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

## TS 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 the questions.
- Each question carries two marks.

Question 1.

If z = 2 – 3i, then show that, z^{2} – 4z + 13 = 0.

Solution:

Given z = 2 – 3i

⇒ (z – 2) = -3i

⇒ (z – 2)^{2} = 9i^{2}

⇒ z^{2} – 4z + 4 = -9

⇒ z^{2} – 4z + 13 = 0

Question 2.

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

Solution:

Given z_{1} = -1

= -1 + i(0)

= cos π + i sin π

∴ Arg z_{1} = π

Given z_{2} = i

= 0 + i(1)

= 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 – \(\frac{\pi}{2}\)

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

Question 3.

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

Solution:

Given 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

∴ \(\left(x^6+\frac{1}{x^6}\right)\) = cos 6θ + i sin 6θ + cos 6θ – i sin 6θ = 2 cos 6θ

Question 4.

Form a quadratic equation whose roots are 7 ± 2√5.

Solution:

Let α = 7 + 2√5 and β = 7 – 2√5

α + β = 7 + 2√5 + 7 – 2√5 = 14

αβ = (7 + 2√5) (7 – 2√5)

= 49 – 20

= 29

Required quadratic equation is x^{2} – (α + β)x + αβ = 0

∴ x^{2} – 14x + 29 = 0

Question 5.

If -1, 2, and α are the roots of 2x^{3} + x^{2} – 7x – 6 = 0, find α.

Solution:

Since -1, 2, α are the roots of the equation 2x^{3} + x^{2} – 7x – 6 = 0

∴ s_{1} = \(\frac{-1}{2}\)

⇒ -1 + 2 + α = \(\frac{-1}{2}\)

⇒ α + 1 = \(\frac{-1}{2}\)

⇒ α = \(\frac{-1}{2}\) – 1

⇒ α = \(\frac{-3}{2}\)

Question 6.

Find the number of ways of arranging the letters of the word MATHEMATICS.

Solution:

The word MATHEMATICS contains 11 letters in which 2M’s, 2A’s, and 2T’s rest are different.

∴ The number of ways of arranging the letters of the MATHEMATICS is \(\frac{11 !}{2 ! 2 ! 2 !}=\frac{11 !}{(2 !)^3}\)

Question 7.

If ^{n}C_{5} = ^{n}C_{6}, then find ^{13}C_{n}.

Solution:

Given ^{n}C_{5} = ^{n}C_{6}

⇒ n = 5 + 6 = 11

∴ ^{13}C_{n} = ^{13}C_{11}

= ^{13}C_{2} [∵ ^{n}C_{r} = ^{n}C_{n-r}]

= \(\frac{13.12}{2}\)

= 78

Question 8.

Prove that C_{0} + 2.C_{1} + 4.C_{2} + 8.C_{3} + ………. + 2^{n} . C_{n} = 3^{n}.

Solution:

Question 9.

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

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

Solution:

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

Now arranging data points in ascending order

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

∴ Median = 6

Absolute values |6 – 2|, |6 – 3|, |6 – 4|, |6 – 6|, |6 – 9|, |6 – 10|, |6 – 13| = 4, 3, 2, 0, 3, 4, 7

∴ Mean deviation about the median = \(\frac{4+3+2+0+3+4+7}{7}\)

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

= 3.28

Question 10.

A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5).

Solution:

X follows a Poisson distribution with parameter λ.

Section – B

(5 × 4 = 20 Marks)

**II. Short Answer Type Questions.**

- Attempt any five questions.
- Each question carries four marks.

Question 11.

Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, -2√3 + 2√3i are the vertices of an equilateral triangle.

Solution:

Let A, B, and C be the points in the Argand plane.

∴ A = (2, 2), B = (-2, -2), C = (-2√3, 2√3)

∴ AB = BC = CA

∴ A, B, and C are the vertices of an equilateral triangle.

Question 12.

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

Solution:

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

⇒ 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

⇒ y^{2} – 5y – 4 ≥ 0

⇒ (y – 1) (y – 4) ≥ 0

⇒ y ≥ 0 (or) y ≥ 4

∴ \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4.

Question 13.

Find the 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, 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 digit 3, 5, 7, 9 appears 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}P3 × 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.

Find the number of ways of selecting a cricket team of 11 players from 7 batsmen and 6 bowlers, such that there will be atleast 5 bowlers in the team.

Solution:

Since the team consists of atleast 5 bowlers, the selection may be of the following types.

Batsmen (7) | Bowlers (6) | |

Type 1 | 6 | 5 |

Type 2 | 5 | 6 |

The no. of selections is first type = ^{7}C_{6} × ^{6}C_{5} = 7 × 6 = 42

The no. of selections in second type = ^{7}C_{5} × ^{6}C_{6} = 21 × 1 = 21

∴ The required number of ways of selecting the cricket team = 42 + 21 = 63

Question 15.

Resolve the fraction \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\) into partial fraction.

Solution:

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

⇒ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A\left(x^2+2\right)+(B x+C)(x-1)}{(x-1)\left(x^2+2\right)}\)

⇒ 2x^{2} + 3x + 4 = A(x^{2} + 2) + (Bx + C) (x – 1) ………(1)

Put x = 1

2 + 3 + 4 = A(1 + 2)

⇒ 9 = 3A

⇒ A = 3

Comparing the co-efficient of x^{2} and constant terms in (1)

2 = A + B

⇒ 2 = 3 + B

⇒ B = -1

4 = 2A – C

⇒ 4 = 2(3) – C

⇒ C = 6 – 4

⇒ C = 2

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

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

Question 16.

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(B/A)

(iv) P(A^{C} ∩ B^{C})

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

= 0.88

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

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

Question 17.

A, B, C are three horses in a race. The probability of A winning the race is twice that of B and the probability of B is twice that of C. What are the probabilities of A, B, and C to win the race?

Solution:

If A, B, C are three independent events of an experiment such that P(A ∩ B^{C} ∩ C^{C}) = \(\frac{1}{4}\), P(A^{C} ∩ B ∩ C^{C}) = \(\frac{1}{8}\), P(A^{C} ∩ B^{C} ∩ C^{C}) = \(\frac{1}{4}\) then find P(A), P(B) and P(C).

Since A, B, C are independent events.

Section – C

(5 × 7 = 35 Marks)

**III. Long Answer Type Questions.**

- Attempt any five questions.
- Each question carries seven marks.

Question 18.

If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ, then prove that cos^{2}α + cos^{2}β + cos^{2}γ = \(\frac{3}{2}\) = sin^{2}α + sin^{2}β + sin^{2}γ.

Solution:

Let a = cos α + i sin α

b = cos β + i sin β

c = cos γ + i sin γ

a + b + c = (cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ)

= (cos α + cos β + cos γ) + i (cos α + cos β + cos γ)

= 0 + i(0)

= 0

∴ a + b + c = 0

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{\cos \alpha+i \sin \alpha}+\frac{1}{\cos \beta+i \sin \beta}+\frac{1}{\cos \gamma+i \sin \gamma}\)

\(\frac{b c+c a+a b}{a b c}\) = cos α – i sin α + cos β – i sin β + cos γ – i sin γ

= (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)

= 0 – i(0)

= 0

∴ ab + bc + ca = 0

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

⇒ 0 = a^{2} + b^{2} + c^{2} + 2(0)

⇒ a^{2} + b^{2} + c^{2} = 0

⇒ (cos α + i sin α)^{2} + (cos β + i sin β)^{2} + (cos γ + i sin γ)^{2} = 0

⇒ cos 2α + i sin 2α + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0

⇒ (cos 2α + cos 2β + cos 2γ) + i (sin 2α + cos 2β + sin 2γ) = 0

Equation real parts on both sides, we have

cos 2α + cos 2β + cos 2γ = 0

⇒ 1 – 2 sin^{2}α + 1 – 2 sin^{2}β + 1 – 2 sin^{2}γ = 0

⇒ 3 = 2(sin^{2}α + sin^{2}β + sin^{2}γ)

⇒ sin^{2}α + sin^{2}β + sin^{2}γ = \(\frac{3}{2}\)

Again cos 2α + cos 2β + cos 2γ = 0

⇒ 2 cos^{2}α – 1 + 2 cos^{2}β – 1 + 2 cos^{2}γ – 1 = 0

⇒ 2(cos^{2}α + cos^{2}β + 2 cos^{2}γ) = 3

⇒ cos^{2}α + cos^{2}β + cos^{2}γ = \(\frac{3}{2}\)

∴ cos^{2}α + cos^{2}β + cos^{2}γ = \(\frac{3}{2}\) = sin^{2}α + sin^{2}β + sin^{2}γ

Question 19.

Solve the equation x^{4} – 10x^{3} + 26x^{2} – 10x + 1 = 0.

Solution:

Given x^{4} – 10x^{3} + 26x^{2} – 10x + 1 = 0

⇒ x^{2} – 10x + 26 – \(\text { 10. } \frac{1}{x}+\frac{1}{x^2}\) = 0

⇒ \(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)\) + 26 = 0

Let x + \(\frac{1}{x}\) = t

Then x^{2} + \(\frac{1}{x^2}\) = t^{2} – 2

∴ (t^{2} – 2) – 10t + 26 = 0

⇒ t^{2} – 2 – 10t + 26 = 0

⇒ t^{2} – 10t + 24 = 0

⇒ t^{2} – 4t – 6t + 24 = 0

⇒ t(t – 4) – 6(t – 4) = 0

⇒ (t – 4) (t – 6) = 0

⇒ t = 4 (or) t = 6

∴ x + \(\frac{1}{x}\) = 4

⇒ x^{2} + 1 = 4x

⇒ x^{2} – 4x + 1 = 0

⇒ x = \(\frac{-(-4) \pm \sqrt{16-4.1 .1}}{2.1}\)

⇒ x = \(\frac{4 \pm \sqrt{12}}{2}\)

⇒ x = \(\frac{4 \pm 2 \sqrt{3}}{2}\)

⇒ x = 2 ± √3

∴ x + \(\frac{1}{x}\) = 6

⇒ x^{2} + 1 = 6x

⇒ x^{2} – 6x + 1 = 0

⇒ x = \(\frac{-(-6) \pm \sqrt{36-4.1 .1}}{2.1}\)

⇒ x = \(\frac{6 \pm \sqrt{32}}{2}\)

⇒ x = \(\frac{6 \pm 4 \sqrt{2}}{2}\)

⇒ x = 3 ± 2√2

∴ The roots are 2 ± √3, 3 ± 2√2

Question 20.

If the coefficients of r^{th}, (r + 1)^{th} and (r + 2)^{nd} terms in the expansion of (1 + x)^{n} are in A.P. Then show that n^{2} – (4r + 1)n + 4r^{2} – 2 = 0.

Solution:

The coefficients of r^{th}, (r + 1)^{th,} and (r + 2)^{th} terms in the expansion of (1 + x)^{3} are ^{n}C_{r-1}, ^{n}C_{r}, and ^{n}C_{r+1} respectively.

Given ^{n}C_{r-1} + ^{n}C_{r+1} = 2 ^{n}C_{r} are in A.P.

∴ ^{n}C_{r-1} + ^{n}C_{r+1} = 2 ^{n}C_{r}

⇒ (n – r) (n – r + 1) = (r + 1) (2n – 3r + 2)

⇒ n^{2} – 2nr + r^{2} + n – r = 2nr – 3r^{2} + 2r + 2n – 3r + 2

⇒ n^{2} – 4nr + 4r^{2} – n – 2 = 0

⇒ n^{2} – (4r + 1)n + 4r^{2} – 2 = 0

Question 21.

If x = \(\frac{1.3}{3.6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6.9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots\), 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 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

\(\bar{x}=\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\mathrm{N}}\right) \mathrm{h}\)

= 25 + (\(\frac{10}{50}\))10

= 27

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

= \(\frac{10}{50}\)(472)

= 9.44

Question 23.

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 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

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_{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})

Question 24.

A random variable ‘X’ has the following probability distribution.

X = x |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

P(X = x) |
0 | K | 2K | 2K | 3K | K^{2} |
2K^{2} |
7K^{2} + K |

Find:

(i) K

(ii) Mean

(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) \(\frac{1}{10}\)

(i) k = \(\frac{1}{10}\) Since k > 0

(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