10th Maths Quarterly Exam 2025-26
Complete Question Paper with Solutions
SECTION - I (Marks: 14)
Answer all of the following:
1. If $n(A \times B) = 6$ and $A = \{1, 3\}$ then $n(B)$ is
Solution:
Given, $n(A \times B) = 6$ and $A = \{1, 3\}$.
We know, $n(A \times B) = n(A) \times n(B)$.
Here, $n(A) = 2$.
So, $6 = 2 \times n(B)$.
$n(B) = \frac{6}{2} = 3$.
Answer: c) 3
2. If there are 1024 relations from a set $A = \{1, 2, 3, 4, 5\}$ to a set $B$, then the number of elements in B is
Solution:
Given, $n(A) = 5$. Let $n(B) = m$.
Number of relations from A to B is $2^{n(A) \times n(B)} = 2^{5m}$.
We are given that the number of relations is 1024.
We know, $1024 = 2^{10}$.
So, $2^{5m} = 2^{10}$.
Equating the powers, $5m = 10 \Rightarrow m = 2$.
Therefore, $n(B) = 2$.
Answer: b) 2
3. Given $f(x) = (-1)^x$ is a function from N to Z, then the range of f is
Solution:
The domain is the set of Natural numbers, N = {1, 2, 3, ...}.
If $x$ is an odd natural number, $f(x) = (-1)^{\text{odd}} = -1$.
If $x$ is an even natural number, $f(x) = (-1)^{\text{even}} = 1$.
So, the range of f is the set of all possible output values, which is $\{-1, 1\}$.
Answer: c) {1, -1}
4. The sum of the exponents of the prime factors in the prime factorization of 1729 is
Solution:
Let's find the prime factors of 1729.
$1729 = 7 \times 247$
$1729 = 7 \times 13 \times 19$
The prime factorization is $1729 = 7^1 \times 13^1 \times 19^1$.
The exponents are 1, 1, and 1.
The sum of the exponents is $1 + 1 + 1 = 3$.
Answer: c) 3
5. $7^{4k} \equiv \text{______} \pmod{100}$
Solution:
$7^1 = 7$
$7^2 = 49$
$7^3 = 343 \equiv 43 \pmod{100}$
$7^4 = 7^3 \times 7 \equiv 43 \times 7 = 301 \equiv 1 \pmod{100}$
So, $7^4 \equiv 1 \pmod{100}$.
Raising both sides to the power of k: $(7^4)^k \equiv 1^k \pmod{100}$.
$7^{4k} \equiv 1 \pmod{100}$.
Answer: a) 1
6. If $(x - 6)$ is the HCF of $x^2 - 2x - 24$ and $x^2 - kx - 6$ then the value of k is
Solution:
If $(x-6)$ is the HCF, it must be a factor of both polynomials.
Let $p(x) = x^2 - kx - 6$.
Since $(x-6)$ is a factor, $p(6) = 0$.
$p(6) = (6)^2 - k(6) - 6 = 0$
$36 - 6k - 6 = 0$
$30 - 6k = 0$
$30 = 6k$
$k = 5$.
Answer: b) 5
7. 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 graph.
Answer: a) straight line
8. If $ax^2 + bx + c = 0$ has equal roots, then c is equal to
Solution:
For a quadratic equation to have equal roots, its discriminant must be zero.
Discriminant, $\Delta = b^2 - 4ac = 0$.
$b^2 = 4ac$
Solving for c, we get: $c = \frac{b^2}{4a}$.
(Note: The options in the paper image are slightly different. Option (d) in the image is $b^2/4a$, which is the correct one. The answer key says (b) but points to $b^2/4a$, so we follow the calculation.)
Answer: $\frac{b^2}{4a}$
9. If in triangle ABC and EDF, $\frac{AB}{DE} = \frac{BC}{FD}$, then they will be similar, when
Solution:
By SAS similarity criterion, if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the triangles are similar.
In $\triangle ABC$, the angle included between sides AB and BC is $\angle B$.
In $\triangle EDF$, the angle included between sides DE and FD is $\angle D$.
Therefore, for the triangles to be similar, we must have $\angle B = \angle D$.
Answer: c) $\angle B = \angle D$
10. In a $\triangle ABC$, AD is the bisector of $\angle BAC$. If AB = 8cm, BD = 6cm and DC = 3cm the length of the side AC is
Solution:
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.
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{8}{AC} = \frac{6}{3}$
$\frac{8}{AC} = 2$
$AC = \frac{8}{2} = 4$ cm.
Answer: b) 4cm
11. The area of triangle formed by the points $(-5,0)$, $(0, -5)$ and $(5,0)$ is
Solution:
The vertices are $A(-5,0)$, $B(0,-5)$, $C(5,0)$.
The base of the triangle can be taken along the x-axis, which is the segment AC.
Length of base AC = $5 - (-5) = 10$ units.
The height of the triangle is the perpendicular distance from vertex B to the x-axis, which is the absolute value of the y-coordinate of B.
Height = $|-5| = 5$ units.
Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 5 = 25$ sq.units.
Answer: b) 25 sq.units
12. If slope of the line PQ is $\frac{1}{\sqrt{3}}$ then slope of the perpendicular bisector of PQ is
Solution:
Let the slope of line PQ be $m_1 = \frac{1}{\sqrt{3}}$.
The perpendicular bisector is a line perpendicular to PQ.
Let its slope be $m_2$.
For perpendicular lines, the product of their slopes is -1.
$m_1 \times m_2 = -1$
$\frac{1}{\sqrt{3}} \times m_2 = -1$
$m_2 = -\sqrt{3}$.
Answer: b) $-\sqrt{3}$
13. $\tan\theta \csc^2\theta - \tan\theta$ is equal to
Solution:
$\tan\theta \csc^2\theta - \tan\theta = \tan\theta (\csc^2\theta - 1)$
We know the identity: $1 + \cot^2\theta = \csc^2\theta \Rightarrow \cot^2\theta = \csc^2\theta - 1$.
So, the expression becomes $\tan\theta (\cot^2\theta)$.
$= \tan\theta \times \frac{1}{\tan^2\theta} = \frac{1}{\tan\theta} = \cot\theta$.
Answer: d) $\cot\theta$
14. If $5x = \sec\theta$ and $\frac{5}{x} = \tan\theta$, then $x^2 - \frac{1}{x^2}$ is equal to
Solution:
Given $5x = \sec\theta$ and $\frac{5}{x} = \tan\theta$.
We know the identity: $\sec^2\theta - \tan^2\theta = 1$.
Substitute the given values into the identity:
$(5x)^2 - (\frac{5}{x})^2 = 1$
$25x^2 - \frac{25}{x^2} = 1$
$25(x^2 - \frac{1}{x^2}) = 1$
$x^2 - \frac{1}{x^2} = \frac{1}{25}$.
Answer: b) $\frac{1}{25}$
SECTION - II (Marks: 20)
Answer any 10 questions. Question No.28 is compulsory.
15. A Relation R is given by the set $\{(x,y) / y = x + 3, x \in \{0,1,2,3,4, 5\}\}$. Determine its domain and range.
Solution:
The relation is defined by $y = x + 3$.
The set of possible values for x (the domain) is given as $\{0, 1, 2, 3, 4, 5\}$.
We find the corresponding y values (the range):
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 $\{(0,3), (1,4), (2,5), (3,6), (4,7), (5,8)\}$.
Domain: The set of all first elements, which is $\{0, 1, 2, 3, 4, 5\}$.
Range: The set of all second elements, which is $\{3, 4, 5, 6, 7, 8\}$.
16. Given the function $f : x \to x^2 - 5x + 6$, evaluate (i) $f(-1)$ (ii) $f(2)$.
Solution:
The function is $f(x) = x^2 - 5x + 6$.
(i) $f(-1)$
$f(-1) = (-1)^2 - 5(-1) + 6$
$f(-1) = 1 + 5 + 6 = 12$.
(ii) $f(2)$
$f(2) = (2)^2 - 5(2) + 6$
$f(2) = 4 - 10 + 6 = 0$.
17. If $13824 = 2^a \times 3^b$ then find a and b.
Solution:
We perform prime factorization of 13824.
$13824 = 2 \times 6912$
$= 2^2 \times 3456$
$= 2^3 \times 1728$
$= 2^4 \times 864$
$= 2^5 \times 432$
$= 2^6 \times 216$
$= 2^7 \times 108$
$= 2^8 \times 54$
$= 2^9 \times 27$
$= 2^9 \times 3^3$
Comparing $2^9 \times 3^3$ with $2^a \times 3^b$, we get $a=9$ and $b=3$.
18. If $1^3+2^3 +3^3 + \dots + k^3 = 44100$ then find $1+2+3+\dots+k$.
Solution:
We know the formula for the sum of the cubes of the first k natural numbers:
$1^3+2^3 + \dots + k^3 = \left(\frac{k(k+1)}{2}\right)^2$.
And the sum of the first k natural numbers is:
$1+2+ \dots + k = \frac{k(k+1)}{2}$.
Given, $\left(\frac{k(k+1)}{2}\right)^2 = 44100$.
Taking the square root on both sides:
$\frac{k(k+1)}{2} = \sqrt{44100} = \sqrt{441 \times 100} = 21 \times 10 = 210$.
Therefore, $1+2+3+\dots+k = 210$.
19. Reduce lowest form of: $\frac{x^2-16}{x^2 + 8x +16}$
Solution:
Factorize the numerator and the denominator.
Numerator: $x^2 - 16 = x^2 - 4^2 = (x-4)(x+4)$ (using $a^2-b^2 = (a-b)(a+b)$).
Denominator: $x^2 + 8x + 16 = x^2 + 2(x)(4) + 4^2 = (x+4)^2 = (x+4)(x+4)$ (using $a^2+2ab+b^2=(a+b)^2$).
So, the expression is $\frac{(x-4)(x+4)}{(x+4)(x+4)}$.
Cancelling the common factor $(x+4)$ (assuming $x \neq -4$), we get:
Lowest form = $\frac{x-4}{x+4}$.
20. If the difference between a number and its reciprocal is $\frac{24}{5}$, find the number.
Solution:
Let the number be $x$. Its reciprocal is $\frac{1}{x}$.
Given, $x - \frac{1}{x} = \frac{24}{5}$.
$\frac{x^2 - 1}{x} = \frac{24}{5}$
$5(x^2 - 1) = 24x$
$5x^2 - 5 = 24x$
$5x^2 - 24x - 5 = 0$
Factoring the quadratic equation:
$5x^2 - 25x + x - 5 = 0$
$5x(x - 5) + 1(x - 5) = 0$
$(5x+1)(x-5) = 0$
So, $x=5$ or $x=-\frac{1}{5}$.
The number can be 5 or -1/5.
21. Determine the nature of roots of: $9x^2 - 24x + 16 = 0$.
Solution:
Compare the equation with $ax^2+bx+c=0$. We have $a=9, b=-24, c=16$.
The nature of the roots is determined by the discriminant, $\Delta = b^2 - 4ac$.
$\Delta = (-24)^2 - 4(9)(16)$
$\Delta = 576 - 576 = 0$.
Since the discriminant is zero ($\Delta=0$), the roots are real and equal.
22. If $\triangle ABC$ is similar to $\triangle DEF$ such that $BC=3$cm, $EF=4$cm and area of $\triangle ABC = 54$cm$^2$. Find the area of $\triangle DEF$.
Solution:
For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2$
$\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2$
$\frac{54}{\text{Area}(\triangle DEF)} = \frac{9}{16}$
$\text{Area}(\triangle DEF) = \frac{54 \times 16}{9} = 6 \times 16 = 96$ cm$^2$.
23. Check whether AD is bisector of $\angle A$ of $\triangle ABC$, AB = 5cm, AC = 10cm, BD = 1.5cm and CD = 3.5 cm.
Solution:
According to the Angle Bisector Theorem, if AD is the bisector of $\angle A$, then it must divide the opposite side BC in the ratio of the other two sides. That is, $\frac{AB}{AC} = \frac{BD}{CD}$.
Let's check the ratios:
LHS: $\frac{AB}{AC} = \frac{5}{10} = \frac{1}{2}$.
RHS: $\frac{BD}{CD} = \frac{1.5}{3.5} = \frac{15}{35} = \frac{3}{7}$.
Since $\frac{1}{2} \neq \frac{3}{7}$, the condition for the theorem is not satisfied.
Therefore, AD is not the bisector of $\angle A$.
24. Find the slope of a line joining the points (14, 10) and (14, -6).
Solution:
Let the points be $(x_1, y_1) = (14, 10)$ and $(x_2, y_2) = (14, -6)$.
The formula for the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
$m = \frac{-6 - 10}{14 - 14} = \frac{-16}{0}$.
Since the denominator is zero, the slope is undefined. This indicates that the line is a vertical line.
25. Find the equation of line passing through the point (3, -4) and having slope : -5/7.
Solution:
Using the point-slope form of a line equation: $y - y_1 = m(x - x_1)$.
Here, $(x_1, y_1) = (3, -4)$ and $m = -\frac{5}{7}$.
$y - (-4) = -\frac{5}{7}(x - 3)$
$y + 4 = -\frac{5}{7}(x - 3)$
$7(y + 4) = -5(x - 3)$
$7y + 28 = -5x + 15$
$5x + 7y + 28 - 15 = 0$
$5x + 7y + 13 = 0$.
26. If the straight lines $12y = -(P+3)x + 12$ and $12x - 7y = 16$ are perpendicular then find P.
Solution:
First, find the slope of each line.
Line 1: $12y = -(P+3)x + 12 \Rightarrow y = -\frac{P+3}{12}x + 1$.
The slope is $m_1 = -\frac{P+3}{12}$.
Line 2: $12x - 7y = 16 \Rightarrow 7y = 12x - 16 \Rightarrow y = \frac{12}{7}x - \frac{16}{7}$.
The slope is $m_2 = \frac{12}{7}$.
For perpendicular lines, the product of their slopes is -1.
$m_1 \times m_2 = -1$
$\left(-\frac{P+3}{12}\right) \times \left(\frac{12}{7}\right) = -1$
$-\frac{P+3}{7} = -1$
$P+3 = 7$
$P = 4$.
27. Prove that $\frac{\sec\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta} = \cot\theta$.
Solution:
LHS = $\frac{\sec\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta}$
$= \frac{1/\cos\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta}$
$= \frac{1}{\sin\theta\cos\theta} - \frac{\sin\theta}{\cos\theta}$
To subtract, find a common denominator, which is $\sin\theta\cos\theta$.
$= \frac{1}{\sin\theta\cos\theta} - \frac{\sin^2\theta}{\sin\theta\cos\theta}$
$= \frac{1 - \sin^2\theta}{\sin\theta\cos\theta}$
Using the identity $\cos^2\theta + \sin^2\theta = 1 \Rightarrow \cos^2\theta = 1 - \sin^2\theta$.
$= \frac{\cos^2\theta}{\sin\theta\cos\theta} = \frac{\cos\theta}{\sin\theta} = \cot\theta$
LHS = RHS. Hence proved.
28. If a, b, c are in A.P, then prove that $\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are also in A.P.
Solution:
If a, b, c are in an Arithmetic Progression (A.P.), then the common difference is constant.
$b - a = c - b$.
To prove that $\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in A.P., we need to show that the difference between consecutive terms is constant.
i.e., we need to prove: $\frac{1}{ca} - \frac{1}{bc} = \frac{1}{ab} - \frac{1}{ca}$.
Consider the LHS: $\frac{1}{ca} - \frac{1}{bc} = \frac{b - a}{abc}$
Consider the RHS: $\frac{1}{ab} - \frac{1}{ca} = \frac{c - b}{abc}$
Since a, b, c are in A.P., we know that $b - a = c - b$.
Therefore, $\frac{b - a}{abc} = \frac{c - b}{abc}$.
LHS = RHS.
Hence, $\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are also in A.P.
29. Let $A = \{ x \in W \mid x < 2 \}$, $B = \{ x \in N \mid 1 < x \le 4 \}$ and $C = \{ 3, 5\}$. Verify that $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
Solution:
First, let's define the sets in roster form:
- $A = \{ x \in W \mid x < 2 \}$: The whole numbers less than 2 are 0 and 1. So, $A = \{0, 1\}$.
- $B = \{ x \in N \mid 1 < x \le 4 \}$: The natural numbers greater than 1 and less than or equal to 4 are 2, 3, and 4. So, $B = \{2, 3, 4\}$.
- $C = \{3, 5\}$.
We need to verify the distributive property of the Cartesian product over union: $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
Step 1: Calculate the Left Hand Side (LHS)
First, find $B \cup C$:
$B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\}$.
Now, find $A \times (B \cup C)$:
$A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\}$
$= \{ (0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5) \}$ --- (1)
Step 2: Calculate the Right Hand Side (RHS)
First, find $A \times B$:
$A \times B = \{0, 1\} \times \{2, 3, 4\}$
$= \{ (0,2), (0,3), (0,4), (1,2), (1,3), (1,4) \}$.
Next, find $A \times C$:
$A \times C = \{0, 1\} \times \{3, 5\}$
$= \{ (0,3), (0,5), (1,3), (1,5) \}$.
Now, find the union $(A \times B) \cup (A \times C)$:
$= \{ (0,2), (0,3), (0,4), (1,2), (1,3), (1,4) \} \cup \{ (0,3), (0,5), (1,3), (1,5) \}$
$= \{ (0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5) \}$ --- (2)
Step 3: Compare LHS and RHS
From equations (1) and (2), we can see that the sets are identical.
LHS = $\{ (0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5) \}$
RHS = $\{ (0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5) \}$
Therefore, LHS = RHS. Hence, verified.
SECTION - III (Marks: 50)
Answer any 10 questions. Q.No.42 is compulsory.
30. Let $A = \{1,2,3,4\}$ and $B = \{2, 5, 8, 11, 14 \}$ be two sets. Let $f : A \to B$ be a function given by $f(x)=3x-1$. Represent this function (i) arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) a graph.
Solution:
Given $f(x) = 3x - 1$.
$f(1) = 3(1) - 1 = 2$
$f(2) = 3(2) - 1 = 5$
$f(3) = 3(3) - 1 = 8$
$f(4) = 3(4) - 1 = 11$
(i) Arrow Diagram:
(An arrow connects each element in A to its corresponding element in B)
(ii) Table Form:
| x | f(x) |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 8 |
| 4 | 11 |
(iii) Set of Ordered Pairs:
$f = \{(1, 2), (2, 5), (3, 8), (4, 11)\}$.
(iv) Graph:
(Plot the points from the set of ordered pairs on a coordinate plane)
31. If $f(x) = x - 1$, $g(x) = 3x+1$ and $h(x) = x^2$, then prove that $fo(goh) = (fog)oh$.
Solution:
We are asked to prove the associative property of function composition, which is $f \circ (g \circ h) = (f \circ g) \circ h$.
Given functions are:
- $f(x) = x - 1$
- $g(x) = 3x + 1$
- $h(x) = x^2$
Step 1: Calculate the Left Hand Side (LHS): $f \circ (g \circ h)$
First, we find the composition of g and h, which is $(g \circ h)(x)$.
$(g \circ h)(x) = g(h(x)) = g(x^2) = 3(x^2) + 1 = 3x^2 + 1$.
Now, we compose f with $(g \circ h)(x)$.
$(f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(3x^2 + 1)$
$= (3x^2 + 1) - 1 = 3x^2$.
So, LHS = $3x^2$. --- (1)
Step 2: Calculate the Right Hand Side (RHS): $(f \circ g) \circ h$
First, we find the composition of f and g, which is $(f \circ g)(x)$.
$(f \circ g)(x) = f(g(x)) = f(3x+1) = (3x+1) - 1 = 3x$.
Now, we compose $(f \circ g)(x)$ with h.
$((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = (f \circ g)(x^2)$
$= 3(x^2) = 3x^2$.
So, RHS = $3x^2$. --- (2)
Step 3: Compare LHS and RHS
From equations (1) and (2), we see that LHS = RHS.
Therefore, $f \circ (g \circ h) = (f \circ g) \circ h$ is proved.
32. Find the HCF of 396, 504 and 636.
Solution:
We will use Euclid's Division Algorithm to find the Highest Common Factor (HCF).
Step 1: Find the HCF of the first two numbers, 396 and 504.
We apply the division lemma to 504 and 396 (where $504 > 396$).
- $504 = 396 \times 1 + 108$
- $396 = 108 \times 3 + 72$
- $108 = 72 \times 1 + 36$
- $72 = 36 \times 2 + 0$
The remainder is now 0. The last non-zero remainder is the HCF. So, HCF(396, 504) = 36.
Step 2: Find the HCF of the result from Step 1 (which is 36) and the third number, 636.
We apply the division lemma to 636 and 36.
- $636 = 36 \times 17 + 24$
- $36 = 24 \times 1 + 12$
- $24 = 12 \times 2 + 0$
The remainder is now 0. The last non-zero remainder is the HCF. So, HCF(36, 636) = 12.
Therefore, the HCF of 396, 504, and 636 is 12.
33. Find the sum of all natural numbers between 300 and 600 which are divisible by 7.
Solution:
The natural numbers between 300 and 600 which are divisible by 7 form an Arithmetic Progression (A.P.).
First term ($a$): The first number greater than 300 divisible by 7 is 301. ($300 \div 7$ gives remainder 6, so $300 + (7-6)=301$).
Last term ($l$): The last number less than 600 divisible by 7 is 595. ($600 \div 7$ gives remainder 5, so $600-5=595$).
Common difference ($d$) = 7.
The A.P. is $301, 308, \dots, 595$.
Number of terms ($n$): $n = \frac{l-a}{d} + 1 = \frac{595-301}{7} + 1 = \frac{294}{7} + 1 = 42 + 1 = 43$.
Sum of the terms ($S_n$): $S_n = \frac{n}{2}(a+l) = \frac{43}{2}(301+595) = \frac{43}{2}(896) = 43 \times 448 = 19264$.
The sum is 19264.
34. Find the sum to n terms of the series $3+33+333+\dots$
Solution:
Let $S_n$ be the sum of the first n terms of the series.
$S_n = 3 + 33 + 333 + \dots + n$ terms
Step 1: Factor out the common term.
$S_n = 3(1 + 11 + 111 + \dots + n \text{ terms})$
Step 2: Multiply and divide by 9 to create a pattern of 9s.
$S_n = \frac{3}{9}(9 + 99 + 999 + \dots + n \text{ terms})$
$S_n = \frac{1}{3}(9 + 99 + 999 + \dots)$
Step 3: Express each term as a difference involving powers of 10.
$S_n = \frac{1}{3}[(10 - 1) + (100 - 1) + (1000 - 1) + \dots + n \text{ terms}]$
$S_n = \frac{1}{3}[(10^1 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$
Step 4: Separate the terms into two groups.
$S_n = \frac{1}{3}[ (10^1 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n \text{ times}) ]$
Step 5: Calculate the sum of each group.
The first group $(10 + 10^2 + \dots + 10^n)$ is a Geometric Progression (G.P.) with:
- First term, $a = 10$
- Common ratio, $r = 10$
- Number of terms = $n$
The second group is the sum of '1' repeated 'n' times, which is simply $n$.
Step 6: Substitute the sums back into the main equation.
$S_n = \frac{1}{3} \left[ \frac{10}{9}(10^n - 1) - n \right]$
This can be further simplified as:
$S_n = \frac{10}{27}(10^n - 1) - \frac{n}{3}$
35. Solve the following system of linear equations in three variables:
$x + y + z = 5$
$2x - y + z = 9$
$x - 2y + 3z = 16$
Solution:
Let the given equations be:
- $x + y + z = 5$ --- (1)
- $2x - y + z = 9$ --- (2)
- $x - 2y + 3z = 16$ --- (3)
Step 1: Eliminate 'y' from equations (1) and (2).
Adding equation (1) and (2):
$(x + y + z) + (2x - y + z) = 5 + 9$
$3x + 2z = 14$ --- (4)
Step 2: Eliminate 'y' from equations (2) and (3).
Multiply equation (2) by 2:
$2(2x - y + z) = 2(9) \Rightarrow 4x - 2y + 2z = 18$ --- (5)
Subtract equation (3) from equation (5):
$(4x - 2y + 2z) - (x - 2y + 3z) = 18 - 16$
$3x - z = 2$ --- (6)
Step 3: Solve the new system of equations (4) and (6).
(4): $3x + 2z = 14$
(6): $3x - z = 2$
Subtract equation (6) from equation (4):
$(3x + 2z) - (3x - z) = 14 - 2$
$3z = 12 \Rightarrow z = 4$.
Step 4: Back-substitute to find x.
Substitute $z=4$ into equation (6):
$3x - 4 = 2 \Rightarrow 3x = 6 \Rightarrow x = 2$.
Step 5: Back-substitute to find y.
Substitute $x=2$ and $z=4$ into the original equation (1):
$2 + y + 4 = 5 \Rightarrow y + 6 = 5 \Rightarrow y = -1$.
The solution is $x=2, y=-1, z=4$.
36. 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 of the polynomial.
Step-by-step Explanation:
- The square root of the first term, $9x^4$, is $3x^2$. Write this as the first term of the root (quotient) and the divisor. Subtract $(3x^2)^2 = 9x^4$ and bring down the next two terms: $12x^3 + 28x^2$.
- Double the current root ($3x^2$) to get $6x^2$. Divide the first term of the new dividend ($12x^3$) by $6x^2$ to get $2x$. This is the second term of the root. The new divisor is $6x^2 + 2x$.
- Multiply the new divisor by $2x$: $(6x^2 + 2x)(2x) = 12x^3 + 4x^2$. Subtract this from the current dividend. The remainder is $(12x^3 + 28x^2) - (12x^3 + 4x^2) = 24x^2$. Bring down the remaining terms: $ax + b$.
- Double the current root ($3x^2 + 2x$) to get $6x^2 + 4x$. Divide the first term of the new dividend ($24x^2$) by $6x^2$ to get $4$. This is the third term of the root. The new divisor is $6x^2 + 4x + 4$.
- Multiply the new divisor by $4$: $(6x^2 + 4x + 4)(4) = 24x^2 + 16x + 16$.
Since the given polynomial is a perfect square, the remainder must be zero.
Therefore, $(24x^2 + ax + b) - (24x^2 + 16x + 16)$ must be 0.
This simplifies to $(a-16)x + (b-16) = 0$.
By comparing the coefficients on both sides:
$a - 16 = 0 \Rightarrow a = 16$
$b - 16 = 0 \Rightarrow b = 16$
The values are $a = 16$ and $b = 16$.
37. The roots of the equation $x^2 + 6x - 4 = 0$ are $\alpha, \beta$. Find the quadratic equation whose roots are $\alpha^2$ and $\beta^2$.
Solution:
For the given equation $x^2 + 6x - 4 = 0$:
Sum of roots: $\alpha + \beta = -6$
Product of roots: $\alpha \beta = -4$
We need to form a new quadratic equation with roots $\alpha^2$ and $\beta^2$.
Sum of new roots: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-6)^2 - 2(-4) = 36 + 8 = 44$.
Product of new roots: $\alpha^2\beta^2 = (\alpha\beta)^2 = (-4)^2 = 16$.
The required quadratic equation is:
$x^2 - (\text{Sum of new roots})x + (\text{Product of new roots}) = 0$
$x^2 - 44x + 16 = 0$.
38. State and prove Thales Theorem.
Solution:
In ∆ABC, D is a point on AB and E is a point on AC.
To prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)
Construction: Draw a line DE || BC
| No. | Statement | Reason |
|---|---|---|
| 1. | ∠ABC = ∠ADE = ∠1 | Corresponding angles are equal because DE || BC |
| 2. | ∠ACB = ∠AED = ∠2 | Corresponding angles are equal because DE || BC |
| 3. | ∠DAE = ∠BAC = ∠3 | Both triangles have a common angle |
| ∆ABC ~ ∆ADE | By AAA similarity | |
| \(\frac{AB}{AD} = \frac{AC}{AE}\) | Corresponding sides are proportional | |
| \(\frac{AD+DB}{AD} = \frac{AE+EC}{AE}\) | Split AB and AC using the points D and E. | |
| \(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\) | On simplification | |
| \(\frac{DB}{AD} = \frac{EC}{AE}\) | Cancelling 1 on both sides | |
| \(\frac{AD}{DB} = \frac{AE}{EC}\) | Taking reciprocals. Hence proved. |
Hence, Thales Theorem is proved.
39. Find the area of the quadrilateral whose vertices are (-9, -2), (-8, -4), (2, 2) and (1, -3).
Solution:
To find the area of a quadrilateral using the shoelace formula, we must list the vertices in counter-clockwise order. A rough plot of the points suggests 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 = $\frac{1}{2} | (x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) |$
Step 1: Calculate the first part $(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1)$
$= (-9)(-4) + (-8)(-3) + (1)(2) + (2)(-2)$
$= 36 + 24 + 2 - 4 = 58$
Step 2: Calculate the second part $(y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)$
$= (-2)(-8) + (-4)(1) + (-3)(2) + (2)(-9)$
$= 16 - 4 - 6 - 18 = -12$
Step 3: Apply the formula
Area = $\frac{1}{2} | 58 - (-12) |$
Area = $\frac{1}{2} | 58 + 12 |$
Area = $\frac{1}{2} | 70 | = 35$
The area of the quadrilateral is 35 square units.
40. Find the equation of the median of $\triangle ABC$ through A where the vertices are A(6, 2), B(-5, -1) and C(1, 9).
Solution:
The median of a triangle through a vertex is the line segment that joins the vertex to the midpoint of the opposite side.
Step 1: Find the midpoint of the side BC.
Let D be the midpoint of BC. The coordinates of B are $(-5, -1)$ and C are $(1, 9)$.
Using the midpoint formula, $D = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$:
$D = \left(\frac{-5+1}{2}, \frac{-1+9}{2}\right) = \left(\frac{-4}{2}, \frac{8}{2}\right) = (-2, 4)$.
Step 2: Find the equation of the line passing through A and D.
We need to find the equation of the median AD, which passes through A(6, 2) and D(-2, 4).
We can use the two-point form of a line: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$.
Let $(x_1, y_1) = (6, 2)$ and $(x_2, y_2) = (-2, 4)$.
$\frac{y - 2}{4 - 2} = \frac{x - 6}{-2 - 6}$
$\frac{y - 2}{2} = \frac{x - 6}{-8}$
Now, cross-multiply:
$-8(y - 2) = 2(x - 6)$
Divide both sides by 2:
$-4(y - 2) = x - 6$
$-4y + 8 = x - 6$
Rearrange the terms to get the standard form:
$x + 4y - 6 - 8 = 0$
$x + 4y - 14 = 0$
The equation of the median through vertex A is $x + 4y - 14 = 0$.
41. If $\frac{\cos\theta}{1+\sin\theta} = \frac{1}{a}$, then prove that $\frac{a^2-1}{a^2+1} = \sin\theta$.
Solution:
We are given the relation: $\frac{\cos\theta}{1+\sin\theta} = \frac{1}{a}$
Step 1: Express 'a' in terms of $\theta$.
From the given equation, by taking the reciprocal on both sides, we get:
$a = \frac{1+\sin\theta}{\cos\theta}$
Step 2: Substitute 'a' into the expression $\frac{a^2-1}{a^2+1}$.
We will evaluate the numerator and the denominator separately.
First, find $a^2$:
$a^2 = \left(\frac{1+\sin\theta}{\cos\theta}\right)^2 = \frac{(1+\sin\theta)^2}{\cos^2\theta} = \frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta}$
Numerator ($a^2-1$):
$a^2 - 1 = \frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta} - 1$
$= \frac{1+2\sin\theta+\sin^2\theta - \cos^2\theta}{\cos^2\theta}$
Using the identity $\cos^2\theta = 1 - \sin^2\theta$:
$= \frac{1+2\sin\theta+\sin^2\theta - (1-\sin^2\theta)}{\cos^2\theta}$
$= \frac{1+2\sin\theta+\sin^2\theta - 1+\sin^2\theta}{\cos^2\theta} = \frac{2\sin\theta+2\sin^2\theta}{\cos^2\theta} = \frac{2\sin\theta(1+\sin\theta)}{\cos^2\theta}$
Denominator ($a^2+1$):
$a^2 + 1 = \frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta} + 1$
$= \frac{1+2\sin\theta+\sin^2\theta + \cos^2\theta}{\cos^2\theta}$
Using the identity $\sin^2\theta + \cos^2\theta = 1$:
$= \frac{1+2\sin\theta+1}{\cos^2\theta} = \frac{2+2\sin\theta}{\cos^2\theta} = \frac{2(1+\sin\theta)}{\cos^2\theta}$
Step 3: Compute the final ratio.
$\frac{a^2-1}{a^2+1} = \frac{\frac{2\sin\theta(1+\sin\theta)}{\cos^2\theta}}{\frac{2(1+\sin\theta)}{\cos^2\theta}}$
By cancelling the common terms $2$, $(1+\sin\theta)$, and $\cos^2\theta$ from the numerator and denominator, we are left with:
$= \sin\theta$
Thus, we have proved that $\frac{a^2-1}{a^2+1} = \sin\theta$.
42. Find the equation of a line passing through the point of intersection of the lines $4x+7y-3=0$ and $2x-3y+1=0$ that has equal intercepts on the axes.
Solution:
Step 1: Find the point of intersection.
(1) $4x + 7y = 3$
(2) $2x - 3y = -1$
Multiply equation (2) by 2: $4x - 6y = -2$ (3)
Subtract equation (3) from (1):
$(4x + 7y) - (4x - 6y) = 3 - (-2)$
$13y = 5 \Rightarrow y = \frac{5}{13}$.
Substitute $y$ in (2): $2x - 3(\frac{5}{13}) = -1 \Rightarrow 2x - \frac{15}{13} = -1 \Rightarrow 2x = \frac{15}{13} - 1 = \frac{2}{13} \Rightarrow x = \frac{1}{13}$.
The point of intersection is $(\frac{1}{13}, \frac{5}{13})$.
Step 2: Find the equation of the required line.
A line with equal intercepts 'a' has the equation $\frac{x}{a} + \frac{y}{a} = 1$, which simplifies to $x + y = a$.
Since this line passes through $(\frac{1}{13}, \frac{5}{13})$, we can substitute these values:
$\frac{1}{13} + \frac{5}{13} = a$
$a = \frac{6}{13}$.
So, the equation is $x + y = \frac{6}{13}$.
Multiplying by 13, we get $13x + 13y = 6$, or $13x + 13y - 6 = 0$.
SECTION - IV (Marks: 16)
Answer the following.
43. 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.
| Diameter (x) cm | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Circumference (y) cm | 3.1 | 6.2 | 9.3 | 12.4 | 15.5 |
(OR)
b) Draw the graph of $xy = 24$, $x, y > 0$. Using the graph find (i) y when x = 3 and (ii) x when y = 6.
Solution:
Solution for part (a)
1. Understanding the Relationship:
The data represents a direct variation between the diameter (x) and the circumference (y). The relationship is given by the formula $y = kx$ (where $k \approx \pi$). From the table, we can see that for every point, $y/x \approx 3.1$. This confirms a linear relationship, and the graph will be a straight line passing through the origin.
2. Drawing the Graph:
- Scale: Choose an appropriate scale. For the X-axis (Diameter), use 1 cm = 1 unit. For the Y-axis (Circumference), use 1 cm = 2 units.
- Plotting Points: Plot the given points on the graph paper: (1, 3.1), (2, 6.2), (3, 9.3), (4, 12.4), and (5, 15.5).
- Drawing the Line: Join the plotted points using a ruler. You will get a straight line that starts from the origin (0,0).
3. Finding the Circumference from the Graph:
- Locate the value 6 on the X-axis (Diameter).
- Draw a vertical dotted line from this point upwards until it intersects the straight line you drew.
- From the point of intersection, draw a horizontal dotted line to the left until it meets the Y-axis.
- Read the value on the Y-axis. This value will be approximately 18.6.
Result: From the graph, when the diameter is 6 cm, the circumference is approximately 18.6 cm.
Solution for part (b)
1. Understanding the Relationship:
The equation $xy = 24$ represents an inverse variation. As x increases, y decreases, and vice-versa. The graph will be a rectangular hyperbola in the first quadrant since $x, y > 0$.
2. Creating a Table of Values:
We need to find pairs of (x, y) such that their product is 24.
| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 |
|---|---|---|---|---|---|---|---|
| y = 24/x | 24 | 12 | 8 | 6 | 4 | 3 | 2 |
3. Drawing the Graph:
- Scale: Choose a suitable scale. For the X-axis, use 1 cm = 2 units. For the Y-axis, use 1 cm = 2 units.
- Plotting Points: Plot the points from the table, such as (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2).
- Drawing the Curve: Join the points with a smooth, continuous curve. This curve is the required graph.
4. Finding Values from the Graph:
- (i) Find y when x = 3:
Locate x = 3 on the X-axis. Draw a vertical line to meet the curve. From that point, draw a horizontal line to the Y-axis. The line meets the Y-axis at 8. - (ii) Find x when y = 6:
Locate y = 6 on the Y-axis. Draw a horizontal line to meet the curve. From that point, draw a vertical line to the X-axis. The line meets the X-axis at 4.
Result:
- (i) When $x = 3$, $y = 8$.
- (ii) When $y = 6$, $x = 4$.
44.
a) Construct a triangle similar to a given triangle ABC with its sides equal to 6/5 of the corresponding sides of the triangle ABC (scale factor 6/5 > 1).
(OR)
b) Construct a triangle $\triangle PQR$ such that $QR = 5$cm, $\angle P = 30^\circ$ and the altitude from P to QR is of length 4.2cm.
Solution:
These are geometric construction problems requiring a compass and a ruler. The following are the steps for construction.
Solution for part (a)
Objective: To construct a triangle A'BC' which is similar to $\triangle ABC$ and whose sides are $\frac{6}{5}$ times the corresponding sides of $\triangle ABC$. Since the scale factor $\frac{6}{5} > 1$, the new triangle will be larger than the original.
Steps of Construction:
- Draw any triangle $\triangle ABC$ with suitable measurements.
- From vertex B, draw a ray BX downwards, making an acute angle with the base BC (i.e., $\angle CBX$ is acute).
- On the ray BX, mark 6 equidistant points using a compass. Let them be $B_1, B_2, B_3, B_4, B_5, B_6$ such that $BB_1 = B_1B_2 = \dots = B_5B_6$.
- Join the 5th point, $B_5$, to the vertex C (since 5 is the denominator of the scale factor).
- Extend the line segment BC. Now, from the 6th point, $B_6$, draw a line parallel to $B_5C$. This parallel line must intersect the extended line segment BC at a point. Call this point C'. ($B_6C' \parallel B_5C$).
- Extend the line segment BA. From C', draw a line parallel to AC. This line must intersect the extended line segment BA at a point. Call this point A'. ($A'C' \parallel AC$).
- The triangle $\triangle A'BC'$ is the required similar triangle.
Solution for part (b)
Objective: To construct $\triangle PQR$ with base $QR = 5$cm, vertical angle $\angle P = 30^\circ$, and altitude from P to QR being 4.2 cm.
Steps of Construction:
- Draw a line segment $QR = 5$ cm.
- At point Q, draw a line QX making an angle of 30° with QR (i.e., $\angle RQX = 30^\circ$).
- At point Q, draw a line QY perpendicular to QX (i.e., $\angle XQY = 90^\circ$).
- Draw the perpendicular bisector of the line segment QR. Let it be MN, and let it intersect QR at point G.
- Let the perpendicular bisector MN and the perpendicular line QY intersect at a point O.
- With O as the center and OQ as the radius, draw a circle. The major arc of this circle contains the vertex P.
- On the perpendicular bisector MN, mark a point H such that $GH = 4.2$ cm (the given altitude).
- Draw a line ST through H, parallel to the base QR.
- This parallel line ST intersects the circle at two points, P and P'.
- Join PQ and PR. The triangle $\triangle PQR$ is the required triangle. (Joining P'Q and P'R would also form a valid triangle).