SECTION A
1. LCM of smallest two digit composite and smallest composite:
Answer: (c) 20
Explanation: Smallest two-digit composite number = 10. Smallest composite number = 4.
LCM (10, 4) = 20.
2. Condition for parallel lines \(3x + 2ky = 2\) and \(2x + 5y + 1 = 0\):
Answer: (c) \(\frac{15}{4}\)
Explanation: For parallel lines, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
\(\frac{3}{2} = \frac{2k}{5} \Rightarrow 4k = 15 \Rightarrow k = \frac{15}{4}\).
3. Zeroes of \(x^2 - 2x - 8\):
Answer: (b) 4, -2
Explanation: Splitting the middle term: \(x^2 - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2) = 0\).
Zeroes are \(x = 4\) and \(x = -2\).
4. Pair of linear equations \(x = 0\) and \(x = 5\):
Answer: (a) No solution
Explanation: \(x = 0\) is the y-axis, and \(x = 5\) is a line parallel to the y-axis. Parallel lines never intersect, hence no solution.
5. Nature of roots of \(2x^2 - 4x + 3 = 0\):
Answer: (c) No real roots
Explanation: Discriminant \(D = b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8\). Since \(D < 0\), the equation has no real roots.
6. 10th term of AP 2, 7, 12...:
Answer: (a) 47
Explanation: \(a = 2\), \(d = 7 - 2 = 5\).
\(a_{10} = a + 9d = 2 + 9(5) = 2 + 45 = 47\).
7. Ratio of areas of similar triangles:
Answer: (c) 36:49
Explanation: Ratio of areas = (Ratio of corresponding sides)\(^2\) = \(\left(\frac{1.2}{1.4}\right)^2 = \left(\frac{6}{7}\right)^2 = \frac{36}{49}\).
8. Distance from x-axis:
Answer: (b) 3 units
Explanation: The distance of a point \((x, y)\) from the x-axis is given by its y-coordinate absolute value, which is \(|3| = 3\).
9. Trig evaluation:
Answer: (d) 0
Explanation: \(\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \frac{1 - (1)^2}{1 + (1)^2} = \frac{0}{2} = 0\).
10. Angle of elevation:
Answer: (b) \(30^\circ\)
Explanation: Let height = \(h\). Shadow = \(h\sqrt{3}\). \(\tan \theta = \frac{\text{Height}}{\text{Shadow}} = \frac{h}{h\sqrt{3}} = \frac{1}{\sqrt{3}}\). Therefore, \(\theta = 30^\circ\).
11. LCM (306, 657):
Answer: (a) 22338
Explanation: \(\text{LCM} \times \text{HCF} = \text{Product of numbers}\).
\(\text{LCM} = \frac{306 \times 657}{9} = 34 \times 657 = 22338\).
12. Zeroes reciprocal to each other:
Answer: (a) 2
Explanation: If zeroes are reciprocal, product of zeroes = 1. \(\frac{c}{a} = 1 \Rightarrow \frac{4k}{k^2 + 4} = 1\).
\(k^2 - 4k + 4 = 0 \Rightarrow (k - 2)^2 = 0 \Rightarrow k = 2\).
13. Next term of AP:
Answer: (c) \(\sqrt{50}\)
Explanation: The AP is \(\sqrt{8} = 2\sqrt{2}\), \(\sqrt{18} = 3\sqrt{2}\), \(\sqrt{32} = 4\sqrt{2}\). Common difference is \(\sqrt{2}\).
Next term = \(4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50}\).
14. BPT application:
Answer: (d) 4
Explanation: By Basic Proportionality Theorem, \(\frac{AD}{DB} = \frac{AE}{EC}\).
\(\frac{x}{x-2} = \frac{x+2}{x-1} \Rightarrow x(x-1) = (x+2)(x-2) \Rightarrow x^2 - x = x^2 - 4 \Rightarrow x = 4\).
15. Mid-point formula:
Answer: (c) 7
Explanation: Mid-point \(P(x, y) = (\frac{3+k}{2}, \frac{4+6}{2}) = (\frac{3+k}{2}, 5)\).
So, \(y = 5\). Since \(x + y - 10 = 0\), \(x + 5 = 10 \Rightarrow x = 5\).
Equating x-coordinates: \(\frac{3+k}{2} = 5 \Rightarrow 3 + k = 10 \Rightarrow k = 7\).
16. Trig ratio evaluation:
Answer: (b) \(\sqrt{3}\)
Explanation: \(\sin A = \frac{1}{2}\) implies \(A = 30^\circ\). \(\cot 30^\circ = \sqrt{3}\).
17. Roots of \(x^2 - p^2 = 0\):
Answer: (c) \(p, -p\)
Explanation: \(x^2 = p^2 \Rightarrow x = \pm p\).
18. Trig identity:
Answer: (b) 9
Explanation: \(9(\sec^2 A - \tan^2 A)\). Using identity \(\sec^2 A - \tan^2 A = 1\), value is \(9 \times 1 = 9\).
19. Assertion & Reason (Triangles):
Answer: (a) Both A and R are true and R is the correct explanation of A.
Explanation: SSS Similarity states that if corresponding sides are proportional, triangles are similar. The reason perfectly explains the assertion.
20. Assertion & Reason (Trigonometry):
Answer: (d) A is false but R is true.
Explanation: The assertion \(\sin(A+B) = \sin A + \sin B\) is a false statement in trigonometry. The counter-example provided in the reason is mathematically correct.
SECTION B (Very Short Answer)
Let \(N = 7 \times 11 \times 13 + 13\).
Taking 13 as a common factor:
\(N = 13 \times (7 \times 11 + 1)\) [1 Mark]
\(N = 13 \times (77 + 1) = 13 \times 78\).
Since \(N\) can be expressed as a product of primes (\(13 \times 13 \times 2 \times 3\)), it has factors other than 1 and itself. Therefore, it is a composite number. [1 Mark]
Let the y-axis divide the segment in the ratio \(k:1\).
The x-coordinate of any point on the y-axis is 0. [0.5 Mark]
Using section formula for x-coordinate:
\(x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}\)
\(0 = \frac{k(-1) + 1(5)}{k + 1}\) [1 Mark]
\(-k + 5 = 0 \Rightarrow k = 5\).
The required ratio is 5:1. [0.5 Mark]
Let the zeroes be \(\alpha = 3 + \sqrt{2}\) and \(\beta = 3 - \sqrt{2}\).
Sum of zeroes (\(S\)) = \(\alpha + \beta = 3 + \sqrt{2} + 3 - \sqrt{2} = 6\). [1 Mark]
Product of zeroes (\(P\)) = \(\alpha \times \beta = (3 + \sqrt{2})(3 - \sqrt{2}) = (3)^2 - (\sqrt{2})^2 = 9 - 2 = 7\). [0.5 Mark]
The quadratic polynomial is \(x^2 - Sx + P = \mathbf{x^2 - 6x + 7}\). [0.5 Mark]
Since \(DE \parallel BC\), by Thales Theorem (BPT):
\(\frac{AD}{BD} = \frac{AE}{CE}\) [0.5 Mark]
\(\frac{4x - 3}{3x - 1} = \frac{8x - 7}{5x - 3}\)
Cross multiplying:
\((4x - 3)(5x - 3) = (8x - 7)(3x - 1)\) [0.5 Mark]
\(20x^2 - 12x - 15x + 9 = 24x^2 - 8x - 21x + 7\)
\(20x^2 - 27x + 9 = 24x^2 - 29x + 7\)
\(4x^2 - 2x - 2 = 0 \Rightarrow 2x^2 - x - 1 = 0\) [0.5 Mark]
\(2x^2 - 2x + x - 1 = 0 \Rightarrow 2x(x - 1) + 1(x - 1) = 0\)
\((2x + 1)(x - 1) = 0\). Since length cannot be negative for \(x = -1/2\), we have \(x = 1\). [0.5 Mark]
LHS = \(\sec A (1 - \sin A) (\sec A + \tan A)\)
Convert to \(\sin\) and \(\cos\):
\(= \frac{1}{\cos A} (1 - \sin A) \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)\) [1 Mark]
\(= \frac{1 - \sin A}{\cos A} \times \frac{1 + \sin A}{\cos A}\)
\(= \frac{1 - \sin^2 A}{\cos^2 A}\) [0.5 Mark]
Using identity \(1 - \sin^2 A = \cos^2 A\):
\(= \frac{\cos^2 A}{\cos^2 A} = \mathbf{1} = \text{RHS}\). Hence proved. [0.5 Mark]
SECTION C (Short Answer)
Let us assume to the contrary that \(\sqrt{3}\) is rational.
Then, \(\sqrt{3} = \frac{a}{b}\), where \(a\) and \(b\) are coprime integers and \(b \neq 0\). [0.5 Mark]
Squaring both sides: \(3 = \frac{a^2}{b^2} \Rightarrow a^2 = 3b^2\).
This implies that 3 divides \(a^2\), and by theorem, 3 divides \(a\).
Let \(a = 3c\) for some integer \(c\). [1 Mark]
Substituting this in our equation: \((3c)^2 = 3b^2 \Rightarrow 9c^2 = 3b^2 \Rightarrow b^2 = 3c^2\).
This implies that 3 divides \(b^2\), which means 3 divides \(b\). [1 Mark]
Therefore, \(a\) and \(b\) have at least 3 as a common factor. This contradicts the fact that \(a\) and \(b\) are coprime.
This contradiction arises due to our incorrect assumption. Hence, \(\sqrt{3}\) is irrational. [0.5 Mark]
Polynomial: \(p(x) = 4\sqrt{3}x^2 + 5x - 2\sqrt{3}\).
Splitting middle term: Product = \(4\sqrt{3} \times (-2\sqrt{3}) = -24\). Sum = 5. Numbers are 8 and -3.
\(4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0\)
\(4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0\) [1 Mark]
\((\sqrt{3}x + 2)(4x - \sqrt{3}) = 0\)
Zeroes are \(\alpha = \frac{-2}{\sqrt{3}}\) and \(\beta = \frac{\sqrt{3}}{4}\). [0.5 Mark]
Verification:
Sum of zeroes = \(\alpha + \beta = \frac{-2}{\sqrt{3}} + \frac{\sqrt{3}}{4} = \frac{-8 + 3}{4\sqrt{3}} = \frac{-5}{4\sqrt{3}}\).
From formula, \(\frac{-b}{a} = \frac{-5}{4\sqrt{3}}\). Hence verified. [1 Mark]
Product of zeroes = \(\alpha \beta = \left(\frac{-2}{\sqrt{3}}\right)\left(\frac{\sqrt{3}}{4}\right) = \frac{-2}{4} = \frac{-1}{2}\).
From formula, \(\frac{c}{a} = \frac{-2\sqrt{3}}{4\sqrt{3}} = \frac{-1}{2}\). Hence verified. [0.5 Mark]
Let the unit digit be \(y\) and tens digit be \(x\). The number is \(10x + y\).
Number obtained by reversing the digits is \(10y + x\).
Given: \((10x + y) + (10y + x) = 66 \Rightarrow 11(x + y) = 66 \Rightarrow x + y = 6\). (Equation 1) [1 Mark]
Given: Digits differ by 2. Thus, \(x - y = 2\) OR \(y - x = 2\). [1 Mark]
Case 1: \(x + y = 6\) and \(x - y = 2\)
Adding both: \(2x = 8 \Rightarrow x = 4\). Then \(y = 2\). Number = 42.
Case 2: \(x + y = 6\) and \(-x + y = 2\)
Adding both: \(2y = 8 \Rightarrow y = 4\). Then \(x = 2\). Number = 24. [0.5 Mark]
There are 2 such numbers. [0.5 Mark]
First 3-digit number divisible by 7 is 105. Last is 994.
The A.P. is 105, 112, 119, ..., 994.
Here \(a = 105\), \(d = 7\), \(a_n = 994\). [1 Mark]
Using \(a_n = a + (n-1)d\):
\(994 = 105 + (n-1)7\)
\(889 = (n-1)7 \Rightarrow n-1 = 127 \Rightarrow n = 128\). [1 Mark]
Sum \(S_n = \frac{n}{2}(a + l) = \frac{128}{2}(105 + 994)\)
\(S_{128} = 64 \times 1099 = \mathbf{70,336}\). [1 Mark]
LHS = \((\sin A + \csc A)^2 + (\cos A + \sec A)^2\)
Expand both squares:
\(= \sin^2 A + \csc^2 A + 2\sin A\csc A + \cos^2 A + \sec^2 A + 2\cos A\sec A\) [1 Mark]
Since \(\sin A \csc A = 1\) and \(\cos A \sec A = 1\):
\(= (\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 2(1) + 2(1)\)
\(= 1 + \csc^2 A + \sec^2 A + 4 = 5 + \csc^2 A + \sec^2 A\) [1 Mark]
Using identities \(\csc^2 A = 1 + \cot^2 A\) and \(\sec^2 A = 1 + \tan^2 A\):
\(= 5 + (1 + \cot^2 A) + (1 + \tan^2 A)\)
\(= \mathbf{7 + \tan^2 A + \cot^2 A} = \text{RHS}\). Hence proved. [1 Mark]
Let the vertices be \(A(6, 1)\), \(B(8, 2)\), \(C(9, 4)\) and \(D(p, 3)\).
We know that the diagonals of a parallelogram bisect each other. Thus, the mid-point of AC is the same as the mid-point of BD. [1 Mark]
Mid-point of AC = \(\left(\frac{6+9}{2}, \frac{1+4}{2}\right) = \left(\frac{15}{2}, \frac{5}{2}\right)\). [1 Mark]
Mid-point of BD = \(\left(\frac{8+p}{2}, \frac{2+3}{2}\right) = \left(\frac{8+p}{2}, \frac{5}{2}\right)\).
Equating the x-coordinates: \(\frac{15}{2} = \frac{8+p}{2}\)
\(15 = 8 + p \Rightarrow \mathbf{p = 7}\). [1 Mark]
SECTION D (Long Answer)
Let the original speed of the train be \(x\) km/h.
Distance = 360 km.
Original time taken \(t_1 = \frac{360}{x}\) hours. [1 Mark]
New speed = \((x + 5)\) km/h.
New time taken \(t_2 = \frac{360}{x+5}\) hours.
According to question: \(t_1 - t_2 = 1\) [1 Mark]
\(\frac{360}{x} - \frac{360}{x+5} = 1\)
\(360 \left[ \frac{x+5-x}{x(x+5)} \right] = 1\) [1 Mark]
\(360 \times 5 = x(x+5) \Rightarrow 1800 = x^2 + 5x\)
\(x^2 + 5x - 1800 = 0\)
Solving by factorization: \(x^2 + 45x - 40x - 1800 = 0\) [1 Mark]
\(x(x + 45) - 40(x + 45) = 0 \Rightarrow (x - 40)(x + 45) = 0\).
\(x = 40\) or \(x = -45\). Since speed cannot be negative, \(x = 40\).
The original speed of the train is 40 km/h. [1 Mark]
Part 1: BPT Theorem
Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. (1 Mark)
Proof steps: Draw \(\triangle ABC\) with \(DE \parallel BC\). Draw perpendiculars from \(D\) to \(AE\) and \(E\) to \(AD\). Join \(BE, CD\). (1 Mark for construction).
Area(\(\triangle ADE\))/Area(\(\triangle BDE\)) = \(\frac{1}{2}AD \times h / \frac{1}{2}BD \times h = AD/BD\). (1 Mark)
Area(\(\triangle ADE\))/Area(\(\triangle CDE\)) = \(AE/CE\).
Since \(\triangle BDE\) and \(\triangle CDE\) are on the same base \(DE\) and between same parallels, their areas are equal. Hence \(AD/BD = AE/CE\). (2 Marks)
OR Part:
In right \(\triangle ABC\) and \(\triangle ADB\):
\(\angle A = \angle A\) (Common)
\(\angle ABC = \angle ADB = 90^\circ\)
\(\triangle ABC \sim \triangle ADB\) (AA similarity) -- (Eq 1) [1.5 Marks]
In \(\triangle ABC\) and \(\triangle BDC\):
\(\angle C = \angle C\) (Common), \(\angle ABC = \angle BDC = 90^\circ\)
\(\triangle ABC \sim \triangle BDC\) (AA similarity) -- (Eq 2) [1.5 Marks]
From (1) and (2), \(\triangle ADB \sim \triangle BDC\). [1 Mark]
Therefore, corresponding sides are proportional: \(\frac{AD}{BD} = \frac{BD}{DC}\).
Cross multiplying gives: \(BD^2 = AD \times DC\). [1 Mark]
Let AB be the building (7m) and CD be the cable tower. Draw BE parallel to ground AC, meeting CD at E. Let CE = \(h\) be the height above building.
Given: \(\angle DBE = 60^\circ\) (elevation of top) and \(\angle EBC = 45^\circ\) (depression of foot, which implies \(\angle BCA = 45^\circ\)). [1 Mark for diagram conceptualization]
In \(\triangle ABC\), \(\tan 45^\circ = \frac{AB}{AC} \Rightarrow 1 = \frac{7}{AC} \Rightarrow AC = 7\) m.
Since AC = BE, horizontal distance \(BE = 7\) m. [1.5 Marks]
In \(\triangle DBE\), \(\tan 60^\circ = \frac{DE}{BE}\).
\(\sqrt{3} = \frac{DE}{7} \Rightarrow DE = 7\sqrt{3}\) m. [1.5 Marks]
Total height of tower CD = CE + DE = AB + DE = \(7 + 7\sqrt{3} = 7(1 + \sqrt{3})\) m.
\(CD = 7(1 + 1.732) = 7 \times 2.732 = \mathbf{19.124 \text{ m}}\). [1 Mark]
Let the speed of boat in still water be \(x\) km/h and stream be \(y\) km/h.
Upstream speed = \((x - y)\) km/h, Downstream speed = \((x + y)\) km/h.
Case 1: \(\frac{30}{x-y} + \frac{44}{x+y} = 10\)
Case 2: \(\frac{40}{x-y} + \frac{55}{x+y} = 13\) [1 Mark for formulating equations]
Let \(\frac{1}{x-y} = u\) and \(\frac{1}{x+y} = v\).
\(30u + 44v = 10 \Rightarrow 15u + 22v = 5\) (Eq 1)
\(40u + 55v = 13\) (Eq 2) [1 Mark]
Multiply (1) by 8 and (2) by 3:
\(120u + 176v = 40\)
\(120u + 165v = 39\)
Subtracting: \(11v = 1 \Rightarrow v = \frac{1}{11}\). [1.5 Marks]
Substitute \(v\) in (1): \(15u + 22(\frac{1}{11}) = 5 \Rightarrow 15u + 2 = 5 \Rightarrow 15u = 3 \Rightarrow u = \frac{1}{5}\).
Now, \(x - y = 5\) and \(x + y = 11\). [1 Mark]
Adding: \(2x = 16 \Rightarrow x = 8\). Subtracting: \(2y = 6 \Rightarrow y = 3\).
Speed of boat = 8 km/h, Speed of stream = 3 km/h. [0.5 Mark]
SECTION E (Case Based)
Let 1st year production be \(a\) and uniform increase be \(d\).
Given: \(a_3 = 6000 \Rightarrow a + 2d = 6000\) (Eq 1)
Given: \(a_7 = 7000 \Rightarrow a + 6d = 7000\) (Eq 2)
Subtracting (1) from (2): \(4d = 1000 \Rightarrow d = 250\).
Substitute \(d\) in (1): \(a + 500 = 6000 \Rightarrow a = 5500\).
(i) Production in 1st year = 5500 cars. [1 Mark]
(ii) 10th year: \(a_{10} = a + 9d = 5500 + 9(250) = 5500 + 2250 =\) 7750 cars. [1 Mark]
(iii) Total in 7 years: \(S_7 = \frac{7}{2}(a + a_7) = \frac{7}{2}(5500 + 7000) = \frac{7}{2}(12500) = 7 \times 6250 =\) 43750 cars. [2 Marks]
OR
Year to reach 10000: \(a_n = 10000 \Rightarrow 5500 + (n-1)250 = 10000\).
\((n-1)250 = 4500 \Rightarrow n-1 = 18 \Rightarrow n = \mathbf{19}\). So in the 19th year. [2 Marks]
(i) Student should draw a ground line, an observer at origin. A balloon at height \(1500\sqrt{3}\) directly above a point \(X\) (angle \(60^\circ\)), moving to a point above \(Y\) (angle \(30^\circ\)). [1 Mark]
(ii) Let \(X\) be horizontal distance for \(60^\circ\) and \(Y\) be total horizontal distance for \(30^\circ\).
\(\tan 60^\circ = \frac{1500\sqrt{3}}{x} \Rightarrow \sqrt{3} = \frac{1500\sqrt{3}}{x} \Rightarrow x = 1500\) m.
\(\tan 30^\circ = \frac{1500\sqrt{3}}{y} \Rightarrow \frac{1}{\sqrt{3}} = \frac{1500\sqrt{3}}{y} \Rightarrow y = 1500 \times 3 = 4500\) m.
Distance traveled = \(y - x = 4500 - 1500 =\) 3000 m. [2 Marks]
OR
Speed = Distance/Time = \(3000 / 15 =\) 200 m/s. [2 Marks]
(iii) Line-of-sight distance (Hypotenuse for 60-degree triangle):
\(\sin 60^\circ = \frac{1500\sqrt{3}}{d} \Rightarrow \frac{\sqrt{3}}{2} = \frac{1500\sqrt{3}}{d} \Rightarrow d = 1500 \times 2 =\) 3000 m. [1 Mark]
Given \(p(x) = x^2 - 2x - 8\).
(i) Zeroes: \(x^2 - 4x + 2x - 8 = 0 \Rightarrow x(x-4) + 2(x-4) = 0 \Rightarrow (x-4)(x+2) = 0\). Zeroes are 4 and -2. [1 Mark]
(ii) Value at \(x = 3\): \(p(3) = (3)^2 - 2(3) - 8 = 9 - 6 - 8 =\) -5. [1 Mark]
(iii) New zeroes are double: \(\alpha' = 4 \times 2 = 8\) and \(\beta' = -2 \times 2 = -4\).
Sum = \(8 - 4 = 4\). Product = \(8 \times (-4) = -32\).
New polynomial = \(k(x^2 - 4x - 32)\). [2 Marks]
OR
Distance between the zeroes on the x-axis = \(4 - (-2) =\) 6 units. [2 Marks]