Thoroughly analyzing AP Inter 2nd Year Maths 2A Model Papers and AP Inter 2nd Year Maths 2A Question Paper March 2018 helps students identify their strengths and weaknesses.

## 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 + 30i^{2}

= -8 – 8i – 30 [∵ i^{2} = -1]

= -38 – 8i

∴ \(\bar{Z}\) = -38 + 8i

Question 2.

If x + iy = cis α . cis β, then find the value of x^{2} + y^{2}.

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)

∴ x^{2} + y^{2} = cos^{2}(α + β) + sin^{2}(α + β) = 1

∴ x^{2} + y^{2} = 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 x^{2} – 5x – 14 is positive?

Solution:

x^{2} – 5x – 14 > 0

⇒ x^{2} – 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 x^{3} – 6x^{2} + 9x – 4 = 0, then find α.

Solution:

Since 1, 1, α are the roots of x^{3} – 6x^{2} + 9x – 4 = 0

s_{1} = \(\frac{-(-6)}{1}\)

⇒ 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 ^{10}C_{5} + 2 . ^{10}C_{4} + ^{10}C_{3}.

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(a^{2} – b^{2}).

Solution:

Given (x – iy)^{1/3} = a – ib

⇒ x – iy = (a – ib)^{3}

= a^{3} – 3a^{2}ib + 3ai^{2}b^{2} + i^{3}b^{3}

= a^{3} – i^{3}a^{2}b – 3ab^{2} + ib^{3}

= (a^{3} – 3ab^{2}) + i(b^{3} – 3a^{2}b)

Equating real and imaginary parts on both sides

x = a^{3} – 3ab^{2}, -y = b^{3} – 3a^{2}b ⇒ y = -(b^{3} – 3a^{2}b)

\(\frac{x}{a}+\frac{y}{b}=\frac{a^3-3 a b^2}{a}+\frac{-\left(b^3-3 a^2 b\right)}{b}\)

= \(\frac{a^3-3 a b^2}{a}+\frac{3 a^2 b-b^3}{b}\)

= a^{2} – 3b^{2} + 3a^{2} – b^{2}

= 4a^{2} – 4b^{2}

= 4(a^{2} – b^{2})

∴ \(\frac{x}{a}+\frac{y}{b}\) = 4(a^{2} – b^{2})

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 = \(\frac{x^2+14 x+9}{x^2+2 x+3}\)

⇒ yx^{2} + 2xy + 3y = x^{2} + 14x + 9

⇒ (1 – y)x^{2} + (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 + y^{2} – (9 – 3y – 9y + 3y2) ≥ 0

⇒ 49 – 14y + y^{2} – 9 + 12y – 3y^{2} ≥ 0

⇒ -2y^{2} – 2y + 40 ≥ 0

⇒ 2y^{2} + 2y – 40 ≤ 0

⇒ y^{2} + 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 ^{5}P_{4} = 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 ^{4}P_{3} ways.

Similarly, each of the digits 3, 5, 7, and 9 appear 24 times in units place.

By adding all these digits, we get

^{4}P_{3} × 1 + ^{4}P_{3} × 3 + ^{4}P_{3} × 5 + ^{4}P_{3} × 7 + ^{4}P_{3} × 9

= ^{4}P_{3} × (1 + 3 + 5 + 7 + 9)

= ^{4}P_{3} × 25

Similarly, the sum of the digits in ten’s place value as ^{4}P_{3} × 25 × 10

Similarly, the values of the sum of the digits 100’s place and 1000’s places are ^{4}P_{3} × 25 × 100 and ^{4}P_{3} × 25 × 1000

Hence the sum of all the 4 digit numbers formed by using the digits 1, 3, 5, 7, 9 is ^{4}P_{3} × 25 × 1 + ^{4}P_{3} × 25 × 10 + ^{4}P_{3} × 25 × 100 + ^{4}P_{3} × 25 × 1000

= ^{4}P_{3} × 25 × (1 + 10 + 100 + 1000)

= ^{4}P_{3} × 25 × 1111

= 24 × 25 × 1111

= 6,66,600

Question 14.

Prove that: ^{25}C_{4} + \(\sum_{r=0}^4{ }^{(29-r)} C_3\) = ^{30}C_{4}

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 E_{1} and E_{2} denote the events that the problem is solved by A and B respectively.

Given P(E_{1}) = \(\frac{1}{3}\) and P(E_{2}) = \(\frac{1}{4}\)

By addition theorem on probability, we have

P(E_{1} ∪ E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

Since E_{1}, E_{2} are independent = P(E_{1}) + P(E_{2}) – P(E_{1}) . P(E_{2})

= \(\frac{1}{3}+\frac{1}{4}-\frac{1}{3} \cdot \frac{1}{4}\)

= \(\frac{4+3-1}{12}\)

= \(\frac{6}{12}\)

= \(\frac{1}{2}\)

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 6x^{6} – 25x^{5} + 31x^{4} – 31x^{2} + 25x – 6 = 0.

Solution:

Given 6×6 – 25x^{5} + 31x^{4} – 31x^{2} + 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) (6x^{4} – 25x^{3} + 37x^{2} – 25x + 6) = 0

Take 6x^{4} – 25x^{3} + 37x^{2} – 25x + 6 = 0

⇒ 6x^{2} – 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 x^{2} + \(\frac{1}{x^2}\) = \(\left(x+\frac{1}{x}\right)^2\) – 2 = t^{2} – 2

∴ 6(t^{2} – 2) – 25t + 37 = 0

⇒ 6t^{2} – 12 – 25t + 37 = 0

⇒ 6t^{2} – 25t + 25 = 0

⇒ 6t^{2} – 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

⇒ 3x^{2} + 3 – 5x = 0

⇒ 3x^{2} – 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

⇒ 2x^{2} + 2 – 5x = 0

⇒ 2x^{2} – 5x + 2 = 0

⇒ 2x^{2} – 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: C_{0}C_{r} + C_{1}C_{r+1} + C_{2}C_{r+2} + ……… + C_{n} = ^{2n}C_{(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) C_{0}C_{1} + C_{1}C_{2} + C_{2}C_{3} + …….. + C_{n-1} . C_{n} = ^{2n}C_{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 E_{1} and E_{2} are any two events of a random experiment and P is the probability function.

Then P(E_{1} ∪ E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

Case (i): When E_{1} ∩ E_{2} = φ

⇒ P(E_{1} ∩ E_{2}) = 0

∴ P(E_{1} ∪ E_{2}) = P(E_{1}) + P(E_{2}) [∵ from the union axiom]

= P (E_{1}) + P(E_{2}) – 0

= P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

Case (ii): When E_{1} ∩ E_{2} = φ

E_{1} ∪ E_{2} can be expressed as union of 2 mutually exclusive events E_{1} – E_{2}, E_{2}.

Hence, E_{1} ∪ E_{2} = (E_{1} – E_{2}) ∪ E_{2} also (E_{1} – E_{2}) ∩ E_{2} = φ

∴ P(E_{1} ∪ E_{2}) = P[(E_{1} – E_{2}) ∪ E_{2}]

= P(E_{1} – E_{2}) + P(E_{2}) ……..(1) [∵ From the union axiom]

Also, E_{1} can be expressed as the union of 2 mutually exclusive events E_{1} – E_{2}, E_{1} ∩ E_{2}.

Hence, E_{1} = (E_{1} – E_{2}) ∪ (E_{1} ∩ E_{2}), also (E_{1} – E_{2}) ∩ (E_{1} ∩ E_{2}) = φ

∴ P(E_{1}) = P(E_{1} – E_{2}) ∪ (E_{1} ∩ E_{2})] = P(E_{1} – E_{2}) + P (E_{1} ∩ E_{2})

⇒ P(E_{1} – E_{2}) = P(E_{1}) – P(E_{1} ∩ E_{2})

From (1),

∴ P(E_{1} ∪ E_{2}) = P(E_{1}) – P(E_{1} ∩ E_{2}) + P(E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2})

(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,……