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

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## TS Inter 2nd Year Maths 2A Question Paper May 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.
Write the conjugate of the complex number (2 + 5i)(-4 + 6i).
Solution:
Let Z = (2 + 5i)(-4 + 6i)
= -8 + 12i – 20i + 30i2
= -8 – 8i – 30
= -38 – 8i
∴ Z = -38 – 8i
∴ The conjugate of (2 + 5i)(-4 + 6i) is -38 – 8i.

Question 2.
Express the complex number -√3 + i in modulus-amplitude form.
Solution:
Let Z = -√3 + i
= $$-2\left[\frac{-\sqrt{3}}{2}+i \frac{1}{2}\right]$$
= $$-2\left[\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right]$$
∴ -√3 + i modulus-amplitude form is $$2\left[\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right]$$.

Question 3.
If 1, w, w2 are the cube roots of unity, then prove that $$\frac{1}{2+w}+\frac{1}{1+2 w}=\frac{1}{1+w}$$.
Solution:
Since 1, w, w2 are the cube roots of units.
∴ 1 + w + w2 = 0 and w3 = 1

Question 4.
If x2 + bx + c = 0, x2 + cx + b = 0 (b ≠ c) have a common root, then show that b + c + 1 = 0.
Solution:
Let α be the common root of x2 + bx + c = 0 and x2 + cx + b = 0
∴ α2 + bα + c = 0
α2 + cα + b = 0
bα – cα + c – b = 0
⇒ (b – c)α = b – c
⇒ α = 1
∴ 12 + b(1) + c = 0
⇒ 1 + b + c = 0
⇒ b + c + 1 = 0

Question 5.
If 1, 1, α are the roots of x3 – 6x2 + 9x – 4 = 0, then find ‘α’.
Solution:
Given 1, 1, α are the roots of x3 – 6x2 + 9x – 4 = 0
∴ S1 = $$\frac{-(-6)}{1}$$
⇒ 1 + 1 + α = 6
⇒ 2 + α = 6
⇒ α = 6 – 2
⇒ α = 4

Question 6.
If nP3 = 1320, find n.
Solution:
Given nP3 = 1320
⇒ n(n – 1)(n – 2) = 10 × 132
⇒ n(n – 1)(n – 2) = 10 × 11 × 12
⇒ n(n – 1)(n – 2) = 12 × 11 × 10
⇒ n = 12

Question 7.
If nC5 = nC6, then find 13Cn.
Solution:
Given nC5 = nC6
∴ n = 5 + 6 = 11
13Cn = 13C11
= 13C2
= $$\frac{13.12}{2}$$
= 78
13Cn = 78

Question 8.
If A and B are coefficients of xn in the expansion of (1 + x)2n and (1 + x)2n-1 respectively, then find the value of $$\frac{A}{B}$$.
Solution:

Question 9.
Find the mean deviation about the mean of 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 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.
A Poisson Variable satisfies P(X = 1) = P(X = 2). Find P(X = 5).
Solution:
We know P(X = r) = $$\frac{\mathrm{e}^{-\lambda} \cdot \lambda^{\mathrm{r}}}{\mathrm{r} !}$$
Given P(X = 1) = P(X = 2)
∴ $$\frac{\mathrm{e}^{-\lambda} \cdot \lambda^1}{1 !}=\frac{\mathrm{e}^{-\lambda} \cdot \lambda^2}{2 !}$$
⇒ λ = $$\frac{\lambda^2}{2}$$
⇒ 2λ = λ2
⇒ λ2 – 2λ = 0
⇒ λ(λ – 2) = 6
Since λ > 0
∴ λ – 2 = 0
⇒ λ = 2

Section – B
(5 × 4 = 20 Marks)

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

Question 11.
Show that the four points in an Argand plane represented by the complex numbers -2 + 7i, $$\frac{-3}{2}+\frac{1}{2} i$$, 4 – 3i, $$\frac{7}{2}$$(1 + i) are the vertices of a rhombus.
Solution:
Let A, B, C, D be the points in the Argand plane.

Here AB = BC = CD = DA and AC ≠ BD
∴ ABCD forms a rhombus.

Question 12.
If x is real, prove that $$\frac{x}{x^2-5 x+9}$$ lies between $$-\frac{1}{11}$$ and 1.
Solution:
Let y = $$\frac{x}{x^2-5 x+9}$$
⇒ yx2 – 5xy + 9y = x
⇒ yx2 + (-5y – 1)x + 9y = 0
x ∈ R ⇒ (-5y – 1)2 – 4 . y . 9y ≥ 0
⇒ 25y2 + 10y + 1 – 36y2 ≥ 0
⇒ -11y2 + 10y + 1 ≥ 0
⇒ 11y2 – 10y – 1 ≤ 0
⇒ 11y2 – 11y + y – 1 ≤ 0
⇒ 11y(y – 1) + 1(y – 1) ≤ 0
⇒ (11y + 1) (y – 1) ≤ 0
⇒ $$-\frac{1}{11}$$ ≤ y ≤ 1
∴ $$\frac{x}{x^2-5 x+9}$$ lies between $$-\frac{1}{11}$$ and 1.

Question 13.
Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8, without repetition.
Solution:
Given digits are 0, 2, 4, 7, 8
Sum of all the 4 digit numbers by the digits 0, 2, 4, 7, 8
= (0 + 2 + 4 + 7 + 8) [5-1P4-1 (1111) – 5-2P4-2 (111)]
= 21[4P3 (1111) – 3P2 (111)]
= 21[24(1111) – 6(111)]
= 21[26664 – 666]
= 21[25998]
= 5,45,958

Question 14.
Prove that $${ }^{25} C_4+\sum_{r=0}^4{ }^{(29-r)} C_3={ }^{30} C_4$$.
Solution:
L.H.S = $${ }^{25} C_4+\sum_{r=0}^4(29-r) C_3$$

Question 15.
Resolve $$\frac{3 x-18}{x^3(x+3)}$$ into partial fractions.
Solution:
Let $$\frac{3 x-18}{x^3(x+3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+3}$$
⇒ $$\frac{3 x-18}{x^3(x+3)}=\frac{A x^2(x+3)+B x(x+3)+C(x+3)+D x^3}{x^3(x+3)}$$
⇒ 3x – 18 = Ax2 (x + 3) + Bx(x + 3) + C(x + 3) + Dx3 ……….(1)
Put x = 0
0 – 18 = 0 + 0 + C(0 + 3) + 0
⇒ -18 = 3C
⇒ C = -6
Put x = -3
3(-3) – 18 = 0 + 0 + 0 + D(-3)3
⇒ -9 – 18 = D(-27)
⇒ -27 = -27D
⇒ D = 1
Comparing the Co-efficient of x3 terms on both sides in (1)
0 = A + D
⇒ A = -D
⇒ A = -1
Comparing the Co-efficient of x2 terms on both sides in (1)
0 = 3A + B
⇒ B = -3A
⇒ B = -3(1)
⇒ B = -3
∴ A = -1, B = -3, C = -6, D = 1
∴ $$\frac{3 x-18}{x^3(x+3)}=\frac{-1}{x}+\frac{-3}{x^2}+\frac{-6}{x^3}+\frac{1}{x+3}$$ = $$\frac{1}{x+3}-\frac{1}{x}-\frac{3}{x^2}-\frac{6}{x^3}$$

Question 16.
If 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($$\overline{\mathrm{A}} \cap \overline{\mathrm{B}}$$)
= 1 – P($$\overline{A \cap B}$$)
= 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 and B are independent events with P(A) = 0.2 and P(B) = 0.5. Find
(i) P(A | B)
(ii) P(B | A)
(iii) P(A ∩ B)
(iv) P(A ∪ B)
Solution:
Given P(A) = 0.2, P(B) = 0.5
Since A and B an independent events
(i) P(A | B) = P(A) = 0.2
(ii) P(B | A) = P(B) = 0.5
(iii) P(A ∩ B) = P(A) . P(B)
= (0.2) (0.5)
= 0.1
(iv) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) . P(B)
= 0.2 + 0.5 – (0.2) (0.5)
= 0.2 + 0.5 – 0.1
= 0.6

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 show that
(i) cos 3α + cos 3β + cos 3γ = 3 cos(α + β + γ)
(ii) sin 3α + sin 3β + sin 3γ = 3 sin(α + β + γ)
Solution:
Given cos α + cos β + cos γ = 0 = sin α + sin β + sin γ
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(sin α + sin β + sin γ)
= 0 + i(0)
= 0
∴ a + b + c = 0
∴ a3 + b3 + c3 = 3abc
⇒ (cos α + i sin α)3 + (cos β + i sin β)3 + (cos γ + i sin γ)3 = 3(cos α + i sin α) (cos β + i sin β) (cos γ + i sin γ)
⇒ (cos 3α + i sin 3α) + (cos 3β + i sin 3β) + (cos 3γ + i sin 3γ) = 3(cis α) (cis β) (cis γ)
⇒ (cos 3α + cos 3β + cos 3γ) + i(sin 3α + sin 3β + sin 3γ) = 3 cis(α + β + γ)
⇒ (cos 3α + cos 3β + cos 3γ) + i(sin 3α + sin 3β + sin 3γ) = 3[cos(α + β + γ) + i sin(α + β + γ)]
Now, equating real and imaginary parts on both sides, we get
∴ cos 3α + cos 3β + cos 3γ = 3 cos(α + β + γ)
∴ sin 3α + sin 3β + sin 3γ = 3 sin(α + β + γ)

Question 19.
Solve 18x3 + 81x2 + 121x + 60 = 0 given that one root is equal to half the sum of the remaining roots.
Solution:
Let α, β, γ be the roots of 18x3 + 81x2 + 121x + 60 = 0.

Question 20.
Prove that $$C_0+C_1 \cdot \frac{x}{2}+C_2 \cdot \frac{x^2}{3}+C_3 \cdot \frac{x^3}{4}+\ldots .+C_n \cdot \frac{x^n}{n+1}=\frac{(1+x)^{n+1}-1}{(n+1) x}$$
Solution:

Question 21.
If x = $$\frac{1.3}{3.6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}$$ + ………, 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 from the mean of the following data, using the step deviation method:

 Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of Students 6 5 8 15 7 6 3

Solution:
Let assumed mean A = 35 Then di = $$\frac{x_i-35}{10}$$

Question 23.
A, B, C are 3 newspapers from a city. 20% of the population read A, 16% read B, 14% read C, 8% both A and B, 5% both A and C, 4% both B and C and 2% all the three. Find the percentage of the population who read atleast one newspaper.
Solution:
Given P(A) = $$\frac{20}{100}$$
P(B) = $$\frac{16}{100}$$
P(C) = $$\frac{14}{100}$$
P(A ∩ B) = $$\frac{8}{100}$$
P(B ∩ C) = $$\frac{4}{100}$$
P(C ∩ A) = $$\frac{5}{100}$$
P(A ∩ B ∩ C) = $$\frac{2}{100}$$
We know P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
= $$\frac{20}{100}+\frac{16}{100}+\frac{14}{100}-\frac{8}{100}-\frac{4}{100}-\frac{5}{100}+\frac{2}{100}$$
= $$\frac{20+16+14-8-4-5+2}{100}$$
= $$\frac{35}{100}$$
∴ The percentage of the population who read atleast one newspaper = $$\frac{35}{100}$$ × 100 = 35%

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) the mean and (iii) P(0 < X < 5).
Solution:
We know ΣP(X = x) = 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
⇒ (10k – 1) (k + 1) = 0
⇒ k = $$\frac{1}{10}$$, -1
(i) k = $$\frac{1}{10}$$ Since k > 0

(ii) Mean = 0 P(X = 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($$\frac{1}{100}$$) + 30($$\frac{1}{10}$$)
= 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