2nd PUC Basic Maths Previous Year Question Paper March 2018

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Karnataka 2nd PUC Basic Maths Previous Year Question Paper March 2018

Time: 3.15 hours
Max Marks: 100

Instructions:

The question paper has 5 parts A, B, C, D, and E. Answer all the parts.
Part – A carries 10 marks, part – B carries 20 marks, part – C carries, part – D carries 30 marks and part – E carries 10 marks.
Write the question number properly as indicated in the question paper.

Part – A

Answer all the ten questions: (10 × 1 = 10)

Question 1.
If A = \(\left[ \begin{matrix} 2 & 4 \\ 3 & -1 \\ 4 & 0 \end{matrix} \right] \) show that (A’)’ = A.
Answer:

Question 2.
How many different 4 digit number can be formed using the digits 1, 2, 4, 5, 7, 8, 9. No digit being repeated.
Answer:
Total ways = ⁷P₄ = 7 × 6 × 5 × 4 = 840 numbers.

Question 3.
Symbolise the propositions “2+ 5 = 6 or all integers are rationals”.
Answer:
P ∨ Q

Question 4.
Find the duplicate ratio of 5 : 3.
Answer:
Duplicate ratio of 5² : 3² = 25 : 9

Question 5.
What rate of interest is obtained by investing in 9% stock at 180?
Answer:

Question 6.
If sin A = \(\frac{3}{5}\), find sin 2A.
Answer:
If sin A = \(\frac{3}{5}\) there cos A = \(\frac{4}{5}\)
∴ sin2 A = 2sin A cos A = \(2 \cdot \frac{3}{5} \cdot \frac{4}{5}=\frac{24}{25}\)

Question 7.
Find the equation of directrix for a given parabola x² = 6y.
Answer:
Compare x² = 6y with x² = 4ay ⇒ 4a – 6 ⇒ a = \(\frac{6}{4}=\frac{3}{2}\)
∴ Equation of dirrectri x is y 2y + 3 = 0 = \(\frac{-3}{2}\) or 2y + 3 = 0

Question 8.

Answer:

Question 9.
If y = 5a^x – log x – 3√x find \(\frac{d y}{d x}\)
Answer:
\(\frac{d y}{d x}\) = 5a^x . log_e a – \(\frac{1}{x}-\frac{3}{2 \sqrt{x}}\)

Question 10.
Evaluate: \(\int \frac{1}{5 e^{-x}} d x\)
Answer:
\(\int \frac{1}{5 e^{-x}} d x\) = \(\frac{1}{5} \int e^{x} d x=\frac{1}{5} e^{x}+C\)

Part – B

II. Answer any ten questions: (10 × 2 = 20)

Question 11.
Solve by crame’s rule 3x + 4y = 7 and 7x – y = 6.
Answer:

Question 12.
Find n if ⁿP₃ = 210.
Answer:
ⁿP₃ = 210
n(n – 1) (n – 2) = 210 ⇒ 7(7 – 1) (7 – 2) = 7.6.5 = 210 ⇒ n = 1

Question 13.
Two dice are thrown at once. What is the probability of getting face upwards with ‘sum equal to 4 or 5″.
Answer:
n(S) = 36
Let A: Sum equal to 4 = {(1,3) (3, 1) (2,2)}
P(A) = \(\frac{3}{36}=\frac{1}{13}\) and B ; Sum equal to 5 = {(1,4) (4, 1) (2, 3) (3, 2)} jo 13
∵ A ∩ B = 0
P(B) = \(\frac{4}{36}\), and P(A ∩ B) = 0
P(Sum 4 or 5) = P(A∪B) = P(A) + P(B) = \(\frac{3}{36}+\frac{4}{36}=\frac{7}{36}\)

Question 14.
If the truth values of propositions p, q, r are T, T, F respectively. Then find the truth values of the compound propositions (p ∨ r)∧ q .
Answer:
Truth value of (p ∨ r) ∧ q
(T ∨ F) ∧ T ⇒ T ∨ T = T ⇒ (P ∨ r) ∧ r is true.

Question 15.
Monthly income of A and B are in the ratio 2 : 3 and their monthly expenditure are in the ratio 3 : 5. If each saves X 100 per month. Find the monthly incomes of A and B.
Answer:
Let their incomes be 2x and 3.x
W.K.T Income – Saving = Expenditure
∴ \(\frac{2 x-100}{3 x-100}=\frac{3}{5}\) ⇒ 5(2x – 100) = 3(3x – 100) ⇒ 10x – 500 = 9x – 300 ⇒ x = 200
∴ Incomes of A and B are 2 × 200 and 3 × 200 = 400 and 600

Question 16.
A bill drawn for 3 month was legally due on 06.07.2018. Find the date of drawing of the bill.
Answer:
L.D.D = D . D + B.P + 3 days
D.D = LDD – B.P -3 days
= (06 – 07 – 2018) – (0.3 – 0) – (3 – 0.0)
= 3 – 4 – 2018

Question 17.
Find the value of sin 15°.
Answer:
Sin 15° = Sin (60 – 45) Sin 60. cos 45 – cos 60. Sin 45

Question 18.
Show that tan (45° + A) tan (45° – A) = 1.
Answer:

Question 19.
Find the equation of parabola given that vertex is origin (0, 0) and passing through the point P (5, 2) and symmetric with respect to the y – axis.
Answer:
Given it is symmetric about y – axis. So it’s equation is x² = 4ay or x² = -4ay, ∵ the parabola passes there the point (5,2) and it is in 1^st quadrant ∴ the parabola is x² = 4ay and 5² = 4.2.a ⇒ 25 = 8a ⇒ a = \(\frac{25}{8}\)
∴ its equation is given by x² = 4 \(\frac{25}{8}\) . y ⇒ x² = \(\frac{25 y}{2}\) or 2x² = 25y

Question 20.

Answer:
Since f(x) is continuous
\(\lim _{x \rightarrow 1}\) f(x) = f(1)
\(\lim _{x \rightarrow 1}\) 4x + 3 = K + 1
4 + 3 = k + 1 ⇒ k = 7 – 1 = 6

Question 21.

Answer:

Question 22.
If S = 5t² + 4t – 8. Find the initial velocity and acceleration.
Answer:
S = 5t² + 4t – 8
V = \(\frac{d s}{d t}\) = 10t + 4 dt
Initial velocity is t = 0 ⇒ V = 4 units/sec dv
acceleration = \(\frac{d v}{d t}\) = 10 sq units/sec

Question 23.
Evaluate: ∫(4x² – 2x + 7)^3/2(4x – 1)dx.
Answer:
Put 4x² – 2x + 7 = t
(8x – 2)dx = dt ⇒ 2(4x – 1 )dx = dt ⇒ (4x – 1 )dx = \(\frac { 1 }{ 2 }\)dt

Question 24.
Evaluate \(\int_{1}^{2} \frac{1}{x} d x\)
Answer:

Part – C

III. Answer any ten questions : (10 × 3 = 30)

Question 25.
If A = \(\left[ \begin{matrix} -1 & 2 \\ 3 & 4 \end{matrix} \right] \) show that A(adj A) = (adj A) A = |A|I.
Answer:
|A| = -4 – 6 = 10

Question 26.
Show that \(\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ab & { -b }^{ 2 } & bc \\ ac & bc & { -c }^{ 2 } \end{matrix} \right| \) = 4a² b² c²
Answer:

= a² b² c²(0 + 2 + 2) = 4a² b² c² = RHS

Question 27.
A team iof 11 is to be chosen from 18 cricketers of whom 6 are bowlers and 3 are wicket keepers. In how many ways can a team be chosen so that.
(i) There are exactly 4 bowlers and one wicket keeper
(ii) There are atleast 4 bowlers and atleast 2 wicket keepers.
Answer:
(i) Exactly 4B and 1 W.K and 6 others

Question 28.
If A and B are event with P(A) = \(\frac{5}{8}\) P(B) = \(\frac{3}{8}\) and P(A ∪ B) = \(\frac{3}{4}\) find (i) p(B/A) (ii) p(A/B)
Answer:

Question 29.
Two taps can separately fill a tank in 12 min and 15 minutes respectively. The tank when full can be emptied by a drain pipe is 20minutes. When the tank was empty, all the three were opened simultaneously. In what time will the tank be filled up?
Answer:
Time taken to fill the tank is 12 and 15 minutes and time taken to drawn the tank is 20 minutes. In one minute the two taps will fill \(\frac{1}{12}, \frac{1}{15}\) of the tank and drain \(\frac{1}{20}\) of the tank.
∴ The required time will be \(\frac{1}{12}+\frac{1}{15}-\frac{1}{20}\)
\(=\frac{20+16-12}{240}\)
\(=\frac{24}{290}=\frac{1}{10}\)
∴ Hence in one minute \(\frac{1}{10}\) of the tank will be full the tank will be filled in minutes.

Question 30.
The banker’s gain on a bill is \(\frac{1}{5}\) th of the banker’s discount and the rate of interest is 20% p.a. Find the unexpired period of the bill.
Answer:
Given BG \(\frac{1}{5}\) BD and r = 20%

Question 31.
Rakshith decides to invest in TCS shares which are selling at ₹ 2,020 per share. How much money is required to purchase 10 shares, if the brokerage in 0.5%.
Answer:
Selling price of 10 shares at 2020 per share = 20,200 ₹
Brokerage = \(\frac{0.5}{100}\) × 20,200 = ₹101
Amount required to purchase = 20,200 + 101 = 20,301 Rs.

Question 32.
The price of a washing machine inclusive of sales tax is ₹ 13,530. If the sales tax is 10%. Find the basic price.
Answer:
Suppose the basic price of the washing machine = ₹x
∴ Total amount paid = MP + ST% of MP
13530 = x + 10% of x 13530 = x + \(\frac{x}{10}=\frac{10 x+x}{10}=\frac{11 x}{10}\)
∴ x = \(\frac{13530 \times 10}{11}\) = 12,300
∴ Basic price = ₹12,300.

Question 33.
Find the length of chord of the circle x² + y² – 6x – 4y – 12 = 0 on the coordinate axes.
Answer:
Given x² + y² – 6x – 4y- 12 = 0
centre = (3, 2) and C = -12
Length of chord intercepted by x – axis = \(2 \sqrt{g^{2}-c}\)
= \(2 \sqrt{3^{2}-(-12)}=2 \sqrt{9+12}=2 \sqrt{21}\) units
Length of chord intercepted byy – axis = \(2 \sqrt{f^{2}-c}=2 \sqrt{4-(-12)}=2 \sqrt{4+12}\) = 8units

Question 34.
If x = a sec θ, y = b tan θ Find \(\frac{d y}{d x}\) at θ = \(\frac{\pi}{4}\).
Answer:
x = a sec θ, y = b tan θ

Question 35.
The volume of a sphere is increasing at the rate 4π cc/sec. Find the rate at which the area of its surface increases when its radius is 10cm.
Answer:

Question 36.
Show that x^x is minimum at x = \(\frac{1}{e}\)
Answer:
Let y = x^x (1)
log y = x log x

Question 37.
Evaluate: ∫x² log x dx.
Answer:
∫udv = uv – ∫v du

Question 38.
Evaluate: ∫cosec x [cosec x + cot x] dx.
Answer:
∫(cosec²x + cosec x.cot x)dx = -cot x – cosec x + C

Part – D

IV. Answer any six questions : (6 × 5 = 30)

Question 39.
Evaluate: \((2+\sqrt{3})^{5}+(2-\sqrt{3})^{5}\)
Answer:
Consider,

Question 40.
Resolve into partial fractions : \(\frac{3 x+5}{(x+2)(x-1)^{2}}\)
Answer:

3x + 5 = A(x – 1)² + B(x + 2)(x – 1) + C(x + 2)
put x = -2
-6 + 5 = A(-2 – 1)² + 0 + 0
-1 = A(-3)² ⇒ A = \(\frac{-1}{9}\)
put;c = 1,
3 + 5 = A(0) + B(0) + C(1 + 2)
8 = 3C ⇒ C = \(\frac{8}{3}\)
put x = 0,
5 = A(-1)² + B(2)(-1) + C(2)
5 = A – 2B + 2C

Question 41.
Verify whether the compound proposition p → (~ p ∨ q) is a tautology or a contradiction or neither.
Answer:

Question 42.
If two men or four women can do a work in 33 days and 3 men and 5 women can do the same work in 24 days ? How long shall 5 men and 2 women take to do the same work ?
Answer:
2 men and 4 boys can work in 33 days
∴ In 1 any 66men and 132 boys can to the same work 3 men and 5 boys can do it in 24 days 72 men and 120 boys can do it in I day
∴ 66 men + 132 boys = 72 men + 120 boys 12 boys = 6 men
5 men & 2 boys are equivalent to 10 + 2 = 12 boys
8 : 12 = x :33

5 men & 2 boys work in 22 days.

Question 43.
Samsung company which manufacture LCD TV. The 1st lot of 10 units was completed in 1,400 laboour hours. Find each subsequent lot, the commutative production was doubled. And it has observed that 90% learning effect applies to all labour related cost. The anticipated production is 320 units of LCD TV find total labour cost required to manufacture 320 unit and also find the total labour cost at ₹ 20/hrs.
Answer:

Totallabour hours for 32 lots =26,453.952. hrs
Labour cost at Rs 20 per hour = 26,453.952 × 20 = 529,07.04

Question 44.
Solve the L.P.P. graphically Z = 3x + 5y
Subject to : x + 3y ≤ 3, x + y ≤ 2, x,y ≥ 0.
Answer:

x + 3y = 3
putx = 0, ⇒ y = 1 ∴ (0, 1)
puty = 0 ⇒ x = 3 ∴ (3, 0)
x + y = 2
put x = o ⇒ y = 2 ∴ (0, 2)
put y = o ⇒ x = 2 ∴ (2, 0)

Question 45.
In any ∆ABC, prove that sin2A + sin2B – sin2C = 4 cos A cos B sin C
Answer:
LHS = sin2A + sin2B – sin2C
= 2sin (A + B). cos (A – B) – 2sin cos C [using trans formation formula]
= 2 sin C cos (A – B) – 2 sin C. cos C
= 2 sin C [cos (A – B) – cos C]
= 2 sin C [cos (A – B) + cos (A + B)] ∴ cos (A + B) = – cos C
= 2 sin C. 2 cos A cos B
= 4 cos A . sin C. cos B [∵ cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= 4 cos A. cos B. Sin A
= RHS
∴ sin 2A + sin 2B – sin 2C = 4 cos A . cos B . sin C.

Question 46.
Find the equation of the circle passing through the point (0, 2) (3, 0) and (3, 2).
Answer:
Let the required equation of the circle is x² + y² + 2yz + 2fy + C = 0
This equation passes through (0,2) (3,0) and (3,2)
(0,2) 0² + 2² + 2g (0) + 2f(2) + C = 0 ⇒ 4f + 4 + C = 0 (1)
(3,0) 3² + 0² + 2g (3) + 2f(0) + C = 0 ⇒ 6g + 9 + C = 0 (2)
(3,2) 3² + 2² + 2g (3) + 2f(2) + C = 0
6g + 4f + 13 + C = 0 (3)
Eqⁿ2 – Eqⁿ1 gives 6g – 4f + 5 = 0 (4)
Eqⁿ3 – Eqⁿ2 gives 4f + 4 = 0 (5)
⇒ f = -1
put f = -1 in Eqⁿ 4, we get 6g + 4 + 5 = 0
6g = -9 ⇒ g = \(\frac{-3}{2}\)
4f + 4 + C = 0
-4 + 4 + C = 0 ⇒ C = 0
∴ the required equation of the circle is x² + y² + 2 \(\left(\frac{-3}{2}\right)\) x + 2(-1)y + 0 = 0
x² + y² – 3x – 2y = 0

Question 47.
If y = a cos (log x) + b sin (log x). P.T x²y₂ + xy₁ + y = 0.
Answer:
If y = acos(log x) + b sin(log x)

xy₁ = -a sin (log x) + b cos (log x)
Again diff w.r.t x.

x²y₂ + xy₁ = -y ⇒ x²y₂ + xy₁ + y = 0

Question 48.
Find the area enclosed between the curves y² = x and x² = y.
Answer:
Given y = x² & y² = x = y = √x
(√x) = x² S.B.S
x = x⁴ ⇒ x⁴ – x = 0 ⇒ x (x³ – 1) = 0 ⇒ x = 0 or x = 1
∴ The required Area bounded = A

Part – E

V. Answer any one question : (1 × 10 = 10)

Question 49.
(a) A school wants to award its students for the values of punctuality good behaviour and hard work with a total cash award ?6,000. Three time the award money for hard work together with the award money for punctuality is ? 11,000. The award money for punctuality and hard work together is double the one given for Good Behaviour. Represent the above situation algebraiclly and also find the award money for each value. Using metrix method.
Answer:
Let the values of punctuality, good behaviour and hard work be denoted by x, y & z respectively
x + y + z = 6000 …… (1)
x + oy + 3z = 11,000 …..(2)
x + z = 2y or x – 2y + z = 0 ……. (3)

(b)
Find the value of (1.01)5 using Binomial upto 4 decimal places. (4)
Answer:
(1.01)⁵ = (1 + 0.1)⁵
= 1⁵ + ⁵C₁(0.01) + ⁵C₂(0.01)² + ⁵C₃.(0.01)³ + ⁵C₄(0.01)⁴ + ⁵C₅(0.01)⁵
= 1 + 5(0.01) + 10(0.001) + 10(0.000001) + Neglect the next terms
= 1 + 0.05 + 0.001 + 0.00001
= 1.0510 correct upto 4 decimals

Question 50.
(a) If n is a rational number and ‘a’ is non-zero real number, then prove that

Answer:
Case 1: Let n be a positive integer.

Let y = x^1/q and a^1/q = b
⇒ x = y² and a = b^q
Also x → a changes to y → b

(b) The angles of depression of two boats as observed fron the mast head of a ship 50m high are 45° and 30°. What is the distance between the boats if they are on the same side of the mast head in line with it ? (4)
Answer:
Let AB be the mast head C and D denote the positions of the boats