
Answer any 10 questions. Q.No.28 is compulsory.
15. If A = B = { p, q}, find i) A x B ii) B x A
Solution:
Given A = {p, q} and B = {p, q}.
i) A x B (Cartesian Product):
\( A \times B = \{(p, p), (p, q), (q, p), (q, q)\} \)
ii) B x A (Cartesian Product):
\( B \times A = \{(p, p), (p, q), (q, p), (q, q)\} \)
In this case, since A = B, A x B = B x A.
16. Find the value of k, such that fog = gof. f(x) = 3x + 2, g(x) = 6x - k
Solution:
Given \( f(x) = 3x + 2 \) and \( g(x) = 6x - k \).
Find fog(x):
\( fog(x) = f(g(x)) = f(6x - k) = 3(6x - k) + 2 = 18x - 3k + 2 \)
Find gof(x):
\( gof(x) = g(f(x)) = g(3x + 2) = 6(3x + 2) - k = 18x + 12 - k \)
Given fog = gof:
\( 18x - 3k + 2 = 18x + 12 - k \)
\( -3k + 2 = 12 - k \)
\( 2 - 12 = -k + 3k \)
\( -10 = 2k \)
\( k = -5 \)
17. If \(a^b \times b^a = 800\). Find a and b.
Solution:
We are given the equation \(a^b \times b^a = 800\).
First, find the prime factorization of 800:
\(800 = 8 \times 100 = 2^3 \times 10^2 = 2^3 \times (2 \times 5)^2 = 2^3 \times 2^2 \times 5^2 = 2^5 \times 5^2\).
So, the equation is \(a^b \times b^a = 2^5 \times 5^2\).
By comparing the terms on both sides, we can match the bases and exponents.
Let's assume \(a = 2\) and \(b = 5\).
Substitute these values back into the expression \(a^b \times b^a\):
\(2^5 \times 5^2 = 32 \times 25 = 800\).
This matches the given value. So, one possible solution is \(a = 2, b = 5\).
Alternatively, if we assume \(a=5\) and \(b=2\), we get \(5^2 \times 2^5 = 25 \times 32 = 800\), which is also correct.
18. Find x, y and z given that the numbers x, 10, y, 24, z are in A.P.
Solution:
The given sequence is x, 10, y, 24, z, which is an Arithmetic Progression (A.P.).
Let the terms be \(t_1 = x, t_2 = 10, t_3 = y, t_4 = 24, t_5 = z\).
In an A.P., the difference between consecutive terms is constant (common difference, d).
We can find 'd' using the known terms \(t_2\) and \(t_4\).
The formula for the nth term is \(t_n = t_1 + (n-1)d\). So, \(t_4 = t_2 + (4-2)d\).
\(24 = 10 + 2d\)
\(2d = 24 - 10 = 14\)
\(d = 7\)
Now we can find x, y, and z:
\(x = t_1 = t_2 - d = 10 - 7 = 3\)
\(y = t_3 = t_2 + d = 10 + 7 = 17\)
\(z = t_5 = t_4 + d = 24 + 7 = 31\)
19. 9, 3, 1, ... are in G.P find 8th term.
Solution:
The given sequence 9, 3, 1, ... is a Geometric Progression (G.P.).
First term, \(a = 9\).
Common ratio, \(r = \frac{t_2}{t_1} = \frac{3}{9} = \frac{1}{3}\).
The formula for the n-th term of a G.P. is \(T_n = a \cdot r^{n-1}\).
We need to find the 8th term, \(T_8\).
\(T_8 = 9 \times \left(\frac{1}{3}\right)^{8-1} = 9 \times \left(\frac{1}{3}\right)^7\)
\(T_8 = 3^2 \times \frac{1}{3^7} = \frac{1}{3^{7-2}} = \frac{1}{3^5}\)
\(3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243\)
\(T_8 = \frac{1}{243}\)
20. Simplify: \( \frac{x^2 - 11x + 18}{x^2 - 4x + 4} \)
Solution:
We need to factorize the numerator and the denominator.
Numerator: \(x^2 - 11x + 18\)
We look for two numbers that multiply to 18 and add up to -11. These are -2 and -9.
So, \(x^2 - 11x + 18 = (x - 2)(x - 9)\).
Denominator: \(x^2 - 4x + 4\)
This is a perfect square of the form \((a-b)^2 = a^2 - 2ab + b^2\).
So, \(x^2 - 4x + 4 = (x - 2)^2\).
Now, the expression becomes:
\( \frac{(x - 2)(x - 9)}{(x - 2)(x - 2)} \)
Cancel the common factor \((x - 2)\) (assuming \(x \neq 2\)).
21. Find the sum and product of the roots for the following quadratic equation \(x^2 + 8x - 65 = 0\).
Solution:
For a quadratic equation of the form \(ax^2 + bx + c = 0\):
Sum of the roots \((\alpha + \beta) = -\frac{b}{a}\)
Product of the roots \((\alpha\beta) = \frac{c}{a}\)
Comparing \(x^2 + 8x - 65 = 0\) with the standard form, we have:
\(a = 1, b = 8, c = -65\)
Sum of the roots = \(-\frac{8}{1} = -8\)
Product of the roots = \(\frac{-65}{1} = -65\)
22. Find the square root of the following rational expression: \( \frac{121(a+b)^8(x+y)^8(b-c)^8}{81(b-c)^4(a-b)^{12}(b-c)^4} \)
Solution:
First, simplify the expression inside. Combine the \((b-c)\) terms in the denominator: \((b-c)^4 \times (b-c)^4 = (b-c)^8\).
Expression = \( \frac{121(a+b)^8(x+y)^8(b-c)^8}{81(a-b)^{12}(b-c)^8} \)
Cancel the common term \((b-c)^8\):
Simplified Expression = \( \frac{121(a+b)^8(x+y)^8}{81(a-b)^{12}} \)
Now, find the square root:
\(\sqrt{\frac{121(a+b)^8(x+y)^8}{81(a-b)^{12}}} = \frac{\sqrt{121}\sqrt{(a+b)^8}\sqrt{(x+y)^8}}{\sqrt{81}\sqrt{(a-b)^{12}}}\)
Using \(\sqrt{x^{2n}} = |x^n|\):
= \( \frac{11 |(a+b)^4| |(x+y)^4|}{9 |(a-b)^6|} \)
Since the powers 4 and 6 are even, the expressions inside the absolute value are always non-negative. We can write the answer with a single modulus sign.
23. If \(\triangle ABC\) is similar to \(\triangle DEF\) such that BC = 3cm, EF = 4cm and area of \(\triangle ABC\) = 54cm². Find the area of \(\triangle DEF\).
Solution:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\(\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2\)
Given: Area(\(\triangle ABC\)) = 54 cm², BC = 3 cm, EF = 4 cm.
\(\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}\)
Rearranging to find the area of \(\triangle DEF\):
\(\text{Area}(\triangle DEF) = 54 \times \frac{16}{9}\)
\(\text{Area}(\triangle DEF) = 6 \times 16 = 96\)
24. Find the value of 'a' for which the given points are collinear: (2, 3), (4, a), (6, -3).
Solution:
If three points are collinear, the slope between any two pairs of points will be equal.
Let the points be P(2, 3), Q(4, a), and R(6, -3).
Slope of PQ = Slope of QR
\(\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_2}{x_3 - x_2}\)
\(\frac{a - 3}{4 - 2} = \frac{-3 - a}{6 - 4}\)
\(\frac{a - 3}{2} = \frac{-3 - a}{2}\)
Since the denominators are equal, the numerators must be equal:
\(a - 3 = -3 - a\)
\(a + a = -3 + 3\)
\(2a = 0\)
\(a = 0\)
25. Find the slope of a line joining the points \((\sqrt{7}, 7)\) with the origin.
Solution:
The given points are \((\sqrt{7}, 7)\) and the origin (0, 0).
Let \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (\sqrt{7}, 7)\).
The formula for the slope (m) is:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\)
\(m = \frac{7 - 0}{\sqrt{7} - 0} = \frac{7}{\sqrt{7}}\)
To simplify, we can write \(7 = \sqrt{7} \times \sqrt{7}\).
\(m = \frac{\sqrt{7} \times \sqrt{7}}{\sqrt{7}} = \sqrt{7}\)
26. Prove that \( \frac{\sec\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta} = \cot\theta \)
Solution:
LHS = \( \frac{\sec\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta} \)
Convert secant to cosine: \(\sec\theta = \frac{1}{\cos\theta}\)
LHS = \( \frac{1/\cos\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta} = \frac{1}{\sin\theta\cos\theta} - \frac{\sin\theta}{\cos\theta} \)
Find a common denominator, which is \(\sin\theta\cos\theta\):
LHS = \( \frac{1}{\sin\theta\cos\theta} - \frac{\sin\theta \times \sin\theta}{\sin\theta\cos\theta} = \frac{1 - \sin^2\theta}{\sin\theta\cos\theta} \)
Using the identity \(\sin^2\theta + \cos^2\theta = 1\), we get \(1 - \sin^2\theta = \cos^2\theta\).
LHS = \( \frac{\cos^2\theta}{\sin\theta\cos\theta} \)
Cancel one \(\cos\theta\) from the numerator and denominator:
LHS = \( \frac{\cos\theta}{\sin\theta} \)
Since \(\cot\theta = \frac{\cos\theta}{\sin\theta}\),
LHS = \(\cot\theta\) = RHS. Hence proved.
27. Prove that \( \sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = \csc\theta + \cot\theta \)
Solution:
LHS = \( \sqrt{\frac{1+\cos\theta}{1-\cos\theta}} \)
Multiply the numerator and denominator inside the square root by the conjugate of the denominator, which is \((1+\cos\theta)\):
LHS = \( \sqrt{\frac{(1+\cos\theta)(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}} = \sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}} \)
Using the identity \(1 - \cos^2\theta = \sin^2\theta\):
LHS = \( \sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}} \)
Take the square root of the numerator and the denominator:
LHS = \( \frac{1+\cos\theta}{\sin\theta} \)
Split the fraction into two parts:
LHS = \( \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} \)
Since \(\csc\theta = \frac{1}{\sin\theta}\) and \(\cot\theta = \frac{\cos\theta}{\sin\theta}\),
LHS = \(\csc\theta + \cot\theta\) = RHS. Hence proved.
28. Find the equation of a straight line passing through (5, -3) and (7, -4).
Solution:
We use the two-point form of the equation of a line: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \)
Here, \( (x_1, y_1) = (5, -3) \) and \( (x_2, y_2) = (7, -4) \).
\( \frac{y - (-3)}{-4 - (-3)} = \frac{x - 5}{7 - 5} \)
\( \frac{y + 3}{-1} = \frac{x - 5}{2} \)
\( 2(y + 3) = -1(x - 5) \)
\( 2y + 6 = -x + 5 \)
\( x + 2y + 6 - 5 = 0 \)
\( x + 2y + 1 = 0 \)
Part - III 10 x 5 = 50
Answer any 10 questions. Q.No.42 is compulsory.
29. Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4}, C = {3, 5}. Verify that \( A \times (B \cup C) = (A \times B) \cup (A \times C) \)
Solution:
First, let's list the elements of the sets:
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}
LHS: \( A \times (B \cup C) \)
\( B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\} \)
\( A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\} \)
\( = \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\} \) --- (1)
RHS: \( (A \times B) \cup (A \times C) \)
\( A \times B = \{0, 1\} \times \{2, 3, 4\} = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\} \)
\( A \times C = \{0, 1\} \times \{3, 5\} = \{(0,3), (0,5), (1,3), (1,5)\} \)
\( (A \times B) \cup (A \times C) = \{(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)\} \cup \{(0,3), (0,5), (1,3), (1,5)\} \)
\( = \{(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)\} \) --- (2)
From (1) and (2), LHS = RHS. Hence, verified.
30. The function 't' which maps temperature in Celsius(C) into temperature in Fahrenheit(F) is defined by \(t(C) = F\) where \(F = \frac{9}{5}C + 32\). Find i) t(0) ii) t(28) iii) t(-10) iv) the value of C when t(C) = 212 v) the temperature when the Celsius value is equal to the Fahrenheit value.
Solution:
The given function is \(t(C) = \frac{9}{5}C + 32\).
i) t(0)
\(t(0) = \frac{9}{5}(0) + 32 = 0 + 32 = 32\). So, \(t(0) = 32^\circ F\).
ii) t(28)
\(t(28) = \frac{9}{5}(28) + 32 = \frac{252}{5} + 32 = 50.4 + 32 = 82.4\). So, \(t(28) = 82.4^\circ F\).
iii) t(-10)
\(t(-10) = \frac{9}{5}(-10) + 32 = 9(-2) + 32 = -18 + 32 = 14\). So, \(t(-10) = 14^\circ F\).
iv) The value of C when t(C) = 212
\(212 = \frac{9}{5}C + 32\)
\(212 - 32 = \frac{9}{5}C\)
\(180 = \frac{9}{5}C\)
\(C = 180 \times \frac{5}{9} = 20 \times 5 = 100\). So, \(C = 100^\circ C\).
v) The temperature when Celsius equals Fahrenheit
Let C = F. We substitute C for F in the formula:
\(C = \frac{9}{5}C + 32\)
\(C - \frac{9}{5}C = 32\)
\(\left(1 - \frac{9}{5}\right)C = 32 \implies \left(\frac{5-9}{5}\right)C = 32\)
\(-\frac{4}{5}C = 32\)
\(C = 32 \times \left(-\frac{5}{4}\right) = -8 \times 5 = -40\). So, the temperature is \(-40^\circ\).
31. If f(x) = x - 4, g(x) = x², h(x) = 3x - 5. Prove that (fog)oh = fo(goh).
Solution:
LHS: (fog)oh
First, find fog(x):
\(fog(x) = f(g(x)) = f(x^2) = x^2 - 4\)
Now, find (fog)oh(x):
\((fog)oh(x) = (fog)(h(x)) = (fog)(3x-5)\)
Substitute (3x-5) into the expression for fog(x):
\(= (3x-5)^2 - 4 = (9x^2 - 30x + 25) - 4 = 9x^2 - 30x + 21\) --- (1)
RHS: fo(goh)
First, find goh(x):
\(goh(x) = g(h(x)) = g(3x-5) = (3x-5)^2\)
Now, find fo(goh)(x):
\(fo(goh)(x) = f(goh(x)) = f((3x-5)^2)\)
Substitute \((3x-5)^2\) into the expression for f(x):
\(= (3x-5)^2 - 4 = (9x^2 - 30x + 25) - 4 = 9x^2 - 30x + 21\) --- (2)
From (1) and (2), LHS = RHS. Hence proved.
32. If 'd' is the highest common factor of 32 and 60. Find x and y satisfying d = 32x + 60y.
Solution:
First, find the HCF (d) of 32 and 60 using Euclid's Division Algorithm.
\(60 = 1 \times 32 + 28\)
\(32 = 1 \times 28 + 4\)
\(28 = 7 \times 4 + 0\)
The last non-zero remainder is 4. So, HCF(32, 60) = d = 4.
Now, we work backwards to express 4 in the form 32x + 60y.
From the second step: \(4 = 32 - 1 \times 28\)
From the first step, we can write \(28 = 60 - 1 \times 32\). Substitute this into the equation for 4:
\(4 = 32 - 1 \times (60 - 1 \times 32)\)
\(4 = 32 - 1 \times 60 + 1 \times 32\)
\(4 = (1+1) \times 32 - 1 \times 60\)
\(4 = 2 \times 32 + (-1) \times 60\)
Comparing this with \(d = 32x + 60y\), we get:
33. Find the middle term of an A.P. The sum of the first three terms is -3, the sum of the last three terms is 69. Total number of terms is 15.
Solution:
Total number of terms, n = 15. The middle term is the \(\left(\frac{15+1}{2}\right)\)th term, which is the 8th term (\(t_8\)).
The first three terms are \(a\), \(a+d\), \(a+2d\). Their sum is -3.
\(a + (a+d) + (a+2d) = 3a + 3d = -3 \implies a+d = -1\) --- (1)
The last three terms are \(t_{13}, t_{14}, t_{15}\). These can be written as \(a+12d\), \(a+13d\), \(a+14d\). Their sum is 69.
\((a+12d) + (a+13d) + (a+14d) = 3a + 39d = 69 \implies a+13d = 23\) --- (2)
Now we solve equations (1) and (2):
Subtract (1) from (2): \((a+13d) - (a+d) = 23 - (-1)\)
\(12d = 24 \implies d = 2\)
Substitute d=2 into (1): \(a + 2 = -1 \implies a = -3\)
The middle term is the 8th term, \(t_8 = a + 7d\).
\(t_8 = -3 + 7(2) = -3 + 14 = 11\)
34. Find the sum of \(15^2 + 16^2 + \dots + 28^2\).
Solution:
We use the formula for the sum of the squares of the first n natural numbers: \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \).
The required sum can be written as:
\( (1^2 + 2^2 + \dots + 28^2) - (1^2 + 2^2 + \dots + 14^2) \)
This is \( \sum_{k=1}^{28} k^2 - \sum_{k=1}^{14} k^2 \).
Part 1: \(\sum_{k=1}^{28} k^2\) (n=28)
\( = \frac{28(28+1)(2 \times 28+1)}{6} = \frac{28 \times 29 \times 57}{6} = 14 \times 29 \times 19 = 7714 \)
Part 2: \(\sum_{k=1}^{14} k^2\) (n=14)
\( = \frac{14(14+1)(2 \times 14+1)}{6} = \frac{14 \times 15 \times 29}{6} = 7 \times 5 \times 29 = 1015 \)
Final Sum:
\(7714 - 1015 = 6699\)
35. Solve: \(3x + y - 3z = 1, -2x - y + 2z = 1, -x - y + z = 2\). (Note: OCR had typos. Corrected based on standard format: 3x+y-3z=1, -2x-y+2z=1, -x-y+z=2)
Solution:
(1) \(3x + y - 3z = 1\)
(2) \(-2x - y + 2z = 1\)
(3) \(-x - y + z = 2\)
Step 1: Eliminate 'y'
Add (1) and (2):
\((3x+y-3z) + (-2x-y+2z) = 1+1 \implies x - z = 2\) --- (4)
Subtract (3) from (2):
\((-2x-y+2z) - (-x-y+z) = 1-2 \implies -x + z = -1\) --- (5)
Step 2: Solve for x and z
Add (4) and (5):
\((x-z) + (-x+z) = 2 + (-1) \implies 0 = 1\)
This is a contradiction, which means the system of equations has no solution. The lines represented by these equations are inconsistent.
36. Simplify: \( \frac{b^2+3b-28}{b^2+4b+4} \div \frac{b^2-49}{b^2-5b-14} \)
Solution:
First, factorize each polynomial:
- \(b^2+3b-28 = (b+7)(b-4)\)
- \(b^2+4b+4 = (b+2)^2 = (b+2)(b+2)\)
- \(b^2-49 = (b+7)(b-7)\)
- \(b^2-5b-14 = (b-7)(b+2)\)
Now, rewrite the expression with the factored forms:
\( \frac{(b+7)(b-4)}{(b+2)(b+2)} \div \frac{(b+7)(b-7)}{(b-7)(b+2)} \)
Change the division to multiplication by inverting the second fraction:
\( \frac{(b+7)(b-4)}{(b+2)(b+2)} \times \frac{(b-7)(b+2)}{(b+7)(b-7)} \)
Now, cancel the common factors: \((b+7)\), \((b-7)\), and one \((b+2)\).
\( \frac{\cancel{(b+7)}(b-4)}{(b+2)\cancel{(b+2)}} \times \frac{\cancel{(b-7)}\cancel{(b+2)}}{\cancel{(b+7)}\cancel{(b-7)}} \)
37. If \(ax^4 + bx^3 + 361x^2 + 220x + 100\) is a perfect square, find the values of a and b.
Solution:
We use the long division method for finding the square root, starting from the constant term since 'a' and 'b' are unknown.
38. State and prove Basic Proportionality theorem.
Solution:
Statement (Thales' Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In \(\triangle ABC\), a line DE is drawn parallel to BC, intersecting AB at D and AC at E.
To Prove: \( \frac{AD}{DB} = \frac{AE}{EC} \)
Construction: Join BE and CD. Draw DM \(\perp\) AC and EN \(\perp\) AB.
Proof:
Area(\(\triangle ADE\)) = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN \)
Area(\(\triangle BDE\)) = \( \frac{1}{2} \times DB \times EN \)
Dividing these, we get: \( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \) --- (1)
Similarly,
Area(\(\triangle ADE\)) = \( \frac{1}{2} \times AE \times DM \)
Area(\(\triangle CDE\)) = \( \frac{1}{2} \times EC \times DM \)
Dividing these, we get: \( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \) --- (2)
Now, \(\triangle BDE\) and \(\triangle CDE\) are on the same base DE and between the same parallel lines DE and BC. Therefore, Area(\(\triangle BDE\)) = Area(\(\triangle CDE\)).
From (1) and (2), using this fact, we can conclude that the left-hand sides are equal:
\( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} \)
Therefore, their right-hand sides must also be equal.
\( \frac{AD}{DB} = \frac{AE}{EC} \). Hence Proved.
39. Prove that \( \tan^2 A - \tan^2 B = \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \)
Solution:
LHS = \( \tan^2 A - \tan^2 B \)
Convert tan to sin/cos:
LHS = \( \frac{\sin^2 A}{\cos^2 A} - \frac{\sin^2 B}{\cos^2 B} \)
Take a common denominator, which is \(\cos^2 A \cos^2 B\):
LHS = \( \frac{\sin^2 A \cos^2 B - \cos^2 A \sin^2 B}{\cos^2 A \cos^2 B} \)
Now, we want the numerator to only have sine terms. Use the identity \(\cos^2 x = 1 - \sin^2 x\).
Numerator = \( \sin^2 A (1 - \sin^2 B) - (1 - \sin^2 A) \sin^2 B \)
\( = \sin^2 A - \sin^2 A \sin^2 B - (\sin^2 B - \sin^2 A \sin^2 B) \)
\( = \sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 A \sin^2 B \)
\( = \sin^2 A - \sin^2 B \)
Substitute this back into the expression:
LHS = \( \frac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \) = RHS. Hence proved.
40. If the vertices of a \(\triangle ABC\) are A(6,2), B(-5, -1) and C(1,9). i) Find the equation of median through A. ii) Find the equation of altitude through A.
Solution:
i) Equation of the median through A
The median from A connects A to the midpoint of the opposite side BC. Let's call the midpoint D.
Midpoint D of BC = \( \left( \frac{-5+1}{2}, \frac{-1+9}{2} \right) = \left( \frac{-4}{2}, \frac{8}{2} \right) = (-2, 4) \)
Now find the equation of the line AD passing through A(6,2) and D(-2,4).
Using the two-point form: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \)
\( \frac{y - 2}{4 - 2} = \frac{x - 6}{-2 - 6} \implies \frac{y - 2}{2} = \frac{x - 6}{-8} \)
\(-8(y-2) = 2(x-6) \implies -4(y-2) = x-6\)
\(-4y + 8 = x - 6 \implies x + 4y - 14 = 0\)
ii) Equation of the altitude through A
The altitude from A is perpendicular to the side BC. First, find the slope of BC.
Slope of BC (\(m_1\)) = \( \frac{9 - (-1)}{1 - (-5)} = \frac{10}{6} = \frac{5}{3} \)
The slope of the altitude AD (\(m_2\)) is the negative reciprocal of \(m_1\).
\(m_2 = -\frac{1}{m_1} = -\frac{3}{5}\)
Now find the equation of the line with slope \(-\frac{3}{5}\) passing through A(6,2) using the point-slope form: \(y - y_1 = m(x - x_1)\).
\(y - 2 = -\frac{3}{5}(x - 6)\)
\(5(y - 2) = -3(x - 6)\)
\(5y - 10 = -3x + 18\)
\(3x + 5y - 28 = 0\)
41. 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 \(2y = x + 3\). (Note: Typo 'z' in the first equation is assumed to be '2').
Solution:
Step 1: Find the first point of intersection (P1)
(1) \(3x + y + 2 = 0 \implies y = -3x - 2\)
(2) \(x - 2y - 4 = 0\)
Substitute (1) into (2): \(x - 2(-3x-2) - 4 = 0 \implies x + 6x + 4 - 4 = 0 \implies 7x = 0 \implies x = 0\).
Substitute x=0 into (1): \(y = -3(0) - 2 = -2\). So, 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 (4) into (3): \(7(2y - 3) - 3y = -12 \implies 14y - 21 - 3y = -12 \implies 11y = 9 \implies y = \frac{9}{11}\).
Substitute y back into (4): \(x = 2\left(\frac{9}{11}\right) - 3 = \frac{18}{11} - \frac{33}{11} = -\frac{15}{11}\). So, P2 = \((-\frac{15}{11}, \frac{9}{11})\).
Step 3: Find the equation of the line joining P1(0, -2) and P2\((-\frac{15}{11}, \frac{9}{11})\)
Using two-point form: \( \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \)
\( \frac{y - (-2)}{\frac{9}{11} - (-2)} = \frac{x - 0}{-\frac{15}{11} - 0} \)
\( \frac{y + 2}{\frac{9+22}{11}} = \frac{x}{-\frac{15}{11}} \implies \frac{y + 2}{31/11} = \frac{x}{-15/11} \)
\(-15(y + 2) = 31x\)
\(-15y - 30 = 31x\)
42. Find the value of k if the area of a quadrilateral is 72 sq. units, whose vertices are taken in the order A(-5, 7), B(-4, k), C(-1, -6) and D(4, 5).
Solution:
The area of a quadrilateral with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) \) is given by:
Area \( = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| \)
Substituting the coordinates:
\( 72 = \frac{1}{2} |((-5)(k) + (-4)(-6) + (-1)(5) + (4)(7)) - ((7)(-4) + (k)(-1) + (-6)(4) + (5)(-5))| \)
\( 144 = |(-5k + 24 - 5 + 28) - (-28 - k - 24 - 25)| \)
\( 144 = |(-5k + 47) - (-k - 77)| \)
\( 144 = |-5k + 47 + k + 77| \)
\( 144 = |-4k + 124| \)
This gives two possibilities:
Case 1: \( -4k + 124 = 144 \implies -4k = 20 \implies k = -5 \)
Case 2: \( -4k + 124 = -144 \implies -4k = -268 \implies k = 67 \)
Part - IV 2 x 8 = 16
Answer all the questions.
43. a) Construct a \(\triangle PQR\) in which PQ = 8cm, \(\angle R = 60^\circ\) and the median RG from R to PQ is 5.8 cm. Find the length of the altitude from R to PQ. (OR) b) Construct a triangle similar to a given triangle LMN with its sides equal to 4/5 of the corresponding sides of the triangle LMN (Scale factor 4/5 < 1).
Solution:
a) Construction of \(\triangle PQR\):
This problem requires geometric construction. The steps are outlined below:
- Draw a line segment PQ = 8 cm.
- At P, draw a line PX such that \(\angle QPX = 60^\circ\).
- Draw a line PY perpendicular to PX at P.
- Draw the perpendicular bisector of PQ, which intersects PQ at G and PY at O.
- With O as the center and OP as the radius, draw a circle. All points on the major arc of this circle will subtend an angle of 60° at PQ.
- With G as the center and radius 5.8 cm (length of the median), draw an arc to intersect the circle at R.
- Join PR and QR. \(\triangle PQR\) is the required triangle.
- To find the altitude, draw a line from R perpendicular to PQ, meeting at H. Measure the length of RH.
(OR)
b) Construction of a similar triangle:
This is a standard construction for similar triangles.
- Construct any triangle \(\triangle LMN\).
- Draw a ray LX from L on the side opposite to M, making an acute angle with LM.
- Mark 5 points (the greater of 4 and 5 in the ratio 4/5) \(L_1, L_2, L_3, L_4, L_5\) on LX such that \(LL_1 = L_1L_2 = \dots = L_4L_5\).
- Join \(L_5\) to M.
- From \(L_4\) (the smaller number in the ratio), draw a line parallel to \(L_5M\) to intersect LM at M'.
- From M', draw a line parallel to MN to intersect LN at N'.
- \(\triangle LM'N'\) is the required triangle similar to \(\triangle LMN\), with sides 4/5 of the corresponding sides.
44. a) Varshika drew 6 circles with different sizes. Draw a graph for the relationship between the diameter and circumference of each circle as shown in the table and use it to find the circumference of a circle when its diameter is 6cm.
(OR)
b) Graph the following linear function \( y = \frac{1}{2}x \). 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:
a) Diameter vs Circumference Graph:
The relationship between diameter (x) and circumference (y) is a direct variation, given by \(y = \pi x\). The table uses an approximation \(k \approx 3.1\).
- Choose a suitable scale for the x-axis (Diameter) and y-axis (Circumference). E.g., x-axis: 1 unit = 1 cm, y-axis: 1 unit = 2 cm.
- Plot the points from the table: (1, 3.1), (2, 6.2), (3, 9.3), (4, 12.4), (5, 15.5).
- Draw a straight line passing through the origin and these points.
- From the graph, locate x = 6 on the x-axis. Draw a vertical line up to the graphed line.
- From that point on the line, draw a horizontal line to the y-axis. The value on the y-axis is the required circumference.
Result from graph: When the diameter is 6 cm, the circumference is approximately 18.6 cm. (Also by calculation: \(y = 3.1 \times 6 = 18.6\))
(OR)
b) Graph of \( y = \frac{1}{2}x \):
This is a linear function representing direct variation, \( y = kx \).
Constant of variation: Comparing with \(y=kx\), the constant of variation is \( k = \frac{1}{2} \).
Graphing:
- Create a table of values:
- If x = 0, y = 0. Point (0,0)
- If x = 2, y = 1. Point (2,1)
- If x = 4, y = 2. Point (4,2)
- If x = -2, y = -1. Point (-2,-1)
- Plot these points on a graph and draw a straight line through them. The line will pass through the origin.
Verification: Take any point on the line, for example (4,2). Calculate \( k = y/x = 2/4 = 1/2 \). This matches the constant of variation. Hence, verified.
i) Find y when x = 9:
\( y = \frac{1}{2}(9) = 4.5 \). This can also be found from the graph.
ii) Find x when y = 7.5:
\( 7.5 = \frac{1}{2}x \implies x = 7.5 \times 2 = 15 \). This can also be found from the graph.