COMMON QUARTERLY EXAMINATION - 2025: Standard X Mathematics - Complete Solutions
Part - I (14 x 1 = 14)
Choose the correct answer.
1. If $n(A \times B) = 6$ and $A = \{1, 3\}$ then $n(B)$ is
- a) 1
- b) 2
- c) 3
- d) 6
Solution:
We know that $n(A \times B) = n(A) \times n(B)$.
Given, $n(A \times B) = 6$ and $A = \{1, 3\}$, so $n(A) = 2$.
Therefore, $6 = 2 \times n(B)$.
$n(B) = \frac{6}{2} = 3$.
2. The range of the relation $R = \{(x, x^2) \mid x \text{ is a prime number less than 13}\}$ is
- a) {2,3,5,7}
- b) {2,3,5,7,11}
- c) {4,9,25,49,121}
- d) {1,4,9,25,49,121}
Solution:
The prime numbers less than 13 are 2, 3, 5, 7, 11.
The relation is $R = \{(x, x^2)\}$. The domain is the set of these prime numbers.
The range is the set of the second elements of the ordered pairs, which are the squares of these primes.
Range = $\{2^2, 3^2, 5^2, 7^2, 11^2\} = \{4, 9, 25, 49, 121\}$.
3. If $f : A \to B$ is a bijective function and if $n(B) = 7$, then $n(A)$ is equal to
- a) 7
- b) 49
- c) 1
- d) 14
Solution:
A bijective function is both one-to-one and onto. In a bijective function, each element of the domain A is paired with exactly one unique element of the codomain B, and every element of B is an image of some element of A.
This implies that the number of elements in the domain must be equal to the number of elements in the codomain.
Therefore, $n(A) = n(B)$.
Given $n(B) = 7$, so $n(A) = 7$.
4. Euclid's division lemma states that for positive integers a and b, there exist unique integers q and r such that $a = bq + r$, where r must satisfy.
- a) $1 < r < b$
- b) $0 < r \le b$
- c) $0 \le r < b$
- d) $0 < r < b$
Solution:
According to Euclid's division lemma, for any two positive integers 'a' and 'b', there exist unique integers 'q' (quotient) and 'r' (remainder) such that $a = bq + r$, where the remainder 'r' is greater than or equal to 0 and strictly less than the divisor 'b'.
The condition is $0 \le r < b$.
5. If 6 times of 6th term of an A.P. is equal to 7 times the 7th term, then the 13th term of the A.P. is
- a) 0
- b) 6
- c) 7
- d) 13
Solution:
Let the A.P. have first term 'a' and common difference 'd'.
The nth term is given by $t_n = a + (n-1)d$.
Given, $6 \times t_6 = 7 \times t_7$.
$6(a + (6-1)d) = 7(a + (7-1)d)$
$6(a + 5d) = 7(a + 6d)$
$6a + 30d = 7a + 42d$
$0 = 7a - 6a + 42d - 30d$
$0 = a + 12d$
The 13th term is $t_{13} = a + (13-1)d = a + 12d$.
Since we found $a + 12d = 0$, the 13th term is 0.
6. If the sequence $t_1, t_2, t_3, \dots$ are in A.P. then the sequence $t_6, t_{12}, t_{18}, \dots$ is
- a) a Geometric Progression
- b) an Arithmetic Progression
- c) neither an Arithmetic Progression nor a Geometric Progression
- d) a constant sequence
Solution:
Let the original A.P. be $a, a+d, a+2d, \dots$. So $t_n = a + (n-1)d$.
The new sequence is $t_6, t_{12}, t_{18}, \dots$.
$t_6 = a + 5d$
$t_{12} = a + 11d$
$t_{18} = a + 17d$
Let's check the difference between consecutive terms:
$t_{12} - t_6 = (a + 11d) - (a + 5d) = 6d$
$t_{18} - t_{12} = (a + 17d) - (a + 11d) = 6d$
Since the difference between consecutive terms is constant (6d), the new sequence is also an Arithmetic Progression.
7. If $(x-6)$ is the HCF of $x^2 - 2x - 24$ and $x^2 - kx - 6$ then the value of k is
- a) 3
- b) 5
- c) 6
- d) 8
Solution:
If $(x-6)$ is the HCF of the two polynomials, then $(x-6)$ is a factor of both.
If $(x-6)$ is a factor of $p(x) = x^2 - kx - 6$, then $p(6)$ must be 0.
$p(6) = (6)^2 - k(6) - 6 = 0$
$36 - 6k - 6 = 0$
$30 - 6k = 0$
$30 = 6k$
$k = \frac{30}{6} = 5$
8. The values of a and b if $4x^4 - 24x^3 + 76x^2 + ax + b$ is a perfect square are
- a) 100, 120
- b) 10, 12
- c) -120, 100
- d) 12, 10
Solution:
We use the long division method to find the square root.
The square root of the first term $4x^4$ is $2x^2$.
Step 1: Divide $4x^4 - 24x^3 + ...$ by $2x^2$. Quotient is $2x^2$. Remainder is $-24x^3 + 76x^2$.
New divisor: $2 \times (2x^2) = 4x^2$.
Step 2: Divide $-24x^3$ by $4x^2$ to get $-6x$. New term in quotient is $-6x$.
$(4x^2 - 6x) \times (-6x) = -24x^3 + 36x^2$.
Subtracting this gives $(76x^2) - (36x^2) = 40x^2$. Bring down $ax+b$.
Step 3: New divisor: $2 \times (2x^2 - 6x) = 4x^2 - 12x$.
Divide $40x^2$ by $4x^2$ to get $10$. New term in quotient is $+10$.
$(4x^2 - 12x + 10) \times (10) = 40x^2 - 120x + 100$.
For the original polynomial to be a perfect square, the remainder must be 0. So, $40x^2 + ax + b = 40x^2 - 120x + 100$.
Comparing coefficients: $a = -120$ and $b = 100$.
9. In $\Delta LMN$, $\angle L = 60^\circ$, $\angle M = 50^\circ$. If $\Delta LMN \sim \Delta PQR$ then the value of $\angle R$ is
- a) 40°
- b) 70°
- c) 30°
- d) 110°
Solution:
In $\Delta LMN$, the sum of angles is $180^\circ$.
$\angle L + \angle M + \angle N = 180^\circ$
$60^\circ + 50^\circ + \angle N = 180^\circ$
$110^\circ + \angle N = 180^\circ$
$\angle N = 180^\circ - 110^\circ = 70^\circ$
Given $\Delta LMN \sim \Delta PQR$. This means corresponding angles are equal.
$\angle L = \angle P$, $\angle M = \angle Q$, and $\angle N = \angle R$.
Therefore, $\angle R = \angle N = 70^\circ$.
10. In a given figure ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of $\Delta PQR$ to the area of $\Delta PST$ is
- a) 25:4
- b) 25:7
- c) 25:11
- d) 25:13
Solution:
Given ST || QR. By the property of similar triangles, $\Delta PST \sim \Delta PQR$.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta PST)} = \left(\frac{PQ}{PS}\right)^2$
We have $PS = 2$ cm and $SQ = 3$ cm. So, $PQ = PS + SQ = 2 + 3 = 5$ cm.
$\frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta PST)} = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$
The ratio is 25:4.
11. If (5,7), (3,p) and (6,6) are collinear, then the value of p is
- a) 3
- b) 6
- c) 9
- d) 12
Solution:
If three points are collinear, the slope between any two pairs of points will be the same.
Let A=(5,7), B=(3,p), C=(6,6).
Slope of AB = Slope of BC
$\frac{p - 7}{3 - 5} = \frac{6 - p}{6 - 3}$
$\frac{p - 7}{-2} = \frac{6 - p}{3}$
$3(p - 7) = -2(6 - p)$
$3p - 21 = -12 + 2p$
$3p - 2p = -12 + 21$
$p = 9$
12. The slope of the line which is perpendicular to a line joining the points (0,0) and (-8, 8) is
- a) -1
- b) 1
- c) 1/3
- d) -8
Solution:
First, find the slope ($m_1$) of the line joining (0,0) and (-8,8).
$m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 0}{-8 - 0} = \frac{8}{-8} = -1$
Let the slope of the perpendicular line be $m_2$.
The condition for perpendicular lines is $m_1 \times m_2 = -1$.
$(-1) \times m_2 = -1$
$m_2 = \frac{-1}{-1} = 1$
13. $\tan\theta \csc^2\theta - \tan\theta$ is equal to
- a) $\sec\theta$
- b) $\cot^2\theta$
- c) $\sin\theta$
- d) $\cot\theta$
Solution:
Factor out $\tan\theta$ from the expression:
$\tan\theta (\csc^2\theta - 1)$
Using the Pythagorean identity $1 + \cot^2\theta = \csc^2\theta$, we get $\csc^2\theta - 1 = \cot^2\theta$.
So the expression becomes $\tan\theta \times \cot^2\theta$.
Since $\cot\theta = \frac{1}{\tan\theta}$, we have:
$\tan\theta \times \frac{1}{\tan^2\theta} = \frac{1}{\tan\theta} = \cot\theta$
14. If f is the identity function, then the value of $f(1) - 2f(2) + f(3)$ is
- a) 1
- b) 0
- c) -1
- d) -3
Solution:
An identity function is defined as $f(x) = x$.
So, $f(1) = 1$, $f(2) = 2$, and $f(3) = 3$.
The expression is $f(1) - 2f(2) + f(3)$.
$= 1 - 2(2) + 3$
$= 1 - 4 + 3$
$= 0$
Part - II (10 x 2 = 20)
Answer any 10 questions. (Q.No. 28 is compulsory)
15. Find $A \times B$ and $B \times A$ if $A = \{2, -2, 3\}$ and $B = \{1, -4\}$.
Solution:
Given $A = \{2, -2, 3\}$ and $B = \{1, -4\}$.
The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.
$A \times B = \{(2, 1), (2, -4), (-2, 1), (-2, -4), (3, 1), (3, -4)\}$.
The Cartesian product $B \times A$ is the set of all ordered pairs $(b, a)$ where $b \in B$ and $a \in A$.
$B \times A = \{(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4, 3)\}$.
16. Let $f = \{(x, y) \mid x, y \in N \text{ and } y = 2x\}$ be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Solution:
The relation is defined as $y = 2x$, where both $x$ and $y$ are natural numbers (N = {1, 2, 3, ...}).
Domain: The set of all possible input values for x. Since $x \in N$, the domain is the set of all natural numbers.
Domain = N = {1, 2, 3, ...}.
Co-domain: The set of all possible output values. Since the relation is on N, the co-domain is N.
Co-domain = N = {1, 2, 3, ...}.
Range: The set of all actual output values (y).
When x=1, y=2. When x=2, y=4. When x=3, y=6, and so on.
Range = {2, 4, 6, 8, ...} = The set of all even natural numbers.
Is this relation a function? Yes. For every element $x$ in the domain (N), there is one and only one image $y = 2x$ in the co-domain (N). Therefore, the relation is a function.
17. If $f(x) = 3x-2$, $g(x) = 2x + k$ and if $fog = gof$, then find the value of k.
Solution:
Given $f(x) = 3x - 2$ and $g(x) = 2x + k$. We are also given $fog = gof$.
$fog(x) = f(g(x)) = f(2x+k) = 3(2x+k) - 2 = 6x + 3k - 2$.
$gof(x) = g(f(x)) = g(3x-2) = 2(3x-2) + k = 6x - 4 + k$.
Since $fog(x) = gof(x)$:
$6x + 3k - 2 = 6x - 4 + k$
$3k - 2 = -4 + k$
$3k - k = -4 + 2$
$2k = -2$
$k = -1$
18. Find the number of terms in the A.P. 3, 6, 9, 12, ..., 111.
Solution:
The given Arithmetic Progression (A.P.) is 3, 6, 9, ..., 111.
First term, $a = 3$.
Common difference, $d = 6 - 3 = 3$.
Last term, $l = t_n = 111$.
The formula for the number of terms is $n = \frac{l - a}{d} + 1$.
$n = \frac{111 - 3}{3} + 1$
$n = \frac{108}{3} + 1$
$n = 36 + 1 = 37$
19. If 3+k, 18-k, 5k+1 are in A.P, then find k.
Solution:
If three terms $a, b, c$ are in A.P., then the middle term is the arithmetic mean of the other two, i.e., $2b = a + c$.
Here, $a = 3+k$, $b = 18-k$, and $c = 5k+1$.
$2(18 - k) = (3 + k) + (5k + 1)$
$36 - 2k = 6k + 4$
$36 - 4 = 6k + 2k$
$32 = 8k$
$k = \frac{32}{8} = 4$
20. Find the sum of 1 + 3 + 5 + ... to 40 terms.
Solution:
The series is an A.P. of odd numbers.
First term, $a = 1$.
Common difference, $d = 3 - 1 = 2$.
Number of terms, $n = 40$.
The sum of an A.P. is $S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_{40} = \frac{40}{2}[2(1) + (40-1)2]$
$S_{40} = 20[2 + (39)2]$
$S_{40} = 20[2 + 78] = 20 = 1600$.
Alternatively, the sum of the first 'n' odd natural numbers is $n^2$.
Here $n=40$, so the sum is $40^2 = 1600$.
21. Find the sum: $3 + 1 + \frac{1}{3} + \dots \infty$
Solution:
The given series is a Geometric Progression (G.P.).
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 G.P. is $S_\infty = \frac{a}{1 - r}$.
$S_\infty = \frac{3}{1 - \frac{1}{3}} = \frac{3}{\frac{2}{3}} = 3 \times \frac{3}{2} = \frac{9}{2}$
22. Find the sum and product of the roots for $x^2 + 3x - 28 = 0$.
Solution:
For a quadratic equation $ax^2 + bx + c = 0$, with roots $\alpha$ and $\beta$:
Sum of roots, $\alpha + \beta = -\frac{b}{a}$.
Product of roots, $\alpha \beta = \frac{c}{a}$.
Comparing $x^2 + 3x - 28 = 0$ with the standard form, we have $a=1, b=3, c=-28$.
Sum of roots = $-\frac{3}{1} = -3$.
Product of roots = $\frac{-28}{1} = -28$.
23. Determine the nature of the roots for $2x^2 - 2x + 9 = 0$.
Solution:
To determine the nature of the roots, we find the discriminant, $\Delta = b^2 - 4ac$.
For the equation $2x^2 - 2x + 9 = 0$, we have $a=2, b=-2, c=9$.
$\Delta = (-2)^2 - 4(2)(9)$
$\Delta = 4 - 72 = -68$.
Since $\Delta < 0$, the roots are not real (they are imaginary and unequal).
24. If the difference between a number and its reciprocal is 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 $x = -1/5$.
25. If $\Delta ABC$ is similar to $\Delta DEF$ such that $BC = 3$ cm, $EF = 4$ cm and area of $\Delta ABC = 54$ cm², find the area of $\Delta DEF$.
Solution:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \left(\frac{BC}{EF}\right)^2$
$\frac{54}{\text{Area}(\Delta DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$
$\text{Area}(\Delta DEF) = 54 \times \frac{16}{9}$
$\text{Area}(\Delta DEF) = 6 \times 16 = 96$ cm².
26. Find the slope of a line joining the points (14,10) and (14,-6).
Solution:
The formula for the slope (m) of a line joining points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
$m = \frac{-6 - 10}{14 - 14} = \frac{-16}{0}$.
Since the denominator is zero, the slope is undefined. This indicates that the line is a vertical line.
27. Prove that $\tan^2\theta - \sin^2\theta = \tan^2\theta \sin^2\theta$.
Solution:
LHS = $\tan^2\theta - \sin^2\theta$
= $\frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta$
= $\sin^2\theta \left( \frac{1}{\cos^2\theta} - 1 \right)$
= $\sin^2\theta \left( \sec^2\theta - 1 \right)$
Using the identity $1 + \tan^2\theta = \sec^2\theta \Rightarrow \sec^2\theta - 1 = \tan^2\theta$.
= $\sin^2\theta (\tan^2\theta)$
= $\tan^2\theta \sin^2\theta$ = RHS.
28. Show that the given points are collinear: (1,2), (2,3) and (3,4).
Solution:
To show that the points are collinear, we can prove that the slope between the first two points is equal to the slope between the last two points.
Let the points be A(1,2), B(2,3), and C(3,4).
Slope of AB = $\frac{3 - 2}{2 - 1} = \frac{1}{1} = 1$.
Slope of BC = $\frac{4 - 3}{3 - 2} = \frac{1}{1} = 1$.
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. Therefore, they are collinear.
Part - III (10 x 5 = 50)
Answer any 10 questions. (Q.No. 42 is compulsory)
29. Let $f : A \to B$ 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
Solution:
First, let's find the images for each element in set 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:
$f = \{(2,0), (4,1), (6,2), (10,4), (12,5)\}$.
ii) A table:
| x | 2 | 4 | 6 | 10 | 12 |
|---|---|---|---|---|---|
| f(x) | 0 | 1 | 2 | 4 | 5 |
iii) An arrow diagram:
(An arrow diagram would show two ovals, one for set A and one for set B. Arrows would be drawn from each element in A to its corresponding image in B: 2→0, 4→1, 6→2, 10→4, 12→5.)
iv) A graph:
(A graph would be plotted on a coordinate plane with the x-axis representing elements of A and the y-axis representing elements of B. The points (2,0), (4,1), (6,2), (10,4), and (12,5) would be plotted as distinct points.)
30. If the function $f : R \to R$ is defined by
$f(x) = \begin{cases} 2x+7 &; x<-2 \\ x^2-2 &; -2 \le x < 3 \\ 3x-2 &; x \ge 3 \end{cases}$
then find the values of
i) f(4) ii) f(-2) iii) f(4) + 2f(1) iv) [f(1) + 3f(4)] / f(-3)
Solution:
i) f(4): Since $4 \ge 3$, we use the definition $f(x) = 3x - 2$.
$f(4) = 3(4) - 2 = 12 - 2 = 10$.
ii) f(-2): Since $-2 \le -2 < 3$, we use the definition $f(x) = x^2 - 2$.
$f(-2) = (-2)^2 - 2 = 4 - 2 = 2$.
iii) f(4) + 2f(1): We know $f(4) = 10$. For $f(1)$, since $-2 \le 1 < 3$, we use $f(x) = x^2 - 2$.
$f(1) = (1)^2 - 2 = 1 - 2 = -1$.
$f(4) + 2f(1) = 10 + 2(-1) = 10 - 2 = 8$.
iv) [f(1) + 3f(4)] / f(-3): We have $f(1) = -1$ and $f(4) = 10$. For $f(-3)$, since $-3 < -2$, we use $f(x) = 2x + 7$.
$f(-3) = 2(-3) + 7 = -6 + 7 = 1$.
$\frac{f(1) + 3f(4)}{f(-3)} = \frac{-1 + 3(10)}{1} = \frac{-1 + 30}{1} = 29$.
31. Find the HCF of 396, 504, 636.
Solution:
We will use Euclid's division algorithm.
Step 1: Find HCF of 504 and 396.
$504 = 396 \times 1 + 108$
$396 = 108 \times 3 + 72$
$108 = 72 \times 1 + 36$
$72 = 36 \times 2 + 0$
The HCF of 504 and 396 is 36.
Step 2: Find HCF of the result (36) and the third number (636).
$636 = 36 \times 17 + 24$
$36 = 24 \times 1 + 12$
$24 = 12 \times 2 + 0$
The HCF of 36 and 636 is 12.
32. Find the sum of all natural numbers between 300 and 600 which are divisible by 7.
Solution:
The numbers form an Arithmetic Progression.
First number > 300 divisible by 7: $300 \div 7$ gives remainder 6. So, the first number is $300 - 6 + 7 = 301$. So, $a = 301$.
Last number < 600 divisible by 7: $600 \div 7$ gives remainder 5. So, the last number is $600 - 5 = 595$. So, $l = 595$.
The common difference is $d=7$.
Number of terms, $n = \frac{l-a}{d} + 1 = \frac{595-301}{7} + 1 = \frac{294}{7} + 1 = 42 + 1 = 43$.
Sum of the terms, $S_n = \frac{n}{2}(a+l)$.
$S_{43} = \frac{43}{2}(301 + 595) = \frac{43}{2}(896) = 43 \times 448 = 19264$.
33. Find the sum to n terms of the series 3 + 33 + 333 + ... to n terms.
Solution:
Let $S_n = 3 + 33 + 333 + \dots$ to n terms.
$S_n = 3(1 + 11 + 111 + \dots)$
Multiply and divide by 9:
$S_n = \frac{3}{9}(9 + 99 + 999 + \dots)$
$S_n = \frac{1}{3}((10-1) + (100-1) + (1000-1) + \dots \text{to n terms})$
$S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots) - (1+1+1+\dots \text{n times})]$
The first part is a G.P. with $a=10, r=10$. Its sum is $\frac{a(r^n-1)}{r-1} = \frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)$.
The second part is simply $n$.
$S_n = \frac{1}{3} \left[ \frac{10}{9}(10^n-1) - n \right]$
$S_n = \frac{10}{27}(10^n-1) - \frac{n}{3}$
34. Find the square root of $64x^4 - 16x^3 + 17x^2 - 2x + 1$.
Solution:
Explanation
Using the long division method for finding the square root:
The square root of the first term $64x^4$ is $8x^2$. This is the first term of our answer.
Step 1: Subtract $(8x^2)^2 = 64x^4$ from the polynomial. Remainder is $-16x^3 + 17x^2 - 2x + 1$.
Step 2: Double the current answer: $2(8x^2) = 16x^2$. Divide the first term of the remainder ($-16x^3$) by this ($16x^2$) to get $-x$. This is the second term of our answer.
Step 3: Multiply the new divisor $(16x^2 - x)$ by the new term $(-x)$: $(16x^2 - x)(-x) = -16x^3 + x^2$. Subtract this from the remainder.
$(-16x^3 + 17x^2) - (-16x^3 + x^2) = 16x^2$. Bring down the remaining terms: $16x^2 - 2x + 1$.
Step 4: Double the current answer: $2(8x^2 - x) = 16x^2 - 2x$. Divide the first term of the new remainder ($16x^2$) by the first term of this ($16x^2$) to get $+1$. This is the third term of our answer.
Step 5: Multiply the new divisor $(16x^2 - 2x + 1)$ by the new term $(+1)$: $(16x^2 - 2x + 1)(1) = 16x^2 - 2x + 1$. Subtracting this gives a remainder of 0.
35. Simplify. $\frac{1}{x^2-5x+6} + \frac{1}{x^2-3x+2} - \frac{1}{x^2-8x+15}$
Solution:
First, factorize the denominators:
- $x^2-5x+6 = (x-2)(x-3)$
- $x^2-3x+2 = (x-1)(x-2)$
- $x^2-8x+15 = (x-3)(x-5)$
The expression becomes: $\frac{1}{(x-2)(x-3)} + \frac{1}{(x-1)(x-2)} - \frac{1}{(x-3)(x-5)}$
Combine the first two terms:
$\frac{(x-1) + (x-3)}{(x-1)(x-2)(x-3)} = \frac{2x-4}{(x-1)(x-2)(x-3)} = \frac{2(x-2)}{(x-1)(x-2)(x-3)} = \frac{2}{(x-1)(x-3)}$
Now, subtract the third term from this result:
$\frac{2}{(x-1)(x-3)} - \frac{1}{(x-3)(x-5)}$
The common denominator is $(x-1)(x-3)(x-5)$.
$\frac{2(x-5) - 1(x-1)}{(x-1)(x-3)(x-5)} = \frac{2x - 10 - x + 1}{(x-1)(x-3)(x-5)} = \frac{x-9}{(x-1)(x-3)(x-5)}$
36. Find the GCD of the following using division algorithm. $2x^4 + 13x^3 + 27x^2 + 23x + 7$, $x^3 + 3x^2 + 3x + 1$, $x^2 + 2x + 1$.
Solution:
Let the polynomials be $p(x), q(x), r(x)$.
$p(x) = 2x^4 + 13x^3 + 27x^2 + 23x + 7$
$q(x) = x^3 + 3x^2 + 3x + 1 = (x+1)^3$
$r(x) = x^2 + 2x + 1 = (x+1)^2$
First, find the GCD of $q(x)$ and $r(x)$. Since $r(x)$ is a factor of $q(x)$, their GCD is $r(x) = (x+1)^2$.
Now, we need to find the GCD of $p(x)$ and $(x+1)^2$. We can do this by dividing $p(x)$ by $(x+1)^2 = x^2 + 2x + 1$.
Using polynomial long division:
Dividing $2x^4 + 13x^3 + 27x^2 + 23x + 7$ by $x^2 + 2x + 1$ gives a quotient of $2x^2 + 9x + 7$ and a remainder of 0.
$(x^2+2x+1)(2x^2+9x+7) = 2x^4+9x^3+7x^2+4x^3+18x^2+14x+2x^2+9x+7 = 2x^4+13x^3+27x^2+23x+7$.
Since the remainder is 0, $x^2+2x+1$ is a factor of $p(x)$.
Therefore, the GCD of all three polynomials is $x^2+2x+1$.
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 $\Delta ABC$, a line DE is parallel to BC, intersecting AB at D and AC at E.
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$
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. |
38. Find the area of the quadrilateral whose vertices are at (-9,0), (-8,6), (-1,-2) and (-6,-3).
Solution:
Let the vertices be A(-9,0), B(-8,6), C(-1,-2), D(-6,-3). We use the shoelace formula for the area of a quadrilateral:
Area = $\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$
Area = $\frac{1}{2} |((-9)(6) + (-8)(-2) + (-1)(-3) + (-6)(0)) - ((0)(-8) + (6)(-1) + (-2)(-6) + (-3)(-9))|$
Area = $\frac{1}{2} |(-54 + 16 + 3 + 0) - (0 - 6 + 12 + 27)|$
Area = $\frac{1}{2} |(-35) - (33)|$
Area = $\frac{1}{2} |-35 - 33| = \frac{1}{2} |-68| = \frac{68}{2} = 34$
39. Without using Pythagoras theorem, show that the points (1,-4), (2,-3) and (4,-7) form a right angled triangle.
Solution:
We can show this by using the concept of slopes. If two lines are perpendicular, the product of their slopes is -1.
Let the points be A(1,-4), B(2,-3), and C(4,-7).
Slope of AB ($m_{AB}$) = $\frac{-3 - (-4)}{2 - 1} = \frac{1}{1} = 1$.
Slope of BC ($m_{BC}$) = $\frac{-7 - (-3)}{4 - 2} = \frac{-4}{2} = -2$.
Slope of AC ($m_{AC}$) = $\frac{-7 - (-4)}{4 - 1} = \frac{-3}{3} = -1$.
Now, let's check the product of the slopes:
$m_{AB} \times m_{AC} = (1) \times (-1) = -1$.
Since the product of the slopes of lines AB and AC is -1, the lines AB and AC are perpendicular. This means the angle at vertex A is $90^\circ$.
40. Find the equation of a straight line passing through (1,-4) and has intercepts which are in the ratio 2:5.
Solution:
Let the x-intercept be $2a$ and the y-intercept be $5a$.
The equation of the line in intercept form is $\frac{x}{x-intercept} + \frac{y}{y-intercept} = 1$.
$\frac{x}{2a} + \frac{y}{5a} = 1$.
The line passes through the point (1, -4). So, we substitute x=1 and y=-4 into the equation:
$\frac{1}{2a} + \frac{-4}{5a} = 1$
To solve for 'a', find a common denominator (10a):
$\frac{5}{10a} - \frac{8}{10a} = 1 \Rightarrow \frac{-3}{10a} = 1 \Rightarrow -3 = 10a \Rightarrow a = -\frac{3}{10}$.
Now find the intercepts:
x-intercept = $2a = 2(-\frac{3}{10}) = -\frac{3}{5}$.
y-intercept = $5a = 5(-\frac{3}{10}) = -\frac{3}{2}$.
The equation is $\frac{x}{-3/5} + \frac{y}{-3/2} = 1$.
$-\frac{5x}{3} - \frac{2y}{3} = 1$.
Multiply by 3: $-5x - 2y = 3$.
$5x + 2y + 3 = 0$.
41. Prove the following identity: $\sqrt{\frac{1+\sin\theta}{1-\sin\theta}} + \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} = 2 \sec \theta$
Solution:
LHS = $\sqrt{\frac{1+\sin\theta}{1-\sin\theta}} + \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}$
Rationalize the denominator for each term:
= $\sqrt{\frac{1+\sin\theta}{1-\sin\theta} \times \frac{1+\sin\theta}{1+\sin\theta}} + \sqrt{\frac{1-\sin\theta}{1+\sin\theta} \times \frac{1-\sin\theta}{1-\sin\theta}}$
= $\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}} + \sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}$
Using the identity $\sin^2\theta + \cos^2\theta = 1 \Rightarrow 1-\sin^2\theta = \cos^2\theta$.
= $\sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}} + \sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}}$
= $\frac{1+\sin\theta}{\cos\theta} + \frac{1-\sin\theta}{\cos\theta}$
= $\frac{1+\sin\theta + 1-\sin\theta}{\cos\theta}$
= $\frac{2}{\cos\theta} = 2 \sec\theta$ = RHS.
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 number. Verify that $(A \cap B) \times C = (A \times C) \cap (B \times C)$.
Solution:
First, let's define the sets:
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2} (The only even prime number is 2)
LHS: $(A \cap B) \times C$
$A \cap B = \{2, 3, 5, 7\}$
$(A \cap B) \times C = \{2, 3, 5, 7\} \times \{2\} = \{(2,2), (3,2), (5,2), (7,2)\}$
RHS: $(A \times C) \cap (B \times C)$
$A \times C = \{(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)\}$
$B \times C = \{(2,2), (3,2), (5,2), (7,2)\}$
Now find the intersection of these two sets:
$(A \times C) \cap (B \times C) = \{(2,2), (3,2), (5,2), (7,2)\}$
Since LHS = RHS, the identity is verified.
Part - IV (2 x 8 = 16)
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:
Since the scale factor is \(\frac{7}{3}\), which is greater than 1, the new triangle will be larger than the original triangle PQR.
- Draw a triangle PQR with any suitable measurements.
- Draw a ray QX making an acute angle with QR, on the side opposite to vertex P.
- Locate 7 points \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6, Q_7\) on the ray QX such that the distances between them are equal (\(QQ_1 = Q_1Q_2 = \dots = Q_6Q_7\)).
- Join \(Q_3\) (the 3rd point, as 3 is the denominator) to R.
- Draw a line through \(Q_7\) parallel to \(Q_3R\). This line will intersect the extended line segment QR at a point R'.
- Draw a line through R' parallel to PR. This line will intersect the extended line segment QP at a point P'.
- \(\triangle P'QR'\) is the required similar triangle, with each side being \(\frac{7}{3}\) times the corresponding side of \(\triangle PQR\).
43. b) (OR) Draw a triangle ABC of base BC = 8 cm, $\angle A = 60^\circ$ and the bisector of $\angle A$ meets BC at D such that BD = 6 cm.
Solution:
Steps of Construction:
44. a) Varshika drew six circles with different sizes. Draw a graph for the relationship between the diameter and circumference (approximately related) of each circle as shown in the table and use it to find the circumference of a circle when its diameter is 6 cm.
| Diameter (x) cm | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Circumference (y) cm | 3.1 | 6.2 | 9.3 | 12.4 | 15.5 |
Solution:
The relationship between diameter (x) and circumference (y) is given by $y = \pi x$. From the table, we can see the constant of proportionality is approximately 3.1 ($y/x \approx 3.1$). So, the relationship is $y = 3.1x$. This is a direct variation, and its graph is a straight line passing through the origin.
Steps to draw the graph and find the solution:
- Choose a suitable scale for the x-axis (Diameter) and y-axis (Circumference). For example, on the x-axis, 1 cm = 1 unit, and on the y-axis, 1 cm = 2 units.
- Plot the given points: (1, 3.1), (2, 6.2), (3, 9.3), (4, 12.4), (5, 15.5).
- Draw a straight line passing through these points and the origin (0,0). This line represents the relationship $y=3.1x$.
- To find the circumference when the diameter is 6 cm, locate the value x = 6 on the x-axis.
- Draw a vertical line from x = 6 upwards until it intersects the graphed line.
- From this point of intersection, draw a horizontal line to the left until it meets the y-axis.
- The value on the y-axis at this point is the required circumference.
From the graph, we will find that when x = 6 cm, y is approximately 18.6 cm.
Calculation Check: Using the equation $y = 3.1x$, for $x=6$, $y = 3.1 \times 6 = 18.6$ cm.
44. b) (OR) Draw the graph of $xy = 24$, $x, y > 0$. Using the graph find
i) y when x = 3 and ii) x when y = 6
Solution:
The equation $xy = 24$ represents indirect variation. Its graph is a rectangular hyperbola. Since $x, y > 0$, we will only draw the curve in the first quadrant.
Step 1: Create a table of values.
From $y = 24/x$, we can find corresponding y values for chosen x values:
| x | 1 | 2 | 3 | 4 | 6 | 8 | 12 |
|---|---|---|---|---|---|---|---|
| y | 24 | 12 | 8 | 6 | 4 | 3 | 2 |
Step 2: Draw the graph.
- Choose a suitable scale. For example, on the x-axis, 1 cm = 2 units, and on the y-axis, 1 cm = 2 units.
- Plot the points from the table: (1, 24), (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2).
- Join the points with a smooth curve. This curve is the graph of $xy=24$.
Step 3: Find the values from the graph.
i) To find y when x = 3:
- Locate x = 3 on the x-axis.
- Draw a vertical line from this point to the curve.
- From the point of intersection on the curve, draw a horizontal line to the y-axis.
- The value on the y-axis is the required value of y.
From the graph, when x = 3, y = 8.
ii) To find x when y = 6:
- Locate y = 6 on the y-axis.
- Draw a horizontal line from this point to the curve.
- From the point of intersection on the curve, draw a vertical line to the x-axis.
- The value on the x-axis is the required value of x.
From the graph, when y = 6, x = 4.
10th Maths EM - Quarterly Exam Paper
