10th Maths Quarterly Exam 2024 - Solutions
Part - I
1. A = {a,b,p}, B = {2,3}, C = {p,q,r,s} then n[(A ∪ C) x B] is
Solution:
Given, A = {a,b,p}, B = {2,3}, C = {p,q,r,s}.
First, find A ∪ C:
A ∪ C = {a,b,p} ∪ {p,q,r,s} = {a,b,p,q,r,s}
Now, find the number of elements in A ∪ C and B:
n(A ∪ C) = 6
n(B) = 2
Then, n[(A ∪ C) x B] = n(A ∪ C) × n(B) = 6 × 2 = 12.
Answer: c) 12
2. If $f(x) = 2x^2$ and $g(x) = \frac{1}{3x}$, then fog is
Solution:
Given $f(x) = 2x^2$ and $g(x) = \frac{1}{3x}$.
fog(x) = f(g(x))
Substitute g(x) into f(x):
$$ f(g(x)) = f\left(\frac{1}{3x}\right) = 2\left(\frac{1}{3x}\right)^2 = 2\left(\frac{1}{9x^2}\right) = \frac{2}{9x^2} $$Answer: c) $\frac{2}{9x^2}$
3. A function f: R→R defined by $f(x) = ax^2 + bx + c, (a \neq 0)$ is called a
Solution:
The expression $ax^2 + bx + c$ with $a \neq 0$ is the standard form of a quadratic polynomial. A function defined by such a polynomial is called a quadratic function.
Answer: d) quadratic function
4. $7^{4k} \equiv$ _____ (mod 100)
Solution:
(Note: The question likely has a typo and should be $7^{4k}$)
Let's find the value of $7^4$ (mod 100):
$7^1 = 7 \equiv 7 \pmod{100}$
$7^2 = 49 \equiv 49 \pmod{100}$
$7^3 = 49 \times 7 = 343 \equiv 43 \pmod{100}$
$7^4 = 43 \times 7 = 301 \equiv 1 \pmod{100}$
Now, for $7^{4k}$:
$$ 7^{4k} = (7^4)^k \equiv (1)^k \equiv 1 \pmod{100} $$Answer: a) 1
5. The sum of first n natural numbers are also called
Solution:
The sum of the first n natural numbers, given by the formula $\frac{n(n+1)}{2}$, generates a sequence of numbers known as triangular numbers.
For n=1, Sum = 1
For n=2, Sum = 1+2 = 3
For n=3, Sum = 1+2+3 = 6
The sequence 1, 3, 6, 10, ... are the triangular numbers.
Answer: c) Triangular numbers
6. The value of $(1^3 + 2^3 + 3^3 + ... + 15^3) - (1 + 2 + 3 + ... + 15)$ 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 natural numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
Here, n = 15.
$1^3 + ... + 15^3 = \left(\frac{15(15+1)}{2}\right)^2 = \left(\frac{15 \times 16}{2}\right)^2 = (15 \times 8)^2 = 120^2 = 14400$.
$1 + ... + 15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120$.
The required value is $14400 - 120 = 14280$.
Answer: c) 14280
7. $\frac{3y-3}{y} \div \frac{7y-7}{3y^2}$ is
(Note: The operator is interpreted as division based on the options.)
Solution:
To divide by a fraction, we multiply by its reciprocal.
$$ \frac{3y-3}{y} \div \frac{7y-7}{3y^2} = \frac{3y-3}{y} \times \frac{3y^2}{7y-7} $$Factor out common terms:
$$ = \frac{3(y-1)}{y} \times \frac{3y^2}{7(y-1)} $$Cancel the $(y-1)$ terms and simplify:
$$ = \frac{3}{y} \times \frac{3y^2}{7} = \frac{9y^2}{7y} = \frac{9y}{7} $$Answer: a) $\frac{9y}{7}$
8. Graph of a linear equation is a ____
Solution:
A linear equation in two variables, of the form $ax + by + c = 0$, always represents a straight line when plotted on a Cartesian coordinate system.
Answer: 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:
$$ \frac{256 x^8y^4z^{10}}{25 x^6y^6z^6} = \frac{256}{25} x^{8-6} y^{4-6} z^{10-6} = \frac{256}{25} x^2 y^{-2} z^4 = \frac{256x^2z^4}{25y^2} $$Now, take the square root:
$$ \sqrt{\frac{256x^2z^4}{25y^2}} = \frac{\sqrt{256}\sqrt{x^2}\sqrt{z^4}}{\sqrt{25}\sqrt{y^2}} = \frac{16|x|z^2}{5|y|} $$Assuming x and y are positive, this simplifies to:
$$ \frac{16xz^2}{5y} $$Answer: d) $\frac{16}{5} \frac{xz^2}{y}$
10. If ΔABC is an isosceles triangle with ∠C = 90° and AC = 5cm, then AB is
Solution:
Since ΔABC is an isosceles triangle with ∠C = 90°, the sides adjacent to the right angle must be equal. Therefore, AC = BC = 5 cm.
By the Pythagorean theorem:
$$ AB^2 = AC^2 + BC^2 $$ $$ AB^2 = 5^2 + 5^2 = 25 + 25 = 50 $$ $$ AB = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \text{ cm} $$Answer: 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 of a triangle divides the opposite side in the ratio of the other two sides.
$$ \frac{AB}{AC} = \frac{BD}{DC} $$Substitute the given values:
$$ \frac{8}{AC} = \frac{6}{3} $$ $$ \frac{8}{AC} = 2 $$ $$ AC = \frac{8}{2} = 4 \text{ cm} $$Answer: b) 4 cm
12. The area of triangle formed by the points (-5,0), (0,-5) and (5,0) is
Solution:
Let the vertices be A(-5,0), B(0,-5), and C(5,0).
Area of triangle = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$$ \text{Area} = \frac{1}{2} |-5(-5 - 0) + 0(0 - 0) + 5(0 - (-5))| $$ $$ = \frac{1}{2} |-5(-5) + 0 + 5(5)| $$ $$ = \frac{1}{2} |25 + 25| = \frac{1}{2} \times 50 = 25 \text{ sq.units} $$Answer: 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 points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Given points are (12,3) and (4,a), and the slope $m = \frac{1}{8}$.
$$ \frac{1}{8} = \frac{a - 3}{4 - 12} $$ $$ \frac{1}{8} = \frac{a - 3}{-8} $$Cross-multiply:
$$ -8 = 8(a - 3) $$ $$ -1 = a - 3 $$ $$ a = 3 - 1 = 2 $$Answer: d) 2
14. $\tan\theta \csc^2\theta - \tan\theta$ is equal to
Solution:
Factor out $\tan\theta$ from the expression:
$$ \tan\theta (\csc^2\theta - 1) $$Using the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$, we get $\csc^2\theta - 1 = \cot^2\theta$.
Substitute this back into the expression:
$$ \tan\theta \times \cot^2\theta $$Since $\cot\theta = \frac{1}{\tan\theta}$:
$$ \tan\theta \times (\cot\theta \times \cot\theta) = (\tan\theta \times \cot\theta) \times \cot\theta = 1 \times \cot\theta = \cot\theta $$Answer: d) $\cot\theta$
Part - II
Answer any 10 questions. (Q.No.28 is compulsory)
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 is defined by y = x + 3 for x in {0,1,2,3,4,5}.
- When x=0, y = 0+3 = 3
- When x=1, y = 1+3 = 4
- When x=2, y = 2+3 = 5
- When x=3, y = 3+3 = 6
- When x=4, y = 4+3 = 7
- When x=5, y = 5+3 = 8
The set of ordered pairs is R = {(0,3), (1,4), (2,5), (3,6), (4,7), (5,8)}.
The domain is the set of all first elements (x-values): Domain = {0, 1, 2, 3, 4, 5}.
The range is the set of all second elements (y-values): Range = {3, 4, 5, 6, 7, 8}.
16. Given the function $f : x \to x^2 - 5x + 6$, evaluate i) f(-1) ii) f(2a)
Solution:
The function is $f(x) = x^2 - 5x + 6$.
i) f(-1)
$$ f(-1) = (-1)^2 - 5(-1) + 6 = 1 + 5 + 6 = 12 $$f(-1) = 12
ii) f(2a)
$$ f(2a) = (2a)^2 - 5(2a) + 6 = 4a^2 - 10a + 6 $$f(2a) = 4a² - 10a + 6
17. Find k if fof(k) = 5 where f(k) = 2k - 1
Solution:
Given $f(k) = 2k - 1$.
First, find the composition f(f(k)):
$$ fof(k) = f(f(k)) = f(2k-1) $$Substitute $(2k-1)$ into the function f:
$$ = 2(2k-1) - 1 = 4k - 2 - 1 = 4k - 3 $$We are given that fof(k) = 5:
$$ 4k - 3 = 5 $$ $$ 4k = 8 $$ $$ k = 2 $$The value of k is 2.
18. Find the HCF of 252525 and 363636.
Solution:
We can factor out 10101 from both numbers:
$252525 = 25 \times 10101$
$363636 = 36 \times 10101$
The HCF of the two numbers is the product of the HCF of (25, 36) and the HCF of (10101, 10101).
HCF(25, 36):
Factors of 25 are 1, 5, 25.
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
The only common factor is 1. So, HCF(25, 36) = 1.
HCF(10101, 10101) = 10101.
Therefore, HCF(252525, 363636) = HCF(25, 36) × 10101 = 1 × 10101 = 10101.
The HCF is 10101.
19. What is the time 15 hours before 11 p.m?
Solution:
11 p.m. in 24-hour format is 23:00.
We need to find the time 15 hours before 23:00.
Time = 23 - 15 = 8.
So, the time is 8:00 in the 24-hour format, which is 8 a.m.
The time is 8 a.m.
20. Find the sum $3+1+\frac{1}{3}+... \infty$
Solution:
This is an infinite geometric progression (GP).
First term, $a = 3$.
Common ratio, $r = \frac{1}{3}$.
Since $|r| = |\frac{1}{3}| < 1$, the sum to infinity exists.
The formula for the sum to infinity of a GP is $S_\infty = \frac{a}{1-r}$.
$$ S_\infty = \frac{3}{1 - \frac{1}{3}} = \frac{3}{\frac{2}{3}} = 3 \times \frac{3}{2} = \frac{9}{2} $$The sum is 9/2 or 4.5.
21. Subtract $\frac{1}{x^2-2}$ from $\frac{2x^3+x^2+3}{x^2-2}$ (Assuming typo correction for common denominator)
Solution:
The problem requires us to calculate:
$$ \frac{2x^3+x^2+3}{x^2-2} - \frac{1}{x^2-2} $$Since the denominators are the same, we can subtract the numerators:
$$ = \frac{(2x^3+x^2+3) - 1}{x^2-2} $$ $$ = \frac{2x^3+x^2+2}{x^2-2} $$Result: $\frac{2x^3+x^2+2}{x^2-2}$
22. Solve $x^2 + 2x - 2 = 0$ by formula method.
Solution:
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
For the equation $x^2 + 2x - 2 = 0$, we have $a=1, b=2, c=-2$.
Substitute these values into the formula:
$$ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-2)}}{2(1)} $$ $$ x = \frac{-2 \pm \sqrt{4 + 8}}{2} $$ $$ x = \frac{-2 \pm \sqrt{12}}{2} $$Simplify the square root: $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.
$$ x = \frac{-2 \pm 2\sqrt{3}}{2} $$Divide by 2:
$$ x = -1 \pm \sqrt{3} $$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:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$$ \frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2 $$Substitute the given values:
$$ \frac{54}{\text{Area}(\Delta DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16} $$Now, solve for the Area(ΔDEF):
$$ \text{Area}(\Delta DEF) = \frac{54 \times 16}{9} $$ $$ \text{Area}(\Delta DEF) = 6 \times 16 = 96 \text{ cm}^2 $$The area of ΔDEF is 96 cm².
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:
By the Basic Proportionality Theorem (Thales' theorem), since DE || BC, we have:
$$ \frac{AD}{DB} = \frac{AE}{EC} $$We are given $\frac{AD}{DB} = \frac{3}{4}$. So, $\frac{AE}{EC} = \frac{3}{4}$.
This means AE = 3k and EC = 4k for some constant k.
We know that AC = AE + EC.
Given AC = 15 cm:
$$ 15 = 3k + 4k = 7k $$ $$ k = \frac{15}{7} $$We need to find AE:
$$ AE = 3k = 3 \times \frac{15}{7} = \frac{45}{7} \text{ cm} $$The length of AE is $\frac{45}{7}$ cm.
25. Show that the points (-2,5), (6,-1) and (2,2) are collinear.
Solution:
Points are collinear if the slope between any two pairs of points is the same.
Let the points be A(-2,5), B(6,-1), and C(2,2).
Slope of AB:
$$ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 5}{6 - (-2)} = \frac{-6}{8} = -\frac{3}{4} $$Slope of BC:
$$ m_{BC} = \frac{2 - (-1)}{2 - 6} = \frac{3}{-4} = -\frac{3}{4} $$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.
The points are 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 rewrite the equation in the slope-intercept form, $y = mx + c$.
$$ (1-\sqrt{3})y = -\sqrt{3}x + 3 $$ $$ y = \frac{-\sqrt{3}}{1-\sqrt{3}}x + \frac{3}{1-\sqrt{3}} $$Slope (m):
$$ m = \frac{-\sqrt{3}}{1-\sqrt{3}} $$Rationalize the denominator:
$$ m = \frac{-\sqrt{3}}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}} = \frac{-\sqrt{3}(1+\sqrt{3})}{1^2 - (\sqrt{3})^2} = \frac{-\sqrt{3}-3}{1-3} = \frac{-(\sqrt{3}+3)}{-2} = \frac{3+\sqrt{3}}{2} $$y-intercept (c):
$$ c = \frac{3}{1-\sqrt{3}} $$Rationalize the denominator:
$$ c = \frac{3}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}} = \frac{3(1+\sqrt{3})}{1-3} = \frac{3(1+\sqrt{3})}{-2} = -\frac{3(1+\sqrt{3})}{2} $$Slope = $\frac{3+\sqrt{3}}{2}$, y-intercept = $-\frac{3(1+\sqrt{3})}{2}$.
27. Prove that $\sec\theta - \cos\theta = \tan\theta \sin\theta$.
Solution:
Starting with the Left Hand Side (LHS):
$$ \text{LHS} = \sec\theta - \cos\theta $$Write $\sec\theta$ as $\frac{1}{\cos\theta}$:
$$ = \frac{1}{\cos\theta} - \cos\theta $$Take a common denominator:
$$ = \frac{1 - \cos^2\theta}{\cos\theta} $$Using the identity $\sin^2\theta + \cos^2\theta = 1$, we have $1 - \cos^2\theta = \sin^2\theta$:
$$ = \frac{\sin^2\theta}{\cos\theta} $$Separate the terms:
$$ = \frac{\sin\theta}{\cos\theta} \times \sin\theta $$Since $\frac{\sin\theta}{\cos\theta} = \tan\theta$:
$$ = \tan\theta \sin\theta = \text{RHS} $$Hence, proved.
28. Find the excluded values of the following expression: $\frac{7P+2}{8P^2+13P+5}$
Solution:
The excluded values are the values of P for which the denominator is zero.
Set the denominator to zero:
$$ 8P^2+13P+5 = 0 $$We can factor this quadratic equation. We need two numbers that multiply to $8 \times 5 = 40$ and add up to 13. The 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 for P:
$8P+5 = 0 \implies P = -\frac{5}{8}$
$P+1 = 0 \implies P = -1$
The excluded values are -1 and -5/8.
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 ∈ W / x < 2} = {0, 1} (W is the set of Whole numbers {0, 1, 2, ...})
- B = {x ∈ N / 1 < x ≤ 4} = {2, 3, 4} (N is the set of Natural numbers {1, 2, 3, ...})
- C = {3, 5}
Left Hand Side (LHS): A x (B ∩ C)
1. Find B ∩ C:
$$ B \cap C = \{2, 3, 4\} \cap \{3, 5\} = \{3\} $$2. Find A x (B ∩ C):
$$ A \times (B \cap C) = \{0, 1\} \times \{3\} = \{(0, 3), (1, 3)\} $$Right Hand Side (RHS): (A x B) ∩ (A x C)
1. Find A x B:
$$ A \times B = \{0, 1\} \times \{2, 3, 4\} = \{(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)\} $$2. Find A x C:
$$ A \times C = \{0, 1\} \times \{3, 5\} = \{(0, 3), (0, 5), (1, 3), (1, 5)\} $$3. Find the intersection (A x B) ∩ (A x C):
$$ (A \times B) \cap (A \times C) = \{(0, 3), (1, 3)\} $$Verification:
LHS = {(0, 3), (1, 3)} and RHS = {(0, 3), (1, 3)}.
Since LHS = RHS, it is 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
- by arrow diagram
- in a table form
- as a set of ordered pairs
- in a graphical form
Solution:
Given A = {1,2,3,4}, B = {2,5,8,11,14} and f(x) = 3x - 1.
- f(1) = 3(1) - 1 = 2
- f(2) = 3(2) - 1 = 5
- f(3) = 3(3) - 1 = 8
- f(4) = 3(4) - 1 = 11
i) By Arrow Diagram:
An arrow diagram would show two ovals. The first oval (Domain A) contains the numbers 1, 2, 3, 4. The second oval (Codomain B) contains 2, 5, 8, 11, 14. Arrows are drawn from each element in A to its corresponding image in B:
1 → 2, 2 → 5, 3 → 8, 4 → 11.
ii) In a Table Form:
| x | f(x) |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 8 |
| 4 | 11 |
iii) As a Set of Ordered Pairs:
The function f can be represented as the set of ordered pairs (x, f(x)).
f = {(1, 2), (2, 5), (3, 8), (4, 11)}
iv) In a Graphical Form:
The function is represented by plotting the ordered pairs as points on a Cartesian coordinate plane. The points to be plotted are (1, 2), (2, 5), (3, 8), and (4, 11).
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:
First, we evaluate the required values from the piecewise function:
- f(-3): Since -5 ≤ -3 < 2, use $f(x) = 6x+1$. f(-3) = 6(-3) + 1 = -17.
- f(2): Since 2 ≤ 2 < 6, use $f(x) = 5x^2-1$. f(2) = 5(2)² - 1 = 19.
- f(7): Since 6 ≤ 7 ≤ 9, use $f(x) = 3x-4$. f(7) = 3(7) - 4 = 17.
- f(1): Since -5 ≤ 1 < 2, use $f(x) = 6x+1$. f(1) = 6(1) + 1 = 7.
- f(4): Since 2 ≤ 4 < 6, use $f(x) = 5x^2-1$. f(4) = 5(4)² - 1 = 79.
- f(8): Since 6 ≤ 8 ≤ 9, use $f(x) = 3x-4$. f(8) = 3(8) - 4 = 20.
- f(-2): Since -5 ≤ -2 < 2, use $f(x) = 6x+1$. f(-2) = 6(-2) + 1 = -11.
- f(6): Since 6 ≤ 6 ≤ 9, use $f(x) = 3x-4$. f(6) = 3(6) - 4 = 14.
Now we perform the calculations:
i) f(-3) + f(2)
= -17 + 19 = 2. Answer: 2
ii) f(7) - f(1)
= 17 - 7 = 10. Answer: 10
iii) 2f(4) + f(8)
= 2(79) + 20 = 158 + 20 = 178. Answer: 178
iv) $\frac{2f(-2)-f(6)}{f(4)+f(-2)}$
$$ = \frac{2(-11) - 14}{79 + (-11)} = \frac{-22 - 14}{79 - 11} = \frac{-36}{68} $$Simplifying by dividing the numerator and denominator by 4:
$$ = -\frac{9}{17} $$Answer: -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:
The formula for the sum of the first 'k' terms of an A.P. is $S_k = \frac{k}{2}[2a + (k-1)d]$, where 'a' is the first term and 'd' is the common difference.
According to the problem:
- $S_1 = S_n = \frac{n}{2}[2a + (n-1)d]$
- $S_2 = S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$
- $S_3 = S_{3n} = \frac{3n}{2}[2a + (3n-1)d]$
We need to prove that $S_3 = 3(S_2 - S_1)$. Let's evaluate the Right Hand Side (RHS).
RHS = 3(S₂ - S₁)
First, calculate $S_2 - S_1$:
$$ S_2 - S_1 = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d] $$Take $\frac{n}{2}$ as a common factor:
$$ = \frac{n}{2} \left( 2[2a + (2n-1)d] - [2a + (n-1)d] \right) $$ $$ = \frac{n}{2} [4a + 2(2n-1)d - 2a - (n-1)d] $$ $$ = \frac{n}{2} [2a + (4n-2 - n+1)d] $$ $$ = \frac{n}{2} [2a + (3n-1)d] $$Now, multiply by 3:
$$ 3(S_2 - S_1) = 3 \times \frac{n}{2} [2a + (3n-1)d] $$ $$ = \frac{3n}{2} [2a + (3n-1)d] $$This expression is exactly the formula for $S_3$.
Thus, RHS = $S_3$ = LHS. Hence, 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 57/2. Find the three terms.
Solution:
Let the three consecutive terms in the G.P. be $\frac{a}{r}$, $a$, and $ar$.
Condition 1: Product of terms is 27.
$$ \left(\frac{a}{r}\right) \times a \times (ar) = 27 $$ $$ a^3 = 27 \implies a = 3 $$Condition 2: Sum of the product of two terms taken at a time is 57/2.
$$ \left(\frac{a}{r}\right)(a) + (a)(ar) + \left(\frac{a}{r}\right)(ar) = \frac{57}{2} $$ $$ \frac{a^2}{r} + a^2r + a^2 = \frac{57}{2} $$Substitute $a=3$ into the equation:
$$ \frac{3^2}{r} + 3^2r + 3^2 = \frac{57}{2} $$ $$ \frac{9}{r} + 9r + 9 = \frac{57}{2} $$Divide the entire equation by 9:
$$ \frac{1}{r} + r + 1 = \frac{57}{18} = \frac{19}{6} $$ $$ \frac{1}{r} + r = \frac{19}{6} - 1 = \frac{19-6}{6} = \frac{13}{6} $$Now, solve for r:
$$ \frac{1+r^2}{r} = \frac{13}{6} $$ $$ 6(1+r^2) = 13r \implies 6 + 6r^2 = 13r $$ $$ 6r^2 - 13r + 6 = 0 $$Factor the quadratic equation:
$$ 6r^2 - 9r - 4r + 6 = 0 $$ $$ 3r(2r - 3) - 2(2r - 3) = 0 $$ $$ (3r-2)(2r-3) = 0 $$This gives two possible values for r: $r = \frac{2}{3}$ or $r = \frac{3}{2}$.
Finding the terms:
- If $r = \frac{2}{3}$, the terms are $\frac{3}{2/3}, 3, 3(\frac{2}{3}) \implies \frac{9}{2}, 3, 2$.
- If $r = \frac{3}{2}$, the terms are $\frac{3}{3/2}, 3, 3(\frac{3}{2}) \implies 2, 3, \frac{9}{2}$.
In both cases, the set of terms is the same.
The three terms are 2, 3, and 9/2.
34. Solve the following system of linear equations in three variables:
3x - 2y + z = 2 ---(1)
2x + 3y - z = 5 ---(2)
x + y + z = 6 ---(3)
Solution:
We can solve this system by elimination.
Step 1: Eliminate 'z' using equations (1) and (2).
Add equation (1) and equation (2):
3x - 2y + z = 2 + (2x + 3y - z = 5) -------------------- 5x + y = 7 ---(4)
Step 2: Eliminate 'z' using equations (2) and (3).
Add equation (2) and equation (3):
2x + 3y - z = 5 + ( x + y + z = 6) -------------------- 3x + 4y = 11 ---(5)
Step 3: Solve the new system of two equations (4) and (5).
From equation (4), we can express y as $y = 7 - 5x$.
Substitute this into equation (5):
$$ 3x + 4(7 - 5x) = 11 $$ $$ 3x + 28 - 20x = 11 $$ $$ -17x = 11 - 28 $$ $$ -17x = -17 $$ $$ x = 1 $$Step 4: Back-substitute to find y and z.
Substitute $x=1$ into equation (4):
$$ 5(1) + y = 7 \implies y = 7 - 5 \implies y = 2 $$Substitute $x=1$ and $y=2$ into equation (3):
$$ (1) + (2) + z = 6 \implies 3 + z = 6 \implies z = 3 $$The solution is x = 1, y = 2, and z = 3.
Part - III
Answer any 10 questions. (Q.No.42 is compulsory)
35. 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.
3x² + 2x + 4
______________________
3x² | 9x⁴ + 12x³ + 28x² + ax + b
|-(9x⁴)
|______________________
6x²+2x | 12x³ + 28x²
| -(12x³ + 4x²)
|______________________
6x²+4x+4| 24x² + ax + b
| -(24x² + 16x + 16)
|______________________
| (a-16)x + (b-16)
For the polynomial to be a perfect square, the remainder must be 0.
Therefore, $(a-16)x + (b-16) = 0$.
This implies the coefficients must be zero:
$a - 16 = 0 \implies a = 16$
$b - 16 = 0 \implies b = 16$
The values are a = 16 and b = 16.
36. If α, β are the roots of $7x^2 + ax + 2 = 0$ and if $β - α = \frac{-13}{7}$, find the value of a.
Solution:
For the quadratic equation $7x^2 + ax + 2 = 0$, the sum and product of the roots (α, β) are:
Sum of roots: $ \alpha + \beta = -\frac{\text{coefficient of x}}{\text{coefficient of } x^2} = -\frac{a}{7} $
Product of roots: $ \alpha\beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{2}{7} $
We are given that $ \beta - \alpha = -\frac{13}{7} $.
We use the algebraic identity: $ (\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta $.
Substitute the known values into the identity:
$$ \left(-\frac{13}{7}\right)^2 = \left(-\frac{a}{7}\right)^2 - 4\left(\frac{2}{7}\right) $$ $$ \frac{169}{49} = \frac{a^2}{49} - \frac{8}{7} $$To eliminate the denominators, multiply the entire equation by 49:
$$ 49 \times \frac{169}{49} = 49 \times \frac{a^2}{49} - 49 \times \frac{8}{7} $$ $$ 169 = a^2 - 7 \times 8 $$ $$ 169 = a^2 - 56 $$ $$ a^2 = 169 + 56 $$ $$ a^2 = 225 $$ $$ a = \pm\sqrt{225} $$a = ±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 triangle ΔABC, a line DE is drawn parallel to BC (DE || BC), intersecting AB at D and AC at E.
To Prove: $ \frac{AD}{DB} = \frac{AE}{EC} $
Construction:
- Join BE and CD.
- Draw DM ⊥ AC and EN ⊥ AB.
Proof:
The area of a triangle is given by the formula: Area = $\frac{1}{2} \times \text{base} \times \text{height}$.
Now, consider the area of ΔADE:
$$ \text{Area}(\Delta ADE) = \frac{1}{2} \times AD \times EN \quad \text{---(1)} $$Consider the area of ΔBDE:
$$ \text{Area}(\Delta BDE) = \frac{1}{2} \times DB \times EN \quad \text{---(2)} $$Divide equation (1) by (2):
$$ \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{---(i)} $$Similarly, considering base AE and EC with height DM:
$$ \text{Area}(\Delta ADE) = \frac{1}{2} \times AE \times DM \quad \text{---(3)} $$ $$ \text{Area}(\Delta CDE) = \frac{1}{2} \times EC \times DM \quad \text{---(4)} $$Divide equation (3) by (4):
$$ \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{---(ii)} $$Now, observe that ΔBDE and ΔCDE are on the same base DE and between the same parallel lines DE and BC. Therefore, their areas are equal.
$$ \text{Area}(\Delta BDE) = \text{Area}(\Delta CDE) \quad \text{---(iii)} $$From equations (i), (ii), and (iii), we can conclude:
$$ \frac{AD}{DB} = \frac{AE}{EC} $$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 of the quadrilateral be A(8,6), B(5,11), C(-5,12), and D(-4,3). We must take the vertices in counter-clockwise order to get a positive area.
The formula for the area of a quadrilateral with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)$ is:
$$ \text{Area} = \frac{1}{2} \left| (x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| $$Let's arrange the points for calculation:
$$ \begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \\ x_3 & y_3 \\ x_4 & y_4 \\ x_1 & y_1 \end{pmatrix} = \begin{pmatrix} 8 & 6 \\ 5 & 11 \\ -5 & 12 \\ -4 & 3 \\ 8 & 6 \end{pmatrix} $$Calculate the first part $(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1)$:
= $(8)(11) + (5)(12) + (-5)(3) + (-4)(6)$
= $88 + 60 - 15 - 24 = 109$
Calculate the second part $(y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)$:
= $(6)(5) + (11)(-5) + (12)(-4) + (3)(8)$
= $30 - 55 - 48 + 24 = -49$
Now, substitute these values into the area formula:
$$ \text{Area} = \frac{1}{2} |109 - (-49)| $$ $$ \text{Area} = \frac{1}{2} |109 + 49| = \frac{1}{2} |158| = 79 $$The area of the quadrilateral is 79 square 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.
- Find the total MB of the song.
- After how many seconds will 75% of the song get downloaded?
- After how many seconds the song will be downloaded completely?
Solution:
The equation given is $y = -0.1x + 1$, where y is the percentage 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 download remaining to time. It does not contain any information about the actual size of the song in megabytes (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 the percentage remaining is 100% - 75% = 25%. In decimal form, the remaining percentage 'y' is 0.25.
Substitute y = 0.25 into the equation:
$$ 0.25 = -0.1x + 1 $$ $$ 0.1x = 1 - 0.25 $$ $$ 0.1x = 0.75 $$ $$ x = \frac{0.75}{0.1} = 7.5 $$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, the percentage remaining is 0%. In decimal form, 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 completely downloaded 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 of a line segment passes through its midpoint and is perpendicular to it.
Step 1: Find the midpoint of the segment AB.
The midpoint M is given by $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
$$ M = \left(\frac{-4+6}{2}, \frac{2+(-4)}{2}\right) = \left(\frac{2}{2}, \frac{-2}{2}\right) = (1, -1) $$Step 2: Find the slope of the line segment AB.
The slope $m_{AB}$ is given by $m = \frac{y_2-y_1}{x_2-x_1}$.
$$ m_{AB} = \frac{-4-2}{6-(-4)} = \frac{-6}{10} = -\frac{3}{5} $$Step 3: Find the slope of the perpendicular bisector.
The slope of the perpendicular bisector ($m_\perp$) is the negative reciprocal of the slope of AB.
$$ m_\perp = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3} $$Step 4: Find the equation of the perpendicular bisector.
Using the point-slope form $y - y_1 = m(x - x_1)$ with the midpoint M(1, -1) and slope $m_\perp = 5/3$.
$$ y - (-1) = \frac{5}{3}(x - 1) $$ $$ y + 1 = \frac{5}{3}(x - 1) $$Multiply by 3 to clear the fraction:
$$ 3(y + 1) = 5(x - 1) $$ $$ 3y + 3 = 5x - 5 $$Rearrange into the standard form $Ax + By + C = 0$:
$$ 5x - 3y - 5 - 3 = 0 $$The equation is 5x - 3y - 8 = 0.
41. If $\cot\theta + \tan\theta = x$ and $\sec\theta - \cos\theta = y$, then prove that $(x^2y)^{2/3} - (xy^2)^{2/3} = 1$.
Solution:
We need to simplify the expressions for x and y first.
Simplifying 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 $\sin^2\theta + \cos^2\theta = 1$:
$$ x = \frac{1}{\sin\theta\cos\theta} $$Simplifying 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 in the expression we need to prove.
First term: $(x^2y)^{2/3}$
$$ x^2y = \left(\frac{1}{\sin\theta\cos\theta}\right)^2 \left(\frac{\sin^2\theta}{\cos\theta}\right) = \frac{1}{\sin^2\theta\cos^2\theta} \cdot \frac{\sin^2\theta}{\cos\theta} = \frac{1}{\cos^3\theta} = \sec^3\theta $$ $$ \implies (x^2y)^{2/3} = (\sec^3\theta)^{2/3} = \sec^2\theta $$Second term: $(xy^2)^{2/3}$
$$ xy^2 = \left(\frac{1}{\sin\theta\cos\theta}\right) \left(\frac{\sin^2\theta}{\cos\theta}\right)^2 = \frac{1}{\sin\theta\cos\theta} \cdot \frac{\sin^4\theta}{\cos^2\theta} = \frac{\sin^3\theta}{\cos^3\theta} = \tan^3\theta $$ $$ \implies (xy^2)^{2/3} = (\tan^3\theta)^{2/3} = \tan^2\theta $$Substitute these back into the Left Hand Side (LHS) of the expression to be proved:
$$ \text{LHS} = (x^2y)^{2/3} - (xy^2)^{2/3} = \sec^2\theta - \tan^2\theta $$Using the Pythagorean identity $1 + \tan^2\theta = \sec^2\theta$, which implies $\sec^2\theta - \tan^2\theta = 1$.
$$ \text{LHS} = 1 = \text{RHS} $$Hence, proved.
42. Find the sum of $10^3 + 11^3 + 12^3 + ... + 20^3$.
Solution:
We can write the series as the difference of two sums:
$$ \sum_{k=10}^{20} k^3 = \sum_{k=1}^{20} k^3 - \sum_{k=1}^{9} k^3 $$Using the formula for the sum of cubes, $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$.
For the first sum (n=20):
$$ \sum_{k=1}^{20} k^3 = \left(\frac{20(20+1)}{2}\right)^2 = \left(\frac{20 \times 21}{2}\right)^2 = (10 \times 21)^2 = 210^2 = 44100 $$For the second sum (n=9):
$$ \sum_{k=1}^{9} k^3 = \left(\frac{9(9+1)}{2}\right)^2 = \left(\frac{9 \times 10}{2}\right)^2 = (9 \times 5)^2 = 45^2 = 2025 $$Now, find the difference:
$$ 44100 - 2025 = 42075 $$The sum is 42075.
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):
Since the scale factor is 7/3, which is greater than 1, the new triangle will be larger than the given triangle PQR.
- Draw the Base Triangle: Draw any triangle PQR with suitable measurements.
- Draw a Ray: Draw a ray QX starting from Q and making an acute angle (e.g., ∠RQX) with the base QR. The ray should be on the side opposite to vertex P.
- Mark Points on the Ray: The scale factor is 7/3. The larger number is 7. Using a compass, mark 7 equal arcs on the ray QX. Label the points as Q₁, Q₂, Q₃, Q₄, Q₅, Q₆, and Q₇ such that Q Q₁ = Q₁Q₂ = ... = Q₆Q₇.
- Join to the Denominator Point: The denominator of the scale factor is 3. Join the third point, Q₃, to the vertex R. This forms the line segment Q₃R.
- Draw a Parallel Line: From the seventh point (the numerator point), Q₇, draw a line parallel to Q₃R. To do this, copy the angle ∠QQ₃R at point Q₇. This parallel line will intersect the extended line segment QR at a new point. Label this point R'.
- Draw the Second Parallel Line: From the new point R', draw a line parallel to the side PR. This line will intersect the extended line segment QP at a new point. Label this point P'.
- The Required Triangle: The triangle ΔP'QR' is the required similar triangle, whose sides are 7/3 times the corresponding sides of ΔPQR.
(OR)
b) 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):
This construction uses the concept of the alternate segment theorem.
- Draw the Base: Draw a line segment QR of length 5 cm.
- Construct the Angle: At point Q, draw a line QE such that ∠RQE = 30°. This angle is equal to the angle in the alternate segment (∠P).
- Draw a Perpendicular: Draw a line QF perpendicular to QE at Q (i.e., ∠EQF = 90°).
- Find the Circumcenter: Draw the perpendicular bisector of the base QR. Let this bisector intersect the line QF at a point O. This point O is the circumcenter of the required triangle.
- Draw the Circumcircle: With O as the center and OQ (or OR) as the radius, draw a circle. The vertex P of the triangle must lie on the major arc of this circle.
- Locate the Altitude Line: The altitude from P to QR is 4.2 cm. Draw a line LM parallel to QR at a distance of 4.2 cm. This line represents all possible locations for vertex P.
- Find Vertex P: The line LM will intersect the circumcircle at two points. Label these points as P and P'.
- Complete the Triangle: Join PQ and PR (or P'Q and P'R).
- The Required Triangle: ΔPQR is the required triangle with QR = 5 cm, ∠P = 30°, and an altitude of 4.2 cm from P to QR.
Part - IV
Answer all the questions.
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:
The relationship between distance (d), speed (s), and time (t) is $d = s \times t$.
Given speed s = 50 km/hr, the equation is $d = 50t$. This is a direct variation.
i) The constant of variation
In the equation $d = kt$, k is the constant of variation. Here, $k=50$.
The constant of variation is 50 km/hr.
Graph Plotting:
Create a table of values for distance vs. time.
| Time (t) in hours | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Distance (d) in km | 0 | 50 | 100 | 150 | 200 | 250 | 300 |
Draw a graph with Time on the x-axis and Distance on the y-axis. Plot the points (0,0), (1,50), (2,100), etc., and draw a straight line through them starting from the origin.
ii) Distance travelled in 90 minutes
90 minutes = 1.5 hours. From the graph, find the y-value corresponding to x = 1.5. A line drawn up from t=1.5 on the x-axis will intersect the graph at d=75 on the y-axis.
Calculation: $d = 50 \times 1.5 = 75$ km.
The bus will travel 75 km in 90 minutes.
iii) Time to cover 300 km
From the graph, find the x-value corresponding to y = 300. A line drawn across from d=300 on the y-axis will intersect the graph at t=6 on the x-axis.
Calculation: $300 = 50 \times t \implies t = 300/50 = 6$ hours.
The time required is 6 hours.
44. b) Draw the graph of xy = 24, x, y > 0. Using the graph find,
i) y when x = 3
ii) x when y = 6
Solution:
The equation is $xy=24$, which represents an indirect variation. The graph will be a rectangular hyperbola in the first quadrant since x, y > 0.
Table of Values:
| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 | 24 |
|---|---|---|---|---|---|---|---|---|
| y = 24/x | 24 | 12 | 8 | 6 | 4 | 3 | 2 | 1 |
Graph Plotting:
Draw a graph with x on the horizontal axis and y on the vertical axis. Plot the points from the table, such as (1,24), (2,12), (3,8), etc. Connect them with a smooth curve.
Using the Graph:
i) Find y when x = 3
On the graph, locate x=3 on the x-axis. Move vertically up to the curve and then horizontally to the y-axis. The corresponding y-value will be 8.
When x = 3, y = 8.
ii) Find x when y = 6
On the graph, locate y=6 on the y-axis. Move horizontally to the curve and then vertically down to the x-axis. The corresponding x-value will be 4.
When y = 6, x = 4.