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Karnataka 2nd PUC Maths Model Question Paper 1 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions:

  1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.
  2. Use the graph sheet for the question on Linear programming in PART E.

Part – A

I. Answer all the following questions ( 10 x 1 = 10 )

Question 1.
Give an example of a relation which is reflexive and symmetric but not transitive.
Answer:
R = {(1,2),(2,3),(2,1),(1,3), (3,2)}

Question 2.
Find the value of cot (tan-1 x + cot-1 x)
Answer:
cot ( tan -1 x cot -1 x = cot ( \(\frac{\pi}{2}\) ) = 0

Question 3.
Define a scalar matrix.
A diagoral matrix is said to be a scalar if the principal diagoral are same.
2nd PUC Maths Model Question Paper 1 with Answers 1

Question 4.
If
2nd PUC Maths Model Question Paper 1 with Answers 2, find the values of x.
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 3
⇒ x2 – 64 = 4 – 64
⇒ x2 = 4
⇒ x = ±2

KSEEB Solutions

Question 5.
Differentiate sin √X with respect to x
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 4

Question 6.
Evaluate : ∫ \(\frac{1-x}{\sqrt{x}} d x\)
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 5

Question 7.
Find the vector components of the vector with initial point {2,1} and terminal point (-5, 7)
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 6

Question 8.
What is the equation of the plane that cuts the coordinate axes at (a, 0,0), (0, b, 0) and (0, 0, c)
Answer:
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)

Question 9.
Define the term corner point in the L.p.p
Answer:
A comer point of a feasible region is a point in the region which is the intersection of the two boundary lines.

Question 10.
If E is an event of a sample space S of an experiment then find P(SIF)
Answer:
P(S/F) = \(\frac{\mathrm{P}(\mathrm{SnF})}{\mathrm{P}(\mathrm{F})}\)

Part -B

Answer any Ten questions: ( 10 x 2 = 20 )

Question 11.
Verify whether the operation * defined on Q by a*b = \(\frac{\mathrm{ab}}{4}\) is associative or not.
Answer:
a*b = \(\frac{\mathrm{ab}}{4}\)
2nd PUC Maths Model Question Paper 1 with Answers 7
from 1 and 2
(a*b)*c = a*(b*c)
⇒ * is associative

Question 12.
Find the value of tan-1(√3) -sec(-2)
Answer:
tan-1(√3) -sec(-2)
2nd PUC Maths Model Question Paper 1 with Answers 8

Question 13.
Write the simplest form of \(\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}})\) , 0 < x < π
2nd PUC Maths Model Question Paper 1 with Answers 9

Question 14.
Let A(1, 3), B(0, 0) and C(K, 0) be the vertices of triangle ABC of area 3 sq. units. Find k using determinant method.
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 10
A = \(\frac { 1 }{ 2 }\) [1(0 – 0) – 3(0 – K) + 1(0 – 0)
A = \(\frac { 1 }{ 2 }\)[3K]
Given , Area = 3 sq. units,
Thus- \(\frac { 3k }{ 2 }\) = +3 \(\frac { 3k }{ 2 }\) = -3
⇒ 3k = 6, 3k = -6
k = 2, k = -2

Question 15.
Prove that greatest integer function defined by f(x) = [x], 0<<3 is not differentiable at x = 1
Answer:
Since fis not continuous at every Integral
point or at x= I
⇒ It follows that [x] cannot differentiate at x = 3.

Question 16.
Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\) ,If x 4t and y = \(\frac{4}{t}\)
Answer:
x = 4t \(\frac{\mathrm{d} x}{\mathrm{d} x}\) = 4
y = \(\frac{4}{t}\) \(\frac{\mathrm{d} y}{\mathrm{d} x}\) = \(\frac{-4}{t^{2}}\)
2nd PUC Maths Model Question Paper 1 with Answers 11

KSEEB Solutions

Question 17.
Find the intervals in which the function fgivenby f(x) = x2 – 4x + 6 is strictly increasing.
Answer:
f(x) = x2 – 4x +6
f1 (x) = 2x -4
f1 (x ) =0 given, 2x – 4=0 ⇒ 2x =4
⇒ x = 2
Thus f1(x ) =0 at x =2
Now, the point x = 2 divides the real line
into two disjoint intervals namely,
(-∞,2) and (2,∞).
At (2,∞),f1 (x) > 0
⇒ f is strictly increasing at(2,∞)

Question 18.
Evaluate : ∫sin3x.cos4xdx
Answer:
Let I = ∫ sin 3x.cos 4x dx
USE:
sin A. cos B = \(\frac{1}{2}\) [sin(A + B)+sin(A + B)]
= \(\frac{1}{2}\)∫ [sin(3x + 4x) + sin(3x – 4x)J dx
= \(\frac{1}{2}\)∫ [sin7x + sin(-x)] dx
= \(\frac{1}{2}\) ∫[sin7x + sin x] dx
2nd PUC Maths Model Question Paper 1 with Answers 12

Question 19.
Evaluate : ∫logx dx.
Answer:
I = ∫logx dx = ∫logx.1 dx
= ∫uv dx =u∫v dx – ∫[\(\frac{d}{d x}\) x∫vdx] dx
Iogx ∫1. dx_∫[\(\frac{d}{d x}\) (logx)x∫1.dx] dx
=1ogx(x)∫( \(\frac{1}{x}x x\) ) dx
=xlogx – ∫1. dx
I = x logx – x + c

Question 20.
Form the differential equation of the family of curves \(\frac{x}{a}+\frac{y}{b}=1\), by eliminating the contants “a” and “b”
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 13

Question 21.
If either \(\overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{0}}\) or then \(\mathbf{a} \cdot\overrightarrow{\mathbf{b}}=\mathbf{0}\) but the converse need not not be true. Justify your answer with an example.
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 15
\(\overrightarrow{\mathrm{a}} . \overrightarrow{\mathrm{b}}\) = 6(4)+2(-4) + (-8)(2)
= 24 – 8 – 16
=0
=> \(\overrightarrow{\mathrm{a}}\) is Δr to b
Take any two non – zero perpendicular vectors \(\overrightarrow{\mathrm{a}}\) & \(\overrightarrow{\mathrm{b}}\)
Hence, the convrse need not be true

Question 22.
Find the angle θ between the vectors \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 16
2nd PUC Maths Model Question Paper 1 with Answers 17

Question 23.
Find the distance between the parallel – lines
\(\vec{r}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) and \overrightarrow{\mathbf{r}}=3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}+\mu(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) )
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 18

KSEEB Solutions

Question 24.
A fair die is rolled. Consider the events E = {1, 3, 5}, F = {2, 3} and G{2, 3, 4, 5}. Find (i) P(E/F) ii) P(E/G)
the set A = {l,2,3,4,5,6}
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 19

Part – C

Answer any Ten Questions : ( 10 x 3 = 30 )

Question 25.
Determine whether the relation R in the set A={I,2,3,4,5,6} as R={x,y}:y is divisible by x} is reflexive, symmetric and transitive.
Answer:
(1) R is reflexive, as (a,a) ∈ R ⇒ a is
divisible by a ∴ R is reflexive.
(2) Let (1,2) ∈ R ⇒ 2 is divisible by 1
But (2,1) ∈ R ⇒ 1 cannot divide by 2
∴ R is not symmetric.
(3) Let (1,2)∈ R ⇒ 1(2, 4)ER
⇒ 2 is divisible by I and 4 is divisibly by 2
Now, (1,4) ∈ R ⇒ 4 is divisible by 1
∴  (1,2)∈ R I(2,4) ∈ R ⇒ (1,4)∈ R
∴ R is transitive.

Question 26.
Find the values of x, if
2nd PUC Maths Model Question Paper 1 with Answers 20
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 21
⇒x2 = 4 – 3 = 1
∴ x = ±1

Question 27.
Answer:
Find the values of x and y.
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 22

Question 28.
If yx + xy = ab, find \(\frac{d y}{d x}\)
yx + xy = ab …..(1)
‘ Let u = yx and v = xy
2nd PUC Maths Model Question Paper 1 with Answers 23
2nd PUC Maths Model Question Paper 1 with Answers 24

Question 29.
Verify Mean value theorem for the
function f(x) = x3 – 5x2 – 3x , in the interval[a,b] where a= 1 and b = 3
(1) f(x) = x3 – 5x2 – 3x is continuous in [1,3]
(2) f1(x) = 3x2 – 10x – 3 exists in [1, 3]
2nd PUC Maths Model Question Paper 1 with Answers 25
3c2 – 10c – 3 +10 = 0
3c2 – 10c + 7 = 0
= 3c2 – 3c – 7c + 7 = 0
= 3c (c – 1) – 7 (c – 1) = 0
=(3c – 7) ((c- 1)) = 0
c = \(\frac{7}{3}\) or c=1
But c ∉ (1,3) ∴c = \(=\frac{7}{3}\)∈ (1 , 3 )

KSEEB Solutions

Question 30.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Answer:
Let one number be x. Then other number is (16 – x)
Let S(x) = x3 +(16 – x)3
S1 (x) = 3x2 +3(16 – x)2 (-1)
S11(x) =6x + 6(16 -x)1(+1)
S1(x) = 0 gives, 3x2 – 3 (16 – x)2 =0
⇒ x2 – (16 – x)2 =0
⇒ x2 = (16 – x)2
⇒ x2 – 256 + x2 – 32x
⇒ 32x = 256
x = \(\frac{256}{32}\)
x =8
Also, S11 (x) = 6(8)+ 6(16 – 8)
= 48 + 6(8)
= 48 + 48
= 96 > 0
∴ By second derivative test, x = 8 is the point of local minima of s.
Hence sum of cubes of numbers is
minimum when the numbers are 8 and 16-8 = 8

Question 31.
Find ∫\(\frac{x}{(x+1)(x+2)} d x\)
Answer:
Let I = \(\frac{x}{(x+1)(x+2)} d x\)
Resolve into partial fractions,
2nd PUC Maths Model Question Paper 1 with Answers 26
2nd PUC Maths Model Question Paper 1 with Answers 27

Question 32.
Evaluate ∫\(\frac{1}{1+\tan x} d x\)
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 28

Question 33.
Find the area of the region bounded by the curve y = x2 and the line y=4
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 29
Given y = x2 and y = 4
To find A and B:
put y = 4 in y = x2
⇒ x2 = 4
⇒ x =±2
put x = 2, y = 4 ⇒(2,4)
x = -2, y = 4 ⇒ (-2,4)
2nd PUC Maths Model Question Paper 1 with Answers 30

Question 34.
Prove that the equation
x2\(\frac{d y}{d x}\) = x2 – 2y2 + xy is a homogeneous differential equation.
Answer:
x2\(\frac{d y}{d x}\) = x2 – 2y2 + xy
2nd PUC Maths Model Question Paper 1 with Answers 31

:. (1) is homogeneous DiffentiaI Equation.
2nd PUC Maths Model Question Paper 1 with Answers 32
2nd PUC Maths Model Question Paper 1 with Answers 33

KSEEB Solutions

Question 35.
Find a vector perpendicular to each of the vectors \(\overrightarrow{\mathbf{a}}=2\) \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
\(\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) which has magnitude 10 units.
Answer:
Let \(\overrightarrow{\mathbf{a}}=2\) \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) and
\(\overrightarrow{\mathbf{b}}=3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
2nd PUC Maths Model Question Paper 1 with Answers 34
= î[- 2 – 15] – ĵ[-4 – 9] + k̂[10 – 3]
= î[-17] – ĵ [-13] + k̂[7]
= -17î + 13 ĵ + 7k̂
2nd PUC Maths Model Question Paper 1 with Answers 35

Question 36.
Show that the points and A (-1, 4, -3), B(3,2,-5) C (-3,8,-5) D(-3,2,1)are coplanar.
Answer:
\(\overrightarrow{\mathrm{AB}}=4 \hat{\mathrm{i}}+(-2) \hat{\mathrm{j}}+(-2) \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{BC}}=-6 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{CD}}=0 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\)
2nd PUC Maths Model Question Paper 1 with Answers 36
= 4[36 + 0] + 2 [- 36 – o] – 2[36 – 0]
= 144 – 72 – 72 = 144 – 144 = 0
∴ A, B, C & D arc coplanar.

Question 37.
Find the cartesian and vector equation of the line that passes through the points (3,-2,-5) and (3,-2,6)
Answer;
2nd PUC Maths Model Question Paper 1 with Answers 37

Question 38.
A die is thrown. If E is the event “the number appearing is a multiple of 3” and F be the event “The number appearing is even”. Then find whether E and F are independent
Answer:
s = {1,2,3,4,5,6}
E={3,6} ∴P(E) = \(\frac{2}{6}=\frac{1}{3}\)
F={2,4,6} ∴ P(F) = \(\frac{3}{6}=\frac{1}{2}\)
we have, P(EnF) = P(E)P(F)
\(\frac{1}{6}\) = \(\frac{1}{3} x \frac{1}{2}\)
\(\frac{1}{6}\) = \(\frac{1}{6}\)
∴ E and F are Independent

Part – D

Answer any sir auestinns: ( 6 x 5 = 30 )

Question 39.
If f:A → A defined by f(x) = \(\frac{4 x+3}{6 x-4}\) where A = R{ \(\frac{2}{3}\) }
show that is invertible and f -1 = f.
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 38

Question 40.
If
2nd PUC Maths Model Question Paper 1 with Answers 39 , Prove that A(B + C) = AB + AC.
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 40
2nd PUC Maths Model Question Paper 1 with Answers 41
2nd PUC Maths Model Question Paper 1 with Answers 42
2nd PUC Maths Model Question Paper 1 with Answers 43

KSEEB Solutions

Question 41.
Solve the following system of equation by
method:x + 2y + 3z = 2; 2x + 3y + z = -1 and x – y – z = -2
Answer:
Consider A x B → (1) ⇒ x = A-1 B – (2)
2nd PUC Maths Model Question Paper 1 with Answers 44
2nd PUC Maths Model Question Paper 1 with Answers 45

Question 42.
ir y = (tan-1x)2, prove that
(x2 +1)2 y2 + 2x(x2 +1)y1 =2
Answer:
y = (tan-1x)2
2nd PUC Maths Model Question Paper 1 with Answers 46

Question 43.
A particle moves along the curve 6y=x3 + 2. Find the points on the curve at which the y – coordinate is changing 8 time as fast as the x- coordinate.
Answer:
6y = x3+2 ……… (1)
2nd PUC Maths Model Question Paper 1 with Answers 47
(or) 3x2 = 48
(or) x2 = 16
∴ x = ±4
from (1): y = 11 and Y = \(\frac{31}{3}\)
Hence , the required points are (4 , 11) and (-4, \(\frac{-31}{3}\))

Question 44.
Find the integral of \(\frac{1}{\sqrt{\mathbf{a}^{2}-\mathbf{x}^{2}}}\) with respect to x, and hence evaluate \(\int \frac{d x}{\sqrt{5-4 x-x^{2}}}\)
Answer:
Let I = ∫\(\frac{1}{\sqrt{a^{2}-x^{2}}} dx\) -(1)
Put x = a sin θ (or) θ = sin -1 ( \(\frac{x}{a}\) )
Then dx = a cosθ dθ
2nd PUC Maths Model Question Paper 1 with Answers 48

Question 45.
Find the area of the of the circle x2 + y2 = a2 by the method of the integration and hence find the area of the circle x2 + y2 = 2.
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 49
2nd PUC Maths Model Question Paper 1 with Answers 50

KSEEB Solutions

Question 46.
Find the general solution of the differential equation
\(\frac{d y}{d x}\) + y.cotx = 2x + x2.cotx
\(\frac{d y}{d x}\) + y.cotx = 2x + x2 .cotx
Let P = cot x, Q = 2x + x2cot x
I.F = e∫pdx = e∫cotxdx = e∫log(sin x ) = sin x
Sol is
y (I.F) = ∫ (QxI.F) dx +C
y sinx = ∫(2x + x2cotx)sin dx + c
= ∫(2x + x2 cotx) dx +c
=2 ∫ x sin xdx + ∫x2 cosx dx + c
= 2[x(-cosx)-(1)(-sinx)] +[x2 sinx – 2x(-cosx) + 2c(-sinx)]
=2xcos x+2sin x +x2sinx
y sin x = x2 sinx + c

This is the Required General solution for the given differential Equation.

Question 47.
Derive the formula to find the shortest distance between the two skew lines \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}_{1}+\lambda \overrightarrow{\mathbf{b}}_{1}\) and
\(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}_{2}+\mu \overrightarrow{\mathbf{b}}_{2}\) in the vector form.
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 51

(1) Let \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}_{1}+\lambda \overrightarrow{\mathrm{b}}_{1}\) and \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}_{2}+\mu \overrightarrow{\mathrm{b}}_{2}\) be two skew lines.
(2) Take any point S on l1, with posotion vector \(\overrightarrow{\mathrm{a}}_{1}\) and T on l2 with the position vector \(\overrightarrow{\mathrm{a}}_{2}\)
(3) clearly \(\overrightarrow{\mathrm{PQ}}=\mathrm{d} \hat{\mathrm{n}}\) where \(\hat{n}=\frac{\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{b}}_{2}}{\left|\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}\right|}\)
where, d – shortest distance
\(\hat{n}\) – units Normal vector
(4) Let θ be angle between \(\overrightarrow{\mathrm{PQ}} and \overrightarrow{\mathrm{ST}}\)
Then, PQ = ST|COSθ| — (1)
(5) But,
2nd PUC Maths Model Question Paper 1 with Answers 52
This is the required shortest Distance between two skew lines.

Question 48.
A fair coin is tossed 8 times. Find the probability of at most 5 Heads.
Answer:
Let p = probability of getting head = \(\frac{1}{2}\)
q = Probability of getting tail = \(\frac{1}{2}\)
n = 8
2nd PUC Maths Model Question Paper 1 with Answers 53

Part – E

Answer any one question: ( 1 x 10 = 10 )

Question 49.
(a). A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and Rs7.OO per package on bolts. How many packages of each should be produced each day so as to maximize his profit if he operates his machines for at the most 12 hours a day?
Let x → nuts, y → bolts
2nd PUC Maths Model Question Paper 1 with Answers 54
Maximimum: Z = Rs. 17.50 x + Rs 7.00 y
x + 3 ≤ 1 2
3x + y ≤ 12 where, xy ≥ 0
2nd PUC Maths Model Question Paper 1 with Answers 55
The shaded portion OAED Represents feasible region and it is bounded.
The feasible solution is obtained at all these comer points.
2nd PUC Maths Model Question Paper 1 with Answers 56
∴ (z)max =73.5atx =3 andy =3.
Thus the manufactures gets maximum prom 01 KS ,j.5 on producing 3 packets of nuts and 3 packets of bolts.

(b).
2nd PUC Maths Model Question Paper 1 with Answers 58
Answer:
2nd PUC Maths Model Question Paper 1 with Answers 59
2nd PUC Maths Model Question Paper 1 with Answers 60
On Expanding along c1
= (b – a) (c – a) {1[(- b)] J
= (b – a) (c – a) (c – b)
= (a – b) (b – c) (c – a)
= RHS

KSEEB Solutions

Question 50.
(a) Prove that \(\int_{2}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) and hence evaluate \(\int_{\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{\mathrm{dx}}{1+\sqrt{\tan x}}\)
Answer:
Let t = a + b – x. Then dt = -dx
when x = a, t = b
and x = b, t=a
2nd PUC Maths Model Question Paper 1 with Answers 61
2nd PUC Maths Model Question Paper 1 with Answers 62

(b). Find all the points of discontinuity on f defined by f(x) =|x| – |x+1|
Answer:
f(x) = |x| – |x+1|
f(c)= |c | – |c+1|
Let c = O, f(0) = |0 |-|1| = 0-1= -1
Lt f(x) = It|x| – |x+1|= |c| – |c+1|
x→ c X → C
let c = 0, Lt f(x)=0 – 1 = – 1
x—+0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)=-1
⇒ f is continuous at x =0
There is no point of Discontinuity.

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