Thoroughly analyzing AP Inter 2nd Year Maths 2A Model Papers Set 2 helps students identify their strengths and weaknesses.

## AP Inter 2nd Year Maths 2A Model Paper Set 2 with Solutions

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 2 marks.

Question 1.

If √3 + i = r(cos θ + i sin θ), then find the value of θ in radian measure.

Solution:

Given that √3 + i = r (cos θ + i sin θ)

⇒ r cos θ = √3 , r sin θ = 1

⇒ r^{2} (cos^{2}θ + sin^{2}θ) = 3 + 1

⇒ r^{2} = 4

⇒ r = 2

∴ cos θ = \(\frac{\sqrt{3}}{2}\) and sin θ = \(\frac{1}{2}\)

Hence θ = \(\frac{\pi}{6}\)

Question 2.

Find the multiplicative inverse of √5 + 3i.

Solution:

The multiplicative inverse of a + ib is \(\frac{a-i b}{a^2+b^2}\)

∴ Multiplicative inverse of √5 + 3i = \(\frac{\sqrt{5}-3 i}{(\sqrt{5})^2+3^2}=\frac{\sqrt{5}-3 i}{5+9}=\frac{\sqrt{5}-3 i}{14}\)

Question 3.

If 1, ω, ω^{2} are the cube roots of unity, then find the value of (1 – ω) (1 – ω^{2}) (1 – ω^{4}) (1 – ω^{8}).

Solution:

Since 1, ω, ω^{2} are the cube roots of unity

⇒ 1 + ω + ω^{2} = 0 and ω^{3} = 1

Now 1 – ω^{4} = 1 – (ω^{3}) ω

= 1 – 1 (ω)

= 1 – ω

1 – ω^{8} = 1 – (ω^{3})^{2} ω^{2}

= 1 – (1) ω^{2}

= 1 – ω^{2}

∴ (1 – ω) (1 – ω^{2}) (1 – ω^{4}) (1 – ω^{8})

= (1 – ω) (1 – ω^{2})(1 – ω) (1 – ω^{2})

= [(1 – ω) (1 – ω^{2})]^{2}

= (1 – ω – ω^{2} + ω^{3})^{2}

= [1 – (ω + ω^{2}) + 1]^{2} (∵ 1 + ω + ω^{2} = 0)

= [1 – (-1) + 1]^{2}

= (3)^{2}

= 9

∴ (1 – ω) (1 – ω^{2}) (1 – ω^{4}) (1 – ω^{8}) = 9

Question 4.

For what values of x, the expression x^{2} – 5x + 14 is positive?

Solution:

Here a = 1, b = -5, c = 14

∆ = b^{2} – 4ac

= 25 – 56

= -31 < 0

∴ ∆ < 0 ∵ a = 1 > 0 and ∆ < 0

⇒ x^{2} – 5x + 14 is positive ∀ x ∈ R.

Question 5.

If α, β, γ are the roots of 4x^{3} – 6x^{2} + 7x + 3 = 0, then find the value of αβ + βγ + γα.

Solution:

α, β, γ are the roots of 4x^{3} – 6x^{2} + 7x + 3 = 0

α + β + γ = \(-\frac{a_1}{a_0}=\frac{6}{4}\)

αβ + βγ + γα = \(\frac{a_2}{a_0}=\frac{7}{4}\)

αβγ = \(-\frac{a_3}{a_0}=-\frac{3}{4}\)

∴ αβ + βγ + γα = \(\frac{7}{4}\)

Question 6.

Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 that are divisible by 2.

Solution:

Take 4 blank places. First, the unit place can be filled by an even digit in 3 ways (2 or 4, or 6).

The remaining three places can be filled with the 6 digits in 6 ways each.

Thus they can be filled in 6 × 6 × 6 = 6^{3} ways.

Therefore, the number of 4 digited numbers divisible by 2 is 3 × 6^{3} = 3 × 216 = 648

Question 7.

Find the number of ways of arranging 4 boys and 3 girls around a circle so that all the girls sit together.

Solution:

Treat all the 3 girls as one unit. Then We have 4 boys and 1 unit of girls. They can be arranged around a circle in 4! ways.

Now, the 3 girls can be arranged among themselves in 3! ways.

∴ The number of required arrangements = 4! × 3!

= 24 × 6

= 144

Question 8.

Find the term independent of x in the expansion of \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\).

Solution:

The general term in \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\) is

For term independent of x, put 42 – 5r = 0

⇒ r = \(\frac{42}{5}\) which is not an integer.

Hence term independent of x in the given expansion is zero.

Question 9.

Find the variance of the following data 5, 12, 3, 18, 6, 8, 2, 10.

Solution:

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 = \(\frac{1}{10}\) = 0.1 and q = 0.9

We have to find P(X = 1)

the probability that exactly one out of 10 is left-handed

P(X = 1) = ^{10}C_{1} p^{1} q^{10-1}

= 10 × 0.1 × (0.9)^{9}

= (0.9)^{9}

Section – B

(5 × 4 = 20 Marks)

**II. Short Answer Type Questions.**

- Answer any five Questions.
- Each Question carries 4 marks.

Question 11.

Find the real values of x and y if \(\frac{x-1}{3+i}-\frac{y-1}{3-i}=i\).

Solution:

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

⇒ \(\frac{(x-1)(3-i)+(y-1)(3+i)}{9-i^2}\) = i

⇒ 3x – xi – 3 + i + 3y – iy – 3 – i = 10i

⇒ (3x + 3y – 6) + i(-x + y) = 0 + 10i

Now equating real and imaginary parts.

3x + 3y – 6 = 0

⇒ x + y- 2 = 0 …….(1)

-x + y = 10

⇒ x – y + 10 = 0 ……..(2)

(1)+(2) ⇒ 2x + 8 = 0

⇒ x = -4

From (1), -4 + y – 2 = 0

⇒ y = 6

∴ x = -4, y = 6

Question 12.

Find the changes in the sign of 4x – 5x^{2} + 2 for x ∈ R and find the extreme value.

Solution:

The roots of 4x – 5x^{2} + 2 = 0

⇒ 5x^{2} – 4x – 2 = 0

Roots are \(\frac{2 \pm \sqrt{14}}{5}\)

∴ \(\frac{2-\sqrt{14}}{5}\) < x < \(\frac{2+\sqrt{14}}{5}\) sign of 4x – 5x^{2} + 2 is positive.

x < \(\frac{2-\sqrt{14}}{5}\) or x > \(\frac{2+\sqrt{14}}{5}\), the sign of 4x – 5x^{2} + 2 is negative.

Since a = -5 < 0, then

Maximum value = \(\frac{4 a c-b^2}{4 a}=\frac{4(-5)(2)-(4)^2}{4(-5)}=\frac{-56}{-20}=\frac{14}{5}\)

Hence extreme value = \(\frac{14}{5}\)

Question 13.

Find the number of 4-digit numbers that can be formed using the digits 2, 3, 5, 6, 8 (without repetition). How many of them are divisible by (i) 2 (ii) 3.

Solution:

The number of 4-digit numbers that can be formed using the 5-digits 2, 3, 5, 6, 8 is ^{5}P_{4} = 120

(i) Divisible by 2: For a number to be divisible by 2, the unit’s place should be filled with an even digit. This can be done in 3 ways (2 or 6 or 8).

Now, the remaining 3 places can be filled with the remaining 4 digits in ^{4}P_{3} = 24 ways.

Hence the number of 4-digit numbers divisible by 2 is 3 × 24 = 72

(ii) Divisible by 3: A number is divisible by 3 if the sum of the digits in it is a multiple of 3.

Since the sum of the given 5 digits is 24, we have to leave either 3 or 6 and use the digits 2, 5, 6, 8 or 2, 3, 5, 8.

In each case, we can permute them in 4! ways. Thus the number of 4-digit numbers divisible by 3 is 2 × 4! = 48

Question 14.

Find the number of 5-letter words that can be formed using the letters of the word NATURE that begin with ‘N’ when repetition is allowed.

Solution:

First, we can fill up the first place with N in one way.

The remaining 4 places can be filled with any one of the 6 letters in 6 × 6 × 6 × 6 = 6^{4} ways.

∴ The number of 5-letter words that can be formed using the letters of the word NATURE that begin with N when repetition is allowed = 1 × 6^{4} = 1296

Question 15.

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

Solution:

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

∴ x^{3} + x^{2} + 1 = (Ax + B) (x^{2} + 3) + (Cx + D) (x^{2} + 2) ………(1)

⇒ x^{3} + x^{2} + 1 = (A + C)x^{3} + (B + D)x^{2} + (3A + 2C)x + (3B + 2D)

Comparing the coefficients of x^{3}, x^{2}, x and constant terms

A + C = 1, B + D = 1, 3A + 2C = 0, 3B + 2D = 1

Solve 3A + 2C = 0 & 2A + 2C = 2

∴ A = -2, C = 3

Solve 3B + 2D = 1 & 2B + 2D = 2

∴ B = – 1, D = 2

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

Question 16.

Three screws are drawn at random from a slot of 50 screws, 5 of which are defective. Find the probability of the event that all 3 screws are non-defective, assuming that the drawing is (a) with replacement and (b) without replacement.

Solution:

Let S be the sample space

∴ The total number of screws = 50

The number of defective screws is 5 and the remaining 45 screws are non-defective.

Let A be the event of getting a drawing of the 3 screws that are non-defective.

(a) With Replacement

(b) Without Replacement

Question 17.

A Bag contains 4 red, 5 black, and 6 blue balls. Find the probability that two balls drawn at random simultaneously from the bag are a red and a black ball.

Solution:

Total number of balls in the bag = 4 + 5 + 6 = 15

From out of these balls, two balls can be drawn in ^{15}C_{2} = \(\frac{15 \times 14}{2}\) = 105 ways.

Out of 4 Red balls, one ball can be drawn in ^{4}C_{1} = 4 ways,

and out of 5 black balls, one ball can be drawn in ^{5}C_{1} = 5 ways.

If E is the event of getting a red and a black ball in a draw,

the number of cases favourable to E = 4 × 5 = 20

∴ The required probability is \(\frac{20}{105}=\frac{4}{21}\)

Section – C

(5 × 7 = 35 Marks)

**III. Long Answer Type Questions.**

- Answer any five Questions.
- Each Question carries 7 marks.

Question 18.

If 1, ω, ω^{2} are the cube roots of unity, then find the value of (a + b)^{3} + (aω + bω^{2})^{3} + (aω^{2} + bω)^{3}

Solution:

Since 1, ω, ω^{2} are the cube root of unity.

⇒ 1 + ω + ω^{2} = 0 and ω^{3} = 1

Now (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} ………(1)

(aω + bω^{2})^{3} = [ω(a + bω)]^{3}

= ω^{3} (a + bω)^{3}

= (1) (a + bω)^{3}

= a^{3} + 3a^{2}bω + 3ab^{2}ω^{2} + b^{3}ω^{3}

= a^{3} + 3a^{2}bω + 3ab^{2}ω^{2} + b^{3} ……..(2)

and (aω^{2} + bω)^{3} = [ω(aω + b)]^{3}

= ω^{3} (aω + b)^{3}

= (1) (aω + b)^{3}

= a^{3}ω^{3} + 3a^{2}bω^{3} + 3ab^{2}ω + b^{3}

= a^{3} (1) + 3a^{2}bω^{2} + 3ab^{2}ω + b^{3}

∴ (aω^{2} + bω)^{3} = a^{3} + 3a^{2}bω^{2} + 3ab^{2}ω + b^{3} ………(3)

adding (1), (2) & (3)

(a + b)^{3} + (aω + bω^{2})^{3} + (aω^{2} + bω)^{3} = 3a^{3} + 3a^{2}b(1 + ω + ω^{2}) + 3ab^{2} (1 + ω + ω^{2}) + 3b^{3}

= 3(a^{3} + b^{3}) + 3a^{2}b (0) + 3ab^{2} (0)

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

∴ (a + b)^{3} + (aω + bω^{2})^{3} + (aω^{2} + bω)^{3} = 3(a^{3} + b^{3})

Question 19.

Solve 3x^{4} + 16x^{3} + 24x^{2} – 16 = 0, given that it has multiple roots.

Solution:

Let f(x) = 3x^{4} + 16x^{3} + 24x^{2} – 16

f'(x) = 12x^{3} + 48x^{2} + 48x

= 12x(x^{2} + 4x + 4)

= 12x(x + 2)^{2}

f”(-2) = 0

f(-2) = 3(16) + 16(-8) + 24(4) – 16 = 0

∴ x + 2 is a factor of f'(x) and f(x)

∴ -2 is multiple root of f(x) = 0

3x^{2} + 4x – 4 = 0

⇒ 3x^{2} + 6x – 2x – 4 = 0

⇒ 3x(x + 2) – 2(x + 2) = 0

⇒ (x + 2) (3x – 2) = 0

⇒ x = -2, \(\frac{2}{3}\)

∴ The roots of the given equation are -2, -2, -2, \(\frac{2}{3}\)

Question 20.

Find the co-efficient of x^{8} in \(\frac{(1-x)^2}{\left(1-\frac{2}{3} x\right)^3}\).

Solution:

Question 21.

Prove that \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\ldots \ldots=\frac{2^n-1}{n+1}\).

Solution:

Question 22.

The following table gives the daily wages of workers. Compute the standard deviation and the coefficient of variance of the wages of the workers.

Wages (Rs.) |
125-175 | 175-225 | 225-275 | 275-325 | 325-375 | 375-425 | 425-475 | 475-525 | 525-575 |

Number of Workers |
2 | 22 | 19 | 14 | 3 | 4 | 6 | 1 | 1 |

Solution:

To solve this problem using the step deviation method.

Here h = 50

Let Assumed mean A = 300 Then y_{i} = \(\frac{x_i-300}{50}\)

Construct Table:

Question 23.

In a certain college, 25% of the boys and 10% of the girls are studying mathematics. The girls constitute 60% of the student’s strength. If a student selected at random is found studying mathematics, find the probability that the student is a girl.

Solution:

The probability that a student selects to be a girl

P(G) = \(\frac{60}{100}=\frac{6}{10}\)

The probability that a student selects to be a boy

P(B) = \(\frac{40}{100}=\frac{4}{10}\)

The probability that a boy studying mathematics

P(M/B) = \(\frac{25}{100}=\frac{1}{4}\)

Similarly, the probability that a girl studying mathematics

P(M/G) = \(\frac{10}{100}=\frac{1}{10}\)

We have to find P(G/M)

By Baye’s theorem

Question 24.

X = x |
-2 | -1 | 0 | 1 | 2 | 3 |

P(X = x) |
0.1 | K | 0.2 | 2K | 0.3 | K |

is the probability distribution of a random variable X. Find the value of K and the variance of X.

Solution:

The Sum of the probabilities = 1

⇒ 0.1 + K + 0.2 + 2K + 0.3 + K = 1

⇒ 4K + 0.6 = 1

⇒ 4K = 1 – 0.6 = 0.4

⇒ K = \(\frac{0.4}{4}\) = 0.1

Mean = (-2) (0.1) + (-1) K + 0 (0.2) + 1 (2K) + 2 (0.3) + 3K

= -0.2 – K + 0 + 2K + 0.6 + 3K

= 4K + 0.4

= 4(0.1) + 0.4

= 0.4 + 0.4

= 0.8

∴ µ = 0.8

Variance (σ^{2}) = \(\sum_{i=1}^n x_i^2 P\left(x=x_i\right)-\mu^2\)

Variance = 4 (0.1) + 1 (K) + 0 (0.2) + 1 (2K) + 4 (0.3) + 9K – µ^{2}

= 0.4 + K + 0 + 2K + 4(0.3) + 9K – (0.8)^{2}

= 12K + 0.4 + 1.2 – (0.8)^{2}

= 12(0.1) + 1.6 – 0.64

= 1.2 + 1.6 – 0.64

= 2.8 – 0.64

= 2.16

∴ σ^{2} = 2.16