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.

• Answer all 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 ¹³C_n.
Solution:
Given ⁿC₅ = ⁿC₆
∴ n = 5 + 6 = 11
¹³C_n = ¹³C₁₁
= ¹³C₂
= \(\frac{13.12}{2}\)
= 78
∴ ¹³C_n = 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)

II. Short Answer Type Questions.

• 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) [^5-1P_4-1 (1111) – ^5-2P_4-2 (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)

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 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 d_i = \(\frac{x_i-35}{10}\)

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