AP Inter 2nd Year Maths 2A Question Paper May 2017

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AP Inter 2nd Year Maths 2A Question Paper May 2017

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.

  • Attempt ALL the questions.
  • Each question carries two marks.

Question 1.
Find the square roots of (-5 + 12i).
Solution:
AP Inter 2nd Year Maths 2A Question Paper May 2017 Q1

Question 2.
If Z1 = -1 and Z2 = -i, then find Arg (Z1Z2).
Solution:
Z1 = -1
= 0 + i(-1)
= cos π + i sin π
∴ Arg Z1 = π
Z2 = -i
= 0 + i(-1)
= \(\cos \left(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)\)
∴ Arg Z2 = \(\frac{-\pi}{2}\)
Arg (Z1Z2) = Arg Z1 + Arg Z2
= π + (\(\frac{-\pi}{2}\))
= π – \(\frac{\pi}{2}\)
= \(\frac{\pi}{2}\)

AP Inter 2nd Year Maths 2A Question Paper May 2017

Question 3.
If x = cos θ, then find the value of \(\left(x^6+\frac{1}{x^6}\right)\).
Solution:
Given x = cis θ
⇒ x = cos θ + i sin θ
⇒ x6 = (cos θ + i sin θ)6
⇒ x6 = cos 6θ + i sin 6θ)
⇒ \(\frac{1}{x^6}=\frac{1}{\cos 6 \theta+i \sin 6 \theta}\)
⇒ \(\frac{1}{x^6}\) = cos 6θ – i sin 6θ
∴ x6 + \(\frac{1}{x^6}\) = cos 6θ + i sin 6θ + cos 6θ – i sin 6θ
∴ x6 + \(\frac{1}{x^6}\) = 2 cos 6θ

Question 4.
Find the values of m for which the equation x2 – 15 – m(2x – 8) = 0 have equal roots.
Solution:
Given quadratic equation is x2 – 15 – m(2x – 8) = 0
⇒ x2 – 15 – 2mx + 8m = 0
⇒ x2 – 2mx + (8m – 15) = 0
Since it has equal roots
∴ (-2m)2 – 4 . 1 (8m – 15) = 0
⇒ 4m2 – 32m + 60 = 0
⇒ m2 – 8m + 15 = 0
⇒ m2 – 3m – 5m + 15 = 0
⇒ m(m – 3) – 5(m – 3) = 0
⇒ (m – 3) (m – 5) = 0
⇒ m = 3, 5

Question 5.
If -1, 2, and α are the roots of 2x3 + x2 – 7x – 6 = 0, then find α.
Solution:
Since -1, 2, α are the roots of 2x3 + x2 – 7x – 6 = 0.
S1 = \(\frac{-1}{2}\)
⇒ -1 + 2 + α = \(\frac{-1}{2}\)
⇒ 1 + α = \(\frac{-1}{2}\)
⇒ α = \(\frac{-1}{2}\) – 1
⇒ α = \(\frac{-3}{2}\)

Question 6.
If nP7 = 42 . nP5, find n.
Solution:
Given nP7 = 42 . nP5
⇒ n(n – 1) (n – 2) (n – 3) (n – 4) (n – 5) (n – 6) = 42 . n(n – 1) (n – 2) (n – 3) (n – 4)
⇒ (n – 5) (n – 6) = 42
⇒ (n – 5) (n – 6) = 7 × 6
⇒ n – 5 = 7
⇒ n = 12

Question 7.
If 17C2t+1 = 17C3t-5, find t.
Solution:
Given 17C2t+1 = 17C3t-5
∴ 2t + 1 = 3t – 5 (or) 17 = 2t + 1 + 3t- 5
⇒ 1 + 5 = 3t – 2t (or) 17 = 5t – 4
⇒ t = 6 (or) 17 + 4 = 5t
⇒ t = 6 (or) t = \(\frac{21}{5}\)
since t is an integer
∴ t = 6

AP Inter 2nd Year Maths 2A Question Paper May 2017

Question 8.
Find the number of terms in the expansion of (2x + 3y + z)7.
Solution:
The number of terms in the expansion of (2x + 3y + z)7 = \(\frac{(7+1)(7+2)}{2}\) = 36

Question 9.
Find the mean deviation from the mean of the following discrete data:
6, 7, 10, 12, 13, 4, 12, 16
Solution:
Given data 6, 7, 10, 12, 13, 4, 12, 16
Mean = \(\frac{6+7+10+12+13+4+12+16}{8}=\frac{80}{8}\) = 10
Absolute values about mean = |6 – 10|, |7 – 10|, |10 – 10|, |12 – 10|, |13 – 10|, |4 – 10|, |12 – 10|, |16 – 10| = 4, 3, 0, 2, 3, 6, 2, 6
∴ Mean deviation = \(\frac{4+3+0+2+3+6+2+6}{8}=\frac{26}{8}\) = 3.25

Question 10.
The mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X ≥ 1).
Solution:
Given mean = 4 ⇒ np = 4
variance = 3 ⇒ npq = 3
\(\frac{n p q}{n p}=\frac{3}{4}\)
⇒ q = \(\frac{3}{4}\)
∴ p = 1 – q
= 1 – \(\frac{3}{4}\)
= \(\frac{1}{4}\)
∴ n(\(\frac{1}{4}\)) = 4
⇒ n = 16
p(x ≥ 1) = 1 – p(x = 0)
= 1 – \(16_{c_o}\left(\frac{1}{4}\right)^0\left(\frac{3}{4}\right)^{16-0}\)
= 1 – \(\left(\frac{3}{4}\right)^{16}\)

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

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

Question 11.
If x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\), then show that 4x2 – 1 = 0.

Question 12.
If x is real, prove that \(\frac{x}{x^2-5 x+9}\) lies between \(\frac{-1}{11}\) and 1.

Question 13.
If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the rank of the word MASTER.
Solution:
The alphabetical order of the letters of the word REMAST is A, E, M, R, S, T
The no.of words beginning with A is 5! = 120
The no.of words begin with E is 5! = 120
The no.of words begin with MAE is 3! = 6
The no.of words begin with MAR is 3! = 6
The no.of words begin with MASE is 2! = 2
The no.of words begin with MASR is 2! = 2
The next word is MASTER = 1
∴ Rank of the word MASTER = 120 + 120 + 6 + 6 + 2 + 2 + 1 = 257

AP Inter 2nd Year Maths 2A Question Paper May 2017

Question 14.
Simplify: \({ }^{34} C_5+\sum_{r=0}^4(38-r) C_4\)
Solution:
AP Inter 2nd Year Maths 2A Question Paper May 2017 Q14
AP Inter 2nd Year Maths 2A Question Paper May 2017 Q14.1

Question 15.
Resolve \(\frac{2 x^2+2 x+1}{x^3+x^2}\) into partial fractions.
Solution:
AP Inter 2nd Year Maths 2A Question Paper May 2017 Q15
⇒ 2x2 + 2x + 1 = Ax(x + 1) + B(x + 1) + Cx2 …….(1)
Put x = 0 in (1)
⇒ 0 + 0 + 1 = 0 + B(0 + 1) + 0
⇒ B = 1
Put x = -1 in (1)
2(-1)2 + 2(-1) + 1 = C(-1)2
⇒ 2 – 2 + 1 = C
⇒ C = 1
Comparing the coefficients of x2 terms in (1)
2 = A + C
⇒ 2 = A + 1
⇒ A = 1
∴ A = 1, B = 1, C = 1
∴ \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\)

Question 16.
State and prove the Addition theorem of Probability.

Question 17.
Suppose A and B are independent events with P(A) = 0.6, P(B) = 0.7, then compute
(i) P(A ∩ B)
(ii) P(A ∪ B)
(iii) P(\(\frac{B}{A}\))
(iv) P(Ac ∩ Bc)

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

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

Question 18.
If α and β are the roots of the equation x2 – 2x + 4 = 0, then for any n ∈ N. Show that \(\alpha^n+\beta^n=2^{n+1} \cos \left(\frac{n \pi}{3}\right)\).

Question 19.
Solve 8x3 – 36x2 – 18x + 81 = 0, given that the roots of this equation are in arithmetic progression.
Solution:
Given 8x3 – 36x2 – 18x + 81 = 0 ……..(1)
Let the roots of (1) be a – d, a, a + d
S1 = \(\frac{-(-36)}{8}\)
⇒ a – d + a + a – d = \(\frac{36}{8}\)
⇒ 3a = \(\frac{36}{8}\)
⇒ a = \(\frac{3}{2}\)
∴ (x – \(\frac{3}{2}\)) is a factor of (1)
AP Inter 2nd Year Maths 2A Question Paper May 2017 Q19
⇒ 8x2 – 24x – 54 = 0
⇒ 4x2 – 12x – 27 = 0
⇒ 4x2 – 18x + 6x – 27 = 0
⇒ 2x(2x – 9) + 3(2x – 9) = 0
⇒ (2x + 3) (2x – 9) = 0
⇒ x = \(\frac{-3}{2}\), \(\frac{9}{2}\)
∴ Roots are \(\frac{-3}{2}\), \(\frac{3}{2}\), \(\frac{9}{2}\).

Question 20.
If n is a positive integer and x is any non-zero real number, then prove that \(C_0+C_1 \cdot \frac{x}{2}+C_2 \cdot \frac{x^2}{3}+C_3 \cdot \frac{x^3}{4}+\ldots . .+C_n \cdot \frac{x^n}{n+1}=\frac{(1+x)^{n+1}-1}{(n+1) x}\).

AP Inter 2nd Year Maths 2A Question Paper May 2017

Question 21.
If x = \(\frac{1}{5}+\frac{1.3}{5.10}+\frac{1.3 .5}{5.10 .15}\) + …….. ∞, then find 3x2 + 6x.

Question 22.
Find the mean deviation about the median for the following continuous distribution:

Age (Years) 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55 55 – 60
No. of Workers (fi) 120 125 175 160 150 140 100 30

Solution:
Construct the table
AP Inter 2nd Year Maths 2A Question Paper May 2017 Q22
AP Inter 2nd Year Maths 2A Question Paper May 2017 Q22.1

Question 23.
Suppose an urn B1 contains 2 white and 3 black balls and another urn B2 contains 3 white and 4 black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found to be black, then find the probability that the urn chosen was B1.
Solution:
Let E1 and E2 denote the events or selecting urns B1 and B2 respectively.
∴ P(E1) = P(E2) = \(\frac{1}{2}\)
AP Inter 2nd Year Maths 2A Question Paper May 2017 Q23

AP Inter 2nd Year Maths 2A Question Paper May 2017

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).

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