Thoroughly analyzing AP Inter 2nd Year Maths 2A Model Papers and AP Inter 2nd Year Maths 2A Question Paper May 2019 helps students identify their strengths and weaknesses.

## 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 i^{2} + i^{4} + i^{6} + …….. up to (2n + 1) terms.

Solution:

i^{2} + i^{4} + i^{6} + ………(2n + 1) terms

= i^{2} + (i^{2})^{2} + (i^{2})^{3} + ……….(2n + 1) terms

= -1 + (-1)^{2} + (-1)^{3} + (-1)^{4} + ………(2n + 1) terms

= -1 + 1 – 1 + 1 – ………(2n + 1) terms

= -1

∴ i^{2} + i^{4} + i^{6} + ……….(2n + 1) terms = -1.

Question 3.

If x = cis θ, then find the value of x^{6} + \(\frac{1}{x^6}\).

Solution:

x = cis θ = cos θ + i sin θ

∴ x^{6} = (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

∴ x^{6} + \(\frac{1}{x^6}\) = cos 6θ + i sin 6θ + cos 6θ – i sin 6θ

∴ x^{6} + \(\frac{1}{x^6}\) = 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 x^{5} – 2x^{4} + 3x^{3} – 2x^{2} + 4x + 3 = 0.

Solution:

Let f(x) = x^{5} – 2x^{4} + 3x^{3} – 2x^{2} + 4x + 3

The algebraic equation whose roots are 2 times the roots of f(x) = 0 is f(\(\frac{x}{2}\)) = 0.

⇒ x^{5} – 4x^{4} + 12x^{3} – 16x^{2} + 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 4^{5}.

Question 7.

If ^{15}C_{2r-1} = ^{15}C_{2r+4}, find r.

Solution:

Given ^{15}C_{2r-1} = ^{15}C_{2r+4}

⇒ 15 = 2r – 1 + 2r + 4

⇒ 15 = 4r + 3

⇒ 4r = 12

⇒ r = 3

Question 8.

If ^{22}C_{r} is the largest binomial coefficient in the expansion of (1 + x)^{22}, find the value of ^{13}C_{r}.

Solution:

Here n = 22 (even integer)

∴ Largest binomial co-efficient = \({ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{\mathrm{n}}{2}\right)}\) = ^{22}C_{11}

∴ r = 11

∴ ^{13}C_{r} = ^{13}C_{11}

= ^{13}C_{2}

= \(\frac{13 \times 12}{2}\)

= 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

⇒ x^{2} + x + y^{2} + y = x^{2} + y^{2} + 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:

⇒ 3yx^{2} + 4yx + 4 = 4x + 1

⇒ 3yx^{2} + (4y – 4)x + (y – 1) = 0

x ∈ R ⇒ (4y – 4)^{2} – 4(3y) (y – 1) ≥ 0

⇒ 16y^{2} – 32y + 16 – 12y^{2} + 12y ≥ 0

⇒ 4y^{2} – 20y + 16 ≥ 0

⇒ 4y^{2} – 20y +16 = 0

⇒ y^{2} – 5y + 4 = 0

⇒ (y – 1) (y – 14) = 0

⇒ y = 1, 4

⇒ 4y^{2} – 20y +16 ≥ 0

⇒ y ≤ 1 (or) y ≥ 4

⇒ y does not lie between 1 and 4.

Since the y^{2} 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 ^{7}p_{4}

= 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 ^{6}p_{3} ways.

∴ The no.of 4 letter words that are begin with vowel = 3 × ^{6}p_{3}

= 3 × 6 × 5 × 4

= 360

(ii) For the required arrangement first and last places can be filled with ^{3}p_{2} ways and the remaining two places can be filled remaining 5 letters it can done in ^{5}p_{2} ways.

∴ The no.of 4-letter words that begin and end with a vowel is ^{3}p_{2} × ^{5}p_{2}

= 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 ^{6}p_{3} ways.

∴ The no.of 4 letter words that are end with consonant = 4 × ^{6}p_{3}

= 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:

x^{4} + x^{2} + 1 = x^{4} + 2x^{2} + 1 – x^{2}

= (x^{2} + 1)^{2} – x^{2}

= (x^{2} + 1 + x) (x^{2} + 1 – x)

= (x^{2} + x + 1) (x^{2} – x + 1)

Let \(\frac{3 x^3-2 x^2-1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}\)

⇒ 3x^{3} – 2x^{2} – 1 = (Ax + B) (x^{2} – x + 1) + (Cx + D) (x^{2} + 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 x^{2} – 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 x^{4} – 5x^{3} + 7x^{2} – 17x + 11 = 0 by -2.

Solution:

Let f(x) ≡ x^{4} – 5x^{3} + 7x^{2} – 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 x^{4} + 3x^{3} + x^{2} – 17x – 19 = 0.

Question 20.

If the coefficients of x^{9}, x^{10}, x^{11} in the expansion of (1 + x)^{n} are in A.P., then prove that n^{2} – 41n + 398 = 0.

Solution:

The co-efficients of x^{9}, x^{10}, x^{11} in the expansion of (1 + x)^{n} are ^{n}C_{9}, ^{n}C_{10}, ^{n}C_{11}.

Given that ^{n}C_{9}, ^{n}C_{10}, and ^{n}C_{11} are in A.P.

∴ ^{n}C_{9} + ^{n}C_{11} = 2 . ^{n}C_{10}

⇒ \(\frac{110+n^2-10 n-9 n+90}{11(n-9)}\) = 2

⇒ n^{2} – 19n + 200 = 22(n – 9)

⇒ n^{2} – 19n + 200 = 22n – 198

⇒ n^{2} – 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 9x^{2} + 24x = 11.

Solution:

⇒ 4 + 3x = 3√3

⇒ (4 + 3x)^{2} = 27

⇒ 16 + 9x^{2} + 24x = 27

⇒ 9x^{2} + 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 B_{1} contains 2 white and 3 black balls and another urn B_{2} 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 B_{1}.

Solution:

Let E_{1} and E_{2} denote the events or selecting urns B_{1} and B_{2} respectively.

∴ P(E_{1}) = P(E_{2}) = \(\frac{1}{2}\)

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 | k^{2} |

6 | 2k^{2} |

7 | 7k^{2} + 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 + k^{2} + 2k^{2} + 7k^{2} + k = 1

⇒ 10k^{2} + 9k = 1

⇒ 10k^{2} + 9k – 1 = 0

⇒ 10k^{2} + 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(k^{2}) + 6(2k^{2}) + 7(7k^{2} + k)

= 0 + k + 4k + 6k + 12k + 5k^{2} + 12k^{2} + 49k^{2} + 7k

= 66k^{2} + 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.