AP Inter 2nd Year Maths 2A Question Paper May 2019

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AP Inter 2nd Year Maths 2A Question Paper May 2019

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 multiplicative inverse of 7 + 24i.
Solution:
The multiplicative inverse of a + ib is \(\frac{a}{a^2+b^2}-i \frac{b}{a^2+b^2}\)
∴ The multiplicative inverse of 7 + 24i is

Question 2.
Simplify i² + i⁴ + i⁶ + …….. up to (2n + 1) terms.
Solution:
i² + i⁴ + i⁶ + ………(2n + 1) terms
= i² + (i²)² + (i²)³ + ……….(2n + 1) terms
= -1 + (-1)² + (-1)³ + (-1)⁴ + ………(2n + 1) terms
= -1 + 1 – 1 + 1 – ………(2n + 1) terms
= -1
∴ i² + i⁴ + i⁶ + ……….(2n + 1) terms = -1.

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

Question 4.
Form the quadratic equation whose roots are \(\frac{p-1}{p+q}\) and \(\frac{-(p+q)}{p-q}\) (p ≠ ±q).
Solution:

Question 5.
Find the algebraic equation whose roots are 2 times the roots of x⁵ – 2x⁴ + 3x³ – 2x² + 4x + 3 = 0.
Solution:
Let f(x) = x⁵ – 2x⁴ + 3x³ – 2x² + 4x + 3
The algebraic equation whose roots are 2 times the roots of f(x) = 0 is f(\(\frac{x}{2}\)) = 0.

⇒ x⁵ – 4x⁴ + 12x³ – 16x² + 64x + 96 = 0.

Question 6.
Find the number of functions from a set A containing 5 elements into a set B containing 4 elements.
Solution:
Here n(A) = 5 and n(B) = 4
The number of functions from a set A into a set B is [n(B)]^n(A).
∴ The number of functions from set A into a set B is 4⁵.

Question 7.
If ¹⁵C_2r-1 = ¹⁵C_2r+4, find r.
Solution:
Given ¹⁵C_2r-1 = ¹⁵C_2r+4
⇒ 15 = 2r – 1 + 2r + 4
⇒ 15 = 4r + 3
⇒ 4r = 12
⇒ r = 3

Question 8.
If ²²C_r is the largest binomial coefficient in the expansion of (1 + x)²², find the value of ¹³C_r.
Solution:
Here n = 22 (even integer)
∴ Largest binomial co-efficient = \({ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{\mathrm{n}}{2}\right)}\) = ²²C₁₁
∴ r = 11
∴ ¹³C_r = ¹³C₁₁
= ¹³C₂
= \(\frac{13 \times 12}{2}\)
= 78

Question 9.
Find the mean deviation from the mean of the following discrete data:
6, 7, 10, 12, 13, 4, 12, 16
Solution:
The arithmetic mean of the given data is
\(\bar{S}\) = \(\frac{6+7+10+12+13+4+12+16}{8}\)
= \(\frac{80}{8}\)
= 10
∴ Absolute values about mean = |6 – 10|, |7 – 10|, |10 – 10|, |10 – 12|, |10 – 13|, |10 – 4|, |10 – 12|, |10 – 16| = 4, 3, 0, 2, 3, 6, 2, 6
∴ Mean deviation = \(\frac{4+3+0+2+3+6+2+6}{8}\)
= \(\frac{26}{8}\)
= 3.25

Question 10.
For a binomial distribution with mean 6 and variance 2, find the first two terms of the distribution.
Solution:
Mean, np = 6
Variance, npq = 2
∴ \(\frac{n p q}{n p}=\frac{2}{6}=\frac{1}{3}\)
⇒ q = \(\frac{1}{3}\)
∴ p = 1 – q
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
∴ n(\(\frac{2}{3}\)) = 6
⇒ n = 9

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

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

Question 11.
If the real part of \(\left(\frac{z+1}{z+i}\right)\) is 1, then find the locus of z where z = x + iy.
Solution:
Given z = x + iy

Since real part of \(\left(\frac{z+1}{z+i}\right)\) is 1
∴ \(\frac{x(x+1)+y(y+1)}{x^2+(y+1)^2}\) = 1
⇒ x² + x + y² + y = x² + y² + 2y + 1
⇒ x – y – 1 = 0
∴ The locus of z is x – y – 1 = 0.

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

⇒ 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
⇒ 4y² – 20y +16 = 0
⇒ y² – 5y + 4 = 0
⇒ (y – 1) (y – 14) = 0
⇒ y = 1, 4
⇒ 4y² – 20y +16 ≥ 0
⇒ y ≤ 1 (or) y ≥ 4
⇒ y does not lie between 1 and 4.
Since the y² coefficient the exp ≥ 0.
∴ \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4.

Question 13.
Find the number of 4-letter words that can be formed using the letters of the word MIRACLE. How many of them:
(i) Begin with a vowel
(ii) Begin and end with vowels
(iii) End with a consonant.
Solution:
The word MIRACLE has 7 letters with 3 vowels (I, A, E) and 4 consonants (M, R, C, L).
The No.of 4-letter words that can be formed by using these letters is ⁷p₄
= 7 × 6 × 5 × 4
= 840

(i) For the required arrangements first place can be filled in 3 ways and the remaining three places can be filled with the remaining 6 letters it can be done in ⁶p₃ ways.
∴ The no.of 4 letter words that are begin with vowel = 3 × ⁶p₃
= 3 × 6 × 5 × 4
= 360
(ii) For the required arrangement first and last places can be filled with ³p₂ ways and the remaining two places can be filled remaining 5 letters it can done in ⁵p₂ ways.
∴ The no.of 4-letter words that begin and end with a vowel is ³p₂ × ⁵p₂
= 3 × 2 × 5 × 4
= 120

(iii) For the required arrangement, first we can fill the last place it can be done in 4 ways.
The remaining three places can be filled with the remaining 6 letters it can be done in ⁶p₃ ways.
∴ The no.of 4 letter words that are end with consonant = 4 × ⁶p₃
= 4 × 6 × 5 × 4
= 480

Question 14.
Prove that \(\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\frac{1.3 .5 \ldots \ldots . .(4 n-1)}{\{1 \cdot 3 \cdot 5 \ldots \ldots \ldots . .(2 n-1)\}^2}\).
Solution:

Question 15.
Resolve \(\frac{3 x^3-2 x^2-1}{x^4+x^2+1}\) into partial fractions.
Solution:
x⁴ + x² + 1 = x⁴ + 2x² + 1 – x²
= (x² + 1)² – x²
= (x² + 1 + x) (x² + 1 – x)
= (x² + x + 1) (x² – x + 1)
Let \(\frac{3 x^3-2 x^2-1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}\)
⇒ 3x³ – 2x² – 1 = (Ax + B) (x² – x + 1) + (Cx + D) (x² + x + 1)
Equating the coefficients of like terms, we have
A + C = 3 ………(1)
⇒ C = 3 – A
-A + B + C + D = -2 ………(2)
A – B + C + D = 0 ……….(3)
B + D = -1 ………(4)
⇒ D = -1 – B
Substitute C, D in (2)
-A + B + 3 – A – 1 – B = -2
⇒ -2A = -4
⇒ A = 2
∴ C = 3 – 2 = 1
Substitute C, D in (3)
A – B + 3 – A – 1 – B = 0
⇒ -2B = -2
⇒ B = 1
∴ D = -1 – 1 = -2
∴ A = 2, B = 1, C = 1, D = -2
∴ \(\frac{3 x^3-2 x^2-1}{x^4+x^2+1}=\frac{2 x+1}{x^2+x+1}+\frac{x-2}{x^2-x+1}\)

Question 16.
The probabilities of three mutually exclusive events are respectively given as \(\frac{1+3 P}{3}, \frac{1-P}{4}, \frac{1-2 P}{2}\). Prove that \(\frac{1}{3}\) ≤ P ≤ \(\frac{1}{2}\).
Solution:
∴ 0 ≤ \(\frac{1+3 p}{3}\) ≤ 1
⇒ 0 ≤ 1 + 3p ≤ 3
⇒ -1 ≤ 3p ≤ 2
⇒ \(\frac{-1}{3} \leq p \leq \frac{2}{3}\) …….(1)
∴ 0 ≤ \(\frac{1-p}{4}\) ≤ 1
⇒ 0 ≤ 1 – p ≤ 4
⇒ -1 ≤ -p ≤ 3
⇒ 1 ≥ p ≥ -3
⇒ -3 ≤ p ≤ 1 ……….(2)
∴ o ≤ \(\frac{1-2 p}{2}\) ≤ 1
⇒ 0 ≤ 1 – 2p ≤ 2
⇒ -1 ≤ – 2p ≤ 1
⇒ 1 ≥ 2p ≥ -1
⇒ -1 ≤ 2p ≤ 1
⇒ \(\frac{-1}{2} \leq p \leq \frac{1}{2}\) ………(3)
From (1), (2), (3) we have
\(\frac{1}{3} \leq p \leq \frac{1}{2} \Rightarrow p \in\left(\frac{1}{3}, \frac{1}{2}\right)\)

Question 17.
If A and B are independent events of a random experiment, show that A^C and B^C are also independent.
Solution:
Given A and B are independent events.
∴ P(A ∩ B) = P(A) . P(B)
P(A^C ∩ B^C) = P[(A ∪ B)^C]
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – [P(A) + P(B) – P(A) + P(B)]
= 1 – P(A) – P(B) + P(A) . P(B)
= [1 – P(A)] – P(B) [1 – P(A)]
= [1 – P(A)] [1 – P(B)]
= P(A^C) . P(B^C)
∴ A^C and B^C are also independent.

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

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

Question 18.
If α, β are the roots of the equation x² – 2x + 4 = 0, then for any n ∈ N show that \(\alpha^n+\beta^n=2^{n+1} \cos \left(\frac{n \pi}{3}\right)\).
Solution:

Question 19.
Find the polynomial equation whose roots are the translates of those of the equation x⁴ – 5x³ + 7x² – 17x + 11 = 0 by -2.
Solution:
Let f(x) ≡ x⁴ – 5x³ + 7x² – 17x + 11 = 0
The polynomial equation whose roots are translates of those of the equation f(x) = 0 by – 2 is f(x + 2) = 0
⇒ (x + 2)⁴ – 5(x + 2)³ + 7(x + 2)² – 17(x + 2) + 11 = 0

∴ Required equation is x⁴ + 3x³ + x² – 17x – 19 = 0.

Question 20.
If the coefficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are in A.P., then prove that n² – 41n + 398 = 0.
Solution:
The co-efficients of x⁹, x¹⁰, x¹¹ in the expansion of (1 + x)ⁿ are ⁿC₉, ⁿC₁₀, ⁿC₁₁.
Given that ⁿC₉, ⁿC₁₀, and ⁿC₁₁ are in A.P.
∴ ⁿC₉ + ⁿC₁₁ = 2 . ⁿC₁₀

⇒ \(\frac{110+n^2-10 n-9 n+90}{11(n-9)}\) = 2
⇒ n² – 19n + 200 = 22(n – 9)
⇒ n² – 19n + 200 = 22n – 198
⇒ n² – 41n + 398 = 0

Question 21.
If x = \(\frac{1.3}{3.6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\)………, 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 median for the following continuous distribution:

Marks Obtained	No. of Boys
0-10	6
10-20	8
20-30	14
30-40	16
40-50	4
50-60	2

Solution:
Construct the table

\(\frac{N}{2}=\frac{50}{2}\) = 25
∴ 20 – 30 is the median class.
∴ Median = L + \(\left(\frac{\frac{N}{2}-\text { Preceding cumulative Frequency }}{f}\right) i\)
= 20 + \(\left(\frac{25-14}{14}\right)\)10
= 20 + 7.857
= 27.857
∴ Mean deviation about median = \(\frac{1}{N} \sum_{i=1}^6 f_i\)|x_i – median|
= \(\frac{1}{50}\)(517.1)
= 10.34

Question 23.
Suppose an urn B₁ contains 2 white and 3 black balls and another urn B₂ contains 3 white and 4 black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found to be black, find the probability that the urn chosen was B₁.
Solution:
Let E₁ and E₂ denote the events or selecting urns B₁ and B₂ respectively.
∴ P(E₁) = P(E₂) = \(\frac{1}{2}\)

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

X = x	P(X = x)
0	0
1	k
2	2k
3	2k
4	3k
5	k2
6	2k2
7	7k2 + k

Find: (i) k (ii) the mean and (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) \(\frac{1}{10}\)
∴ k = \(\frac{1}{10}\) Since k > 0
(i) k = \(\frac{1}{10}\)

(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.