# TS Inter 2nd Year Maths 2A Question Paper March 2018

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## 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.

• Each question carries two marks.

Question 1.
If z = 2 – 3i, then show that, z2 – 4z + 13 = 0.
Solution:
Given z = 2 – 3i
⇒ (z – 2) = -3i
⇒ (z – 2)2 = 9i2
⇒ z2 – 4z + 4 = -9
⇒ z2 – 4z + 13 = 0

Question 2.
If z1 = -1 and z2 = i, then find Arg($$\frac{z_1}{z_2}$$).
Solution:
Given z1 = -1
= -1 + i(0)
= cos π + i sin π
∴ Arg z1 = π
Given z2 = i
= 0 + i(1)
= cos $$\frac{\pi}{2}$$ + i sin $$\frac{\pi}{2}$$
∴ Arg z2 = $$\frac{\pi}{2}$$
Arg($$\frac{z_1}{z_2}$$) = Arg z1 – Arg z2
= 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
⇒ x6 = (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 x2 – (α + β)x + αβ = 0
∴ x2 – 14x + 29 = 0

Question 5.
If -1, 2, and α are the roots of 2x3 + x2 – 7x – 6 = 0, find α.
Solution:
Since -1, 2, α are the roots of the equation 2x3 + x2 – 7x – 6 = 0
∴ s1 = $$\frac{-1}{2}$$
⇒ -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 nC5 = nC6, then find 13Cn.
Solution:
Given nC5 = nC6
⇒ n = 5 + 6 = 11
13Cn = 13C11
= 13C2 [∵ nCr = nCn-r]
= $$\frac{13.12}{2}$$
= 78

Question 8.
Prove that C0 + 2.C1 + 4.C2 + 8.C3 + ………. + 2n . Cn = 3n.
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)

• 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:

= $$\frac{4 x+1}{3 x^2+4 x+1}$$
⇒ 3yx2 + 4yx + 4 = 4x + 1
⇒ 3yx2 + (4y – 4)x + (y – 1) = 0
x ∈ R ⇒ (4y – 4)2 – 4(3y)(y – 1) ≥ 0
⇒ 16y2 – 32y + 16 – 12y2 + 12y ≥ 0
⇒ 4y2 – 20y + 16 ≥ 0
⇒ y2 – 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 5P4 = 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 4P3 ways.
Similarly, each digit 3, 5, 7, 9 appears 24 times in units place.
By adding all these digits, we get
4P3 × 1 + 4P3 × 3 + 4P3 × 5 + 4P3 × 7 + 4P3 × 9
= 4P3 × (1 + 3 + 5 + 7 + 9)
= 4P3 × 25
Similarly, the sum of the digits in ten’s place value as 4P3 × 25 × 10
Similarly, the values of the sum of the digits 100’s place and 1000’s places are 4P3 × 25 × 100 and 4P3 × 25 × 1000
Hence the sum of all the 4-digit numbers formed by using the digits 1, 3, 5, 7, 9 is 4P3 × 25 × 1 + 4P3 × 25 × 10 + 4P3 × 25 × 100 + 4P3 × 25 × 1000
= 4P3 × 25 × (1 + 10 + 100 + 1000)
= 4P3 × 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 = 7C6 × 6C5 = 7 × 6 = 42
The no. of selections in second type = 7C5 × 6C6 = 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 $$\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+2}$$
⇒ $$\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A\left(x^2+2\right)+(B x+C)(x-1)}{(x-1)\left(x^2+2\right)}$$
⇒ 2x2 + 3x + 4 = A(x2 + 2) + (Bx + C) (x – 1) ………(1)
Put x = 1
2 + 3 + 4 = A(1 + 2)
⇒ 9 = 3A
⇒ A = 3
Comparing the co-efficient of x2 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(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
= 0.88

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

(iv) P(AC ∩ BC) = P(AC) . P(BC) (∵ Since AC, BC 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 ∩ BC ∩ CC) = $$\frac{1}{4}$$, P(AC ∩ B ∩ CC) = $$\frac{1}{8}$$, P(AC ∩ BC ∩ CC) = $$\frac{1}{4}$$ then find P(A), P(B) and P(C).
Since A, B, C are independent events.

Section – C
(5 × 7 = 35 Marks)

• 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 cos2α + cos2β + cos2γ = $$\frac{3}{2}$$ = sin2α + sin2β + sin2γ.
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 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ 0 = a2 + b2 + c2 + 2(0)
⇒ a2 + b2 + c2 = 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 sin2α + 1 – 2 sin2β + 1 – 2 sin2γ = 0
⇒ 3 = 2(sin2α + sin2β + sin2γ)
⇒ sin2α + sin2β + sin2γ = $$\frac{3}{2}$$
Again cos 2α + cos 2β + cos 2γ = 0
⇒ 2 cos2α – 1 + 2 cos2β – 1 + 2 cos2γ – 1 = 0
⇒ 2(cos2α + cos2β + 2 cos2γ) = 3
⇒ cos2α + cos2β + cos2γ = $$\frac{3}{2}$$
∴ cos2α + cos2β + cos2γ = $$\frac{3}{2}$$ = sin2α + sin2β + sin2γ

Question 19.
Solve the equation x4 – 10x3 + 26x2 – 10x + 1 = 0.
Solution:
Given x4 – 10x3 + 26x2 – 10x + 1 = 0
⇒ x2 – 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 x2 + $$\frac{1}{x^2}$$ = t2 – 2
∴ (t2 – 2) – 10t + 26 = 0
⇒ t2 – 2 – 10t + 26 = 0
⇒ t2 – 10t + 24 = 0
⇒ t2 – 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
⇒ x2 + 1 = 4x
⇒ x2 – 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
⇒ x2 + 1 = 6x
⇒ x2 – 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 rth, (r + 1)th and (r + 2)nd terms in the expansion of (1 + x)n are in A.P. Then show that n2 – (4r + 1)n + 4r2 – 2 = 0.
Solution:
The coefficients of rth, (r + 1)th, and (r + 2)th terms in the expansion of (1 + x)3 are nCr-1, nCr, and nCr+1 respectively.
Given nCr-1 + nCr+1 = 2 nCr are in A.P.
nCr-1 + nCr+1 = 2 nCr

⇒ (n – r) (n – r + 1) = (r + 1) (2n – 3r + 2)
⇒ n2 – 2nr + r2 + n – r = 2nr – 3r2 + 2r + 2n – 3r + 2
⇒ n2 – 4nr + 4r2 – n – 2 = 0
⇒ n2 – (4r + 1)n + 4r2 – 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 9x2 + 24x = 11.
Solution:

⇒ 4 + 3x = 3√3
⇒ (4 + 3x)2 = 27
⇒ 16 + 9x2 + 24x = 27
∴ 9x2 + 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 E1 and E2 are any two events of a random experiment and P is the probability function.
Then P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2)
Case (i): When E1 ∩ E2 = φ
E1 ∩ E2 = φ
⇒ P (E1 ∩ E2) = 0
∴ P(E1 ∪ E2) = P(E1) + P(E2) [∵ from the union axiom]
= P(E1) + P(E2) – 0
= P(E1) + P(E2) – P(E1 ∩ E2)
Case (ii): When E1 ∩ E2 = φ
∴ E1 ∪ E2 can be expressed as union of 2 mutually exclusive events E1 – E2, E2.
Hence, E1 ∪ E2 = (E1 – E2) ∪ E2
Also (E1 – E2) ∩ E2 = φ
∴ P(E1 ∪ E2) = P[(E1 – E2) ∪ E2]
P(E1 ∪ E2) = P(E1 – E2) + P(E2) ………(1) [∵ from the union axiom]
Also, E1 can be expressed as the union of 2 mutually exclusive events E1 – E2, E1 ∩ E2.
Hence, E1 = (E1 – E2) ∪ (E1 ∩ E2)
Also (E1 – E2) ∩ (E1 ∩ E2) = φ
∴ P(E1) = P(E1 – E2) ∪ (E1 ∩ E2)] = P(E1 – E2) + P(E1 ∩ E2)
⇒ P(E1 – E2) = P(E1) – P(E1 ∩ E2)
From (1 ),
P(E1 ∪ E2) = P(E1) – P(E1 ∩ E2) + P(E2) = P(E1) + P(E2) – P(E1 ∩ E2)

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 K2 2K2 7K2 + 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 + k2 + 2k2 + 7k2 + k = 1
⇒ 10k2 + 9k = 1
⇒ 10k2 + 9k – 1 = 0
⇒ 10k2 + 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(k2) + 6(2k2) + 7(7k2 + k)
= 0 + k + 4k + 6k + 12k + 5k2 + 12k2 + 49k2 + 7k
= 66k2 + 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