2nd PUC Basic Maths Model Question Paper 1 with Answers

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Karnataka 2nd PUC Basic Maths Model Question Paper 1 with Answers

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 = \(\left[ \begin{matrix} 5 & -6 & 3 \end{matrix} \right] \), B = \(\left[ \begin{matrix} -3 \\ 1 \\ 0 \end{matrix} \right] \)
Answer:

= [-15 – 6 + 10]
AB = [-21]_{1 × 1}

Question 2.
If ⁿP₄ = 360 find n.
Answer:
ⁿP₄ = 360
\(\frac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4 !)}\) = 360
n(n – 1) (n – 2) (n – 3) = 360
Assume n = 6
6 × 5 × 4 × 3 = 360 n = 6

Question 3.
Write the truth value of the proposition “If 2 is not a prim numbers then √2 is an irrational number”.
Answer:
Consider, p : 2 is not a prime number = F
q : √2 is an irrational number = T
Then the symbolic represantation is p → q
F → T
= T.

Question 4.
Find the mean proportion of 9 and 16
Answer:
Let x be the mean proportion
9 : x : : x : 16
x² = 144
x = 12

Question 5.
Define the “Index of learning”.
Answer:

Question 6.
If sin A = \(\frac { 3 }{ 5 }\), find sin 2A.
Answer:
Sin 2A = 2. sin A. cos A = \(2 \cdot \frac{3}{5} \cdot \frac{4}{5}\) cos A = \(\sqrt{1-\sin ^{2} A}\)
= \(2 \cdot \frac{3}{5} \cdot \frac{4}{5}\) cos A = \(\frac{4}{5}\)
∴ Sin 2A = \(\frac{24}{25}\)

Question 7.
Find the radius of the circle x² + y² – 4x – 5 = 0
Answer:
The equation of the circle is given by
x² + y² – 4x + 0y – 5 = 0
Here, 2g = – 4; 2f = 0; c = -5
g = -2; f = 0; c = -5.
radius = \(\sqrt{g^{2}+f^{2}-c}\) = \(\sqrt{4+0+5}=\sqrt{9}\)
∴ radius = 3 units

Question 8.

Answer:

= 3.3^3-1 = 27

Question 9.
If y = 2√x – cos 2x + 2, find \(\frac{d y}{d x}\)
Answer:

Question 10.
Evaluate : ∫\(\left(\frac{1}{x}-\sin x+3\right)\) dx.
Answer:
log x + cos x + 3x + c.

Part – B

II. Answer any 10 Questions (2 × 10 = 20)

Question 11.
If A = \(\left[ \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right]\) show that A² – 4A + 3I = 0
Answer:

Question 12.
There are 15 points in a plane of which 5 are collinear. Find the number of straightlines can be formed.
Answer:
By Joining any two points we get a straight line
∴ Total lines from 15 points = ¹⁵C₂
But 5 points are collinear, if both points are selected from these 5 points they give rise to one single line.
∴ Total lines = ¹⁵C₂ – ⁵C₂ + 1 = 105 – 10 + 1 = 96.

Question 13.
A boy drawn at random 3 balls from a bag containing 9 red and 5 white balls. What is ‘ the probability of getting (i) all red balls (ii) 2 red and 1 white.
Answer:
There are totally 9 + 5 = 14 balls
Total ways of choosing 3 balls from 14 balls is = n = ¹⁴C₃

(i) Total ways of choosing 3 balls from 9 red balls is ⁹C₃
favourable cases = ⁹C₃
∴ P(all red balls) = \(\frac{^{9} \mathrm{C}_{3}}{^{14} \mathrm{C}_{3}}=\frac{84}{364}\)
∴ P(all red balls) = 0.2307
∴ The probability that the boy draws 3 reds balls randomly is 0.2307.

(ii) Total ways of choosing 2 red and 1 white ball is ⁹C₂ × ⁵C₁
∴ favourable cases = ⁹C₂ × ⁵C₁
P(2 red and 1 white ball) = \(\frac{^{9} C_{2} \times^{5} C_{1}}{^{14} C_{3}}\) = \(\frac{36 \times 5}{364}\)
P(2 red and 1 white ball) = 0.4945
∴ The probability that the boy draws 2 red and 1 white ball is 0.4945.

Question 14.
Write the converse and contrapositive of the compound proposition “If prashanth got first class in mathematics then prof. John will gift him a watch”.
Answer:
Consider, p: Prashanth got first class in mathematics
q: Prof John will gift him a watch
The symbolic representation is q → p
Converse of is p → q is q → p
“If Prof. John gifted him a watch then Prashanth will get first class in mathematics”.
Contrapositive of p → q is ~q → ~p
“If Prof. John doesnot gift a watch then Prashanth will not get first class in mathematics.”

Question 15.
A certain number is subtracted from each of the two term o,” the ratio 21: 35 to give a new ratio 3 : 10. Find the number which is subtracted?
Answer:
Given: If x is subtracted from 21 : 35 then it results to be 3 : 10
⇒ 21 – x : 35 – x : 3 : 10
⇒ \(\frac{21-x}{35-x}=\frac{3}{10}\)
⇒ 210 – 10x = 105 – 3x
210 – 105 = 10x – 3x
105 = 7x
∴ x = 15
∴ The number subtracted from 21:35 to give 3 : 10 is 15.

Question 16.
Find the Banker’s discount on ₹ 1000 due 6 months hence at 10% per annum.
Answer:
r = 10 % = 0.1, F = ₹ 1000, t = 1/2 yrs, BD = ? .
BD = Ftr = 1000 × 0.5 × 0.1 = ₹ 50

Question 17.
Derive the value for cos 105°.
Answer:
cos 105° = cos (180° – 75°)
= -cos (75°)
cos 75° = cos(45° + 30°)
= [cos 45, cos 30 – sin 45, sin 30]

∴ cos(105)° = \(\frac{1-\sqrt{3}}{2 \sqrt{2}}\)

Question 18.
Transform 2sin 40° × cos 20° into sum.
Answer:
Consider, sinA. cosB = \(\frac { 1 }{ 2 }\)[sin(A + B) + sin(A – B)]
2 sin 40°. cos 20° = [sin (40° + 20°) + sin (40° – 20°)]
2 sin 40°. cos 20° = [sin 60° + sin 20°] = \(\frac{\sqrt{3}}{2}\) + sin 20°.

Question 19.
Find the focus and the equation of Directrix of the parabola x² – 16y = 0.
Answer:
x² = I6y
This curve turns upwards and it is of the form x² = 4ay
⇒ 4a = 16 ⇒ a = 4
∴ Equation of directrix is : y = -a, y = – 4 ⇒ y + 4 = 0.

Question 20.

Answer:

Question 21.
If y = X^x, find \(\frac{d y}{d x}\)
Answer:
y = X^x
Taking log^m on both sides we get
log y = log X^x
= x log x
diff w.r.t x

Question 22.
If S = 4t³ – 6t² + t – 7 (S = distance in mt, t = time sec). Find the velocity and acceleration at t = 2 sec.
Answer:
Given: S = 4t³ – 6t² + t – 7
V = \(\frac{d s}{d t}\) = 12t² – 12t + 1
acceleration = \(\frac{d v}{d t}\) = 24t – 12
∴ accenⁿ = 24 × 2 – 12 = 48 – 12 = 36 m/sec²
V_{t = 2 sec} = 12 × 2² – 12 × 2 + 1 = 48 – 24 + 1 = 25 mts/sec.

Question 23.
Evaluate : \(\int \frac{x}{x^{2}+4} d x\)
Answer:

Question 24.
Evaluate : \(\int_{0}^{x / 4} \sec ^{2} 3 x d x\)
Answer:

Part – C

III. Answer any 10 Questions (3 × 10 = 30)

Question 25.
If A = \(\left[ \begin{matrix} -1 & 2 \\ 3 & 4 \end{matrix} \right] \) verify A. (adj A) = (adj A). A = |A|. I and (I = Identity Matrix of 2^nd order).
Answer:

Question 26.
prove that \(\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ bc & ca & ab \end{matrix} \right| \) = (a – b) (b – c) (c – a).
Answer:

⇒ [Taking (a – b) and (b – c) as common factor from c₁ and c₂ respectively]

(a – b) (b – c) [1 (- a – (-c) ] = (a – b)(b – c)(c – a) = RHS
Hence, proved.

Question 27.
Find the number of permutation of the letters of the word ‘COMMISSION’. If the word
(i) Start with M and end with M
(ii) 2S’s are together
(iii) 2O’s are not together
Answer:
There are 10 letters out of which O, M, I and S are repeated twice.
∴ The Total no. of permutations = \(\frac{10 !}{2 ! 2 ! 2 ! 2 !}\)
(i) The number of permutations which begin with m and end with m = \(\frac{8 !}{2 ! 2 ! 2 !}\)
(ii) 2S^S are together can be taken as 1 unit
∴ Total no. of letters = 10 – 2 + 1 = 9
∴ No of permutations = \(\frac{9 !}{2 ! 2 ! 2 !}\)
(iii) 2 ‘O’s are not together = Total – 2 ‘O’s are together = \(\frac{9 !}{2 ! 2 ! 2 !}=\frac{10 !}{(2 !)^{4}}-\frac{9 !}{(2 !)^{3}}\)

Question 28.
In a class of 80 student 40 are taking Mathematics 25 are taking Statistics. If each student has taken at least one of these subject. Find the probability that the student is taking
(i) both Mathematics and Statistics
(ii) only Mathematics
(iii) Only Statistics.
Answer:
Let M = Mathematics, S = Statistics
(i) P(M ∩ S) = P(M).P(S) = \(\frac{1}{2} \cdot \frac{5}{16}=\frac{5}{32}\)
(ii) P(M) = \(\frac{40}{80}=\frac{1}{2}\)(iii) p(s) \(\frac{25}{80}=\frac{5}{16}\)

Question 29.
A sum of ₹. 2415 has to be divided among three person A, B and C in such proportion that A’s share to B’s as 4 : 5 and B’s share to C’s share as 9 : 16. How much does each get?
Answer:
Given A: B = 4 : 5 × 9,
B : C = 9 : 16 × 5
A : B = 36 : 45
B : C = 45 : 80
⇒ A : B : C = 36 : 45 : 80
Sum of the ratios 161
A^s share = \(\frac{36}{161}\) × 2415 = ₹540
B^s share = \(\frac{45}{161}\) × 2415 = ₹675
C^s share = \(\frac{80}{161}\) × 2415 = ₹161

Question 30.
A bill of ₹ 5000 was drawn on 10/4/2013 at 3 months and was discounted on 1/5/2013 at 12% p.a. For what sum was the bill Discounted and also find the Banker’s gain.
Answer:
Here F = Rs. 5,000, Bill period = 3 Months
legally due date = 10/4/2012 (Date of drawing) +
(+) 0 – 3 – 0 (Bill period)
(+) 3 – 0 – 0 (Grace period)
L.D.D = 13 – 7 – 13
Date of discount = 1 – 5 – 13
This means, the number of days from 1/5/13 to 13/7/13 becomes the unexpired period. This may be calculated as follows:
1/5/13 to 13/7/13
In June = (+) 30 days
In July = (+) 13
Total = 73 days
i.e., + = 73 days = \(\frac{73}{365}\) year
t = \(\frac{1}{5}\)
Given r = 12% ∴ BD = Ftr = 5000 × \(\frac{1}{5}\) × 0.12
Discount value = F – BD BD = ₹120
= 5000 – 120 = 4880
Discount value = 4880 ₹ TD = Ftr = 4882.81 × \(\frac{1}{5}\) × 0.12
P = \(\frac{\mathrm{F}}{1+\mathrm{tr}}\) BG = BD – TD
= \(\frac{5000}{1+\frac{1}{5} \times 0.12}\) = 120 – 117.18
i.e., p = ₹4882.81 i.e. BG = ₹2.82

Question 31.
Find the Interest earned on ₹ 4897.50 cash invested in 15% stock at 81.50, given that brokerage is 0.125%.
Answer:
Amount invested is Rs 4897.50
Cost of 1 share = 81.5 + 0.125 = 81.625

Question 32.
The price of a washing machine, inclusive of sales tax is ₹ 13,530. If the sales tax is 10%, find its basic price?
Answer:
Suppose the basic price (MP) of the washing machine = ₹x
∴ total amount paid = MP + ST% of x
13530 = x + 10% of x

Question 33.
Find the equation of the parabola given that the ends of the latus rectum are L (3, 6) and L¹ (-5, 6).

Question 34.
If x = a cos⁴t, y = b sin⁴t. find \(\frac{d y}{d x}\) at t = \(\pi / 4\)
Answer:
We have x = a cos⁴ t, y = b sin⁴t
Given x = a cos⁴t, y = b sin⁴t
differentiate both w.r.t t we get

Question 35.
The height of a cone is 30 cm and it is constant the radius of the base is increasing at the rate 0.25cm / sec. Find the rate of increase of volume of the cone when the radius is 10 cm?
Answer:
Given, h = 30 cm (constent)
\(\frac{d r}{d t}\) = 0.25 cm/sec, \(\frac{d v}{d t}\) = ?, r = 10 cm
W.K.T Volume of the cone = V = \(\frac{1}{3}\) πr² h

∴ The volume of the cone is increasing at the rate 50 π cm²/sec

Question 36.
Find the maximum and minimum value of the function f(x) = x⁵ – 5x⁴ + 5x³ – 1.
Answer:
Given f(x) = x⁵ – 5x⁴ + 5x³ – 1 _____ 1
f¹(x) = 5x⁴ – 20x³ + 15x² _____ 2 ….. (1)
f¹¹(x) = 20x² – 60x² + 30x ______ 3
for a function to be max^m of min^m f¹(x⁴) = 0
5x⁴ – 20x² + 15x² = 0
x²(5x² – 20x + 15) = 0
x²(x² – 4x + 3) = 0
x²(x – 3) (x – 1) = 0
x = 0, x = 1 & x = 3
Case 1: Put x = 0 in equation(3) we get f¹¹(x) = 0 the fun^x is neither maximum nor minimum at the point x = 0. This point is called the point of inflexion.

Case 2: Put x = 1 in equⁿ(3) we get f¹¹(x) = 20 – 60 + 30 = -10 < 0 the function is maximum at x = 1 and maximum value is

Case 3: Put x = 3 in equ<sn(3) we get

The function is minimum at the point x = 3 and minimum value is f(3) = 3⁵ – 5.3⁴ + 5.3³ – 1243 – 405 + 135 – 1 = -28.

Question 37.
Evaluate \(\int \frac{5 x}{(x-3)(x+4)} d x\)
Answer:

By partial fractions find A and B
∴ 5x = A(x + 4) + B(x – 3)
put x = 3 15 = A(3 + 4) + B(0)
15 = 7A ⇒ A = \(\frac{15}{7}\)
put x = 4
-20 = A(0) + B(-4 – 3)
-20 = 7B ⇒ B = \(\frac{20}{7}\)

= \(\frac{15}{7}\)log(x – 3) + \(\frac{20}{7}\)log(x + 4) + C

Question 38.
Evaluate ∫x^2, .cos3x dx.
Answer:
∫u.dv = u.v – ∫v.du

Part – D

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

Question 39.
Find the term Independent of ‘x’ in the expansion of \(\left(\sqrt{x}+\frac{1}{3 x^{2}}\right)^{10}\)
Answer:

To find the term Independent of x, we get

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

⇒ 2x + 1 = A(x + 2)(3x- 1) + B(3x – 1)(x – 1) + C(x + 2)(x – 1)
put x = -2
2(-2) + 1 = A(-2 + 2) + (3(2-)-l) + B(3(-2) -1) (-2 – 1) + C(- 2 + 2) (-2 – 1)
-4 + 1 = A(0) + B(21) + C(0)
-3 = B.21
∴ B = \(\frac{-3}{21}\)
put x = 1
⇒ 2(1) + 1 = A(1 + 2) (3.1 – 1) + B(3(1) – 1) (1 – 1) + C (1 + 2) (1 – 1)
3 = A(6)
A = \(\frac{1}{2}\)
put x = \(\frac{1}{3}\)

Question 41.
Verify: (p → 9) ∨ [(p ∧ q) ↔ q] is a Tautology or contradiction or neither.
Answer:

∴ (p → 9) ∨ [(p ∧ q) ↔ q] is a Tautology

Question 42.
8 men and 16 boys can finish a job in 6 days, but 12 men and 24 boys can finish the same job in 8 days. How many days will 16 men and 20 boys take to finish the same Job?
Answer:
8 men + 16 boys can do the job in 6 days Multiply by 6
48men + 96 boys can do the × 6 job in 1 day
Next 12 men and 24 boys can do the same job in 8 days
12 men + 24 boys can do it is 8 days Multiply by 8
96 men + 192 boys can do the same job is 1 day
Hence in doing one days work it may be started that
48men + 96boys = 96men + 192 boys
1 man = 2 boys
∴ 16 men + 20 boys = 32 boys + 20 boys = 52 boys can do the job in how many days
This is the case of inverse proportion
Let the unknown quantity be x days

32 : 52 : : x : 6
32 × 6 = 52x
x = 3.7 days.

Question 43.
Samsung company which manufacture LCD TV. The 1st lot of 10 unit was completed in 1400 labour hrs. Find each subsequent lot, the commutative production was doubled. And it has observed that 90% tearing effect applies to all labour related cost. The anticipated production is 320 unit of LCD TV find total labour cost required to manufacture 320 unit and also find the total labour cost at ₹ 20/per hrs.
Answer:

Total labour hour is 26,453.76 hrs and the labour cost at
₹ 20/per hr = 26,453.76 × 20 = ₹ 5,29,075.2

Question 44.
Using the graphical method, solve the following. LPP:-
Objective function : minimize z = 1.5Ax + 2.5y
Subject to constraints : x + 3y ≥ 3
x + y ≥ 2
and x ≥ 0, y ≥ 0.
Answer:
Consider the in equalities as equalities
x + 3y = 3 x + y = 2
If x = 0 then y = 1, (0, 1) If x = 0, y = 2, (0, 2)
If y = 0 then A: = 3, (3, 0) If y = 0, x = 2, (2, 0)
Plot the points on the graph

Question 45.

Answer:

= 3 – 3 sin² θ + 3 – 3 cos² θ
= 3 [1 – sin² θ + 1 – cos² θ]
= 3 [2 – (sin² θ + cos² θ)]
= 3[2 – 1]
= 3 = RHS.

Question 46.
A circle has its centre on x-axis and passess through (5,1) and (3,4). Find its equation.
Answer:
Let the equation of the required circle be x² + y² + 2gx + 2fy + c = 0
By data, 25 + 1 + 2g (5) + 2f(1) + c = 0
⇒ 10g + 2f + c = -26 ____ (1)
9 + 16 + 2g(3) + 2f(4) + c = 0
6g + 8f + c = -25 _____ (2)
The centre (-g, -f) of (1) lies on x-axis
⇒ y-coordinate of the centre = 0 ⇒ f = 0
By putting f in equation (1) and (2) we get,
10g + c = -26
6g + c = -25
(-) (-) (+)
4g = -1
g = \(\frac { -1 }{ 4 }\)
puting g = \(\frac { -1 }{ 4 }\) in eq (1)

puting g = \(\frac { -1 }{ 4 }\) ; f = 0. c = \(\frac { -47 }{ 2 }\) in the equation, we get
x² + y² + 2\(\left(\frac{-1}{4}\right)\)x + 2(0)y + \(\left(\frac{-47}{2}\right)\) = 0
x² + y² \(\frac { -1 }{ 2 }\) x – \(\frac { 47 }{ 2 }\) = 0
⇒ 2x² + 2y² – x – 47 = 0
2x² + 2y² – 1x – 47 = 0 is the required circle’s equation.

Question 47.
If y = (x² + a²)⁶, show that (x² + a²)\(\frac{d^{2} y}{d x^{2}}\) – 10x \(\frac{d y}{d x}\) – 12y = 0
Answer:
y = (a² + x²)⁶
DifF w.r.t x
y₁ = 6(a² + x²)⁵.2x
= 12x (a² + x²)5 …….. (1)
Again diffn. w.r.t x
y₂ = 12x.5 (a² + x²)⁴.2x + (a² + x²)⁵.12
= 120x²(a² + x²)⁴ + 12(a² + x²)⁵ …… (2)
∴ LHS = (a² + x²)\(\frac{d^{2} y}{d x^{2}}\) – 10xy₁ – 12y
= (x² + a²) {120x² (a² + x²)⁴ + 12(a² + x²)⁵ – 10x {12x (a² + x²)⁵} – 12(a² + x²)⁶
∴ LHS = RHS

Question 48.
Find the area enclosed between the curves y = 11x – 24 – x² and the line y = x.
Answer:
Given y = 11x – 24 – x² but y = x
∴ x = 11x – 24 – x²

(x – 4)(x – 6) = 0
x = 4 or x = 6

Part – E

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

Question 49.
(a) Transport corporation operate bus service between two villages. Data regarding the passenger traffic during the 1st three days of the week is given below along with the total revenue.

Find the bus fare charged per children, senior citizen and per adult. By using matrix method.
Answer:
Let x be the bus fare charged per children
Let y be the bus fare charged per senior children
Let z be the bus fare charged per adult
Consider
10x + 10y + 20z = 90 ⇒ x + y + 2z = 9 (By dividing 10)
30x + 20y + 10y = 100 ⇒ 3x + 2y + z = 10
1ox + 20y + 30z = 140 ⇒ x + 2y + 3z = 14

(b) The angle of elevations of the top of an unfinished tower at a point distance 120 mt from its base in 45°. How much higher must the tower be raised so that the angle of elevation at the same point many be 60° ?
Answer:
From triangle ABD

Question 50.

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) Find the total Revenue obtained by raising the output from 10 to 20 units. Where the marginal revenue function in given by MR = 3\(\left(\frac{x^{2}}{20}\right)\) – 10x + 100 (x = out put).
Answer: