2nd PUC Basic Maths Previous Year Question Paper March 2019

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Karnataka 2nd PUC Basic Maths Previous Year Question Paper March 2019

Time: 3.15 hours
Max Marks: 100

Instructions:

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

Part – A

I. Answer all the questions (1 × 10 =10)

Question 1.
If A = \(\left[ \begin{matrix} 1 & -2 \\ 3 & 4 \end{matrix} \right]\) then find 2A’.
Answer:
2A’ = \(\left[ \begin{matrix} 2 & 6 \\ -4 & 8 \end{matrix} \right]\)

Question 2.
In how many ways can 10 people be seated around a table?
Answer:

Question 3.
Symbolise the proposition: “If Oxygen is a gas then gold is a compound”.
Answer:
p → q

Question 4.
Find the duplicate ratio of 5 : 3.
Answer:
5³ : 4³ or 125 : 64

Question 5.
Find the income obtained by investing ₹3,600 in 5% stock at 90.
Answer:

Question 6.
Express sin5A cos3A as sum or difference of two trignometric functions.
Answer:
\(\frac { 1 }{ 2 }\)[sin8A + sin2A]

Question 7.
Find the centre of the circle x² + y² – 4x – y – 5 = 0.
Answer:
(2 + \(\frac { 1 }{ 2 }\))

Question 8.

Answer:
– 3

Question 9.
Differentiate (5e^x – logg x – 3√x) with respect to x.
Answer:
5e^x – \(\frac{1}{x}\) – \(\frac{3}{2 \sqrt{x}}\)

Question 10.
Integrate (x² – \(\frac{6}{x}\) + 5e^x)with respect to x.
Answer:
\(\frac{x^{3}}{3}\) – 6 log_ex + 5e^x + C

Part – B

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

Question 11.
If A = \(\left[ \begin{matrix} 1 & 3 & -1 \\ -1 & 0 & 2 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} 4 & -1 & 2 \\ 1 & 3 & -2 \end{matrix} \right] \) then, find A – 3B
Answer:
3B = \(\left[ \begin{matrix} 12 & -3 & 6 \\ 3 & 9 & -6 \end{matrix} \right] \) ∴ A = 3B = \(\left[ \begin{matrix} -11 & 6 & -7 \\ -4 & -9 & 8 \end{matrix} \right] \)

Question 12.
In how many ways the word “CARROM” be arranged such that 2R’S are always together?
Answer:
₹2 as 1 unit
∴ No. of letters = 6 – 2 + 1 = 5

Question 13.
A box contain 8 red marbles, 6 green marbles and 10 pink marbles. One marble is drawn at random from the box. What is the probability that the marble drawn is either red or green?
Answer:
Writing 8R + 6G + 10P = 24 OR No. of R & G = 14
∴ Probability = \(\frac{14}{24}=\frac{7}{12}\)

Question 14.
If the truth values of the propositions p, q, r are T, T, F respectively, then find the truth values of p → (q ∧ r).
Answer:

Question 15.
What must be added to each term in the ratio 5:6 so that it becomes 8:9?
Answer:
Let x be added ∴ \(\frac{5+x}{6+x}=\frac{8}{9}\)
Solving to get x = 3

Question 16.
True discount on a bill was ?900 and Banker’s gain was ₹27. What is the face value of the bill?
Answer:
Getting BD = ₹927. ₹ F = Rs. 30,900

Question 17.
Find the value of cos 15°.
Answer:
cos 15° = cos (45° – 30°) = cos 45°.cos 30° + sin 45°.sin 30°
Getting answer = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

Question 18.
If tan A = \(\frac { 1 }{ 2 }\), tan B = \(\frac { 1 }{ 3 }\), then find tan(A + B).
Answer:

Question 19.
Find the equation of parabola whose vertex is (0,0) and focus is (3,0).
Answer:
Identifying y² = 4ax .
Getting a = 3 and hence y² = 12x

Question 20.

Answer:
Getting LHL = 0 and RHL = -2
Concluding f(x) is not continuous at x = 1

Question 21.
If y = (a² – x²)¹⁰, find \(\frac{d y}{d x}\)
Answer:
\(\frac{d y}{d x}\) = 10(a² – x²)⁹ (0 – 2x) = -20(a² – x²)⁹

Question 22.
If the Cost function of a firm is given by C(x) = x³ – 3x + 7. Find the average cost and marginal cost.
Answer:
AC = x² – 3 + \(\frac{7}{x}\)
MC = 3x² – 3

Question 23.
Evaluate: \(\int \frac{4 x+3}{2 x^{2}+3 x+5} d x\)
Answer:
Substituting 2x² + 3x + 5 = t
∴ (4x + 3)dx = dt
∴ Given integral = \(\frac{1}{t}\)dt = log t + C = log_e (2x² + 3x + 5) + C

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

Simplifying to get answer = e²

Part – C

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

Question 25.
If A = \(\left[ \begin{matrix} 2 & 3 \\ 4 & -1 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} -1 & 5 \\ 6 & 2 \end{matrix} \right] \) show that (AB)’ = B’A’
Answer:

Question 26.
prove that \(\left| \begin{matrix} 1+a & b & c \\ a & 1+b & c \\ a & b & 1+c \end{matrix} \right| \) = 1 + a + b + c.
Answer:
Using C¹₁ → C₁ + C₂ + C₃ to get

Question 27.
From a class of 9 boys and 7 girls 12 students are to be chosen for a competition which includes atleast 6 boys and atleast 4 girls. In how many ways can this be done if a particular boy is always chosen?
Answer:
No. of ways = ⁸C₅ × ⁷C₆ × ⁸C₆ × ⁷C₅ × ⁸C₇ × ⁷C₄ = 56 × 7 + 28 × 21 + 8 × 35 = 1260

Question 28.
A couple appears in an interview for two vacancies in the same post. The probability of husband getting selected is \(\frac{1}{7}\) and the probability of wife getting selected is \(\frac{1}{5}\). What is the probability that
(a) both of them will be selected ?
(b) only one of them will be selected ?
(c) none of them will be selected ?
Answer:
Writing: P (husband not selected) = \(\frac{6}{7}\) and p (wife not selected) = \(\frac{4}{5}\)
(OR) Getting: (a) P (both selected) = \(\frac{1}{35}\)
(b) P (only one selected) = \(\frac{10}{35}=\frac{2}{7}\)
(c) P (none selected) = \(\frac{24}{35}\)

Question 29.
In a Fort there was ration for 560 soldiers that would last the soldiers for 70 days after 20 days, 60 soldiers left the fort. For how many days the remaining ration can support the remaining soldiers ?
Answer:
Writing: remaining no. of soldiers = 560 – 60 = 500
remaining no. of days = 70 – 20 = 50

Writing: 560 : 500 = x : 50 ⇒ x = 56 days

Question 30.
For ₹ 512.50 due 6 months at 15% p.a. Find the true present value and discounted cash value.
Answer:
p = \(\frac{\mathrm{F}}{1+t r}\) = ₹ 476.74 ⇒ Discounted value = F (1 – tr) = ₹ 474.06
Getting P = ₹ 476.74
Getting BD = F.tr = ₹ 38.43. ∴ Getting Discounted Value = F – BD = ₹ 474.06

Question 31.
Sanjana invests ₹ 3240 in a stock at 108 and sells when price falls to 104. How much stock at 130 can sanjana now buy?
Answer:
Case 1: (Buying) stock at 108
Worth of stock purchased = \(\frac{100 \times 3240}{108}\) = ₹ 3000

Case 2: Stock sold at 104
Amount of money received by selling = \(\frac{104 \times 3000}{100}\) = ₹ 3120

Case 3: Stock sold at 130
Amount of stock bought = \(\frac{3120 \times 100}{130}\) = ₹ 2400

Question 32.
Sanju goes to a shop to buy a bicycle quoted at ₹ 2000. The rate of sales tax is 12% on it. He asks the shopkeeper for a rebate on the price of the bicycle to such an extent that has to pay ₹ 2016 inclusive of sales tax. Find the rebate percentage on the price of the bicycle.
Answer:
SP = 2016, ST = 12%. Let MP = x
SP = MP + 12% MP ∴ 2016 = x + 0.12x
⇒ x = 1800 ⇒ Rebate = 2000 – 1800 = ₹ 200 Rebate % = \(\frac{200}{2000} \times 100\) = 10%

Question 33.
Find the equation of parabola when vertex is (0,0) and axis is y – axis and passes through the point (-1, -3).
Answer:
Identifying x² = 4ay [or x² = -4ay]
Substituting x = -1, y = -3
to get a = –\(\frac{1}{12}\) [or a = \(\frac{1}{12}\) ]
∴ Required equation : x² = –\(\frac{1}{3}\)y OR 3x² + y = 0

Question 34.
Differentiate y = \(\frac{e^{x}-1}{e^{x}+1}\) with respect to x.
Answer:

Question 35.
Find the maximum and minimum value of y = x³ – 9x⁴ 15x – 1.
Answer:

Question 36.
A square plate is expanding uniformly, the side is increasing at the rate of 5 cm/sec. what is the rate at which the area and its perimeter is increasing when the side is 20 cm long?
Answer:

Question 37.
Integrate x² sinx with respect to x.
Answer:
∫ x² sin x dx = -x² cosx- ∫ -cosx.2x dx
= -x² cosx + 2[x sinx – ∫sinx.1dx
= -x² cosx + 2xsinx + 2cosx + C

Question 38.
Evaluate: ∫(2x + 3)(x² +3x + 5)^3/2 dx.
Answer:
Substituting x² – 3x + 5 = t to get (2x + 3) dx = dt
∴ Given integral becomes ∫t^3/2 dt

Part – D

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

Question 39.
Find the middle term in the expansion of \(\left(\frac{2 x^{2}}{3}-\frac{3}{2 x}\right)^{10}\)
Answer:
Identifying middle term as T₆

Substituting r = 5 and simplifying to get T₆ = -252x⁵

Question 40.
Resolve \(\frac{x-1}{x(x+2)(x+4)}\) into partial fractions.
Answer:

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

Question 42.
Walking 4 kmph a student reaches his college 5 minutes late and if he walks at 5 kmph he reaches \(2 \frac{1}{2}\) minutes early. What is the distance from his house to the college?
Answer:
Let required distance = x kms
at 4 kmph, time taken = \(\frac{x}{4}\) ⇒ at 5 kmph, time taken = \(\frac{x}{5}\)
Given = \(\frac{x}{4}=\frac{5}{60}=\frac{x}{5}+\frac{2.5}{60}\) ⇒ Getting x = 2.5 Km

Question 43.
A motor company Ltd. has observed that a 90% learning effect applies to all labour related costs whenver a new product is taken up for production the anticipated production 320 units for the coming year. The production is done in lots of 10 units each. Each lot requires 1000 hours at ₹ 15/hour. Calculate the total labour hours and labour cost to manufacture 320 units.
Answer:
10 units = 1 lot

∴ Total hours = 18,895.68 ⇒ Total labour cost = 18,895.68 × 15 = ₹ 2,83,435.20

Question 44.
Solve the following L.P.P. graphically
Maximise Z = 60x + 15y
Subject to the constraints : x + y ≤ 50,
3x + y ≤ 90,
and x ,y ≥ 0.
Answer:

Question 45.
If A + B + C = 180°, then prove that sin2A + sin2B + sin2C = 4 sinA sinB sinC.
Answer:
sin 2A + sin 2B + sin 2C
= 2 sin\(\left(\frac{2 A+2 B}{2}\right)\).cos \(\left(\frac{2 A-2 B}{2}\right)\) + 2 sin C.cos C
= 2sin C.cos (A – B) + 2 sin C.cos C
= 2 sin C [cos(A – B) + cos C]
= 2 sin C[cos(A – B) + cos {180 – (A + B)}]
= 2 sin C[cos(A – B) – cos (A + B)]
= 2 sin C[-2 sin A. sin (-B)]
= 4 sin A. sin B . sin C

Question 46.
Find the equation of the circle passing through the points (5,3), (1, 5) and (3, -1).
Answer:
Getting 3 equations as
10 g + 6f + C = -34
2 g +10 f + C = -26
6 g – 2f + C = -10
Solving to get g = -2, f = -2, C = -2
Final equation: x² + y² – 4x – 4y – 2 = 0

Question 47.
If y = log \((x+\sqrt{x^{2}+1})\) show that (x² + 1)y₂ + xy₁ = 0.
Answer:

Question 48.
Find the area bounded by the parabola y² = 4x and the line x=y.
Answer:
y² = 4x ……… (1) x = y ………. (2)
(1) & (2) ⇒ x = 0, x = 4

Part – E

Answer any one of the following questions: (1 × 10 = 10)

Question 49.
(a) A sales person Samanth has the following record of sales. He is paid a commission at fixed rate per unit but varying rates for products P, Q, and R. (6)

Find the rate of commission payable on P, Q and R per unit sold
Answer:
Let rates of commission per unit of sales be x, y and z for P, Q and R respectively
Writing 9x + 10y + 2z = 780
15x + 5y + 4z = 900
6x + 10y + 3 z = 820

Note: Proportionate marks should be given to any other appropriate method.

(b) Find the value of (1.2)⁵ using Binomial theorem upto 4 places of decimals. (4)
Answer:
(1.2)⁵ = (1 + 0.2)⁵
= 1⁵ + ⁵C₁ (0.2) + ⁵C₂ (0.2)2 + ⁵C₃ (0.2)3 + ⁵C₄ (0.2)4 + (0.2)5
= 1 + 5(0.2) + 10 (0.04) + 10 (0.008) + 5(00004) + (0.00032)
= 2.4883

Question 50.
(a) If ‘n’ is a rational number and ‘a’ is non-zero real number, then prove that (6)

Answer:
Proof for n positive integer
Proof for n negative integer
Proof for n rational number

(b) A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he returns 80 mts from the bank he finds the angle to be 30°. Find the height of the tree and the width of the river. (4)
Answer: