TS Inter 2nd Year Maths 2A Question Paper March 2018

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TS Inter 2nd Year Maths 2A Question Paper March 2018

Time: 3 Hours
Maximum Marks: 75

Note: This question paper consists of three sections A, B, and C.

Section – A
(10 × 2 = 20 Marks)

I. Very Short Answer Type Questions.

• Answer all the questions.
• Each question carries two marks.

Question 1.
If z = 2 – 3i, then show that, z² – 4z + 13 = 0.
Solution:
Given z = 2 – 3i
⇒ (z – 2) = -3i
⇒ (z – 2)² = 9i²
⇒ z² – 4z + 4 = -9
⇒ z² – 4z + 13 = 0

Question 2.
If z₁ = -1 and z₂ = i, then find Arg(\(\frac{z_1}{z_2}\)).
Solution:
Given z₁ = -1
= -1 + i(0)
= cos π + i sin π
∴ Arg z₁ = π
Given z₂ = i
= 0 + i(1)
= cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
∴ Arg z₂ = \(\frac{\pi}{2}\)
Arg(\(\frac{z_1}{z_2}\)) = Arg z₁ – Arg z₂
= π – \(\frac{\pi}{2}\)
= \(\frac{\pi}{2}\)

Question 3.
If x = cis θ, then find the value of \(\left(x^6+\frac{1}{x^6}\right)\).
Solution:
Given x = cis θ = cos θ + i sin θ
⇒ x⁶ = (cos θ + i sin θ)⁶ = cos 6θ + i sin 6θ
\(\frac{1}{x^6}=\frac{1}{\cos 6 \theta+i \sin 6 \theta}\) = cos 6θ – i sin 6θ
∴ \(\left(x^6+\frac{1}{x^6}\right)\) = cos 6θ + i sin 6θ + cos 6θ – i sin 6θ = 2 cos 6θ

Question 4.
Form a quadratic equation whose roots are 7 ± 2√5.
Solution:
Let α = 7 + 2√5 and β = 7 – 2√5
α + β = 7 + 2√5 + 7 – 2√5 = 14
αβ = (7 + 2√5) (7 – 2√5)
= 49 – 20
= 29
Required quadratic equation is x² – (α + β)x + αβ = 0
∴ x² – 14x + 29 = 0

Question 5.
If -1, 2, and α are the roots of 2x³ + x² – 7x – 6 = 0, find α.
Solution:
Since -1, 2, α are the roots of the equation 2x³ + x² – 7x – 6 = 0
∴ s₁ = \(\frac{-1}{2}\)
⇒ -1 + 2 + α = \(\frac{-1}{2}\)
⇒ α + 1 = \(\frac{-1}{2}\)
⇒ α = \(\frac{-1}{2}\) – 1
⇒ α = \(\frac{-3}{2}\)

Question 6.
Find the number of ways of arranging the letters of the word MATHEMATICS.
Solution:
The word MATHEMATICS contains 11 letters in which 2M’s, 2A’s, and 2T’s rest are different.
∴ The number of ways of arranging the letters of the MATHEMATICS is \(\frac{11 !}{2 ! 2 ! 2 !}=\frac{11 !}{(2 !)^3}\)

Question 7.
If ⁿC₅ = ⁿC₆, then find ¹³C_n.
Solution:
Given ⁿC₅ = ⁿC₆
⇒ n = 5 + 6 = 11
∴ ¹³C_n = ¹³C₁₁
= ¹³C₂ [∵ ⁿC_r = ⁿC_n-r]
= \(\frac{13.12}{2}\)
= 78

Question 8.
Prove that C₀ + 2.C₁ + 4.C₂ + 8.C₃ + ………. + 2ⁿ . C_n = 3ⁿ.
Solution:

Question 9.
Find the mean deviation about the median for the following data:
4, 6, 9, 3, 10, 13, 2
Solution:
Given data points are 4, 6, 9, 3, 10, 13, 2
Now arranging data points in ascending order
2, 3, 4, 6, 9, 10, 13
∴ Median = 6
Absolute values |6 – 2|, |6 – 3|, |6 – 4|, |6 – 6|, |6 – 9|, |6 – 10|, |6 – 13| = 4, 3, 2, 0, 3, 4, 7
∴ Mean deviation about the median = \(\frac{4+3+2+0+3+4+7}{7}\)
= \(\frac{23}{7}\)
= 3.28

Question 10.
A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5).
Solution:
X follows a Poisson distribution with parameter λ.

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

• Attempt any five questions.
• Each question carries four marks.

Question 11.
Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, -2√3 + 2√3i are the vertices of an equilateral triangle.
Solution:
Let A, B, and C be the points in the Argand plane.
∴ A = (2, 2), B = (-2, -2), C = (-2√3, 2√3)

∴ AB = BC = CA
∴ A, B, and C are the vertices of an equilateral triangle.

Question 12.
Prove that \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\) does not lie in between 1 and 4, if x is real.
Solution:

= \(\frac{4 x+1}{3 x^2+4 x+1}\)
⇒ 3yx² + 4yx + 4 = 4x + 1
⇒ 3yx² + (4y – 4)x + (y – 1) = 0
x ∈ R ⇒ (4y – 4)² – 4(3y)(y – 1) ≥ 0
⇒ 16y² – 32y + 16 – 12y² + 12y ≥ 0
⇒ 4y² – 20y + 16 ≥ 0
⇒ y² – 5y – 4 ≥ 0
⇒ (y – 1) (y – 4) ≥ 0
⇒ y ≥ 0 (or) y ≥ 4
∴ \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4.

Question 13.
Find the sum of all 4 digit numbers that can be formed using the digits 1, 3, 5, 7, 9.
Solution:
Given digits are 1, 3, 5, 7, 9
The no.of four-digit numbers that can be formed using the digits 1, 3, 5, 7, 9 is ⁵P₄ = 120.
We first find the sum of the digits in the unit place of all these 120 numbers.
If we fill the unit place with 1 then the remaining 3 places can be filled with the remaining 4 digits it can be done in ⁴P₃ ways.
Similarly, each digit 3, 5, 7, 9 appears 24 times in units place.
By adding all these digits, we get
⁴P₃ × 1 + ⁴P₃ × 3 + ⁴P₃ × 5 + ⁴P₃ × 7 + ⁴P₃ × 9
= ⁴P₃ × (1 + 3 + 5 + 7 + 9)
= ⁴P₃ × 25
Similarly, the sum of the digits in ten’s place value as ⁴P3 × 25 × 10
Similarly, the values of the sum of the digits 100’s place and 1000’s places are ⁴P₃ × 25 × 100 and ⁴P₃ × 25 × 1000
Hence the sum of all the 4-digit numbers formed by using the digits 1, 3, 5, 7, 9 is ⁴P₃ × 25 × 1 + ⁴P₃ × 25 × 10 + ⁴P₃ × 25 × 100 + ⁴P₃ × 25 × 1000
= ⁴P₃ × 25 × (1 + 10 + 100 + 1000)
= ⁴P₃ × 25 × 1111
= 24 × 25 × 1111
= 6,66,600

Question 14.
Find the number of ways of selecting a cricket team of 11 players from 7 batsmen and 6 bowlers, such that there will be atleast 5 bowlers in the team.
Solution:
Since the team consists of atleast 5 bowlers, the selection may be of the following types.

	Batsmen (7)	Bowlers (6)
Type 1	6	5
Type 2	5	6

The no. of selections is first type = ⁷C₆ × ⁶C₅ = 7 × 6 = 42
The no. of selections in second type = ⁷C₅ × ⁶C₆ = 21 × 1 = 21
∴ The required number of ways of selecting the cricket team = 42 + 21 = 63

Question 15.
Resolve the fraction \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\) into partial fraction.
Solution:
Let \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+2}\)
⇒ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A\left(x^2+2\right)+(B x+C)(x-1)}{(x-1)\left(x^2+2\right)}\)
⇒ 2x² + 3x + 4 = A(x² + 2) + (Bx + C) (x – 1) ………(1)
Put x = 1
2 + 3 + 4 = A(1 + 2)
⇒ 9 = 3A
⇒ A = 3
Comparing the co-efficient of x² and constant terms in (1)
2 = A + B
⇒ 2 = 3 + B
⇒ B = -1
4 = 2A – C
⇒ 4 = 2(3) – C
⇒ C = 6 – 4
⇒ C = 2
∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}\)
∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{2-x}{x^2+2}\)

Question 16.
Suppose A and B are independent events with P(A) = 0.6 and P(B) = 0.7. Then compute:
(i) P(A ∩ B)
(ii) P(A ∪ B)
(iii) P(B/A)
(iv) P(A^C ∩ B^C)
Solution:
Given A, B are independent events and P(A) = 0.6, P(B) = 0.7
(i) P(A ∩ B) = P(A) P(B)
= (0.6) (0.7)
= 0.42

(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.6 + 0.7 – 0.42
= 0.88

(iii) P(B/A) = P(B) = 0.7

(iv) P(A^C ∩ B^C) = P(A^C) . P(B^C) (∵ Since A^C, B^C are also independent events)
= [1 – P(A)] [1 – P(B)]
= [1 – 0.6] [1 – 0.7]
= (0.4)(0.3)
= 0.12

Question 17.
A, B, C are three horses in a race. The probability of A winning the race is twice that of B and the probability of B is twice that of C. What are the probabilities of A, B, and C to win the race?
Solution:
If A, B, C are three independent events of an experiment such that P(A ∩ B^C ∩ C^C) = \(\frac{1}{4}\), P(A^C ∩ B ∩ C^C) = \(\frac{1}{8}\), P(A^C ∩ B^C ∩ C^C) = \(\frac{1}{4}\) then find P(A), P(B) and P(C).
Since A, B, C are independent events.

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

• Attempt any five questions.
• Each question carries seven marks.

Question 18.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ, then prove that cos²α + cos²β + cos²γ = \(\frac{3}{2}\) = sin²α + sin²β + sin²γ.
Solution:
Let a = cos α + i sin α
b = cos β + i sin β
c = cos γ + i sin γ
a + b + c = (cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ)
= (cos α + cos β + cos γ) + i (cos α + cos β + cos γ)
= 0 + i(0)
= 0
∴ a + b + c = 0
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{\cos \alpha+i \sin \alpha}+\frac{1}{\cos \beta+i \sin \beta}+\frac{1}{\cos \gamma+i \sin \gamma}\)
\(\frac{b c+c a+a b}{a b c}\) = cos α – i sin α + cos β – i sin β + cos γ – i sin γ
= (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 – i(0)
= 0
∴ ab + bc + ca = 0
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ 0 = a² + b² + c² + 2(0)
⇒ a² + b² + c² = 0
⇒ (cos α + i sin α)² + (cos β + i sin β)² + (cos γ + i sin γ)² = 0
⇒ cos 2α + i sin 2α + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0
⇒ (cos 2α + cos 2β + cos 2γ) + i (sin 2α + cos 2β + sin 2γ) = 0
Equation real parts on both sides, we have
cos 2α + cos 2β + cos 2γ = 0
⇒ 1 – 2 sin²α + 1 – 2 sin²β + 1 – 2 sin²γ = 0
⇒ 3 = 2(sin²α + sin²β + sin²γ)
⇒ sin²α + sin²β + sin²γ = \(\frac{3}{2}\)
Again cos 2α + cos 2β + cos 2γ = 0
⇒ 2 cos²α – 1 + 2 cos²β – 1 + 2 cos²γ – 1 = 0
⇒ 2(cos²α + cos²β + 2 cos²γ) = 3
⇒ cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
∴ cos²α + cos²β + cos²γ = \(\frac{3}{2}\) = sin²α + sin²β + sin²γ

Question 19.
Solve the equation x⁴ – 10x³ + 26x² – 10x + 1 = 0.
Solution:
Given x⁴ – 10x³ + 26x² – 10x + 1 = 0
⇒ x² – 10x + 26 – \(\text { 10. } \frac{1}{x}+\frac{1}{x^2}\) = 0
⇒ \(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)\) + 26 = 0
Let x + \(\frac{1}{x}\) = t
Then x² + \(\frac{1}{x^2}\) = t² – 2
∴ (t² – 2) – 10t + 26 = 0
⇒ t² – 2 – 10t + 26 = 0
⇒ t² – 10t + 24 = 0
⇒ t² – 4t – 6t + 24 = 0
⇒ t(t – 4) – 6(t – 4) = 0
⇒ (t – 4) (t – 6) = 0
⇒ t = 4 (or) t = 6
∴ x + \(\frac{1}{x}\) = 4
⇒ x² + 1 = 4x
⇒ x² – 4x + 1 = 0
⇒ x = \(\frac{-(-4) \pm \sqrt{16-4.1 .1}}{2.1}\)
⇒ x = \(\frac{4 \pm \sqrt{12}}{2}\)
⇒ x = \(\frac{4 \pm 2 \sqrt{3}}{2}\)
⇒ x = 2 ± √3
∴ x + \(\frac{1}{x}\) = 6
⇒ x² + 1 = 6x
⇒ x² – 6x + 1 = 0
⇒ x = \(\frac{-(-6) \pm \sqrt{36-4.1 .1}}{2.1}\)
⇒ x = \(\frac{6 \pm \sqrt{32}}{2}\)
⇒ x = \(\frac{6 \pm 4 \sqrt{2}}{2}\)
⇒ x = 3 ± 2√2
∴ The roots are 2 ± √3, 3 ± 2√2

Question 20.
If the coefficients of r^th, (r + 1)^th and (r + 2)^nd terms in the expansion of (1 + x)ⁿ are in A.P. Then show that n² – (4r + 1)n + 4r² – 2 = 0.
Solution:
The coefficients of r^th, (r + 1)^th, and (r + 2)^th terms in the expansion of (1 + x)³ are ⁿC_r-1, ⁿC_r, and ⁿC_r+1 respectively.
Given ⁿC_r-1 + ⁿC_r+1 = 2 ⁿC_r are in A.P.
∴ ⁿC_r-1 + ⁿC_r+1 = 2 ⁿC_r

⇒ (n – r) (n – r + 1) = (r + 1) (2n – 3r + 2)
⇒ n² – 2nr + r² + n – r = 2nr – 3r² + 2r + 2n – 3r + 2
⇒ n² – 4nr + 4r² – n – 2 = 0
⇒ n² – (4r + 1)n + 4r² – 2 = 0

Question 21.
If x = \(\frac{1.3}{3.6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6.9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots\), then prove that 9x² + 24x = 11.
Solution:


⇒ 4 + 3x = 3√3
⇒ (4 + 3x)² = 27
⇒ 16 + 9x² + 24x = 27
∴ 9x² + 24x = 11

Question 22.
Find the mean deviation about the mean for the following data:

Marks Obtained	0-10	10-20	20-30	30-40	40-50
No. of Students	5	8	15	16	6

Solution:
Taking the assumed mean a = 25 and h = 10
Construct the table

\(\bar{x}=\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\mathrm{N}}\right) \mathrm{h}\)
= 25 + (\(\frac{10}{50}\))10
= 27
∴ Mean deviation from the mean = \(\frac{1}{N} \Sigma f_i\left|x_i-\bar{x}\right|\)
= \(\frac{10}{50}\)(472)
= 9.44

Question 23.
State and prove the addition theorem on probability.
Solution:
Statement: If E₁ and E₂ are any two events of a random experiment and P is the probability function.
Then P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)
Case (i): When E₁ ∩ E₂ = φ
E₁ ∩ E₂ = φ
⇒ P (E₁ ∩ E₂) = 0
∴ P(E₁ ∪ E₂) = P(E₁) + P(E₂) [∵ from the union axiom]
= P(E₁) + P(E₂) – 0
= P(E₁) + P(E₂) – P(E₁ ∩ E₂)
Case (ii): When E₁ ∩ E₂ = φ
∴ E₁ ∪ E₂ can be expressed as union of 2 mutually exclusive events E₁ – E₂, E₂.
Hence, E₁ ∪ E₂ = (E₁ – E₂) ∪ E₂
Also (E₁ – E₂) ∩ E₂ = φ
∴ P(E₁ ∪ E₂) = P[(E₁ – E₂) ∪ E₂]
P(E₁ ∪ E₂) = P(E₁ – E₂) + P(E₂) ………(1) [∵ from the union axiom]
Also, E₁ can be expressed as the union of 2 mutually exclusive events E₁ – E₂, E₁ ∩ E₂.
Hence, E₁ = (E₁ – E₂) ∪ (E₁ ∩ E₂)
Also (E₁ – E₂) ∩ (E₁ ∩ E₂) = φ
∴ P(E₁) = P(E₁ – E₂) ∪ (E₁ ∩ E₂)] = P(E₁ – E₂) + P(E₁ ∩ E₂)
⇒ P(E₁ – E₂) = P(E₁) – P(E₁ ∩ E₂)
From (1 ),
P(E₁ ∪ E₂) = P(E₁) – P(E₁ ∩ E₂) + P(E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)

Question 24.
A random variable ‘X’ has the following probability distribution.

X = x	0	1	2	3	4	5	6	7
P(X = x)	0	K	2K	2K	3K	K2	2K2	7K2 + K

Find:
(i) K
(ii) Mean
(iii) P(0 < x < 5)
Solution:
We know P(X = x) = 1
⇒ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 1
⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
⇒ 10k² + 9k = 1
⇒ 10k² + 9k – 1 = 0
⇒ 10k² + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (k + 1) (10k – 1) = 0
⇒ k = -1 (or) \(\frac{1}{10}\)

(i) k = \(\frac{1}{10}\) Since k > 0

(ii) Mean = 0 (P = 0) + 1 P(X = 1) + 2 P(X = 2) + 3 P(X = 3) + 4 P(X = 4) + 5 P(X = 5) + 6 P(X = 6) + 7 P(X = 7)
= 0(0) + 1(k) + 2(2k) + 3(2k) + 4(3k) + 5(k²) + 6(2k²) + 7(7k² + k)
= 0 + k + 4k + 6k + 12k + 5k² + 12k² + 49k² + 7k
= 66k² + 30k
= \(66\left(\frac{1}{10}\right)^2+30\left(\frac{1}{10}\right)\)
= 66(\(\frac{1}{100}\)) + 3
= 0.66 + 3
= 3.66

(iii) P(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 2k + 3k
= 8k
= 8(\(\frac{1}{10}\))
=0.8
∴ P(0 < x < 5) = 0.8