10th Maths Quarterly Exam Key Answer September 2023 | Chennai District

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10th Maths Quarterly Exam Key Answer - September 2023 (Chennai)

QUARTERLY SEPTEMBER 2023
10^TH MATHS KEY ANSWER ( CHENNAI )

I. CHOOSE THE CORRECT ANSWER

• 1. c. 3
• 2. a. (8, 6)
• 3. d. Quadratic
• 4. b. 2
• 5. a. 1
• 6. b. \(k^2\)
• 7. b. \((y + \frac{1}{2})^2\)
• 8. c. -1, 2
• 9. d. \(5\sqrt{2}\)
• 10. a. 1.4 cm
• 11. b. Parallel to Y-axis
• 12. d. \(-\frac{a}{b}\)
• 13. b. \(x + y = 3, 3x + y = 7\)
• 14. d. \(\cot \theta\)

II. ANSWER ANY 10 ONLY

• 15.
  Domain = \( \{ 0, 1, 2, 3, 4, 5 \} \)
  Range = \( \{ 3, 4, 5, 6, 7, 8 \} \)
• 16.
  \( \Rightarrow f(x^2) = (f(x))^2 \)
  \( \Rightarrow 3 - 2x^2 = (3 – 2x)^2 \)
  \( \Rightarrow x^2 - 2x + 1 = 0 \)
  \( \Rightarrow (x-1)(x-1) = 0 \)
  \( \therefore x = 1 \)
• 17.
  Image of 1 = 5 & Image of 2 = 8
  Preimage of 29 is 9 & Preimage of 53 is 17
• 18.
  \( 2^a \times 3^b = 13824 \)
  \( 2^a \times 3^b = 2^9 \times 3^3 \)
  \( \therefore a = 9 \text{ and } b = 3 \)
• 19.
  \( a_8 = \frac{63}{11} \) & \( a_{15} = \frac{225}{31} \)
• 20.
  Here \( a = 3 \) and \( r = \frac{t_2}{t_1} = -\frac{1}{3} \) \[ S_\infty = \frac{a}{1-r} \] \[ \Rightarrow S_\infty = \frac{3}{1 - (-\frac{1}{3})} = \frac{3}{1 + \frac{1}{3}} = \frac{3}{\frac{4}{3}} = \frac{9}{2} \]
• 21.
  \( x(x + 3)(x-2) = 0 \)
  Therefore excluded values are \( x = 0, 2 \text{ & } -3 \)
• 22.
  \( x^2 - (\text{SOR})x + \text{POR} = 0 \)
  \( x^2 + 9x + 20 = 0 \)
• 23.
  \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \frac{BC^2}{EF^2} \)
  \( \frac{54}{\text{Area}(\triangle DEF)} = \frac{3^2}{4^2} \)
  \( \text{Area}(\triangle DEF) = \frac{16 \times 54}{9} = 96 \, \text{cm}^2 \)
• 24.
  From Thales' Theorem, assuming \( \frac{AD}{DB} = \frac{AE}{EC} \). Given calculation implies \( \frac{3}{4} = \frac{x}{15-x} \)
  \( 3(15-x) = 4x \Rightarrow 45 - 3x = 4x \Rightarrow 7x = 45 \)
  \( x = \frac{45}{7} \approx 6.43 \, \text{cm} \)
• 25.
  \( \frac{3-a}{9+2} = -\frac{1}{2} \Rightarrow \frac{3-a}{11} = -\frac{1}{2} \)
  \( 2(3-a) = -11 \Rightarrow 6 - 2a = -11 \Rightarrow 2a = 17 \)
  \( a = \frac{17}{2} \)
• 26.
  x-intercept = -9
  y-intercept = 4
• 27.
  \[ \sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = \sqrt{\frac{(1+\cos\theta)(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}} \] \[ = \sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}} = \sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}} \] \[ = \frac{1+\cos\theta}{\sin\theta} = \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} \] \[ = \csc\theta + \cot\theta = \text{RHS} \]
• 28.
  The given expression leads to the condition: \[ \frac{1}{p^2} + \frac{1}{q^2} = 1 \]

III. ANSWER ANY 10 ONLY

• 29.
  \( (B \cup C) = \{ 2, 3, 4 \} \cup \{3, 5\} = \{2, 3, 4, 5\} \)
  \( A \times (B \cup C) = \{ 0, 1 \} \times \{ 2, 3, 4, 5 \} \)
  \( = \{ (0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5) \} \)
  \( (A \times B) = \{ (0,2), (0,3), (0,4), (1,2), (1,3), (1,4) \} \)
  \( (A \times C) = \{ (0,3), (0,5), (1,3), (1,5) \} \)
  \( (A \times B) \cup (A \times C) = \{ (0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5) \} \)
• 30.
  (i) \( f(4) = 3x - 2 = 3(4) - 2 = 10 \)
  (ii) \( f(-2) = x^2 - 2 = (-2)^2 - 2 = 4 - 2 = 2 \)
  (iii) To find \( f(4) + 2f(1) \):
  \( 2f(1) = 2(x^2 - 1) = 2(1^2 - 2) = 2(-1) = -2 \)
  \( f(4) + 2f(1) = 10 - 2 = 8 \)
  (iv) \( \frac{f(1) - 3f(4)}{f(-3)} = \frac{-1 - 3(10)}{1} = -31 \) (Assuming \(f(-3)=1\))
• 31.
  \( \text{fog} = f(g(x)) = (3x+1)-1 = 3x \)
  \( (\text{fog}) \circ h = (3x) \circ (x^2) = 3x^2 \)
  \( \text{goh} = g(h(x)) = 3(x^2)+1 = 3x^2+1 \)
  \( f \circ (\text{goh}) = f(3x^2+1) = (3x^2+1)-1 = 3x^2 \)
• 32.
  Let the numbers be \(a-3d, a-d, a+d, a+3d\). Given \(a=7, d=\pm2\).
  Numbers are 1, 5, 9, 13.
• 33.
  \( S_n = 3 + 33 + 333 + \dots \)
  \( S_n = \frac{3}{9} (9 + 99 + 999 + \dots) \)
  \( S_n = \frac{1}{3} [(10-1) + (10^2-1) + (10^3-1) + \dots] \)
  \( S_n = \frac{1}{3} [ (10 + 10^2 + \dots + 10^n) - n ] \)
  \( S_n = \frac{1}{3} \left[ \frac{10(10^n-1)}{10-1} - n \right] \)
  \( S_n = \frac{10(10^n-1)}{27} - \frac{n}{3} \)
• 34.
  Assuming this means \( \sum_{k=10}^{24} k^2 \):
  \( \sum_{k=1}^{24} k^2 - \sum_{k=1}^{9} k^2 = 4900 - 285 = 4615 \)
• 35.
  \( x = 2, y = -1, z = 4 \)
• 36.
  \( m = 30 \) and \( n = 9 \)
• 37.
  Discriminant, \( \Delta = b^2 - 4ac \)
  \( = [2(pr + qs)]^2 - 4(p^2 + q^2)(r^2 + s^2) \)
  \( = 4(p^2r^2+q^2s^2+2prqs) - 4(p^2r^2+p^2s^2+q^2r^2+q^2s^2) \)
  \( = -4(p^2s^2+q^2r^2-2prqs) = -4(ps - qr)^2 \)
  Since \( (ps-qr)^2 \ge 0 \), then \( \Delta \le 0 \). The roots are not real unless \( \Delta=0 \).
  If \( ps = qr \), then \( \Delta = 0 \), and the roots will be real and equal.
• 39.
  Area of quadrilateral by Shoelace formula: \[ A = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| \] \[ A = \frac{1}{2} |(8 \cdot 11 + 5 \cdot 12 + (-5) \cdot 3 + (-4) \cdot 6) - (6 \cdot 5 + 11 \cdot (-5) + 12 \cdot (-4) + 3 \cdot 8)| \] \[ = \frac{1}{2} |(88 + 60 - 15 - 24) - (30 - 55 - 48 + 24)| \] \[ = \frac{1}{2} |109 - (-49)| = \frac{1}{2}(158) = 79 \text{ Sq. Units.} \]
• 40.
  Midpoint of AB = \((\frac{6-4}{2}, \frac{-4+2}{2}) = (1, -1)\)
  Slope of AB (\(m_1\)) = \( \frac{2-(-4)}{-2-6} = \frac{6}{-8} = -\frac{3}{4} \) (Corrected value)
  Slope of perpendicular to AB (\(m_2\)) = \( \frac{4}{3} \)
  Equation of the line: \( y - y_1 = m_2(x - x_1) \)
  \( y - (-1) = \frac{4}{3}(x-1) \Rightarrow y+1 = \frac{4}{3}(x-1) \)
  \( 3(y+1) = 4(x-1) \Rightarrow 3y+3=4x-4 \)
  \( 4x - 3y - 7 = 0 \)
• 41.
  Given \(a = \frac{1+\sin\theta}{\cos\theta}\). Find \( \frac{a^2 - 1}{a^2 + 1} \). \[ a^2 = \frac{(1+\sin\theta)^2}{\cos^2\theta} = \frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta} \] \[ \frac{a^2 - 1}{a^2 + 1} = \frac{\frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta} - 1}{\frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta} + 1} = \frac{1+2\sin\theta+\sin^2\theta - \cos^2\theta}{1+2\sin\theta+\sin^2\theta + \cos^2\theta} \] Using \( \cos^2\theta = 1-\sin^2\theta \) and \( \sin^2\theta+\cos^2\theta = 1 \): \[ = \frac{1+2\sin\theta+\sin^2\theta - (1-\sin^2\theta)}{1+2\sin\theta+1} = \frac{2\sin\theta+2\sin^2\theta}{2+2\sin\theta} \] \[ = \frac{2\sin\theta(1+\sin\theta)}{2(1+\sin\theta)} = \sin\theta \]
• 42.
  \( \frac{1}{x+2} + \frac{1}{x-2} = \frac{6}{4x+7} \)
  \( \frac{(x-2) + (x+2)}{(x+2)(x-2)} = \frac{6}{4x+7} \)
  \( \frac{2x}{x^2-4} = \frac{6}{4x+7} \)
  \( 2x(4x+7) = 6(x^2-4) \)
  \( 8x^2 + 14x = 6x^2 - 24 \)
  \( 2x^2 + 14x + 24 = 0 \)
  \( x^2 + 7x + 12 = 0 \)
  \( (x+3)(x+4) = 0 \)
  \( \therefore x = -3 \text{ and } x = -4 \)

GEOMETRY & GRAPH

43. (a) Direct Variation

Time taken (hr) - x	1	2	3	4	5	6
Distance (km) - y	50	100	150	200	250	300

Equation: \(y = kx \Rightarrow k = \frac{y}{x}\)
(i) Constant of variation \(k = \frac{50}{1} = 50 \)
(ii) The distance travelled for \(1\frac{1}{2}\) hours = \(50 \times 1.5 = 75\) km
(iii) The time taken to cover 300 km = \(300/50 = 6\) hours.

(b) Indirect Variation

X	1	2	3	4	6	8	12	24
Y	24	12	8	6	4	3	2	1

Variation: Indirect Variation
Equation: \(xy = k \Rightarrow k = 1 \times 24 = 24 \)
Solution from the graph:
(i) If \( x = 3 \), then \( y = 8 \)
(ii) If \( y = 6 \), then \( x = 4 \)

44. (a) & (b) Geometric Constructions

Students are required to perform geometric constructions as per the question's instructions.

Prepared By:

D.S.GNANADASON – MATHS TEACHER (MCCHSS) 9940359935