10-ஆம் வகுப்பு கணிதம்: 5 மதிப்பெண் வினாக்கள் மற்றும் முழுமையான விடைகள் (Long Answer Questions)
தமிழகப் பள்ளிகளில் நடைபெறும் 10-ஆம் வகுப்பு மாணவர்களுக்கான கணிதத் தேர்வில் (ஆங்கில வழி) கேட்கப்படும் முக்கிய 5 மதிப்பெண் வினாக்கள் மற்றும் அதற்கான படிநிலை விடைகள் (Step-by-Step Solutions) கீழே கொடுக்கப்பட்டுள்ளன. இதனை மாணவர்கள் எளிதில் படிக்கும் வகையிலும், மொபைல் போனில் பார்ப்பதற்கு வசதியாகவும் வடிவமைத்துள்ளோம்.
PART III: LONG ANSWER QUESTIONS (5 MARKS EACH)
1. If $A = \{x \in W \mid x < 2\}$, $B = \{x \in \mathbb{N} \mid 1 < x \le 4\}$ and $C = \{3, 5\}$, verify that: $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
Solution:
Given sets are in descriptive form. First, list the elements:
$W$ is the set of Whole numbers $\{0, 1, 2, \dots\}$. So, $A = \{0, 1\}$
$\mathbb{N}$ is the set of Natural numbers $\{1, 2, 3, \dots\}$. So, $B = \{2, 3, 4\}$
$C = \{3, 5\}$
LHS: $A \times (B \cup C)$
First, find $B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\}$
Now, $A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\}$
$A \times (B \cup C) = \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\} \quad \dots \text{(Equation 1)}$
RHS: $(A \times B) \cup (A \times C)$
$A \times B = \{0, 1\} \times \{2, 3, 4\}$
$A \times B = \{(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)\}$
$A \times C = \{0, 1\} \times \{3, 5\}$
$A \times C = \{(0, 3), (0, 5), (1, 3), (1, 5)\}$
Now, $(A \times B) \cup (A \times C) = \{(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)\} \cup \{(0, 3), (0, 5), (1, 3), (1, 5)\}$
$(A \times B) \cup (A \times C) = \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\} \quad \dots \text{(Equation 2)}$
From Equations (1) and (2), LHS = RHS. Hence verified.
2. If $f(x) = x - 1$, $g(x) = 3x + 1$, and $h(x) = x^2$, prove that $(f \circ g) \circ h = f \circ (g \circ h)$.
Solution:
LHS: $(f \circ g) \circ h$
First, find $f \circ g$
$(f \circ g)(x) = f(g(x)) = f(3x + 1)$
Since $f(x) = x - 1$, substitute $x$ with $(3x + 1)$
$= (3x + 1) - 1 = 3x$
Now find $(f \circ g) \circ h$
$((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = (f \circ g)(x^2)$
Since $(f \circ g)(x) = 3x$, substitute $x$ with $x^2$
$= 3(x^2) = 3x^2 \quad \dots \text{(Equation 1)}$
RHS: $f \circ (g \circ h)$
First, find $g \circ h$
$(g \circ h)(x) = g(h(x)) = g(x^2)$
Since $g(x) = 3x + 1$, substitute $x$ with $x^2$
$= 3(x^2) + 1 = 3x^2 + 1$
Now find $f \circ (g \circ h)$
$(f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(3x^2 + 1)$
Since $f(x) = x - 1$, substitute $x$ with $(3x^2 + 1)$
$= (3x^2 + 1) - 1 = 3x^2 \quad \dots \text{(Equation 2)}$
From (1) and (2), $(f \circ g) \circ h = f \circ (g \circ h)$. Hence proved.
3. Find the sum to $n$ terms of the sequence: $5 + 55 + 555 + \dots$
Solution:
Let $S_n = 5 + 55 + 555 + \dots \text{ to } n \text{ terms}$
Take 5 common from all terms:
$S_n = 5(1 + 11 + 111 + \dots \text{ to } n \text{ terms})$
Multiply and divide by 9:
$S_n = \frac{5}{9} (9 + 99 + 999 + \dots \text{ to } n \text{ terms})$
Rewrite 9, 99, 999 in powers of 10:
$S_n = \frac{5}{9} [(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$
Group the powers of 10 and the 1s:
$S_n = \frac{5}{9} [(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ to } n \text{ terms})]$
The first part is a Geometric Progression (G.P.) with $a = 10$, $r = 10$, and number of terms $= n$.
Sum of G.P. formula: $S_n = \frac{a(r^n - 1)}{r - 1}$
$S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{10 - 1} - n \right]$
$S_n = \frac{5}{9} \left[ \frac{10(10^n - 1)}{9} - n \right]$
$S_n = \frac{50(10^n - 1)}{81} - \frac{5n}{9}$
4. Rekha has 15 square colour papers of sizes $10\text{ cm}, 11\text{ cm}, 12\text{ cm}, \dots, 24\text{ cm}$. How much total area can be decorated using all these colour papers?
Solution:
Area of a square = $\text{side}^2$
Total area to be decorated $= 10^2 + 11^2 + 12^2 + \dots + 24^2$
This can be written as the difference of sum of squares of first 24 natural numbers and first 9 natural numbers:
$= (1^2 + 2^2 + \dots + 24^2) - (1^2 + 2^2 + \dots + 9^2)$
Formula for sum of squares of first $n$ natural numbers: $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$
Here, substitute $n = 24$ for the first part and $n = 9$ for the second part.
$= \left[ \frac{24 \times (24+1) \times (2(24)+1)}{6} \right] - \left[ \frac{9 \times (9+1) \times (2(9)+1)}{6} \right]$
$= \left[ \frac{24 \times 25 \times 49}{6} \right] - \left[ \frac{9 \times 10 \times 19}{6} \right]$
$= (4 \times 25 \times 49) - (3 \times 5 \times 19)$
$= (100 \times 49) - (15 \times 19)$
$= 4900 - 285$
$= 4615 \text{ cm}^2$
Total area that can be decorated is $4615 \text{ cm}^2$.
5. Find the area of the quadrilateral whose vertices are $(8, 6), (5, 11), (-5, 12)$, and $(-4, 3)$.
Solution:
Let the given vertices be plotted in a rough graph and taken in counter-clockwise order to ensure positive area.
Let $A(x_1, y_1) = (8, 6)$
Let $B(x_2, y_2) = (5, 11)$
Let $C(x_3, y_3) = (-5, 12)$
Let $D(x_4, y_4) = (-4, 3)$
Area of quadrilateral ABCD $= \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)|$
$= \frac{1}{2} | (8 \times 11 + 5 \times 12 + (-5) \times 3 + (-4) \times 6) - (6 \times 5 + 11 \times (-5) + 12 \times (-4) + 3 \times 8) |$
$= \frac{1}{2} | (88 + 60 - 15 - 24) - (30 - 55 - 48 + 24) |$
$= \frac{1}{2} | (148 - 39) - (54 - 103) |$
$= \frac{1}{2} | (109) - (-49) |$
$= \frac{1}{2} | 109 + 49 |$
$= \frac{1}{2} (158)$
$= 79 \text{ sq. units}$.
6. State and prove the Basic Proportionality Theorem (Thales Theorem).
Solution:
Statement:
A straight line drawn parallel to a side of a triangle intersecting the other two sides, divides the sides in the same ratio.
Proof:
Given: In $\Delta ABC$, $D$ is a point on $AB$ and $E$ is a point on $AC$. Line $DE \parallel BC$.
To prove: $\frac{AD}{DB} = \frac{AE}{EC}$
Construction: Draw a line $DE$ parallel to $BC$.
Step 1: In $\Delta ABC$ and $\Delta ADE$,
$\angle ABC = \angle ADE$ (Corresponding angles are equal because $DE \parallel BC$)
$\angle ACB = \angle AED$ (Corresponding angles are equal because $DE \parallel BC$)
$\angle DAE = \angle BAC$ (Common angle)
Step 2: Therefore, $\Delta ABC \sim \Delta ADE$ (By AAA similarity criterion)
Step 3: Since the triangles are similar, their corresponding sides are proportional.
$\frac{AB}{AD} = \frac{AC}{AE}$
Step 4: Subtract 1 from both sides.
$\frac{AB}{AD} - 1 = \frac{AC}{AE} - 1$
$\frac{AB - AD}{AD} = \frac{AC - AE}{AE}$
Step 5: From the figure, $AB - AD = DB$ and $AC - AE = EC$
$\frac{DB}{AD} = \frac{EC}{AE}$
Step 6: Taking reciprocals on both sides,
$\frac{AD}{DB} = \frac{AE}{EC}$
Hence the theorem is proved.
7. Show that the points $(-2, -1), (4, 0), (3, 3)$, and $(-3, 2)$ form the vertices of a parallelogram.
Solution:
Let the given points be $A(-2, -1), B(4, 0), C(3, 3)$ and $D(-3, 2)$.
A quadrilateral is a parallelogram if its opposite sides are parallel. We can prove this by showing slopes of opposite sides are equal.
Slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$
Slope of $AB = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6}$
Slope of $CD = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}$
Since Slope of $AB$ = Slope of $CD$, line $AB \parallel \text{line } CD$.
Now, Slope of $BC = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$
Slope of $AD = \frac{2 - (-1)}{-3 - (-2)} = \frac{3}{-3 + 2} = \frac{3}{-1} = -3$
Since Slope of $BC$ = Slope of $AD$, line $BC \parallel \text{line } AD$.
Since both pairs of opposite sides are parallel, $ABCD$ is a parallelogram.
8. Find the equation of the straight line passing through $(-3, 8)$ such that the sum of its positive intercepts on the coordinate axes is 7.
Solution:
Let the $x$-intercept be '$a$' and $y$-intercept be '$b$'.
Given that sum of intercepts is 7: $a + b = 7 \Rightarrow b = 7 - a$
The intercept form of the line equation is: $\frac{x}{a} + \frac{y}{b} = 1$
Substitute $b = 7 - a$:
$\frac{x}{a} + \frac{y}{7 - a} = 1$
Since the line passes through the point $(-3, 8)$, substitute $x = -3$ and $y = 8$:
$\frac{-3}{a} + \frac{8}{7 - a} = 1$
Take LCM:
$\frac{-3(7 - a) + 8a}{a(7 - a)} = 1$
$-21 + 3a + 8a = a(7 - a)$
$-21 + 11a = 7a - a^2$
Bring all terms to one side to form a quadratic equation:
$a^2 + 11a - 7a - 21 = 0$
$a^2 + 4a - 21 = 0$
Factorizing the quadratic equation:
$(a + 7)(a - 3) = 0$
So, $a = -7$ or $a = 3$.
The question states "positive intercepts", so we take $a = 3$.
If $a = 3$, then $b = 7 - 3 = 4$.
Now, substitute $a = 3$ and $b = 4$ in the intercept form:
$\frac{x}{3} + \frac{y}{4} = 1$
Multiply the entire equation by 12 (LCM of 3 and 4):
$4x + 3y = 12$
$4x + 3y - 12 = 0$
This is the required equation of the straight line.
9. Let $f: A \rightarrow B$ be defined by $f(x) = \frac{x}{2} - 1$ where $A = \{2, 4, 6, 10, 12\}$ and $B = \{0, 1, 2, 4, 5, 9\}$. Represent $f$ by: (i) an arrow diagram (ii) a set of ordered pairs (iii) a table (iv) a graph.
Solution:
Given $f(x) = \frac{x}{2} - 1$. Calculate $f(x)$ for all values in set $A$:
If $x = 2$, $f(2) = \frac{2}{2} - 1 = 1 - 1 = 0$
If $x = 4$, $f(4) = \frac{4}{2} - 1 = 2 - 1 = 1$
If $x = 6$, $f(6) = \frac{6}{2} - 1 = 3 - 1 = 2$
If $x = 10$, $f(10) = \frac{10}{2} - 1 = 5 - 1 = 4$
If $x = 12$, $f(12) = \frac{12}{2} - 1 = 6 - 1 = 5$
(i) Arrow Diagram:
Draw two ovals representing sets A and B. Draw arrows connecting 2 to 0, 4 to 1, 6 to 2, 10 to 4, and 12 to 5. Element 9 in set B will not have a pre-image.
(ii) A set of ordered pairs:
$f = \{(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)\}$
(iii) A table:
| $x$ | 2 | 4 | 6 | 10 | 12 |
|---|---|---|---|---|---|
| $f(x)$ | 0 | 1 | 2 | 4 | 5 |
(iv) A graph:
Plot the points $(2, 0), (4, 1), (6, 2), (10, 4), \text{ and } (12, 5)$ on the X-Y coordinate plane.
10. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by:
$f(x) = \begin{cases} 2x + 7, & \text{if } x < -2 \\ x^2 - 2, & \text{if } -2 \le x < 3 \\ 3x - 2, & \text{if } x \ge 3 \end{cases}$
Find: (i) $f(4)$ (ii) $f(-2)$ (iii) $f(4) + 2f(1)$ (iv) $\frac{f(1) - 3f(4)}{f(-3)}$
Solution:
Identify the correct interval for each value of $x$:
(i) To find $f(4)$, $x = 4$ lies in the interval $x \ge 3$.
So, use $f(x) = 3x - 2$
$f(4) = 3(4) - 2 = 12 - 2 = 10$
(ii) To find $f(-2)$, $x = -2$ lies in the interval $-2 \le x < 3$.
So, use $f(x) = x^2 - 2$
$f(-2) = (-2)^2 - 2 = 4 - 2 = 2$
(iii) To find $f(4) + 2f(1)$:
We already have $f(4) = 10$.
To find $f(1)$, $x = 1$ lies in the interval $-2 \le x < 3$.
So, use $f(x) = x^2 - 2 \Rightarrow f(1) = (1)^2 - 2 = 1 - 2 = -1$
$f(4) + 2f(1) = 10 + 2(-1) = 10 - 2 = 8$
(iv) To find $\frac{f(1) - 3f(4)}{f(-3)}$:
We already have $f(1) = -1$ and $f(4) = 10$.
To find $f(-3)$, $x = -3$ lies in the interval $x < -2$.
So, use $f(x) = 2x + 7 \Rightarrow f(-3) = 2(-3) + 7 = -6 + 7 = 1$
Now substitute the values in the expression:
$= \frac{(-1) - 3(10)}{1}$
$= \frac{-1 - 30}{1}$
$= -31$
11. The vertices of $\Delta ABC$ are $A(-3, 0), B(10, -2)$, and $C(12, 3)$. Find the equations of the altitudes drawn through $A$ and $B$.
Solution:
Altitude drawn through A:
The altitude through A is a line passing through A and perpendicular to the opposite side BC.
First, find the slope of BC. Points are $B(10, -2)$ and $C(12, 3)$.
Slope of $BC (m_1) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{12 - 10} = \frac{5}{2}$
Since the altitude is perpendicular to BC, its slope ($m$) will be the negative reciprocal of $m_1$.
Slope of altitude through $A (m) = -\frac{2}{5}$
Equation of the altitude passing through $A(-3, 0)$ with slope $m = -\frac{2}{5}$:
$y - y_1 = m(x - x_1)$
$y - 0 = -\frac{2}{5}(x - (-3))$
$5y = -2(x + 3)$
$5y = -2x - 6$
$2x + 5y + 6 = 0$ (This is the equation of altitude through A)
Altitude drawn through B:
The altitude through B is a line passing through B and perpendicular to the opposite side AC.
First, find the slope of AC. Points are $A(-3, 0)$ and $C(12, 3)$.
Slope of $AC (m_2) = \frac{3 - 0}{12 - (-3)} = \frac{3}{15} = \frac{1}{5}$
Slope of altitude through $B (m) = -5$ (negative reciprocal)
Equation of the altitude passing through $B(10, -2)$ with slope $m = -5$:
$y - (-2) = -5(x - 10)$
$y + 2 = -5x + 50$
$5x + y + 2 - 50 = 0$
$5x + y - 48 = 0$ (This is the equation of altitude through B)
12. If $S_1, S_2$, and $S_3$ denote the sum of the first $n, 2n$, and $3n$ terms of an A.P. respectively, prove that $S_3 = 3(S_2 - S_1)$.
Solution:
Sum of first $n$ terms of an A.P is $S_n = \frac{n}{2}[2a + (n-1)d]$
Given:
$S_1 = \text{Sum of } n \text{ terms} = \frac{n}{2}[2a + (n-1)d] \quad \dots \text{(1)}$
$S_2 = \text{Sum of } 2n \text{ terms} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d] \quad \dots \text{(2)}$
$S_3 = \text{Sum of } 3n \text{ terms} = \frac{3n}{2}[2a + (3n-1)d] \quad \dots \text{(3)}$
RHS: $3(S_2 - S_1)$
First, find $S_2 - S_1$:
$= \frac{2n}{2}[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$
Take $\frac{n}{2}$ as common:
$= \frac{n}{2} \left[ 2[2a + (2n-1)d] - [2a + (n-1)d] \right]$
$= \frac{n}{2} \left[ 4a + 2(2n-1)d - 2a - (n-1)d \right]$
$= \frac{n}{2} \left[ 2a + d(4n - 2 - n + 1) \right]$
$= \frac{n}{2} \left[ 2a + d(3n - 1) \right]$
Now multiply by 3:
$3(S_2 - S_1) = 3 \times \frac{n}{2} [2a + (3n-1)d]$
$= \frac{3n}{2} [2a + (3n-1)d]$
From equation (3), this is exactly $S_3$.
Therefore, $S_3 = 3(S_2 - S_1)$. Hence proved.
13. State and prove the Angle Bisector Theorem.
Solution:
Statement:
The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.
Proof:
Given: In $\Delta ABC$, $AD$ is the internal angle bisector of $\angle A$, which meets $BC$ at $D$.
To prove: $\frac{AB}{AC} = \frac{BD}{CD}$
Construction: Draw a line through $C$ parallel to $AB$. Extend $AD$ to meet this line at $E$.
Step 1: Since $CE \parallel AB$ and $AE$ is the transversal, the alternate interior angles are equal.
$\angle BAE = \angle CEA$ ($\dots \text{Equation 1}$)
Step 2: Since $AD$ is the angle bisector of $\angle A$:
$\angle BAE = \angle CAE$ ($\dots \text{Equation 2}$)
Step 3: From Equations 1 and 2, we get:
$\angle CEA = \angle CAE$
In $\Delta ACE$, since two angles are equal, the sides opposite to them are equal (Isosceles triangle property).
Therefore, $AC = CE$ ($\dots \text{Equation 3}$)
Step 4: Now, consider $\Delta ABD$ and $\Delta ECD$.
$\angle ABD = \angle ECD$ (Alternate interior angles, since $AB \parallel CE$)
$\angle ADB = \angle EDC$ (Vertically opposite angles)
Therefore, $\Delta ABD \sim \Delta ECD$ (By AA similarity criterion)
Step 5: Since the triangles are similar, their corresponding sides are proportional.
$\frac{AB}{CE} = \frac{BD}{CD}$
Step 6: Substitute $CE = AC$ (from Equation 3) in the above ratio.
$\frac{AB}{AC} = \frac{BD}{CD}$
Hence the theorem is proved.
14. If $36x^4 - 60x^3 + 61x^2 - mx + n$ is a perfect square, find the values of $m$ and $n$.
Solution:
We use the square root method (long division for polynomials) to find the coefficients.
$\quad \quad \quad \ \ \ 6x^2 \quad - 5x \quad + 3$
$\quad \quad $ -------------------------------------------
$6x^2$ $\quad \quad | \ 36x^4 - 60x^3 + 61x^2 - mx + n$
$\quad \quad \quad \ \ \ \ 36x^4$
$\quad \quad $ -------------------------------------------
$12x^2 - 5x \ \ | \quad \quad \ - 60x^3 + 61x^2$
$\quad \quad \quad \quad \quad \quad - 60x^3 + 25x^2$
$\quad \quad $ -------------------------------------------
$12x^2 - 10x + 3 \ | \quad \quad \quad \quad \quad \ 36x^2 - mx + n$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ 36x^2 - 30x + 9$
$\quad \quad $ -------------------------------------------
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ 0$ (Since it is a perfect square, the remainder is 0)
Comparing the terms in the last step to make the remainder zero:
$-m = -30 \Rightarrow m = 30$
$n = 9$
Therefore, the values are $m = 30$ and $n = 9$.
