10th Maths Quarterly Exam 2024 - Solved Question Paper
COMMON QUARTERLY EXAMINATION - 2024
Standard: X
Subject: Mathematics
Marks: 100
Time: 3.00 hrs
Part - I (14 x 1 = 14)
I. Choose the correct answer:
1. If A = {a,b,p}, B = {2,3}, C = {p,q,r,s} then \(n[(A \cup C) \times B]\) is
Given, A = {a, b, p}, B = {2, 3}, C = {p, q, r, s}.
\(A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\}\).
\(n(A \cup C) = 6\).
\(n(B) = 2\).
\(n[(A \cup C) \times B] = n(A \cup C) \times n(B) = 6 \times 2 = 12\).
Answer: c) 12
2. \(f(x) = (x + 1)^3 – (x - 1)^3\) represents a function which is
\(f(x) = (x + 1)^3 – (x - 1)^3\)
Using \((a+b)^3 = a^3+3a^2b+3ab^2+b^3\) and \((a-b)^3 = a^3-3a^2b+3ab^2-b^3\).
\(f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)\)
\(f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1\)
\(f(x) = 6x^2 + 2\).
The highest power of x is 2, so it is a quadratic function.
Answer: d) quadratic
3. If n(A) = p, n(B) = q, then the total number of relations that exist from A to B is
The total number of relations from set A to set B is the number of possible subsets of the Cartesian product A x B.
n(A) = p, n(B) = q.
\(n(A \times B) = n(A) \times n(B) = pq\).
The total number of subsets of A x B is \(2^{n(A \times B)} = 2^{pq}\).
(Note: Option d in the paper seems to be a typo '2pq', it should be \(2^{pq}\)).
Answer: d) \(2^{pq}\)
4. The sum of the exponents of the prime factors in the prime factorisation of 1729 is
Prime factorisation of 1729:
1729 is divisible by 7: \(1729 = 7 \times 247\).
247 is divisible by 13: \(247 = 13 \times 19\).
So, \(1729 = 7^1 \times 13^1 \times 19^1\).
The exponents of the prime factors are 1, 1, and 1.
Sum of the exponents = 1 + 1 + 1 = 3.
Answer: c) 3
5. An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P is
Given, number of terms n = 31 (an odd number).
The middle term of the A.P is the \((\frac{n+1}{2})^{th}\) term.
Middle term = \(\frac{31+1}{2} = 16^{th}\) term.
It is given that the \(16^{th}\) term is m.
For an A.P with an odd number of terms, the sum of all terms is given by: Sum = n × (middle term).
Sum = 31 × m = 31m.
Answer: c) 31 m
6. The value of \((1^3 + 2^3 + 3^3 + \dots + 15^3) - (1 + 2 + 3 + \dots + 15)\) is
We use the formulas for the sum of the first n natural numbers and the sum of their cubes.
Sum of cubes: \(\Sigma n^3 = [\frac{n(n+1)}{2}]^2\).
Sum of natural numbers: \(\Sigma n = \frac{n(n+1)}{2}\).
Here, n = 15.
\(1^3 + 2^3 + \dots + 15^3 = [\frac{15(15+1)}{2}]^2 = [\frac{15 \times 16}{2}]^2 = (15 \times 8)^2 = 120^2 = 14400\).
\(1 + 2 + \dots + 15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120\).
The required value = 14400 - 120 = 14280.
Answer: c) 14280
7. \(\frac{3y-3}{y} \div \frac{7y-7}{3y^2}\) is
\(\frac{3y-3}{y} \div \frac{7y-7}{3y^2}\)
\( = [\frac{3(y-1)}{y}] \times [\frac{3y^2}{7(y-1)}] \)
Cancel out the common term (y-1).
\( = \frac{3}{y} \times \frac{3y^2}{7} \)
\( = \frac{3 \times 3y^2}{y \times 7} \)
\( = \frac{9y^2}{7y} \)
\( = \frac{9y}{7} \)
Answer: a) 9y/7
8. The square root of \(\frac{256x^8y^4z^{10}}{25x^6y^6z^6}\) is equal to
First, simplify the expression inside the square root:
\(\frac{256x^8y^4z^{10}}{25x^6y^6z^6} = \frac{256}{25} \cdot x^{8-6} \cdot y^{4-6} \cdot z^{10-6}\)
\( = \frac{256}{25} \cdot x^2 \cdot y^{-2} \cdot z^4 \)
\( = \frac{256x^2z^4}{25y^2} \)
Now, take the square root:
\(\sqrt{\frac{256x^2z^4}{25y^2}} = \frac{\sqrt{256} \cdot \sqrt{x^2} \cdot \sqrt{z^4}}{\sqrt{25} \cdot \sqrt{y^2}}\)
\( = \frac{16 |x| z^2}{5 |y|} \)
We use the modulus for generality.
\( = |\frac{16 xz^2}{5 y}| \)
Answer: d) \(|\frac{16 xz^2}{5 y}|\)
9. The solution of \((2x – 1)^2 = 9\) is equal to
\((2x - 1)^2 = 9\)
Take the square root on both sides:
\(2x - 1 = \pm\sqrt{9}\)
\(2x - 1 = \pm3\)
Case 1: \(2x - 1 = 3 \Rightarrow 2x = 4 \Rightarrow x = 2\).
Case 2: \(2x - 1 = -3 \Rightarrow 2x = -2 \Rightarrow x = -1\).
The solutions are -1 and 2.
Answer: c) -1, 2
10. If in \(\triangle ABC\), DE||BC, AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm then the length of AE is
Since DE || BC, by the Basic Proportionality Theorem (Thales' Theorem), we have:
\(\frac{AD}{AB} = \frac{AE}{AC}\)
Given: AD = 2.1 cm, AB = 3.6 cm, AC = 2.4 cm.
\(\frac{2.1}{3.6} = \frac{AE}{2.4}\)
\(AE = \frac{2.1 \times 2.4}{3.6}\)
\(AE = \frac{2.1 \times 24}{36}\)
\(AE = \frac{2.1 \times 2}{3}\)
\(AE = \frac{4.2}{3} = 1.4\) cm.
Answer: a) 1.4 cm
11. The point of intersection of 3x - y = 4 and x + y = 8 is
We have two equations:
1) \(3x - y = 4\)
2) \(x + y = 8\)
Add equation (1) and (2):
\((3x - y) + (x + y) = 4 + 8\)
\(4x = 12 \Rightarrow x = 3\).
Substitute x = 3 into equation (2):
\(3 + y = 8 \Rightarrow y = 5\).
The point of intersection is (3, 5).
Answer: c) (3,5)
12. When proving that a quadrilateral is a parallelogram by using slopes you must find
A quadrilateral is a parallelogram if and only if both pairs of opposite sides are parallel. Two lines are parallel if and only if their slopes are equal. Therefore, to prove a quadrilateral is a parallelogram using slopes, one must show that the slopes of two pairs of opposite sides are equal.
Answer: b) the slopes of two pair of opposite sides
13. The area of triangle formed by the points (-2,0), (0,-2) and (2,0) is
Using the formula for the area of a triangle with vertices \((x_1,y_1), (x_2,y_2), (x_3,y_3)\):
Area = \(\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|\)
Area = \(\frac{1}{2} |(-2)(-2-0) + 0(0-0) + 2(0-(-2))|\)
Area = \(\frac{1}{2} |(-2)(-2) + 0 + 2(2)|\)
Area = \(\frac{1}{2} |4 + 4| = \frac{1}{2} \times 8 = 4\) sq.units.
Answer: b) 4 sq.units
14. If \((\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = k + \tan^2\alpha + \cot^2\alpha\), then the value of k is equal to
Expand the Left Hand Side (LHS):
LHS = \((\sin^2\alpha + \csc^2\alpha + 2\sin\alpha\csc\alpha) + (\cos^2\alpha + \sec^2\alpha + 2\cos\alpha\sec\alpha)\)
Since \(\sin\alpha\csc\alpha = 1\) and \(\cos\alpha\sec\alpha = 1\):
LHS = \((\sin^2\alpha + \cos^2\alpha) + \csc^2\alpha + 2 + \sec^2\alpha + 2\)
Using identities \(\sin^2\alpha+\cos^2\alpha=1\), \(\csc^2\alpha=1+\cot^2\alpha\), \(\sec^2\alpha=1+\tan^2\alpha\):
LHS = \(1 + (1 + \cot^2\alpha) + 2 + (1 + \tan^2\alpha) + 2\)
LHS = \(1 + 1 + \cot^2\alpha + 2 + 1 + \tan^2\alpha + 2\)
LHS = \(7 + \tan^2\alpha + \cot^2\alpha\).
Comparing with RHS = \(k + \tan^2\alpha + \cot^2\alpha\), we get k = 7.
Answer: b) 7
Part - II (10 x 2 = 20)
II. Answer any 10 questions. (Q.No.28 is compulsory)
15. If A x B = {(3,2), (3,4), (5,2), (5,4)}, then find A and B.
The set A is the collection of all first elements in the ordered pairs of A x B.
A = {3, 5}.
The set B is the collection of all second elements in the ordered pairs of A x B.
B = {2, 4}.
16. Let X = {1,2,3,4} and Y = {2,4,6,8,10} and R = {(1,2), (2,4), (3,6), (4,8)}, show that R is a function and find its domain, co-domain and range.
To show R is a function from X to Y, we must verify that every element in X has exactly one image in Y.
- The image of 1 is 2.
- The image of 2 is 4.
- The image of 3 is 6.
- The image of 4 is 8.
Since each element in X has a unique image in Y, R is a function.
Domain: The set of all first elements in R. Domain = {1, 2, 3, 4} = X.
Co-domain: The entire set Y. Co-domain = {2, 4, 6, 8, 10}.
Range: The set of all second elements in R. Range = {2, 4, 6, 8}.
17. Find the \(8^{th}\) term of the G.P. 9, 3, 1, ...
The given Geometric Progression (G.P) is 9, 3, 1, ...
First term (a) = 9.
Common ratio (r) = \(T_2/T_1 = 3/9 = 1/3\).
The \(n^{th}\) term of a G.P is given by \(T_n = a \cdot r^{n-1}\).
The \(8^{th}\) term \(T_8 = 9 \cdot (1/3)^{8-1} = 9 \cdot (1/3)^7\)
\( = 3^2 \cdot (1/3^7) = 3^2 / 3^7 = 3^{2-7} = 3^{-5} = 1/3^5 = 1/243\).
The 8th term is 1/243.
18. Find the LCM of \(5x - 10, 5x^2 – 20\)
First, factorize each expression:
1) \(5x - 10 = 5(x - 2)\)
2) \(5x^2 - 20 = 5(x^2 - 4) = 5(x - 2)(x + 2)\)
The LCM is the product of the highest powers of all factors that appear in the expressions.
LCM = \(5 \times (x - 2) \times (x + 2) = 5(x^2 - 4) = 5x^2 - 20\).
LCM is 5(x-2)(x+2) or \(5x^2 - 20\).
19. Simplify: \( \frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4} \)
\( \frac{4x^2y}{2z^2} \times \frac{6xz^3}{20y^4} \)
\( = (\frac{4 \times 6}{2 \times 20}) \times (x^2 \times x) \times (\frac{y}{y^4}) \times (\frac{z^3}{z^2}) \)
\( = (\frac{24}{40}) \times x^{2+1} \times y^{1-4} \times z^{3-2} \)
\( = (\frac{3}{5}) \times x^3 \times y^{-3} \times z^1 \)
\( = \frac{3x^3z}{5y^3} \)
Simplified form: \( \frac{3x^3z}{5y^3} \)
20. In the figure, AD is the bisector of ∠A. If BD = 4 cm, DC = 3 cm and AB = 6 cm, find AC.
By the Angle Bisector Theorem, the bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides.
Therefore, \(\frac{AB}{AC} = \frac{BD}{DC}\).
Given: AB = 6 cm, BD = 4 cm, DC = 3 cm.
\(\frac{6}{AC} = \frac{4}{3}\)
\(4 \times AC = 6 \times 3\)
\(4 \times AC = 18\)
\(AC = 18 / 4 = 4.5\) cm.
The length of AC is 4.5 cm.
21. A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.
Solution:
The stick and the tower form similar triangles with their shadows, as the angle of elevation of the sun is the same for both.
Height of stick = 6 m.
Length of stick's shadow = 400 cm = 4 m.
Length of tower's shadow = 28 m.
Let the height of the tower be 'h' m.
By similarity, the ratio of corresponding sides is equal:
\(\frac{\text{Height of tower}}{\text{Height of stick}} = \frac{\text{Shadow of tower}}{\text{Shadow of stick}}\)
\(\frac{h}{6} = \frac{28}{4}\)
\(\frac{h}{6} = 7\)
\(h = 7 \times 6 = 42\) m.
The height of the tower is 42 m.
22. Show that the points P(-1.5, 3), Q(6, -2), R(-3, 4) are collinear.
Three points are collinear if the slope of the line joining the first two points is equal to the slope of the line joining the last two points.
Slope of PQ = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 3}{6 - (-1.5)} = \frac{-5}{7.5} = \frac{-50}{75} = -\frac{2}{3}\).
Slope of QR = \(\frac{y_3 - y_2}{x_3 - x_2} = \frac{4 - (-2)}{-3 - 6} = \frac{6}{-9} = -\frac{2}{3}\).
Since the slope of PQ is equal to the slope of QR, the points P, Q, and R are collinear.
23. Find the slope of a line joining the given points (-6,1) and (14,10).
The slope (m) of a line joining points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\)
Here, \((x_1, y_1) = (-6, 1)\) and \((x_2, y_2) = (14, 10)\).
\(m = \frac{10 - 1}{14 - (-6)} = \frac{9}{14 + 6} = \frac{9}{20}\).
The slope of the line is 9/20.
24. Show that the straight lines 2x + 3y - 8 = 0 and 4x + 6y + 18 = 0 are parallel.
Two lines ax + by + c = 0 and dx + ey + f = 0 are parallel if their slopes are equal. The slope of a line in this form is -a/b.
For the first line, 2x + 3y - 8 = 0:
Slope \(m_1 = -\frac{\text{coefficient of x}}{\text{coefficient of y}} = -\frac{2}{3}\).
For the second line, 4x + 6y + 18 = 0:
Slope \(m_2 = -\frac{4}{6} = -\frac{2}{3}\).
Since \(m_1 = m_2\), the two straight lines are parallel.
25. Given the function f : \(x \rightarrow x^2 – 5x + 6\), evaluate i) f(-1) and ii) f(2).
The function is \(f(x) = x^2 – 5x + 6\).
i) f(-1):
\(f(-1) = (-1)^2 - 5(-1) + 6 = 1 + 5 + 6 = 12\).
ii) f(2):
\(f(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0\).
f(-1) = 12 and f(2) = 0.
26. If \(13824 = 2^a \times 3^b\), then find a and b.
We need to find the prime factorization of 13824.
\(13824 = 2 \times 6912\)
\( = 2^2 \times 3456\)
\( = 2^3 \times 1728\)
\( = 2^3 \times 12^3\)
\( = 2^3 \times (4 \times 3)^3 = 2^3 \times (2^2 \times 3)^3\)
\( = 2^3 \times (2^2)^3 \times 3^3 = 2^3 \times 2^6 \times 3^3\)
\( = 2^{3+6} \times 3^3 = 2^9 \times 3^3\).
Comparing with \(2^a \times 3^b\), we get a = 9 and b = 3.
a = 9, b = 3.
27. Prove that \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = \csc\theta + \cot\theta\)
LHS = \(\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}\)
Multiply numerator and denominator inside the root by \((1+\cos\theta)\):
\( = \sqrt{\frac{(1+\cos\theta)(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}} \)
\( = \sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}} \)
Using the identity \(\sin^2\theta + \cos^2\theta = 1\), we have \(1 - \cos^2\theta = \sin^2\theta\).
\( = \sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}} \)
\( = \frac{1+\cos\theta}{\sin\theta} \)
\( = \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} \)
\( = \csc\theta + \cot\theta\) = RHS.
Hence Proved.
28. Find the sum of 1 + 3 + 5 + ... + 51. (Compulsory Question)
This is an Arithmetic Progression (A.P.) of odd numbers.
First term (a) = 1.
Common difference (d) = 3 - 1 = 2.
Last term (l) = 51.
First, find the number of terms (n):
\(l = a + (n-1)d\)
\(51 = 1 + (n-1)2\)
\(50 = (n-1)2\)
\(25 = n - 1 \Rightarrow n = 26\).
Now, find the sum (\(S_n\)):
\(S_n = \frac{n}{2}(a + l)\)
\(S_{26} = \frac{26}{2}(1 + 51) = 13 \times 52 = 676\).
The sum is 676.
Part - III (10 x 5 = 50)
III. Answer any 10 questions. (Q.No.42 is compulsory)
29. Let f : A→B be a function defined by \(f(x) = x/2 - 1\) where A = {2,4,6,10,12}, B = {0,1,2,4,5,9}. Represent f by i) Set of ordered pairs ii) A table iii) An arrow diagram iv) A graph
Given \(f(x) = \frac{x}{2} - 1\). We find the image for each element in A:
f(2) = 2/2 - 1 = 1 - 1 = 0
f(4) = 4/2 - 1 = 2 - 1 = 1
f(6) = 6/2 - 1 = 3 - 1 = 2
f(10) = 10/2 - 1 = 5 - 1 = 4
f(12) = 12/2 - 1 = 6 - 1 = 5
i) Set of ordered pairs:
f = {(2,0), (4,1), (6,2), (10,4), (12,5)}
| x | f(x) |
|---|---|
| 2 | 0 |
| 4 | 1 |
| 6 | 2 |
| 10 | 4 |
| 12 | 5 |
(Draw two ovals for sets A and B. Write elements. Draw arrows from 2 to 0, 4 to 1, 6 to 2, 10 to 4, 12 to 5).
iv) A graph:
(Plot the points (2,0), (4,1), (6,2), (10,4), (12,5) on a graph paper).
30. Find x if gff(x) = fgg(x), given f(x) = 3x + 1 and g(x) = x + 3.
Given f(x) = 3x + 1 and g(x) = x + 3.
First, find gff(x):
f(x) = 3x + 1
ff(x) = f(f(x)) = f(3x+1) = 3(3x+1) + 1 = 9x + 3 + 1 = 9x + 4.
gff(x) = g(ff(x)) = g(9x+4) = (9x+4) + 3 = 9x + 7.
Next, find fgg(x):
g(x) = x + 3
gg(x) = g(g(x)) = g(x+3) = (x+3) + 3 = x + 6.
fgg(x) = f(gg(x)) = f(x+6) = 3(x+6) + 1 = 3x + 18 + 1 = 3x + 19.
Given gff(x) = fgg(x):
9x + 7 = 3x + 19
9x - 3x = 19 - 7
6x = 12
x = 2.
The value of x is 2.
31. If \(p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} = 113400\) where \(p_1, p_2, p_3, p_4\) are primes in ascending order and \(x_1, x_2, x_3, x_4\) are integers, find the values of p's and x's.
We perform prime factorization of 113400.
\(113400 = 1134 \times 100 = 1134 \times 10^2\)
\( = 1134 \times (2 \times 5)^2 = 1134 \times 2^2 \times 5^2\)
Now factorize 1134:
\(1134 = 2 \times 567 = 2 \times 9 \times 63 = 2 \times 3^2 \times 9 \times 7 = 2 \times 3^2 \times 3^2 \times 7 = 2^1 \times 3^4 \times 7^1\).
So, \(113400 = (2^1 \times 3^4 \times 7^1) \times (2^2 \times 5^2) = 2^{1+2} \times 3^4 \times 5^2 \times 7^1 = 2^3 \times 3^4 \times 5^2 \times 7^1\).
The primes in ascending order are \(p_1=2, p_2=3, p_3=5, p_4=7\).
The corresponding exponents are \(x_1=3, x_2=4, x_3=2, x_4=1\).
\(p_1=2, p_2=3, p_3=5, p_4=7\) and \(x_1=3, x_2=4, x_3=2, x_4=1\).
32. Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these colour papers?
The sides of the square papers are 10 cm, 11 cm, ..., 24 cm.
The areas of the squares are \(10^2, 11^2, \dots, 24^2\).
Total area = \(10^2 + 11^2 + \dots + 24^2\).
This can be calculated as \((1^2 + 2^2 + \dots + 24^2) - (1^2 + 2^2 + \dots + 9^2)\).
Using the formula for the sum of squares of first n natural numbers, \(S_n = \frac{n(n+1)(2n+1)}{6}\).
Sum of squares up to 24: \(S_{24} = \frac{24(25)(49)}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900\).
Sum of squares up to 9: \(S_9 = \frac{9(10)(19)}{6} = 3 \times 5 \times 19 = 285\).
Total area = 4900 - 285 = 4615 cm².
The total area that can be decorated is 4615 cm².
33. In an A.P, sum of 4 consecutive terms is 28 and the sum of their squares is 276. Find the four numbers.
Let the four consecutive terms of the A.P. be (a - 3d), (a - d), (a + d), (a + 3d).
Sum of the terms = (a-3d) + (a-d) + (a+d) + (a+3d) = 28.
4a = 28 \(\Rightarrow\) a = 7.
Sum of their squares = \((a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 276\).
\((a^2-6ad+9d^2) + (a^2-2ad+d^2) + (a^2+2ad+d^2) + (a^2+6ad+9d^2) = 276\).
\(4a^2 + 20d^2 = 276\).
Substitute a = 7:
\(4(7)^2 + 20d^2 = 276 \Rightarrow 4(49) + 20d^2 = 276 \Rightarrow 196 + 20d^2 = 276\).
\(20d^2 = 276 - 196 = 80\).
\(d^2 = 4 \Rightarrow d = \pm2\).
If d = 2, the terms are (7-6), (7-2), (7+2), (7+6) \(\Rightarrow\) 1, 5, 9, 13.
If d = -2, the terms are (7+6), (7+2), (7-2), (7-6) \(\Rightarrow\) 13, 9, 5, 1.
The four numbers are 1, 5, 9, 13.
34. If \(9x^4 + 12x^3 + 28x^2 + ax + b\) 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 \(9x^4\) is \(3x^2\). So, the first term of the root is \(3x^2\).
\((3x^2)^2 = 9x^4\). Subtracting this leaves \(12x^3 + 28x^2 + ax + b\).
New divisor is \(2(3x^2) = 6x^2\). Divide \(12x^3\) by \(6x^2\) to get \(2x\). This is the second term of the root.
New divisor is \(6x^2 + 2x\). Multiply by \(2x\): \((6x^2+2x)(2x) = 12x^3 + 4x^2\).
Subtract this from \(12x^3 + 28x^2\): \((12x^3+28x^2) - (12x^3+4x^2) = 24x^2\).
Bring down ax+b. We have \(24x^2 + ax + b\).
New divisor is \(2(3x^2+2x) = 6x^2 + 4x\). Divide \(24x^2\) by \(6x^2\) to get 4. This is the third term.
New divisor is \(6x^2+4x+4\). Multiply by 4: \((6x^2+4x+4)(4) = 24x^2 + 16x + 16\).
For the polynomial to be a perfect square, the remainder must be zero.
\((24x^2 + ax + b) - (24x^2 + 16x + 16) = 0\).
\((a - 16)x + (b - 16) = 0\).
Comparing coefficients, a - 16 = 0 \(\Rightarrow\) a = 16, and b - 16 = 0 \(\Rightarrow\) b = 16.
a = 16 and b = 16.
35. Solve the equation: \(\frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}\)
Given the equation: \(\frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}\)
Rearrange the terms to group variables with x on one side.
\(\frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b}\)
Find a common denominator for each side.
LHS: \(\frac{x - (a+b+x)}{x(a+b+x)} = \frac{x - a - b - x}{x(a+b+x)} = \frac{-(a+b)}{x(a+b+x)}\)
RHS: \(\frac{b+a}{ab} = \frac{a+b}{ab}\)
Now equate the simplified sides:
\(\frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab}\)
Assuming \(a+b \neq 0\), we can divide both sides by (a+b):
\(\frac{-1}{x(a+b+x)} = \frac{1}{ab}\)
Cross-multiply:
\(-ab = x(a+b+x)\)
\(-ab = ax + bx + x^2\)
Rearrange into a standard quadratic equation form: \(x^2 + ax + bx + ab = 0\)
Factor by grouping:
\(x(x+a) + b(x+a) = 0\)
\((x+a)(x+b) = 0\)
This gives two possible solutions for x.
The solutions are x = -a and x = -b.
36. State and prove Angle Bisector Theorem.
Statement: The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.
Given: In \(\triangle ABC\), AD is the internal bisector of \(\angle A\) which meets the side BC at D.
To Prove: \(\frac{AB}{AC} = \frac{BD}{DC}\)
Construction: Draw a line through C parallel to AB. Extend AD to meet this line at E.
Proof:
In \(\triangle ABC\), since CE is parallel to AB (by construction), we can consider AE and BC as transversals.
1. \(\angle BAE = \angle AEC\) (Alternate interior angles, since AB || CE and AE is transversal).
2. \(\angle BAD = \angle CAD\) (Given, as AD is the angle bisector of \(\angle A\)).
From (1) and (2), we get \(\angle CAD = \angle AEC\).
3. \(\angle DAC = \angle ACE\) (This is incorrect. The correct relation is \(\angle PAB = \angle ACE\) which is \(\angle BAD = \angle AEC\)... Wait let's use corresponding angles).
Let's re-state the proof using a different standard construction for clarity.
Alternative Standard Proof:
Construction: Through C, draw a line CE parallel to DA, meeting the line BA extended at E.
Proof:
Since DA || CE, by Basic Proportionality Theorem (Thales' Theorem) on \(\triangle BCE\), we have:
\(\frac{BD}{DC} = \frac{BA}{AE}\) --- (1)
Now, since DA || CE and AC is a transversal:
\(\angle DAC = \angle ACE\) (Alternate interior angles) --- (2)
And since DA || CE and BE is a transversal:
\(\angle BAD = \angle AEC\) (Corresponding angles) --- (3)
It is given that AD is the angle bisector, so:
\(\angle BAD = \angle DAC\) --- (4)
From equations (2), (3), and (4), we get:
\(\angle ACE = \angle AEC\)
In \(\triangle ACE\), since the angles opposite to sides AE and AC are equal, the sides themselves are equal.
Therefore, AE = AC --- (5)
Now, substitute (5) into (1):
\(\frac{BD}{DC} = \frac{BA}{AC}\)
Hence, \(\frac{AB}{AC} = \frac{BD}{DC}\). The theorem is proved.
37. Find the area of the quadrilateral whose vertices are at (-9,-2), (-8,-4), (2,2) and (1,-3).
Let the vertices be A(-9,-2), B(-8,-4), C(2,2) and D(1,-3). To find the area of the quadrilateral, we must take the vertices in counter-clockwise order. A rough sketch shows the order should be A(-9, -2), B(-8, -4), D(1, -3), and C(2, 2).
Let the vertices be \((x_1, y_1) = (-9,-2)\), \((x_2, y_2) = (-8,-4)\), \((x_3, y_3) = (1,-3)\), \((x_4, y_4) = (2,2)\).
The formula for the area of a quadrilateral is:
Area = \(\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|\)
First part: \(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1\)
= \((-9)(-4) + (-8)(-3) + (1)(2) + (2)(-2)\)
= \(36 + 24 + 2 - 4\)
= 58
Second part: \(y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1\)
= \((-2)(-8) + (-4)(1) + (-3)(2) + (2)(-9)\)
= \(16 - 4 - 6 - 18\)
= \(16 - 28 = -12\)
Area = \(\frac{1}{2} |58 - (-12)|\)
= \(\frac{1}{2} |58 + 12|\)
= \(\frac{1}{2} |70| = 35\)
The area of the quadrilateral is 35 sq. units.
38. Find the equation of a straight line passing through (1,4) and has intercepts which are in the ratio 2:5.
Let the equation of the line in intercept form be \(\frac{x}{a} + \frac{y}{b} = 1\), where 'a' is the x-intercept and 'b' is the y-intercept.
Given that the intercepts are in the ratio 2:5. This means \(\frac{a}{b} = \frac{2}{5}\).
We can write the intercepts in terms of a constant k: \(a = 2k\) and \(b = 5k\).
Substitute these into the intercept form equation:
\(\frac{x}{2k} + \frac{y}{5k} = 1\)
The line passes through the point (1, 4). Substitute x = 1 and y = 4 into the equation to find k:
\(\frac{1}{2k} + \frac{4}{5k} = 1\)
To solve for k, find a common denominator (10k):
\(\frac{5(1) + 2(4)}{10k} = 1\)
\(\frac{5 + 8}{10k} = 1\)
\(\frac{13}{10k} = 1 \implies 13 = 10k \implies k = \frac{13}{10}\)
Now, substitute the value of k back into the equation of the line:
\(\frac{x}{2(\frac{13}{10})} + \frac{y}{5(\frac{13}{10})} = 1\)
\(\frac{x}{\frac{13}{5}} + \frac{y}{\frac{13}{2}} = 1\)
\(\frac{5x}{13} + \frac{2y}{13} = 1\)
Multiply the entire equation by 13 to clear the denominator:
\(5x + 2y = 13\)
The equation of the straight line is 5x + 2y - 13 = 0.
39. Let A(3,-4), B(9,-4), C(5,-7) and D(7,-7). Show that ABCD is a trapezium.
A trapezium is a quadrilateral with at least one pair of parallel sides. We find the slopes of all four sides.
Slope of AB = \(\frac{-4 - (-4)}{9 - 3} = \frac{0}{6} = 0\).
Slope of BC = \(\frac{-7 - (-4)}{5 - 9} = \frac{-3}{-4} = \frac{3}{4}\).
Slope of CD = \(\frac{-7 - (-7)}{7 - 5} = \frac{0}{2} = 0\).
Slope of DA = \(\frac{-4 - (-7)}{3 - 7} = \frac{3}{-4} = -\frac{3}{4}\).
Since Slope(AB) = Slope(CD) = 0, the side AB is parallel to the side CD.
Since Slope(BC) ≠ Slope(DA), the sides BC and DA are not parallel.
Because ABCD has exactly one pair of parallel sides, it is a trapezium.
40. Simplify: \(\frac{b^2+3b-28}{b^2+4b+4} \div \frac{b^2-49}{b^2-5b-14}\)
First, factorize all the quadratic expressions.
\(b^2 + 3b - 28 = (b+7)(b-4)\)
\(b^2 + 4b + 4 = (b+2)^2\)
\(b^2 - 49 = (b+7)(b-7)\)
\(b^2 - 5b - 14 = (b-7)(b+2)\)
The expression becomes:
\(\frac{(b+7)(b-4)}{(b+2)^2} \div \frac{(b+7)(b-7)}{(b-7)(b+2)}\)
Invert the second fraction and multiply:
\( = \frac{(b+7)(b-4)}{(b+2)^2} \times \frac{(b-7)(b+2)}{(b+7)(b-7)}\)
Cancel out common factors: (b+7), (b-7), and one (b+2).
\( = \frac{b-4}{b+2} \).
The simplified expression is \(\frac{b-4}{b+2}\).
41. Prove the following identity: \(\frac{\sin^3A + \cos^3A}{\sin A + \cos A} + \frac{\sin^3A - \cos^3A}{\sin A - \cos A} = 2\)
We use the sum and difference of cubes formulas:
\(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)
\(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)
For the first term:
\(\frac{\sin^3A + \cos^3A}{\sin A + \cos A} = \frac{(\sin A + \cos A)(\sin^2A - \sin A\cos A + \cos^2A)}{\sin A + \cos A}\)
\( = \sin^2A - \sin A\cos A + \cos^2A\). Since \(\sin^2A + \cos^2A = 1\), this simplifies to \(1 - \sin A\cos A\).
For the second term:
\(\frac{\sin^3A - \cos^3A}{\sin A - \cos A} = \frac{(\sin A - \cos A)(\sin^2A + \sin A\cos A + \cos^2A)}{\sin A - \cos A}\)
\( = \sin^2A + \sin A\cos A + \cos^2A\). This simplifies to \(1 + \sin A\cos A\).
Adding the two simplified terms:
LHS = \((1 - \sin A\cos A) + (1 + \sin A\cos A) = 1 + 1 = 2\) = RHS.
Hence Proved.
42. Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime numbers. Verify that \(A \times (B - C) = (A \times B) - (A \times C)\). (Compulsory Question)
First, define the sets:
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
Calculate the LHS: \(A \times (B - C)\)
B - C = {2, 3, 5, 7} - {2} = {3, 5, 7}.
\(A \times (B - C) = \{1, 2, 3, 4, 5, 6, 7\} \times \{3, 5, 7\}\)
= {(1,3), (1,5), (1,7), (2,3), (2,5), (2,7), (3,3), (3,5), (3,7), (4,3), (4,5), (4,7), (5,3), (5,5), (5,7), (6,3), (6,5), (6,7), (7,3), (7,5), (7,7)}.
Calculate the RHS: \((A \times B) - (A \times C)\)
\(A \times B = \{(a, b) | a \in A, b \in B\}\). This has 7x4 = 28 elements.
\(A \times C = \{(a, c) | a \in A, c \in C\} = \{(a, 2) | a \in A\}\)
= {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)}.
\((A \times B) - (A \times C)\) is the set of all elements in \(A \times B\) that are not in \(A \times C\). This means we remove all pairs from \(A \times B\) where the second element is 2.
This leaves us with pairs (a, b) where \(a \in A\) and \(b \in \{3, 5, 7\}\), which is exactly \(A \times (B - C)\).
LHS = RHS. Hence, verified.
Part - IV (2 x 8 = 16)
IV. Answer all the questions.
43. a) Construct a triangle similar to a given triangle PQR with its sides equal to 7/3 of the corresponding sides of the triangle PQR (Scale factor 7/3 > 1).
Solution:
Steps of Construction:
1. Construct any triangle PQR.
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 7 points (the greater of 7 and 3 in 7/3) Q₁, Q₂, ..., Q₇ on QX so that QQ₁ = Q₁Q₂ = ... = Q₆Q₇.
4. Join Q₃ (the 3rd point, 3 being the denominator of 7/3) to R.
5. Draw a line through Q₇ parallel to Q₃R to intersect the extended line segment QR at R'.
6. Draw a line through R' parallel to PR to intersect the extended line segment QP at P'.
7. Then, ΔP'QR' is the required triangle whose sides are 7/3 of 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.
Solution:
Steps of Construction:
1. Draw a line segment QR = 5 cm.
2. At Q, draw a line QE such that ∠RQE = 30°.
3. At Q, draw a line QF perpendicular to QE.
4. Draw the perpendicular bisector of QR, which intersects QF at O and QR at M.
5. With O as center and OQ as radius, draw a circle. The points on the major arc of this circle will subtend an angle of 30° at the circumference.
6. Draw a line GH parallel to QR at a distance of 4.2 cm.
7. This line GH intersects the circle at two points, P and P'.
8. Join PQ and PR (or P'Q and P'R).
9. 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.
Let time be x (in hours) and distance be y (in km).
The relationship is y = 50x. This is a direct variation.
i) The constant of variation (k) is 50.
Table of values:
| Time x (hr) | 60 | 120 | 180 | 240 |
|---|---|---|---|---|
| Distance y (km) | 50 | 100 | 150 | 200 |
ii) How far will it travel in 90 minutes?90 minutes = 1.5 hours. From the graph, draw a vertical line from x= 90 minutes or 1.5 hours to the graph line, then a horizontal line to the y-axis. The value on the y-axis will be 75.
Answer: 75 km. iii) The time required to cover a distance of 300 km from the graph.
From the graph, draw a horizontal line from y=300 to the graph line, then a vertical line to the x-axis. The value on the x-axis will be 360 minutes or 6 hours..
Answer: 6 hours.
44. b) (OR) A school announces that for a certain competition, the cash prize will be distributed for all the participants equally as shown below.
i) Find the constant of variation
ii) Graph the above data and hence, find how much will each participant get if the number of participants are 12.
Let X be the number of participants and y be the amount for each participant.
Let's check the product xy for the given data:
2 × 180 = 360
4 × 90 = 360
6 × 60 = 360
8 × 45 = 360
10 × 36 = 360
Since xy is constant, this is an inverse variation.
i) The constant of variation (k) is xy = 360.
ii) Graph and solution for 12 participants:
Graph: Plot the points (2, 180), (4, 90), (6, 60), (8, 45), (10, 36) on a graph. Join them with a smooth curve (a hyperbola).
To find the amount for 12 participants, we can use the equation xy = 360.
12 × y = 360
y = 360 / 12 = 30.
From the graph, locate 12 on the X-axis, move up to the curve, and then move left to the Y-axis. The corresponding value will be 30.
Answer: If there are 12 participants, each will get Rs. 30.