Thoroughly analyzing AP Inter 2nd Year Maths 2A Model Papers and AP Inter 2nd Year Maths 2A Question Paper March 2017 helps students identify their strengths and weaknesses.

## AP Inter 2nd Year Maths 2A Question Paper March 2017

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.

Write the complex, number (2 – 3i) (3 + 4i) in the form A + iB.

Solution:

(2 – 3i) (3 + 4i) = 6 + 8i – 9i – 12i^{2}

= 6 – i – 12(-1)

= 6 – i + 12

= 18 – i

= 18 + i(-1)

= A + iB where A = 18, B = -1

Question 2.

If z_{1} = -1 and z_{2} = i, then find Arg(\(\frac{z_1}{z_2}\)).

Solution:

z_{1} = -1 cos π + i sin π

∴ Arg (z_{1}) = π

z_{2} = i = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)

∴ Arg z_{2} = \(\frac{\pi}{2}\)

Arg(\(\frac{z_1}{z_2}\)) = Arg z_{1} – Arg z_{2}

= π – \(\frac{\pi}{2}\)

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

∴ \({Arg}\left(\frac{z_1}{z_2}\right)=\frac{\pi}{2}\)

Question 3.

If 1, ω, ω^{6} are the cube roots of unity, prove that: (a + b) (aω + bω^{2}) (aω^{2} + bω) = a^{3} + b^{3}.

Solution:

(a + b) (aω + bω^{2}) (aω^{2} + bω)

= (a + b) [a^{2} ω^{3} + abω^{2} + abω^{2} + b^{2}ω^{3}]

= (a + b) [a^{2} + ab(ω^{2} + ω^{4}) + b^{2}]

= (a + b) [a^{2} + ab(ω^{2} + ω) + b^{2}]

= (a + b) (a^{2} – ab + b^{2})

= a^{3} + b^{3}

Question 4.

Find the values of m for which the equation x^{2} – 15 – m(2x – 8) = 0 have equal roots.

Solution:

Given equation is x^{2} – 15 – m(2x – 8) = 0

⇒ x^{2} – 2mx + 8m – 15 = 0

If the equation ax^{2} + bx + c = 0 has equal roots then its discriminant is zero.

∵ The roots are equal ⇒ b^{2} – 4ac = 0

a = 1, b = -2m, c = 8m – 15

b^{2} – 4ac = (-2m)^{2} – 4(1) (8m – 15)

= 4m^{2} – 32m + 60

= 4(m^{2} – 8m + 15)

= 4(m – 3) (m – 5)

∴ b^{2} – 4ac = 0

⇒ 4(m – 3) (m – 5) = 0

⇒ m – 3 = 0 or m – 5 = 0

⇒ m = 3 or 5

Question 5.

If the product of the roots of 4x^{3} + 16x^{2} – 9x – a = 0 is 9, then find a.

Solution:

α, β, γ are the roots of 4x^{3} + 16x^{2} – 9x – a = 0

αβγ = 9

⇒ \(\frac{a}{4}\) = 9

⇒ a = 36

Question 6.

Find the number of different chains that can be prepared using 7 different coloured beads.

Solution:

We know that the number of circular permutations of hanging type that can be formed using n things is \(\frac{1}{2}\) {(n – 1)!}.

Hence the number of chains is \(\frac{1}{2}\) {(7 – 1)!} = \(\frac{1}{2}\) (6!) = 360.

Question 7.

If ^{n}P_{r} = 5040 and ^{n}C_{r} = 210, find n and r.

Solution:

Hint: ^{n}P_{r} = r! ^{n}C_{r} and ^{n}P_{r} = n(n – 1) (n – 2), …… [n – (r – 1)]

^{n}P_{r} = 5040, ^{n}C_{r} = 210

r! = \(\frac{{ }^n P_r}{{ }^n C_r}=\frac{5040}{210}=\frac{504}{21}\) = 24 = 4!

∴ r = 4

^{n}P_{r} = 5040

⇒ ^{n}P_{4} = 5040

⇒ ^{n}P_{4} = 10 × 504

⇒ ^{n}P_{4} = 10 × 9 × 56

⇒ ^{n}P_{4} = 10 × 9 × 8 × 7

⇒ ^{n}P_{4} = ^{10}P_{4}

∴ n = 10

∴ n = 10, r = 4

Question 8.

Find the set E of the values of x for which the binomial expansion of \((3-4 x)^{\frac{3}{4}}\) is valid.

Solution:

\((3-4 x)^{3 / 4}=3^{3 / 4}\left(1-\frac{4 x}{3}\right)^{3 / 4}\)

The binomial expansion of \((3-4 x)^{\frac{3}{4}}\) is valid, when \(\left|\frac{4 x}{3}\right|\) < 1

i.e., |x| < \(\frac{3}{4}\)

i.e., E = \(\left(\frac{-3}{4}, \frac{3}{4}\right)\)

Question 9.

Find the mean deviation about the median for the following data:

4, 6, 9, 3, 10, 13, 2.

Solution:

Given ungrouped data are 4, 6, 9, 3, 10, 13, 2

Expressing the data in ascending order of magnitude,

we have 2, 3, 4, 6, 9, 10, 13

∴ Median = 6 = b (say)

The absolute values are |6 – 2|, |6 – 3|, |6 – 4|, |6 – 6|, |6 – 9|, |6 – 10|, |6 – 13| = 4, 3, 2, 0, 3, 4, 7

∴ Mean deviation from the median = \(\frac{\sum_{i=1}^7\left|x_i-b\right|}{7}\)

= \(\frac{4+3+2+0+3+4+7}{7}\)

= \(\frac{23}{7}\)

= 3.285

Question 10.

The probability that a person chosen at random is left-handed (in handwriting) is 0.1. What is the probability that in a group of 10 people, there is one who is left-handed?

Solution:

Here n = 10, p = 0.1

q = 1 – p = 1 – 0.1 = 0.9

P(X = 1) = ^{10}C_{1} (0.1)^{1} (0.9)^{10-1}

= 10 × 0.1 × (0.9)^{9}

= 1 × (0.9)^{9}

= (0.9)^{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 point P denotes the complex number z = x + iy in the Argand plane and if \(\frac{z-i}{z-1}\) is a purely imaginary number, find the locus of P.

Solution:

We note that the quotient \(\frac{z-i}{z-1}\) is not defined if z = 1.

Since z = x + iy

∴ The locus of P is the circle x^{2} + y^{2} – x – y = 0 excluding the point (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:

Let y = \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\)

y = \(\frac{x+1+3 x+1-1}{(3 x+1)(x+1)}\)

y = \(\frac{4 x+1}{3 x^2+3 x+x+1}\)

y = \(\frac{4 x+1}{3 x^2+4 x+1}\)

⇒ 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 – 4) = 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.

If the letters of the word PRISON are permuted in all possible ways and the words thus formed are arranged in dictionary order, find the rank of the word PRISON.

Solution:

The letters of the given word in dictionary order are I N O P R S

In the dictionary order, first, all the words that begin with I come.

If I occupy the first place then the remaining 5 places can be filled with the remaining 5 letters in 5! ways.

Thus, there are 5! number of words that begin with I.

On proceeding like this we get

I → 5! ways

N → 5! ways

O → 5! ways

P I → 4! ways

P N → 4! ways

P O → 4! ways

P R I N → 2! ways

P R I O → 2! ways

P R I S N O → 1 way

P R I S O N → 1 way

Hence the rank of PRISON is 3 × 5! + 3 × 4! + 2 × 2! + 1 × 1 = 360 + 72 + 4 + 1 + 1 = 438.

Question 14.

Simplify \({ }^{34} C_5+\sum_{r=0}^4{ }^{(38-r)} C_4\).

Solution:

Question 15.

Resolve \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\) into partial fractions.

Solution:

Let \(\frac{x^2-3}{(x+2)\left(\dot{x}^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1}\)

⇒ \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A\left(x^2+1\right)+(B x+C)(x+2)}{(x+2)\left(x^2+9\right)}\)

⇒ x^{2} – 3 = A(x^{2} + 1) + (Bx + C) (x + 2) ………(1)

Put x = -2 in (1)

(-2)^{2} – 3 = A[(-2)^{2} + 1]

⇒ 4 – 3 = A(4 + 1)

⇒ 1 = 5A

⇒ A = \(\frac{1}{5}\)

In (1), Comparing the coefficients of x^{2} on both sides, we have

1 = A + B

⇒ B = 1 – A

⇒ B = 1 – \(\frac{1}{5}\)

⇒ B = \(\frac{4}{5}\)

In (1), Comparing constant terms on both sides, we have

-3 = A + 2C

⇒ 2C = -3 – A

⇒ 2C = -3 – \(\frac{1}{5}\)

⇒ 2C = \(\frac{-16}{5}\)

⇒ C = \(\frac{-8}{5}\)

∴ A = \(\frac{1}{5}\), B = \(\frac{4}{5}\), C = \(\frac{-8}{5}\)

∴ \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{\frac{1}{5}}{x+2}+\frac{\frac{4}{5} x-\frac{8}{5}}{x^2+1}\) = \(\frac{1}{5(x+2)}+\frac{4 x-8}{5\left(x^2+1\right)}\)

Question 16.

State and prove the Addition theorem on probability.

Solution:

Statement: If E_{1} and E_{2} are any two events of a random experiment and P is a probability function, then P(E_{1} ∪ E_{2}) = P(E_{1}) + P(E_{2}) – P(E_{1} ∩ E_{2}).

Proof: Case (i): E_{1} ∩ E_{2} = φ

Then P(E_{1} ∩ E_{2}) = 0

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

= P(E_{1}) + P(E_{2}) – 0

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

Case (ii): Suppose E_{1} ∩ E_{2} ≠ φ

Then E_{1} ∪ E_{2} = E_{1} ∪ (E_{2} – E_{1}) and E_{1} ∩ (E_{2} – E_{1}) = φ

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

= P(E_{1}) + P(E_{2} – E_{1})

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

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

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

Question 17.

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(\(\frac{B}{A}\))

(iv) P(Ac ∩ Bc)

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

= 1.3 – 0.42

= 0.88

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

(iv) P(A^{C} ∩ B^{C}) = P(A^{C}) . P(B^{C}) (∵ 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

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 an integer then show that: (1 + cos θ + i sin θ)^{n} + (1 + cos θ – i sin θ)^{n} = 2^{n+1} cos(\(\frac{\theta}{2}\)) cos(\(\frac{n \theta}{2}\)).

Solution:

Question 19.

Solve the equation: 2x^{5} + x^{4} – 12x^{3} – 12x^{2} + x + 2 = 0.

Solution:

Given f(x) = 2x^{5} + x^{4} – 12x^{3} – 12x^{2} + x + 2 = 0

This is an odd-degree reciprocal equation of the first type.

∴ -1 is a root.

Dividing f(x) by (x + 1), we get 2x^{4} – x^{3} – 11x^{2} – 11x + 2 = 0

Dividing by x^{2},

2x^{2} – x – 11 – \(\frac{1}{x}+\frac{2}{x^2}\) = 0

\(2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)-11=0\) …….(1)

Put a = x + \(\frac{1}{x}\) so that x2 + \(\frac{1}{x^2}\) = a^{2} – 2

Substituting in (1), the required equation is

2(a^{2} – 2) – a – 11 = 0

⇒ 2a^{2} – 4 – a – 11 = 0

⇒ 2a^{2} – a – 15 = 0

⇒ (a – 3) (2a + 5) = 0

⇒ a = 3 (or) a = \(\frac{-5}{2}\)

Case (1): a = 3

x + \(\frac{1}{x}\) = 3

⇒ x^{2} + 1 = 3x

⇒ x^{2} – 3x + 1 = 0

⇒ x = \(\frac{3 \pm \sqrt{9-4.1 .1}}{2}\)

⇒ x = \(\frac{3 \pm \sqrt{9-4}}{2}\)

⇒ x = \(\frac{3 \pm \sqrt{5}}{2}\)

Case (2): a = \(\frac{-5}{2}\)

x + \(\frac{1}{x}\) = \(\frac{-5}{2}\)

⇒ \(\frac{x^2+1}{x}=\frac{-5}{2}\)

⇒ 2x^{2} + 2 = -5x

⇒ 2x^{2} + 5x + 2 = 0

⇒ (2x + 1) (x + 2) = 0

⇒ x = \(\frac{-1}{2}\), -2

∴ The roots of the given equation are -1, \(\frac{3 \pm \sqrt{5}}{2}\), -2, \(\frac{-1}{2}\).

Question 20.

If the coefficients of 4 consecutive terms in the expansion of (1 + x)^{n} are a_{1}, a_{2}, a_{3}, a_{4} respectively, then show that: \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\)

Solution:

Given a_{1}, a_{2}, a_{3}, a_{4} are the coefficients of 4 consecutive terms in (1 + x)^{n} respectively.

Question 21.

If x = \(\frac{1.3}{3.6}+\frac{1 \cdot 3.5}{3 \cdot 6.9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}\) + …….., then Prove that 9x^{2} + 24x = 11.

Solution:

= \(\left(\frac{1}{3}\right)^{-1 / 2}-\frac{4}{3}\)

= \(\sqrt{3}-\frac{4}{3}\)

⇒ 3x + 4 = 3√3

Squaring on both sides

(3x + 4)^{2} = (3√3)^{2}

⇒ 9x^{2} + 24x + 16 = 27

⇒ 9x^{2} + 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

∴ Mean deviation from the mean = \(\frac{1}{N} \Sigma f_i\left|x_i-\bar{x}\right|\)

= \(\frac{1}{50}\)(477)

= 9.54

Question 23.

Three urns have the following composition of balls:

Urn I: 1 white, 2 black

Urn II: 2 white, 1 black

Urn III: 2 white, 2 black

One of the urns is selected at random and a ball is drawn. It turns out to be white. Find the probability that it came from urn III.

Solution:

Let E_{i} be the event of choosing the Urn i = 1, 2, 3 and P(E_{i}) be the probability of choosing the Urn i = 1, 2, 3.

Then P(E_{1}) = P(E_{2}) = P(E_{3}) = \(\frac{1}{3}\)

Having choosen the Urn i, the probability of drawing a white ball, P(W/E_{i}), is given by

P(W/E_{1}) = \(\frac{1}{3}\); P(W/E_{2}) = \(\frac{2}{3}\); P(W/E_{3}) = \(\frac{2}{4}\)

We have to find the probability P(E_{3}/W) by Baye’s theorem.

Question 24.

The probability distribution of a random variable X is given below:

X = x_{i} |
P(X = x_{i}) |

1 | k |

2 | 2k |

3 | 3k |

4 | 4k |

5 | 5k |

Find the value of k and the mean and variance of X.

Solution:

We have \(\sum_{i=1}^5 P\left(X=x_i\right)\) = 1

⇒ k + 2k + 3k + 4k + 5k = 1

⇒ k = \(\frac{1}{15}\)

Mean μ of x = \(\sum_{r=1}^5 r \cdot p\left(X=x_i\right)=\sum_{r=1}^5 r(r k)\)

= 1.(k) + 2.(2k) + 3.(3k) + 4.(4k) + 5.(5k)

= 55k

= 55 × \(\frac{1}{15}\)

= \(\frac{11}{3}\)

variance (σ^{2}) = (1)^{2} . k + (2)^{2} . 2k + (3)^{2} . 3k + (4)^{2} . 4k + (5)^{2} (5k) – µ^{2}

= k + 8k + 27k + 64k + 125k – \(\left(\frac{11}{3}\right)^2\)

= 225k – \(\frac{121}{9}\)

= 225 × \(\frac{1}{15}\) – \(\frac{121}{9}\)

= \(\frac{135-121}{9}\)

= \(\frac{14}{9}\)