10th Maths Quarterly Exam 2024 Question Paper with Answer Key
PART - A (14 x 1 = 14)
Answer all the questions. Choose the correct answer from the given four alternatives.
1. 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 .................
Answer: b) 2
Solution:
Given, $n(A) = 5$.
Number of relations from A to B is $2^{n(A) \times n(B)}$.
$2^{n(A) \times n(B)} = 1024$
We know that $1024 = 2^{10}$.
So, $2^{5 \times n(B)} = 2^{10}$.
Equating the powers, $5 \times n(B) = 10$.
$n(B) = 10 / 5 = 2$.
Given, $n(A) = 5$.
Number of relations from A to B is $2^{n(A) \times n(B)}$.
$2^{n(A) \times n(B)} = 1024$
We know that $1024 = 2^{10}$.
So, $2^{5 \times n(B)} = 2^{10}$.
Equating the powers, $5 \times n(B) = 10$.
$n(B) = 10 / 5 = 2$.
2. If the ordered pairs (a, -1) and (5, b) belong to {(x,y) / y = 2x + 3}, then the values of ‘a’ and ‘b’ are
Answer: d) (-2, 13)
Solution:
Given relation: $y = 2x + 3$.
For (a, -1): $x = a, y = -1$.
$-1 = 2a + 3 \implies 2a = -4 \implies a = -2$.
For (5, b): $x = 5, y = b$.
$b = 2(5) + 3 \implies b = 10 + 3 \implies b = 13$.
So, the values are $(a, b) = (-2, 13)$.
Given relation: $y = 2x + 3$.
For (a, -1): $x = a, y = -1$.
$-1 = 2a + 3 \implies 2a = -4 \implies a = -2$.
For (5, b): $x = 5, y = b$.
$b = 2(5) + 3 \implies b = 10 + 3 \implies b = 13$.
So, the values are $(a, b) = (-2, 13)$.
3. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
Answer: d) 2520
Solution:
We need to find the LCM of numbers from 1 to 10.
Prime factorization of numbers: 1=1, 2=2, 3=3, 4=2², 5=5, 6=2×3, 7=7, 8=2³, 9=3², 10=2×5.
LCM = Highest power of all prime factors involved.
LCM = $2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7 = 72 \times 35 = 2520$.
We need to find the LCM of numbers from 1 to 10.
Prime factorization of numbers: 1=1, 2=2, 3=3, 4=2², 5=5, 6=2×3, 7=7, 8=2³, 9=3², 10=2×5.
LCM = Highest power of all prime factors involved.
LCM = $2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7 = 72 \times 35 = 2520$.
4. If $t_n$ is the $n^{th}$ term of an A.P. then $t_{2n} - t_n$ is
Answer: a) nd
Solution:
For an A.P., the $n^{th}$ term is $t_n = a + (n-1)d$.
$t_{2n} = a + (2n-1)d$.
$t_{2n} - t_n = [a + (2n-1)d] - [a + (n-1)d]$
$= a + 2nd - d - a - nd + d$
$= (2nd - nd) = nd$.
For an A.P., the $n^{th}$ term is $t_n = a + (n-1)d$.
$t_{2n} = a + (2n-1)d$.
$t_{2n} - t_n = [a + (2n-1)d] - [a + (n-1)d]$
$= a + 2nd - d - a - nd + d$
$= (2nd - nd) = nd$.
5. The sequence $\sqrt{11}, \sqrt{55}, 5\sqrt{11}, 5\sqrt{55}, 25\sqrt{11}, \dots$ represents
Answer: b) a G.P. only
Solution:
Let's check the ratio of consecutive terms.
$t_2/t_1 = \sqrt{55}/\sqrt{11} = \sqrt{5 \times 11}/\sqrt{11} = \sqrt{5}$.
$t_3/t_2 = 5\sqrt{11}/\sqrt{55} = 5\sqrt{11}/(\sqrt{5}\sqrt{11}) = 5/\sqrt{5} = \sqrt{5}$.
Since the common ratio is constant ($r = \sqrt{5}$), the sequence is a G.P. The difference is not constant, so it's not an A.P.
Let's check the ratio of consecutive terms.
$t_2/t_1 = \sqrt{55}/\sqrt{11} = \sqrt{5 \times 11}/\sqrt{11} = \sqrt{5}$.
$t_3/t_2 = 5\sqrt{11}/\sqrt{55} = 5\sqrt{11}/(\sqrt{5}\sqrt{11}) = 5/\sqrt{5} = \sqrt{5}$.
Since the common ratio is constant ($r = \sqrt{5}$), the sequence is a G.P. The difference is not constant, so it's not an A.P.
6. If $(x - 6)$ is the HCF of $x^2 - 2x - 24$ and $x^2 - kx - 6$ then the value of k is
Answer: b) 5
Solution:
If $(x-6)$ is the HCF, then $x=6$ is a root of both polynomials.
Let $P(x) = x^2 - kx - 6$.
Since $x=6$ is a root, $P(6) = 0$.
$(6)^2 - k(6) - 6 = 0$
$36 - 6k - 6 = 0$
$30 - 6k = 0 \implies 6k = 30 \implies k = 5$.
If $(x-6)$ is the HCF, then $x=6$ is a root of both polynomials.
Let $P(x) = x^2 - kx - 6$.
Since $x=6$ is a root, $P(6) = 0$.
$(6)^2 - k(6) - 6 = 0$
$36 - 6k - 6 = 0$
$30 - 6k = 0 \implies 6k = 30 \implies k = 5$.
7. Which of the following should be added to make $x^4 + 64$ a perfect square?
Answer: b) $16x^2$
Solution:
$x^4 + 64 = (x^2)^2 + (8)^2$.
To make it a perfect square of the form $(a+b)^2 = a^2 + 2ab + b^2$, we need to add the term $2ab$.
Here, $a = x^2$ and $b = 8$.
$2ab = 2(x^2)(8) = 16x^2$.
So, $x^4 + 16x^2 + 64 = (x^2 + 8)^2$.
$x^4 + 64 = (x^2)^2 + (8)^2$.
To make it a perfect square of the form $(a+b)^2 = a^2 + 2ab + b^2$, we need to add the term $2ab$.
Here, $a = x^2$ and $b = 8$.
$2ab = 2(x^2)(8) = 16x^2$.
So, $x^4 + 16x^2 + 64 = (x^2 + 8)^2$.
8. A quadratic equation whose one zero is 5 and the sum of the zeroes is 0 is given by the equation
Answer: c) $x^2 - 25 = 0$
Solution:
Let the zeroes be $\alpha$ and $\beta$.
Given $\alpha = 5$ and sum of zeroes $\alpha + \beta = 0$.
$5 + \beta = 0 \implies \beta = -5$.
Product of zeroes, $\alpha\beta = 5 \times (-5) = -25$.
The quadratic equation is $x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) = 0$.
$x^2 - (0)x + (-25) = 0 \implies x^2 - 25 = 0$.
Let the zeroes be $\alpha$ and $\beta$.
Given $\alpha = 5$ and sum of zeroes $\alpha + \beta = 0$.
$5 + \beta = 0 \implies \beta = -5$.
Product of zeroes, $\alpha\beta = 5 \times (-5) = -25$.
The quadratic equation is $x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) = 0$.
$x^2 - (0)x + (-25) = 0 \implies x^2 - 25 = 0$.
9. The perimeters of two similar triangles $\triangle ABC$ and $\triangle PQR$ are 36cm and 24cm respectively. If PQ = 10 cm, the length of AB is ...............
Answer: d) 15 cm
Solution:
For similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.
$\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle PQR)} = \frac{AB}{PQ}$
$\frac{36}{24} = \frac{AB}{10}$
$\frac{3}{2} = \frac{AB}{10} \implies AB = \frac{3 \times 10}{2} = 15$ cm.
For similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.
$\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle PQR)} = \frac{AB}{PQ}$
$\frac{36}{24} = \frac{AB}{10}$
$\frac{3}{2} = \frac{AB}{10} \implies AB = \frac{3 \times 10}{2} = 15$ cm.
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
Answer: b) 4 cm
Solution:
By Angle Bisector Theorem,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{8}{AC} = \frac{6}{3}$
$\frac{8}{AC} = 2 \implies AC = \frac{8}{2} = 4$ cm.
By Angle Bisector Theorem,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{8}{AC} = \frac{6}{3}$
$\frac{8}{AC} = 2 \implies AC = \frac{8}{2} = 4$ cm.
11. The straight line given by the equation x = 11 is
Answer: b) parallel to y axis
Solution:
The equation $x = 11$ represents a vertical line where the x-coordinate is always 11, regardless of the y-coordinate. A vertical line is always parallel to the y-axis.
The equation $x = 11$ represents a vertical line where the x-coordinate is always 11, regardless of the y-coordinate. A vertical line is always parallel to the y-axis.
12. If (5,7), (3, p) and (6, 6) are collinear then the value of p is
Answer: c) 9
Solution:
If three points are collinear, the area of the triangle formed by them is zero.
Area = $\frac{1}{2} [x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)] = 0$
$5(p-6) + 3(6-7) + 6(7-p) = 0$
$5p - 30 + 3(-1) + 42 - 6p = 0$
$-p + 9 = 0 \implies p = 9$.
If three points are collinear, the area of the triangle formed by them is zero.
Area = $\frac{1}{2} [x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)] = 0$
$5(p-6) + 3(6-7) + 6(7-p) = 0$
$5p - 30 + 3(-1) + 42 - 6p = 0$
$-p + 9 = 0 \implies p = 9$.
13. (2,1) is the point of intersection of two lines
Answer: b) x + y = 3; 3x + y = 7
Solution:
Substitute the point (2,1) into the equations of each option.
Option b):
For $x+y=3$: $2+1=3$. (True)
For $3x+y=7$: $3(2)+1 = 6+1 = 7$. (True)
Since the point (2,1) satisfies both equations, it is their point of intersection.
Substitute the point (2,1) into the equations of each option.
Option b):
For $x+y=3$: $2+1=3$. (True)
For $3x+y=7$: $3(2)+1 = 6+1 = 7$. (True)
Since the point (2,1) satisfies both equations, it is their point of intersection.
14. If $5x = \sec\theta$ and $\frac{5}{y} = \tan\theta$, then $x^2 - \frac{1}{y^2}$ is equal to
Answer: b) $\frac{1}{25}$
Solution:
Given $5x = \sec\theta \implies x = \frac{\sec\theta}{5}$.
Given $\frac{5}{y} = \tan\theta \implies \frac{1}{y} = \frac{\tan\theta}{5}$.
We need to find $x^2 - \frac{1}{y^2}$.
$x^2 - \frac{1}{y^2} = \left(\frac{\sec\theta}{5}\right)^2 - \left(\frac{\tan\theta}{5}\right)^2$
$= \frac{\sec^2\theta}{25} - \frac{\tan^2\theta}{25} = \frac{\sec^2\theta - \tan^2\theta}{25}$
Since $\sec^2\theta - \tan^2\theta = 1$, the expression equals $\frac{1}{25}$.
Given $5x = \sec\theta \implies x = \frac{\sec\theta}{5}$.
Given $\frac{5}{y} = \tan\theta \implies \frac{1}{y} = \frac{\tan\theta}{5}$.
We need to find $x^2 - \frac{1}{y^2}$.
$x^2 - \frac{1}{y^2} = \left(\frac{\sec\theta}{5}\right)^2 - \left(\frac{\tan\theta}{5}\right)^2$
$= \frac{\sec^2\theta}{25} - \frac{\tan^2\theta}{25} = \frac{\sec^2\theta - \tan^2\theta}{25}$
Since $\sec^2\theta - \tan^2\theta = 1$, the expression equals $\frac{1}{25}$.
PART - B (10 x 2 = 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:
Given $y = x + 3$ and $x \in \{0,1,2,3,4,5\}$.
If $x=0, y=0+3=3$
If $x=1, y=1+3=4$
If $x=2, y=2+3=5$
If $x=3, y=3+3=6$
If $x=4, y=4+3=7$
If $x=5, y=5+3=8$
The relation R as a set of ordered pairs is R = {(0,3), (1,4), (2,5), (3,6), (4,7), (5,8)}.
Domain (set of all first elements) = {0, 1, 2, 3, 4, 5}.
Range (set of all second elements) = {3, 4, 5, 6, 7, 8}.
Given $y = x + 3$ and $x \in \{0,1,2,3,4,5\}$.
If $x=0, y=0+3=3$
If $x=1, y=1+3=4$
If $x=2, y=2+3=5$
If $x=3, y=3+3=6$
If $x=4, y=4+3=7$
If $x=5, y=5+3=8$
The relation R as a set of ordered pairs is R = {(0,3), (1,4), (2,5), (3,6), (4,7), (5,8)}.
Domain (set of all first elements) = {0, 1, 2, 3, 4, 5}.
Range (set of all second elements) = {3, 4, 5, 6, 7, 8}.
16. Let f be a function from R to R defined by $f(x) = 3x - 5$. Find the values of a and b given that (a, 4) and (1, b) belong to f.
Solution:
Given $f(x) = 3x - 5$.
If (a, 4) belongs to f, then $f(a) = 4$.
$3a - 5 = 4 \implies 3a = 9 \implies a = 3$.
If (1, b) belongs to f, then $f(1) = b$.
$b = 3(1) - 5 \implies b = 3 - 5 \implies b = -2$.
Thus, $a = 3$ and $b = -2$.
Given $f(x) = 3x - 5$.
If (a, 4) belongs to f, then $f(a) = 4$.
$3a - 5 = 4 \implies 3a = 9 \implies a = 3$.
If (1, b) belongs to f, then $f(1) = b$.
$b = 3(1) - 5 \implies b = 3 - 5 \implies b = -2$.
Thus, $a = 3$ and $b = -2$.
17. If $f(x) = x^2 - 1$, $g(x) = x - 2$ find a, if $g \circ f(a) = 1$.
Solution:
First, find the composite function $g \circ f(x)$.
$g \circ f(x) = g(f(x)) = g(x^2 - 1) = (x^2 - 1) - 2 = x^2 - 3$.
Given $g \circ f(a) = 1$.
$a^2 - 3 = 1$
$a^2 = 4$
$a = \pm\sqrt{4} \implies a = 2$ or $a = -2$.
First, find the composite function $g \circ f(x)$.
$g \circ f(x) = g(f(x)) = g(x^2 - 1) = (x^2 - 1) - 2 = x^2 - 3$.
Given $g \circ f(a) = 1$.
$a^2 - 3 = 1$
$a^2 = 4$
$a = \pm\sqrt{4} \implies a = 2$ or $a = -2$.
18. Solve: $5x \equiv 4 \pmod{6}$
Solution:
$5x \equiv 4 \pmod{6}$ means $5x - 4$ is a multiple of 6.
$5x - 4 = 6k$ for some integer $k$.
$5x = 6k + 4$.
We can test values for $x$:
If $x=1, 5(1) = 5 \equiv 5 \pmod{6}$.
If $x=2, 5(2) = 10 \equiv 4 \pmod{6}$.
So, a solution is $x=2$.
$5x \equiv 4 \pmod{6}$ means $5x - 4$ is a multiple of 6.
$5x - 4 = 6k$ for some integer $k$.
$5x = 6k + 4$.
We can test values for $x$:
If $x=1, 5(1) = 5 \equiv 5 \pmod{6}$.
If $x=2, 5(2) = 10 \equiv 4 \pmod{6}$.
So, a solution is $x=2$.
19. In a G.P. 729, 243, 81, ......... find $t_7$.
Solution:
The given G.P is 729, 243, 81, ...
First term, $a = 729$.
Common ratio, $r = \frac{243}{729} = \frac{1}{3}$.
The $n^{th}$ term of a G.P. is $t_n = ar^{n-1}$.
For the $7^{th}$ term, $n=7$.
$t_7 = 729 \times (\frac{1}{3})^{7-1} = 729 \times (\frac{1}{3})^6$
Since $729 = 3^6$,
$t_7 = 3^6 \times \frac{1}{3^6} = 1$.
The given G.P is 729, 243, 81, ...
First term, $a = 729$.
Common ratio, $r = \frac{243}{729} = \frac{1}{3}$.
The $n^{th}$ term of a G.P. is $t_n = ar^{n-1}$.
For the $7^{th}$ term, $n=7$.
$t_7 = 729 \times (\frac{1}{3})^{7-1} = 729 \times (\frac{1}{3})^6$
Since $729 = 3^6$,
$t_7 = 3^6 \times \frac{1}{3^6} = 1$.
20. If $1+2+3+...+k = 325$ then find $1^3+2^3+3^3 +...+ k^3$.
Solution:
We know the formula for the sum of the first k natural numbers: $1+2+3+...+k = \frac{k(k+1)}{2}$.
Given $\frac{k(k+1)}{2} = 325$.
The formula for the sum of the cubes of the first k natural numbers is: $1^3+2^3+3^3 +...+ k^3 = \left(\frac{k(k+1)}{2}\right)^2$.
Substituting the given value:
$1^3+2^3+3^3 +...+ k^3 = (325)^2 = 105625$.
We know the formula for the sum of the first k natural numbers: $1+2+3+...+k = \frac{k(k+1)}{2}$.
Given $\frac{k(k+1)}{2} = 325$.
The formula for the sum of the cubes of the first k natural numbers is: $1^3+2^3+3^3 +...+ k^3 = \left(\frac{k(k+1)}{2}\right)^2$.
Substituting the given value:
$1^3+2^3+3^3 +...+ k^3 = (325)^2 = 105625$.
21. Find the excluded value of the rational expression $\frac{t}{t^2-5t+6}$
Solution:
The excluded values are the values of $t$ that make the denominator zero.
Set the denominator to zero: $t^2 - 5t + 6 = 0$.
Factor the quadratic equation: $(t-2)(t-3) = 0$.
This gives $t-2=0$ or $t-3=0$.
So, $t=2$ and $t=3$.
The excluded values are 2 and 3.
The excluded values are the values of $t$ that make the denominator zero.
Set the denominator to zero: $t^2 - 5t + 6 = 0$.
Factor the quadratic equation: $(t-2)(t-3) = 0$.
This gives $t-2=0$ or $t-3=0$.
So, $t=2$ and $t=3$.
The excluded values are 2 and 3.
22. Simplify : $\frac{x^3}{x-y} + \frac{y^3}{y-x}$
Solution:
$\frac{x^3}{x-y} + \frac{y^3}{y-x} = \frac{x^3}{x-y} + \frac{y^3}{-(x-y)}$
$= \frac{x^3}{x-y} - \frac{y^3}{x-y}$
$= \frac{x^3 - y^3}{x-y}$
Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:
$= \frac{(x-y)(x^2+xy+y^2)}{x-y}$
$= x^2+xy+y^2$.
$\frac{x^3}{x-y} + \frac{y^3}{y-x} = \frac{x^3}{x-y} + \frac{y^3}{-(x-y)}$
$= \frac{x^3}{x-y} - \frac{y^3}{x-y}$
$= \frac{x^3 - y^3}{x-y}$
Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:
$= \frac{(x-y)(x^2+xy+y^2)}{x-y}$
$= x^2+xy+y^2$.
23. Determine the nature of the roots of the quadratic equation $15x^2 + 11x + 2 = 0$.
Solution:
The nature of the roots is determined by the discriminant, $\Delta = b^2 - 4ac$.
For $15x^2 + 11x + 2 = 0$, we have $a=15, b=11, c=2$.
$\Delta = (11)^2 - 4(15)(2) = 121 - 120 = 1$.
Since $\Delta = 1 > 0$ and is a perfect square, the roots are real, unequal, and rational.
The nature of the roots is determined by the discriminant, $\Delta = b^2 - 4ac$.
For $15x^2 + 11x + 2 = 0$, we have $a=15, b=11, c=2$.
$\Delta = (11)^2 - 4(15)(2) = 121 - 120 = 1$.
Since $\Delta = 1 > 0$ and is a perfect square, the roots are real, unequal, and rational.
24. If $\triangle ABC$ is similar to $\triangle DEF$ such that BC = 3cm, EF = 4cm and area of $\triangle ABC = 54cm^2$. Find the area of $\triangle DEF$.
Solution:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2$
$\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$
$\text{Area}(\triangle DEF) = \frac{54 \times 16}{9} = 6 \times 16 = 96$ cm$^2$.
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left(\frac{BC}{EF}\right)^2$
$\frac{54}{\text{Area}(\triangle DEF)} = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$
$\text{Area}(\triangle DEF) = \frac{54 \times 16}{9} = 6 \times 16 = 96$ cm$^2$.
25. Find the slope of a line joining the points (-6, 1) and (-3, 2).
Solution:
The slope $m$ of a line joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Here, $(x_1, y_1) = (-6, 1)$ and $(x_2, y_2) = (-3, 2)$.
$m = \frac{2 - 1}{-3 - (-6)} = \frac{1}{-3 + 6} = \frac{1}{3}$.
The slope is $\frac{1}{3}$.
The slope $m$ of a line joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Here, $(x_1, y_1) = (-6, 1)$ and $(x_2, y_2) = (-3, 2)$.
$m = \frac{2 - 1}{-3 - (-6)} = \frac{1}{-3 + 6} = \frac{1}{3}$.
The slope is $\frac{1}{3}$.
26. Show that the straight lines $x - 2y + 3 = 0$ and $6x + 3y + 8 = 0$ are perpendicular.
Solution:
Two lines are perpendicular if the product of their slopes is -1 ($m_1 \times m_2 = -1$).
Slope of the first line, $m_1 = -\frac{\text{coeff. of x}}{\text{coeff. of y}} = -\frac{1}{-2} = \frac{1}{2}$.
Slope of the second line, $m_2 = -\frac{6}{3} = -2$.
Product of slopes, $m_1 \times m_2 = \frac{1}{2} \times (-2) = -1$.
Since the product of their slopes is -1, the lines are perpendicular.
Two lines are perpendicular if the product of their slopes is -1 ($m_1 \times m_2 = -1$).
Slope of the first line, $m_1 = -\frac{\text{coeff. of x}}{\text{coeff. of y}} = -\frac{1}{-2} = \frac{1}{2}$.
Slope of the second line, $m_2 = -\frac{6}{3} = -2$.
Product of slopes, $m_1 \times m_2 = \frac{1}{2} \times (-2) = -1$.
Since the product of their slopes is -1, the lines are perpendicular.
27. Prove that $\sqrt{\frac{1+\sin\theta}{1-\sin\theta}} = \sec\theta + \tan\theta$.
Solution:
LHS = $\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}$
Multiply numerator and denominator by the conjugate of the denominator, $(1+\sin\theta)$: $= \sqrt{\frac{(1+\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}} = \sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}$
Using the identity $\cos^2\theta = 1-\sin^2\theta$:
$= \sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}} = \frac{1+\sin\theta}{\cos\theta}$
$= \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \sec\theta + \tan\theta$ = RHS.
Hence proved.
LHS = $\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}$
Multiply numerator and denominator by the conjugate of the denominator, $(1+\sin\theta)$: $= \sqrt{\frac{(1+\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}} = \sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}$
Using the identity $\cos^2\theta = 1-\sin^2\theta$:
$= \sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}} = \frac{1+\sin\theta}{\cos\theta}$
$= \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \sec\theta + \tan\theta$ = RHS.
Hence proved.
28. (Compulsory) Find the equation of straight line whose slope is -4 and passing through the point (1,2).
Solution:
The point-slope form of a line is $y - y_1 = m(x - x_1)$.
Given slope $m = -4$ and point $(x_1, y_1) = (1, 2)$.
Substituting the values:
$y - 2 = -4(x - 1)$
$y - 2 = -4x + 4$
$4x + y - 2 - 4 = 0$
The required equation is $4x + y - 6 = 0$.
The point-slope form of a line is $y - y_1 = m(x - x_1)$.
Given slope $m = -4$ and point $(x_1, y_1) = (1, 2)$.
Substituting the values:
$y - 2 = -4(x - 1)$
$y - 2 = -4x + 4$
$4x + y - 2 - 4 = 0$
The required equation is $4x + y - 6 = 0$.
PART - C (10 x 5 = 50) - Solutions
Answer any 10 questions. Question No. 42 is compulsory.
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:
Given sets:
A = {x ∈ W / x < 2} = {0, 1} (W is Whole Numbers)
B = {x ∈ N / 1 < x ≤ 4} = {2, 3, 4} (N is Natural Numbers)
C = {3, 5}
LHS: A X (B ∩ C)
First, find B ∩ C:
B ∩ C = {2, 3, 4} ∩ {3, 5} = {3}
Now, find A X (B ∩ C):
A X (B ∩ C) = {0, 1} X {3} = {(0, 3), (1, 3)} --- (1)
RHS: (A X B) ∩ (A X C)
First, find A X B:
A X B = {0, 1} X {2, 3, 4} = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
Next, find A X C:
A X C = {0, 1} X {3, 5} = {(0, 3), (0, 5), (1, 3), (1, 5)}
Now, find the intersection:
(A X B) ∩ (A X C) = {(0, 3), (1, 3)} --- (2)
From (1) and (2), LHS = RHS. Hence verified.
Given sets:
A = {x ∈ W / x < 2} = {0, 1} (W is Whole Numbers)
B = {x ∈ N / 1 < x ≤ 4} = {2, 3, 4} (N is Natural Numbers)
C = {3, 5}
LHS: A X (B ∩ C)
First, find B ∩ C:
B ∩ C = {2, 3, 4} ∩ {3, 5} = {3}
Now, find A X (B ∩ C):
A X (B ∩ C) = {0, 1} X {3} = {(0, 3), (1, 3)} --- (1)
RHS: (A X B) ∩ (A X C)
First, find A X B:
A X B = {0, 1} X {2, 3, 4} = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
Next, find A X C:
A X C = {0, 1} X {3, 5} = {(0, 3), (0, 5), (1, 3), (1, 5)}
Now, find the intersection:
(A X B) ∩ (A X C) = {(0, 3), (1, 3)} --- (2)
From (1) and (2), LHS = RHS. Hence verified.
30. Let f : A→B be a function defined by $f(x) = \frac{x}{2} - 1$, where A = {2,4,6,10,12}, B = {0,1,2,4,5,9}. Represent f by i) a set of ordered pairs ii) a table iii) an arrow diagram iv) a graph.
Solution:
Given $f(x) = \frac{x}{2} - 1$ and A = {2, 4, 6, 10, 12}.
$f(2) = \frac{2}{2} - 1 = 1 - 1 = 0$
$f(4) = \frac{4}{2} - 1 = 2 - 1 = 1$
$f(6) = \frac{6}{2} - 1 = 3 - 1 = 2$
$f(10) = \frac{10}{2} - 1 = 5 - 1 = 4$
$f(12) = \frac{12}{2} - 1 = 6 - 1 = 5$
i) Set of ordered pairs:
f = {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
ii) A table:
iii) An arrow diagram:
iv) A graph:
Given $f(x) = \frac{x}{2} - 1$ and A = {2, 4, 6, 10, 12}.
$f(2) = \frac{2}{2} - 1 = 1 - 1 = 0$
$f(4) = \frac{4}{2} - 1 = 2 - 1 = 1$
$f(6) = \frac{6}{2} - 1 = 3 - 1 = 2$
$f(10) = \frac{10}{2} - 1 = 5 - 1 = 4$
$f(12) = \frac{12}{2} - 1 = 6 - 1 = 5$
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:
iv) A graph:
31. If f(x) = x², g(x) = 3x and h(x) = x - 2, prove that (fog)oh = fo(goh).
Solution:
LHS: (fog)oh
First, find $fog(x)$: $fog(x) = f(g(x)) = f(3x) = (3x)^2 = 9x^2$.
Now, find $(fog)oh(x)$: $(fog)oh(x) = (fog)(h(x)) = (fog)(x-2) = 9(x-2)^2$. --- (1)
RHS: fo(goh)
First, find $goh(x)$: $goh(x) = g(h(x)) = g(x-2) = 3(x-2)$.
Now, find $fo(goh)(x)$: $fo(goh)(x) = f(goh(x)) = f(3(x-2)) = (3(x-2))^2 = 9(x-2)^2$. --- (2)
From (1) and (2), LHS = RHS. Hence proved.
LHS: (fog)oh
First, find $fog(x)$: $fog(x) = f(g(x)) = f(3x) = (3x)^2 = 9x^2$.
Now, find $(fog)oh(x)$: $(fog)oh(x) = (fog)(h(x)) = (fog)(x-2) = 9(x-2)^2$. --- (1)
RHS: fo(goh)
First, find $goh(x)$: $goh(x) = g(h(x)) = g(x-2) = 3(x-2)$.
Now, find $fo(goh)(x)$: $fo(goh)(x) = f(goh(x)) = f(3(x-2)) = (3(x-2))^2 = 9(x-2)^2$. --- (2)
From (1) and (2), LHS = RHS. Hence proved.
32. The 13th term of an A.P is 3 and the sum of first 13 terms is 234. Find the common difference and the sum of first 21 terms.
Solution:
Given, $t_{13} = 3$ and $S_{13} = 234$.
We know $t_n = a + (n-1)d$.
$t_{13} = a + (13-1)d = a + 12d = 3$. --- (1)
We know $S_n = \frac{n}{2}(2a + (n-1)d)$.
$S_{13} = \frac{13}{2}(2a + (13-1)d) = \frac{13}{2}(2a + 12d) = 234$.
$13(a+6d) = 234 \implies a + 6d = \frac{234}{13} = 18$. --- (2)
Subtracting (2) from (1):
$(a + 12d) - (a + 6d) = 3 - 18$
$6d = -15 \implies d = -\frac{15}{6} = -\frac{5}{2}$.
Substitute $d = -5/2$ into (2):
$a + 6(-\frac{5}{2}) = 18 \implies a - 15 = 18 \implies a = 33$.
Now, find the sum of the first 21 terms, $S_{21}$.
$S_{21} = \frac{21}{2}(2a + (21-1)d) = \frac{21}{2}(2(33) + 20(-\frac{5}{2}))$.
$S_{21} = \frac{21}{2}(66 - 50) = \frac{21}{2}(16) = 21 \times 8 = 168$.
Common difference $d = -5/2$. Sum of first 21 terms is 168.
Given, $t_{13} = 3$ and $S_{13} = 234$.
We know $t_n = a + (n-1)d$.
$t_{13} = a + (13-1)d = a + 12d = 3$. --- (1)
We know $S_n = \frac{n}{2}(2a + (n-1)d)$.
$S_{13} = \frac{13}{2}(2a + (13-1)d) = \frac{13}{2}(2a + 12d) = 234$.
$13(a+6d) = 234 \implies a + 6d = \frac{234}{13} = 18$. --- (2)
Subtracting (2) from (1):
$(a + 12d) - (a + 6d) = 3 - 18$
$6d = -15 \implies d = -\frac{15}{6} = -\frac{5}{2}$.
Substitute $d = -5/2$ into (2):
$a + 6(-\frac{5}{2}) = 18 \implies a - 15 = 18 \implies a = 33$.
Now, find the sum of the first 21 terms, $S_{21}$.
$S_{21} = \frac{21}{2}(2a + (21-1)d) = \frac{21}{2}(2(33) + 20(-\frac{5}{2}))$.
$S_{21} = \frac{21}{2}(66 - 50) = \frac{21}{2}(16) = 21 \times 8 = 168$.
Common difference $d = -5/2$. Sum of first 21 terms is 168.
33. Find the sum of n terms of the series 3 + 33 + 333 + ... to n terms.
Solution:
Let $S_n = 3 + 33 + 333 + \dots + n$ terms.
$S_n = 3(1 + 11 + 111 + \dots + n$ terms).
Multiply and divide by 9:
$S_n = \frac{3}{9}(9 + 99 + 999 + \dots + n$ terms).
$S_n = \frac{1}{3}[(10-1) + (10^2-1) + (10^3-1) + \dots + (10^n-1)]$.
$S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1+1+1+\dots n \text{ times})]$.
The first part is a G.P. with first term $a=10$ and common ratio $r=10$.
Sum of G.P. = $a\frac{r^n-1}{r-1} = 10\frac{10^n-1}{10-1} = \frac{10}{9}(10^n-1)$.
So, $S_n = \frac{1}{3}\left[\frac{10}{9}(10^n-1) - n\right]$.
$S_n = \frac{10}{27}(10^n-1) - \frac{n}{3}$.
Let $S_n = 3 + 33 + 333 + \dots + n$ terms.
$S_n = 3(1 + 11 + 111 + \dots + n$ terms).
Multiply and divide by 9:
$S_n = \frac{3}{9}(9 + 99 + 999 + \dots + n$ terms).
$S_n = \frac{1}{3}[(10-1) + (10^2-1) + (10^3-1) + \dots + (10^n-1)]$.
$S_n = \frac{1}{3}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1+1+1+\dots n \text{ times})]$.
The first part is a G.P. with first term $a=10$ and common ratio $r=10$.
Sum of G.P. = $a\frac{r^n-1}{r-1} = 10\frac{10^n-1}{10-1} = \frac{10}{9}(10^n-1)$.
So, $S_n = \frac{1}{3}\left[\frac{10}{9}(10^n-1) - n\right]$.
$S_n = \frac{10}{27}(10^n-1) - \frac{n}{3}$.
34. There are 12 pieces of five, ten and twenty rupee currencies whose total value is Rs. 105. When first 2 sorts are interchanged in their numbers its value will be increased by Rs. 20. Find the number of currencies in each sort.
Solution:
Let the number of Rs. 5, Rs. 10, and Rs. 20 currencies be x, y, and z respectively.
From the problem statement:
1. Total number of currencies: $x + y + z = 12$ --- (1)
2. Total value: $5x + 10y + 20z = 105 \implies x + 2y + 4z = 21$ --- (2)
3. After interchanging x and y, the value is Rs. 125: $5y + 10x + 20z = 125 \implies 2x + y + 4z = 25$ --- (3)
Subtracting (2) from (3):
$(2x + y + 4z) - (x + 2y + 4z) = 25 - 21 \implies x - y = 4$ --- (4)
Subtracting (1) from (2):
$(x + 2y + 4z) - (x + y + z) = 21 - 12 \implies y + 3z = 9$ --- (5)
From (4), $x = y + 4$. Substitute this into (1):
$(y + 4) + y + z = 12 \implies 2y + z = 8 \implies z = 8 - 2y$.
Substitute this expression for z into (5):
$y + 3(8 - 2y) = 9$
$y + 24 - 6y = 9$
$-5y = -15 \implies y = 3$.
Now find x and z:
$x = y + 4 = 3 + 4 = 7$.
$z = 8 - 2y = 8 - 2(3) = 8 - 6 = 2$.
Number of Rs. 5 currencies = 7.
Number of Rs. 10 currencies = 3.
Number of Rs. 20 currencies = 2.
Let the number of Rs. 5, Rs. 10, and Rs. 20 currencies be x, y, and z respectively.
From the problem statement:
1. Total number of currencies: $x + y + z = 12$ --- (1)
2. Total value: $5x + 10y + 20z = 105 \implies x + 2y + 4z = 21$ --- (2)
3. After interchanging x and y, the value is Rs. 125: $5y + 10x + 20z = 125 \implies 2x + y + 4z = 25$ --- (3)
Subtracting (2) from (3):
$(2x + y + 4z) - (x + 2y + 4z) = 25 - 21 \implies x - y = 4$ --- (4)
Subtracting (1) from (2):
$(x + 2y + 4z) - (x + y + z) = 21 - 12 \implies y + 3z = 9$ --- (5)
From (4), $x = y + 4$. Substitute this into (1):
$(y + 4) + y + z = 12 \implies 2y + z = 8 \implies z = 8 - 2y$.
Substitute this expression for z into (5):
$y + 3(8 - 2y) = 9$
$y + 24 - 6y = 9$
$-5y = -15 \implies y = 3$.
Now find x and z:
$x = y + 4 = 3 + 4 = 7$.
$z = 8 - 2y = 8 - 2(3) = 8 - 6 = 2$.
Number of Rs. 5 currencies = 7.
Number of Rs. 10 currencies = 3.
Number of Rs. 20 currencies = 2.
35. Find the square root of the polynomial $4x^4 - 28x^3 + 37x^2 + 42x + 9$ by division method.
Solution:
Using the long division method for square root:
The square root of the first term $4x^4$ is $2x^2$.
Using the long division method for square root:
The square root of the first term $4x^4$ is $2x^2$.
$2x^2 - 7x - 3$
____________________
$2x^2$ | $4x^4 - 28x^3 + 37x^2 + 42x + 9$
$4x^4$
____________________
$4x^2 - 7x$ | $-28x^3 + 37x^2$
$-28x^3 + 49x^2$
____________________
$4x^2 - 14x - 3$ | $-12x^2 + 42x + 9$
$-12x^2 + 42x + 9$
____________________
0
The square root is $|2x^2 - 7x - 3|$.
____________________
$2x^2$ | $4x^4 - 28x^3 + 37x^2 + 42x + 9$
$4x^4$
____________________
$4x^2 - 7x$ | $-28x^3 + 37x^2$
$-28x^3 + 49x^2$
____________________
$4x^2 - 14x - 3$ | $-12x^2 + 42x + 9$
$-12x^2 + 42x + 9$
____________________
0
36. The roots of the equation $x^2 + 6x - 4 = 0$ are α, β. Find the quadratic equation whose roots are α² and β².
Solution:
For the equation $x^2 + 6x - 4 = 0$, the roots are $\alpha$ and $\beta$.
Sum of roots: $\alpha + \beta = -b/a = -6/1 = -6$.
Product of roots: $\alpha\beta = c/a = -4/1 = -4$.
We need to find 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 $x^2 - (\text{Sum of new roots})x + (\text{Product of new roots}) = 0$.
$x^2 - 44x + 16 = 0$.
For the equation $x^2 + 6x - 4 = 0$, the roots are $\alpha$ and $\beta$.
Sum of roots: $\alpha + \beta = -b/a = -6/1 = -6$.
Product of roots: $\alpha\beta = c/a = -4/1 = -4$.
We need to find 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 $x^2 - (\text{Sum of new roots})x + (\text{Product of new roots}) = 0$.
$x^2 - 44x + 16 = 0$.
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, the other two sides are divided in the same ratio.
Given: In $\triangle ABC$, a line DE is parallel to BC, intersecting AB at D and AC at E.
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.
Construction: Join BE and CD. Draw $DM \perp AC$ and $EN \perp AB$.
Proof:
Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height}$.
In $\triangle ADE$, Area($\triangle ADE$) = $\frac{1}{2} \times AD \times EN$.
In $\triangle BDE$, Area($\triangle BDE$) = $\frac{1}{2} \times DB \times EN$.
$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}$ --- (1)
Similarly, in $\triangle ADE$, Area($\triangle ADE$) = $\frac{1}{2} \times AE \times DM$.
In $\triangle CDE$, Area($\triangle CDE$) = $\frac{1}{2} \times EC \times DM$.
$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}$ --- (2)
Triangles $\triangle BDE$ and $\triangle CDE$ are on the same base DE and between the same parallel lines DE and BC. Therefore, Area($\triangle BDE$) = Area($\triangle CDE$).
From (1) and (2), we get:
$\frac{AD}{DB} = \frac{AE}{EC}$. Hence proved.
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, the other two sides are divided in the same ratio.
Given: In $\triangle ABC$, a line DE is parallel to BC, intersecting AB at D and AC at E.
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.
Construction: Join BE and CD. Draw $DM \perp AC$ and $EN \perp AB$.
Proof:Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height}$.
In $\triangle ADE$, Area($\triangle ADE$) = $\frac{1}{2} \times AD \times EN$.
In $\triangle BDE$, Area($\triangle BDE$) = $\frac{1}{2} \times DB \times EN$.
$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}$ --- (1)
Similarly, in $\triangle ADE$, Area($\triangle ADE$) = $\frac{1}{2} \times AE \times DM$.
In $\triangle CDE$, Area($\triangle CDE$) = $\frac{1}{2} \times EC \times DM$.
$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}$ --- (2)
Triangles $\triangle BDE$ and $\triangle CDE$ are on the same base DE and between the same parallel lines DE and BC. Therefore, Area($\triangle BDE$) = Area($\triangle CDE$).
From (1) and (2), we get:
$\frac{AD}{DB} = \frac{AE}{EC}$. Hence 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 be A(8, 6), B(5, 11), C(-5, 12), and D(-4, 3). Area of quadrilateral = $\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$.
$= \frac{1}{2} |((8)(11) + (5)(12) + (-5)(3) + (-4)(6)) - ((6)(5) + (11)(-5) + (12)(-4) + (3)(8))|$.
$= \frac{1}{2} |(88 + 60 - 15 - 24) - (30 - 55 - 48 + 24)|$.
$= \frac{1}{2} |(109) - (-49)|$.
$= \frac{1}{2} |109 + 49| = \frac{1}{2} |158|$.
$= 79$ square units.
Let the vertices be A(8, 6), B(5, 11), C(-5, 12), and D(-4, 3). Area of quadrilateral = $\frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$.
$= \frac{1}{2} |((8)(11) + (5)(12) + (-5)(3) + (-4)(6)) - ((6)(5) + (11)(-5) + (12)(-4) + (3)(8))|$.
$= \frac{1}{2} |(88 + 60 - 15 - 24) - (30 - 55 - 48 + 24)|$.
$= \frac{1}{2} |(109) - (-49)|$.
$= \frac{1}{2} |109 + 49| = \frac{1}{2} |158|$.
$= 79$ square units.
39. 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 'a' and the y-intercept be 'b'. Given ratio of intercepts is 2:5. So, $a/b = 2/5 \implies a = (2/5)b$. Or let $a=2k, b=5k$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
$\frac{x}{2k} + \frac{y}{5k} = 1$.
To clear the denominators, multiply by 10k: $5x + 2y = 10k$.
The line passes through the point (1, -4). Substitute these values for x and y:
$5(1) + 2(-4) = 10k$
$5 - 8 = 10k \implies -3 = 10k \implies k = -3/10$.
Substitute the value of k back into the equation $5x + 2y = 10k$:
$5x + 2y = 10(-3/10) \implies 5x + 2y = -3$.
The required equation is $5x + 2y + 3 = 0$.
Let the x-intercept be 'a' and the y-intercept be 'b'. Given ratio of intercepts is 2:5. So, $a/b = 2/5 \implies a = (2/5)b$. Or let $a=2k, b=5k$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
$\frac{x}{2k} + \frac{y}{5k} = 1$.
To clear the denominators, multiply by 10k: $5x + 2y = 10k$.
The line passes through the point (1, -4). Substitute these values for x and y:
$5(1) + 2(-4) = 10k$
$5 - 8 = 10k \implies -3 = 10k \implies k = -3/10$.
Substitute the value of k back into the equation $5x + 2y = 10k$:
$5x + 2y = 10(-3/10) \implies 5x + 2y = -3$.
The required equation is $5x + 2y + 3 = 0$.
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 passes through the midpoint of AB and is perpendicular to AB.
1. Find the midpoint of AB:
Midpoint M = $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{-4+6}{2}, \frac{2-4}{2}) = (\frac{2}{2}, \frac{-2}{2}) = (1, -1)$.
2. Find the slope of AB:
Slope $m_{AB} = \frac{y_2-y_1}{x_2-x_1} = \frac{-4-2}{6-(-4)} = \frac{-6}{10} = -\frac{3}{5}$.
3. Find the slope of the perpendicular bisector:
Slope $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3}$.
4. Find the equation of the line:
Using point-slope form $y - y_1 = m(x - x_1)$ with point M(1, -1) and slope $m=5/3$.
$y - (-1) = \frac{5}{3}(x - 1)$
$y + 1 = \frac{5}{3}(x - 1)$
$3(y+1) = 5(x-1) \implies 3y + 3 = 5x - 5$.
$5x - 3y - 8 = 0$.
The perpendicular bisector passes through the midpoint of AB and is perpendicular to AB.
1. Find the midpoint of AB:
Midpoint M = $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{-4+6}{2}, \frac{2-4}{2}) = (\frac{2}{2}, \frac{-2}{2}) = (1, -1)$.
2. Find the slope of AB:
Slope $m_{AB} = \frac{y_2-y_1}{x_2-x_1} = \frac{-4-2}{6-(-4)} = \frac{-6}{10} = -\frac{3}{5}$.
3. Find the slope of the perpendicular bisector:
Slope $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-3/5} = \frac{5}{3}$.
4. Find the equation of the line:
Using point-slope form $y - y_1 = m(x - x_1)$ with point M(1, -1) and slope $m=5/3$.
$y - (-1) = \frac{5}{3}(x - 1)$
$y + 1 = \frac{5}{3}(x - 1)$
$3(y+1) = 5(x-1) \implies 3y + 3 = 5x - 5$.
$5x - 3y - 8 = 0$.
41. If cotθ + tanθ = x and secθ - cosθ = y then prove that $(x^2y)^{2/3} - (xy^2)^{2/3} = 1$.
Solution:
Simplify x and y:
$x = \cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$.
$y = \sec\theta - \cos\theta = \frac{1}{\cos\theta} - \cos\theta = \frac{1-\cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$.
Now, calculate the terms in the expression to prove:
$x^2y = \left(\frac{1}{\sin\theta\cos\theta}\right)^2 \left(\frac{\sin^2\theta}{\cos\theta}\right) = \frac{1}{\sin^2\theta\cos^2\theta} \cdot \frac{\sin^2\theta}{\cos\theta} = \frac{1}{\cos^3\theta} = \sec^3\theta$.
$(x^2y)^{2/3} = (\sec^3\theta)^{2/3} = \sec^2\theta$.
$xy^2 = \left(\frac{1}{\sin\theta\cos\theta}\right) \left(\frac{\sin^2\theta}{\cos\theta}\right)^2 = \frac{1}{\sin\theta\cos\theta} \cdot \frac{\sin^4\theta}{\cos^2\theta} = \frac{\sin^3\theta}{\cos^3\theta} = \tan^3\theta$.
$(xy^2)^{2/3} = (\tan^3\theta)^{2/3} = \tan^2\theta$.
Substitute back into the expression:
LHS = $(x^2y)^{2/3} - (xy^2)^{2/3} = \sec^2\theta - \tan^2\theta$.
Using the trigonometric identity $\sec^2\theta - \tan^2\theta = 1$.
LHS = 1 = RHS. Hence proved.
Simplify x and y:
$x = \cot\theta + \tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$.
$y = \sec\theta - \cos\theta = \frac{1}{\cos\theta} - \cos\theta = \frac{1-\cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$.
Now, calculate the terms in the expression to prove:
$x^2y = \left(\frac{1}{\sin\theta\cos\theta}\right)^2 \left(\frac{\sin^2\theta}{\cos\theta}\right) = \frac{1}{\sin^2\theta\cos^2\theta} \cdot \frac{\sin^2\theta}{\cos\theta} = \frac{1}{\cos^3\theta} = \sec^3\theta$.
$(x^2y)^{2/3} = (\sec^3\theta)^{2/3} = \sec^2\theta$.
$xy^2 = \left(\frac{1}{\sin\theta\cos\theta}\right) \left(\frac{\sin^2\theta}{\cos\theta}\right)^2 = \frac{1}{\sin\theta\cos\theta} \cdot \frac{\sin^4\theta}{\cos^2\theta} = \frac{\sin^3\theta}{\cos^3\theta} = \tan^3\theta$.
$(xy^2)^{2/3} = (\tan^3\theta)^{2/3} = \tan^2\theta$.
Substitute back into the expression:
LHS = $(x^2y)^{2/3} - (xy^2)^{2/3} = \sec^2\theta - \tan^2\theta$.
Using the trigonometric identity $\sec^2\theta - \tan^2\theta = 1$.
LHS = 1 = RHS. Hence proved.
42. (Compulsory) Swathi has 15 ice cubes of different sizes 9cm, 10cm, 11cm, ..., 23cm. How much volume of ice cubes can be used to prepare some fruit juice with these ice cubes?
Solution:
The sizes of the cubes are the side lengths. The volume of a cube with side 'a' is $a^3$.
We need to find the total volume, which is the sum of the volumes of all cubes.
Total Volume = $9^3 + 10^3 + 11^3 + \dots + 23^3$.
This can be calculated as the difference of two sums of cubes:
Total Volume = $(1^3 + 2^3 + \dots + 23^3) - (1^3 + 2^3 + \dots + 8^3)$.
Using the formula for the sum of cubes of first n natural numbers, $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$.
For n=23: Sum = $\left(\frac{23(23+1)}{2}\right)^2 = \left(\frac{23 \times 24}{2}\right)^2 = (23 \times 12)^2 = 276^2 = 76176$.
For n=8: Sum = $\left(\frac{8(8+1)}{2}\right)^2 = \left(\frac{8 \times 9}{2}\right)^2 = (4 \times 9)^2 = 36^2 = 1296$.
Total Volume = $76176 - 1296 = 74880$ cm$^3$.
The total volume of the ice cubes is 74,880 cm$^3$.
The sizes of the cubes are the side lengths. The volume of a cube with side 'a' is $a^3$.
We need to find the total volume, which is the sum of the volumes of all cubes.
Total Volume = $9^3 + 10^3 + 11^3 + \dots + 23^3$.
This can be calculated as the difference of two sums of cubes:
Total Volume = $(1^3 + 2^3 + \dots + 23^3) - (1^3 + 2^3 + \dots + 8^3)$.
Using the formula for the sum of cubes of first n natural numbers, $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$.
For n=23: Sum = $\left(\frac{23(23+1)}{2}\right)^2 = \left(\frac{23 \times 24}{2}\right)^2 = (23 \times 12)^2 = 276^2 = 76176$.
For n=8: Sum = $\left(\frac{8(8+1)}{2}\right)^2 = \left(\frac{8 \times 9}{2}\right)^2 = (4 \times 9)^2 = 36^2 = 1296$.
Total Volume = $76176 - 1296 = 74880$ cm$^3$.
The total volume of the ice cubes is 74,880 cm$^3$.
PART - D (2 x 8 = 16) - Solutions
Answer all the questions.
43. 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 APQR such that QR = 5cm, ∠P = 30° and the altitude from P to QR is of length 4.2cm.
Solution a) Steps of Construction (Similar Triangle):
1. Draw any triangle ABC.
2. Draw a ray AX making an acute angle with AB on the side opposite to vertex C.
3. Since the scale factor is 6/5, locate 6 points (the greater of 6 and 5) $A_1, A_2, \dots, A_6$ on AX so that $AA_1 = A_1A_2 = \dots = A_5A_6$.
4. Join $A_5$ (the smaller number in the ratio) to B.
5. Draw a line through $A_6$ parallel to $A_5B$ to intersect the extended line segment AB at B'.
6. Draw a line through B' parallel to BC to intersect the extended line segment AC at C'.
7. The triangle $\triangle$AB'C' is the required triangle similar to $\triangle$ABC.
Solution b) Steps of Construction (Triangle PQR):
1. Draw a line segment QR = 5cm.
2. At Q, draw a line QX such that $\angle RQX = 30^\circ$.
3. Draw a line QY perpendicular to QX at Q.
4. Draw the perpendicular bisector of QR. Let it intersect QR at M and the line QY at O.
5. With O as the center and OQ as the radius, draw a circle. All angles subtended by the arc QR on the major segment will be 30°.
6. Draw a line segment GH parallel to QR at a distance of 4.2 cm.
7. This line GH intersects the circle at two points, P and P'.
8. Join PQ and PR. $\triangle$PQR is the required triangle.
1. Draw any triangle ABC.
2. Draw a ray AX making an acute angle with AB on the side opposite to vertex C.
3. Since the scale factor is 6/5, locate 6 points (the greater of 6 and 5) $A_1, A_2, \dots, A_6$ on AX so that $AA_1 = A_1A_2 = \dots = A_5A_6$.
4. Join $A_5$ (the smaller number in the ratio) to B.
5. Draw a line through $A_6$ parallel to $A_5B$ to intersect the extended line segment AB at B'.
6. Draw a line through B' parallel to BC to intersect the extended line segment AC at C'.
7. The triangle $\triangle$AB'C' is the required triangle similar to $\triangle$ABC.
Diagram showing construction of a similar triangle with a scale factor greater than 1.
Solution b) Steps of Construction (Triangle PQR):
1. Draw a line segment QR = 5cm.
2. At Q, draw a line QX such that $\angle RQX = 30^\circ$.
3. Draw a line QY perpendicular to QX at Q.
4. Draw the perpendicular bisector of QR. Let it intersect QR at M and the line QY at O.
5. With O as the center and OQ as the radius, draw a circle. All angles subtended by the arc QR on the major segment will be 30°.
6. Draw a line segment GH parallel to QR at a distance of 4.2 cm.
7. This line GH intersects the circle at two points, P and P'.
8. Join PQ and PR. $\triangle$PQR is the required triangle.
Diagram showing construction of triangle PQR with given base, vertex angle, and altitude.
44. a) A bus is travelling at a uniform speed of 50km/hr. Draw the distance-time graph and hence find. i) the constant of variation. ii) how far it will travel in 90 minutes? iii) the time required to cover a distance of 300km from the graph. (OR)
b) The following table shows the data about the number of pipes and the time taken to fill the same tank.
No. of pipes (X): 2, 3, 6, 9
Time taken (Y) (in mts): 45, 30, 15, 10
Draw the graph for the above data and hence. i) Find the time taken to fill the tank when five pipes are used. ii) Find the number of pipes when the time is 9 minutes.
Solution a) Distance-Time Graph (Direct Variation):
The relationship is Distance = Speed × Time, so $y = 50x$, where y is distance (km) and x is time (hr).
Table of values:
Plot these points and draw a straight line through the origin (0,0).
From the graph:
i) The constant of variation is the speed, k = 50 km/hr.
ii) For 90 minutes (1.5 hours), find x=1.5 on the graph. The corresponding y-value will be 75. The bus will travel 75 km.
iii) To cover 300 km, find y=300 on the graph. The corresponding x-value will be 6. The time required is 6 hours.
Solution b) Pipes-Time Graph (Inverse Variation):
Check the type of variation: $XY = 2 \times 45 = 90$; $3 \times 30 = 90$; $6 \times 15 = 90$; $9 \times 10 = 90$.
Since XY is constant, this is an inverse variation. The constant of variation $k = 90$. The equation is $xy=90$.
Plot the points (2, 45), (3, 30), (6, 15), (9, 10) and draw a smooth curve (a hyperbola).
From the graph:
i) When 5 pipes are used (x=5), find the corresponding y-value on the curve. From the equation $5y=90 \implies y=18$. The time taken will be 18 minutes.
ii) When the time is 9 minutes (y=9), find the corresponding x-value on the curve. From the equation $x(9)=90 \implies x=10$. The number of pipes required is 10.
The relationship is Distance = Speed × Time, so $y = 50x$, where y is distance (km) and x is time (hr).
Table of values:
| Time (x) in hr | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Distance (y) in km | 50 | 100 | 150 | 200 |
Graph showing a straight line passing through the origin, representing direct variation (y=50x).
From the graph:i) The constant of variation is the speed, k = 50 km/hr.
ii) For 90 minutes (1.5 hours), find x=1.5 on the graph. The corresponding y-value will be 75. The bus will travel 75 km.
iii) To cover 300 km, find y=300 on the graph. The corresponding x-value will be 6. The time required is 6 hours.
Solution b) Pipes-Time Graph (Inverse Variation):
Check the type of variation: $XY = 2 \times 45 = 90$; $3 \times 30 = 90$; $6 \times 15 = 90$; $9 \times 10 = 90$.
Since XY is constant, this is an inverse variation. The constant of variation $k = 90$. The equation is $xy=90$.
Plot the points (2, 45), (3, 30), (6, 15), (9, 10) and draw a smooth curve (a hyperbola).
Graph showing a hyperbola in the first quadrant, representing inverse variation (xy=90).
From the graph:i) When 5 pipes are used (x=5), find the corresponding y-value on the curve. From the equation $5y=90 \implies y=18$. The time taken will be 18 minutes.
ii) When the time is 9 minutes (y=9), find the corresponding x-value on the curve. From the equation $x(9)=90 \implies x=10$. The number of pipes required is 10.