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

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AP 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 questions.
• Each question carries TWO marks.

Question 1.
Find the complex conjugate of (2 + 5i) (-4 + 6i).
Solution:
Let Z = (2 + 5i) (-4 + 6i)
= -8 + 12i – 20i + 30i²
= -8 – 8i – 30 [∵ i² = -1]
= -38 – 8i
∴ \(\bar{Z}\) = -38 + 8i

Question 2.
If x + iy = cis α . cis β, then find the value of x² + y².
Solution:
Given x + iy = cis α . cis β
= cis(α + β)
= cos(α + β) + i sin(α + β)
Equating real and imaginary parts on bothsides
x = cos(α + β), y = sin(α + β)
∴ x² + y² = cos²(α + β) + sin²(α + β) = 1
∴ x² + y² = 1

Question 3.
If A, B, and C are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find the value of XYZ.
Solution:
Since A, B, C are angles of a triangle.
∴ A + B + C = 180°
Given x = cis A, y = cis B, z = cis C
XYZ = (cis A) (cis B) (cis C)
= cis (A + B + C)
= cis 180°
= cos 180° + i sin 180°
= -1 + i(0)
= -1
∴ XYZ = -1

Question 4.
For what values of x the expression x² – 5x – 14 is positive?
Solution:
x² – 5x – 14 > 0
⇒ x² – 7x + 2x – 14 > 6
⇒ x(x – 7) + 2(x – 7) > 0
⇒ (x + 2) (x – 7) > 0
⇒ x < -2 (or) x > 7

Question 5.
If 1, 1, α are the roots of x³ – 6x² + 9x – 4 = 0, then find α.
Solution:
Since 1, 1, α are the roots of x³ – 6x² + 9x – 4 = 0
s₁ = \(\frac{-(-6)}{1}\)
⇒ 1 + 1 + α = 6
⇒ α = 6 – 2
⇒ α = 4

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

Question 7.
Find the value of ¹⁰C₅ + 2 . ¹⁰C₄ + ¹⁰C₃.
Solution:

Question 8.
Find the 8th term of \(\left(1-\frac{5 x}{2}\right)^{-3 / 5}\).
Solution:

Question 9.
Find the mean deviation about the mean for the following data:
3, 6, 10, 4, 9, 10
Solution:
Given data 3, 6, 10, 4, 9, 10
Mean = \(\frac{3+6+10+4+9+10}{6}=\frac{42}{6}\) = 7
The absolute values of the deviations are |3 – 7|, |6 – 7|, |10 – 7|, |4 – 7|, |9 – 7|, |10 – 7| = 4, 1, 3, 3, 2, 3
∴ The mean deviation about the mean = \(\frac{4+1+3+3+2+3}{6}\)
= \(\frac{16}{6}\)
= 2.66

Question 10.
If the mean and variance of a binomial variable X are 2.4 and 1.44 respectively, find the parameters of the distribution X. (Binomial).
Solution:
X following Binomial distribution with parameters n and p.
mean np = 2.4 ………(1)
variance npq = 1.44 …………(2)
\(\frac{n p q}{n p}=\frac{1.44}{2.4}\)
⇒ q = 0.6
∴ p = 1 – q
= 1 – 0.6
= 0.4
from (1), n(0.4) = 2.4
⇒ n = 6
∴ n = 6, p = 0.4

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

• Answer ANY FIVE questions.
• Each question carries FOUR marks.

Question 11.
If (x – iy)^1/3 = a – ib, then show that \(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²).
Solution:
Given (x – iy)^1/3 = a – ib
⇒ x – iy = (a – ib)³
= a³ – 3a²ib + 3ai²b² + i³b³
= a³ – i³a²b – 3ab² + ib³
= (a³ – 3ab²) + i(b³ – 3a²b)
Equating real and imaginary parts on both sides
x = a³ – 3ab², -y = b³ – 3a²b ⇒ y = -(b³ – 3a²b)
\(\frac{x}{a}+\frac{y}{b}=\frac{a^3-3 a b^2}{a}+\frac{-\left(b^3-3 a^2 b\right)}{b}\)
= \(\frac{a^3-3 a b^2}{a}+\frac{3 a^2 b-b^3}{b}\)
= a² – 3b² + 3a² – b²
= 4a² – 4b²
= 4(a² – b²)
∴ \(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²)

Question 12.
Find the maximum value of the function \(\frac{x^2+14 x+9}{x^2+2 x+3}\) over R.
Solution:
Let y = \(\frac{x^2+14 x+9}{x^2+2 x+3}\)
⇒ yx² + 2xy + 3y = x² + 14x + 9
⇒ (1 – y)x² + (14 – 2y)x + (9 – 3y) = 0
x ∈ R ⇒ (14 – 2y)² – 4(1 – y)(9 – 3y) ≥ 0
⇒ 4(7 – y)² – 4(1 – y)(9 – 3y) ≥ 0
⇒ (7 – y)² – (1 – y)(9 – 3y) ≥ 0
⇒ 49 – 14y + y² – (9 – 3y – 9y + 3y2) ≥ 0
⇒ 49 – 14y + y² – 9 + 12y – 3y² ≥ 0
⇒ -2y² – 2y + 40 ≥ 0
⇒ 2y² + 2y – 40 ≤ 0
⇒ y² + y – 20 ≤ 0
⇒ (y + 5)(y – 4) ≤ 0
⇒ -5 ≤ y ≤ 4
∴ Maximum value = 4, Minimum value = -5

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, and 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 of the digits 3, 5, 7, and 9 appear 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.
Prove that: ²⁵C₄ + \(\sum_{r=0}^4{ }^{(29-r)} C_3\) = ³⁰C₄
Solution:

Question 15.
Resolve \(\frac{x^2+5 x+7}{(x-3)^3}\) into partial fractions.
Solution:

Question 16.
A and B are events with P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.3. Find the probability that:
(i) A does not occur
(ii) Neither A nor B occurs
Solution:
Given P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.3
(i) The probability that A does not occur = P(\(\bar{A}\))
= 1 – P(A)
= 1 – 0.5
= 0.5
(ii) The probability that neither A nor B occurs = P(\(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\))
= P(\(\overline{A \cup B}\))
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – [0.5 + 0.4 – 0.3]
= 1 – 0.6
= 0.4

Question 17.
A problem in calculus is given to two students A and B whose chances of solving it are \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively. Find the probability of the problem being solved if both of them try independently.
Solution:
Let E₁ and E₂ denote the events that the problem is solved by A and B respectively.
Given P(E₁) = \(\frac{1}{3}\) and P(E₂) = \(\frac{1}{4}\)
By addition theorem on probability, we have
P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)
Since E₁, E₂ are independent = P(E₁) + P(E₂) – P(E₁) . P(E₂)
= \(\frac{1}{3}+\frac{1}{4}-\frac{1}{3} \cdot \frac{1}{4}\)
= \(\frac{4+3-1}{12}\)
= \(\frac{6}{12}\)
= \(\frac{1}{2}\)

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

• Answer ANY FIVE questions.
• Each question carries SEVEN marks.

Question 18.
If n is a positive integer, show that \((P+i Q)^{\frac{1}{n}}+(P-i Q)^{\frac{1}{n}}=2\left(P^2+Q^2\right)^{\frac{1}{2 n}} \cdot \cos \left[\frac{1}{n} \tan ^{-1} \frac{Q}{P}\right]\)
Solution:

Question 19.
Solve the equation 6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0.
Solution:
Given 6×6 – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0
∴ Given equation is an even degree reciprocal equation of class two.
∴ 1 and -1 are the roots of the given equation.
∴ (x – 1) and (x + 1) are the factors of the given equation.

(x – 1) (x + 1) (6x⁴ – 25x³ + 37x² – 25x + 6) = 0
Take 6x⁴ – 25x³ + 37x² – 25x + 6 = 0
⇒ 6x² – 25x + 37x – 25 . \(\frac{1}{x}\) + 6 . \(\frac{1}{x^2}\) = 0
⇒ \(6\left(x^2+\frac{1}{x^2}\right)-25\left(x+\frac{1}{x}\right)\) + 37 = 0
Let x + \(\frac{1}{x}\) = t
Then x² + \(\frac{1}{x^2}\) = \(\left(x+\frac{1}{x}\right)^2\) – 2 = t² – 2
∴ 6(t² – 2) – 25t + 37 = 0
⇒ 6t² – 12 – 25t + 37 = 0
⇒ 6t² – 25t + 25 = 0
⇒ 6t² – 10t – 15t + 25 = 0
⇒ 2t(3t – 5) – 5(3t – 5) = 0
⇒ (3t – 5) (2t – 5) = 0
⇒ 3t – 5 = 0 (or) 2t – 5 = 0
∴ 3t – 5 = 0
3(x + \(\frac{1}{x}\)) – 5 = 0
⇒ 3x² + 3 – 5x = 0
⇒ 3x² – 5x + 3 = 0
⇒ x = \(\frac{5 \pm \sqrt{25-4.3 .3}}{2-3}\)
⇒ x = \(\frac{5 \pm i \sqrt{11}}{6}\)
∴ 2t – 5 = 0
⇒ 2(x + \(\frac{1}{x}\)) – 5 = 0
⇒ 2x² + 2 – 5x = 0
⇒ 2x² – 5x + 2 = 0
⇒ 2x² – 4x – x + 2 = 0
⇒ 2x(x – 2) – 1(x – 2) = 0
⇒ (x – 2) (2x – 1) = 0
⇒ x = 2, \(\frac{1}{2}\)
Hence the roots of the given equation are 1, -1, 2, \(\frac{1}{2}\), \(\frac{5 \pm i \sqrt{11}}{6}\)

Question 20.
For r = 0, 1, 2, ……n, prove that: C₀C_r + C₁C_r+1 + C₂C_r+2 + ……… + C_n = ²ⁿC_(n+r) and hence deduce that:
(i) \(C_0^2+C_1^2+C_2^2+\ldots \ldots \ldots \ldots \ldots+C_n^2={ }^{2 n} C_n\)
(ii) C₀C₁ + C₁C₂ + C₂C₃ + …….. + C_n-1 . C_n = ²ⁿC_n+1.
Solution:

Question 21.
Find the sum to infinite terms of the series: \(\frac{7}{.5}\left(1+\frac{1}{10^2}+\frac{1.3}{1.2} \cdot \frac{1}{10^4}+\frac{1.3 .5}{1.2 .3} \cdot \frac{1}{10^6}+\ldots . . \infty\right)\)
Solution:

Question 22.
Find the mean deviation from the mean of the following data, using the step deviation method:

Marks	No. of Students
0-10	6
10-20	5
20-30	8
30-40	15
40-50	7
50-60	6
60-70	3

Solution:
Let assumed mean A = 35 then d_i = \(\frac{x_i-35}{10}\)

Mean \(\overline{\mathbf{x}}=\mathrm{A}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\mathrm{N}}\right)_h\)
= 35 + (\(\frac{-8}{50}\)) 10
= 35 – 1.6
= 33.4
Mean deviation from Mean = \(\frac{1}{N} \Sigma f_i\left|x_i-\bar{x}\right|\)
= \(\frac{1}{50}\) (659.2)
= 13.18

Question 23.
(a) State and prove the Addition theorem on probability.
(b) Find the probability of drawing an ace or a spade from a well-shuffled pack of 52 playing cards.
Solution:
(a) 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
⇒ 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(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)

(b) Let A be the event that gets an ace and B be the event that gets a spade card from a well-shuffled pack of 52 playing cards S is the sample space.

Question 24.
If X is a random variable with probability distribution P(X = k) = \(\frac{(k+1) C}{2^k}\), k = 0, 1, 2, 3,………. then find C.
Solution:
Given X is a random variable with a probability distribution
P(X = k) = \(\frac{(k+1) C}{2^k}\), k = 0, 1, 2, 3,……