Câu hỏi mô hình toán học 2023 Lớp 12 Ban Karnataka

Students can Download 2nd PUC Maths Model Question Paper 3 with Answers, Karnataka 2nd PUC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations

Karnataka 2nd PUC Maths Model Question Paper 3 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 questions ( 10 × 1 =10 )

Question 1
Let * be a binary operation defined on the set of Rational numbers Q defined by a*b = ab + 1 prove that * is a commutative
Answer
If * be a commutative binary operation
a*b = b*a
⇒ a*b = ab + 1 = ba + 1= b*a
∴ * is commutative binary operation on *

Question 2
Find the principle value of cos-1(-1/2)
Answer

Question 3
Define a diagonal matrix
Answer
A square matrix in which except the principle diagonal elements all the elements are zero

Question 4
If A = \(\left[ \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} \right]\) then find the value of . 2A
Answer

Question 5
Find \(\frac{\mathrm{d} y}{\mathrm{d} x}\) if y = cos(1 – x)
Answer
y = cos(1 – x)
Dijf w. rl. x
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) = cox(1 – x)
= -sin(1 – x) x \(\frac{d}{d x}\) (1 – x) dx
= -sin(1 – x)x – 1 = sin(1 – x)

Question 6
Evaluate ∫(2x – 3 cos x + ex ). dx
=> ∫(2x – 3cosx +ex ). dx
= ∫ 2 x. dx – 3 ∫ cos x. dx + ∫ ex . dx
= 2 x \(\frac{x^{2}}{2}\) – 3 x sinx + e +c
= x2 – 3sin x + ex + c

Question 7
Define a Unit vector
Answer
A vector having Unit magnitude along any vector \(\vec{a}\) is Called a unit vector in the direction of \(\vec{a}\) add is denoted \(\vec{a}. \vec{a}. \) = 1

Question 8
If a line makes angle 90°, 60° and 30° with positive direction of x, y and z axis respectively. Find its direction cosines
Answer
The Direction cosines are given by
=(cos90°, cos60°, cos30°)
= (0, 1/2, √3/2 )

Question 9
In linear programming problems, define linear objectives function
Answer
The linear function z = ax + by where a, b are constants which has to be maximised or minimized is called linear objective function

Question 10
If P(A) = 0. 6 P(B) = 0. 3 and P(A ∩B ) = 0. 2 find P. A. B)
Answer
P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0. 2}{0. 3}=\frac{2}{3}\)

Part – B

II. Answer any Ten questions

Question 11
S. T. function f . N → N by f (1) = f(2) = 1 and f(x) = x -1 for every x > 2, is on to but not one-one
Answer
f . N → N by f(1) = f(2) = 1 and f(x) = x – 1
f is not one-one because
f (1) = 1 and f(2) = 1
∴ f(1) = f(2)
but
1≠2
∴ f is not one-one
for every y ∈ N then
f(x) = y – x – 1 then y = x – 1
⇒ x ∈ N
∴ y ∈ N ∋ x ∈ N
∴ f is onto

Question 12
P. T. tan -1 – cot -1 x = \(\frac{\pi}{2}\) ∀ x ∈R
Answer
Let tan -1 x = y
x = tan y = cot ( \(\pi / 2\) – y)
cot -1 x = \(\pi / 2\) – y
cot -1 x + y = \(\pi / 2\)
cot -1 x – tan -1 x = \(\pi / 2\)

Question 13
Write the simplest form of \(\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}})\) 0 < x < π
Answer

Question 14
Find the equation of a line passing through (3,1) (9,3) using determinants
Let P(x,y) be the line passing through (3,1) and (9,3)
∴ Let A(3,1) B (9,3)
∴ If P is a point on the line AB then Area of the Δle is zero

∴x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0
-2x – y (-6) + (0) = 0
-2x + 6y = 0
6y = 2x
x = 3 y
∴ Equation of line is x = 3y

Question 15
if \(\sqrt{x}+\sqrt{y}=\sqrt{10}\) S. T \frac{d y}{d x}+\sqrt{\frac{y}{x}}=0[/latex]
Answer

Question 16
Find \(\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} \mathbf{x}}\) If y = (log x)cos x
Answer
y = (log x)cos x
Take log on both side
log y – log(log x)cos x
log y – cos x. log(log x)
Diff w. r. t x, we get

Question 17
Approximate √36. 6 by using differential
Answer
y = √x Let x = 36 and Ax = 0. 6 then
Δy = \(\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{36. 6}-\sqrt{36}=\sqrt{36. 6}-6\)
√36. 6 = 6 + Δy
Now dy is approximately equal to Ay and is given by

The approximate value of √36. 6 is 6+0. 05 = 6. 05

Question 18
Integrate sinx. sin(cosx) w. r. t. x
∫sinx. sin(cosx). dx
Take cos x = t
Diff w. r. t x, we have
-sinx = \(\frac{d t}{d x}\)
-sin x. dx = dt
sin x. dx = -dt
⇒ ∫sin(cosx). sinx. dx
= ∫sin(t)λ – dt = – ∫ sin t. dt
= -x – cos t. + c =cost + c = cos(cosx) + c

Question 19
Evaluate \(\int_{0}^{1} \frac{1}{1+x^{2}} \cdot d x\)
Answer

Question 20
Find the order and degree of differential equation

Answer
order – 3 degree – 2

Question 21
Find the area of parallelogram whose adjacent sides determine by the vectors
\(\overrightarrow{\mathrm{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Answer
Area of the parallelogram whose adjacent sides are given by

Question 22
Obtain the projection of the vector \(\vec{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}\) on the vector \(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\)
Answer

Question 23
Find the equation of the plane through the inter section of planes 3x – y + 2z – 4 = 0 and x + y – z – 2 = 0 and the point (2,2,1)
Answer
Equation of any plane through the intersection of the given planes, is of the form
3x – y + 2z – 4 + λ (x + y – z – 2) = 0
(3 + λ)x + (λ – 1)y + (2 – λ) z – (4 + 2λ) = 0
By Data it is passes through (2,2,1)
⇒ (3 + λ)2 + (λ -1)2 + (2 – λ)1 – (4 + 2λ) = 0
⇒ 6 + 2λ + 2λ – 2 + 2 – λ – 4 – 2λ = 0
⇒ 2 + λ = 0
λ = – 2
Thus the required equation is
(3 – 2)x + (-2 -1 )y + (+2 – 2)1 – (4 + 2(-2) = 0
x – 3y + (0)(1) – (4 – 4) = 0
x – 3y = 0

Question 24
A die is thrown. If E is the event the number appearing is a multiple of 3 are F be the event the number appearing is even, then prove that E and F are independent events
Answer
A die is thrown the sample space are {1,2,3,4,5,6}
E is the event “the number appearing is a multiple of 3”
E = {3,6} P(E) = \(\frac{n(E)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)
F be the event the number appearing is even
F = {2, 4, 6} P(F) = \(\frac{n(F)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)
⇒ (E∩F) = {6} =P(E∩F) =  \(\frac{n(E \cap F)}{n(s)}=\frac{1}{6}\)
∴P(E)
∴ P(E) × P(F) = P(E∩F)
∴ \(\frac{1}{3} \times \frac{1}{2}=\frac{1}{6} \Rightarrow \frac{1}{6}=\frac{1}{6}\)
∴ E and F are independent Events

Part – C

III. Answer any Ten questions

Question 25
Show that the Relation R in the set Z of integers given by R= {(x y) . 2 divides (x-y)} is an Equivalence Relation
Answer
→ R – {(xy). 2 divides (x – y) }
→ R is Reflexive as 2 divides (a – a) ∀ a ∈ z
If (a b) ∈ R then 2 divides a – b
2 divids b – a Hence (ba)∈R
⇒R is Symmetric
(ab) ∈ R and (be) ∈ R
⇒ a – b and b-c are divisible by 2
Now a – c = (a – b) + (b – c)
= is even
’. a – c is divisible by 2
⇒ R is Transitive
R is Reflexive, Transitive, Symmetric
’. R is an Equivalence Relation in Z

Question 26
Prove that \(\tan ^{-1} x \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\) = \(\tan ^{-1}\left[\frac{3 x-x^{3}}{1-3 x^{2}}\right] \cdot. x. <\frac{1}{\sqrt{3}}\)
Answer
→ Take tan-1 x = θ ⇒ x = tanθ

Question 27
For any square matrix A with Real numbers. Prove that A + A1 is a symmetric and A – A1 is a skew symmetric
Answer
Let B = A + A1 then
B1 = (A + A1)1 = A1 + (A1)1
= A1 + A= A + A1 =B
∴ B1 =B ∴B is symmetric matrix
∴ A + A1 is a symmetric matrix
C = A – A1
C1 = (A – A1 )1 = A1 – (A1)1
= A1 – A = -(A – A1) = -C
C1 = -C is a skew-symmetric matrix
∴ A – A1 is a skew-symmetric matrix

Question 28
If x = a(θ + sinθ) and y = a(1 – cosθ). Prove that \(\frac{d y}{d x}=\tan \frac{\theta}{2}\)
Answer
⇒ x = a(θ + sinθ)
Diff. w. r. t θ

Question 29
Verify mean value theorem if f(x) = x2 – 4x – 3 in the interval [ab]
where a = 1 and b = 4
→ Given f(x) = x2 – 4x – 3
⇒ f1 (x) = 2x – 4 which exists ∀ x in R
⇒f is derivable for all x in particular it is derivable in
[a b] = [1 4] and hence continuous in[a b]= [ 1 4] Since f is derivable in [1 4 ] ⇒ f is derivable in (1 4)
∴ Both the condition of L. M. V. Theorem as satisfied
∃ at least one real C in (1 4) ∋

⇒ MVT is verified for f(x) in [14],

Question 30
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum
→ x + y = 60
y = 60 – x
P = xy3
P = x(60 – x)3
P = x(60 – x)3
Diffw. r. t x

= x. 3(60 – x)2 x (-1) + (60 – x)2. 1
= -3x(60 – x)2 + (60 – x)3
= (60 – x)3 – 3x(60 – x)2
=(60 – x)2 [60-x – 3x)
= (60 – x)2 (60 – 4x)
xy3 is to be maximum
\(\frac{d p}{d x}\) = 0
0 = (60-x)2 (60-4x)
60 – 4x = 0
+4x = -60
x = \(\frac{60}{4}\)
x = 15
∴ x = 15 y = 45
x + y = 60
y = 60 – x
= 60 – 15
= 45

Question 31
Evaluate ∫sin3x. cos4x. dx
∫cos4x. sin3x. dx
= ∫ \(\frac{1}{2}\) sin(4x + 3x) – sin(4x -3x)]. dx
= ∫(sin 7x – sin x) dx
= \(\frac{1}{2}\) ∫ sin 7x – ∫\(\frac{1}{2}\) sin x. dx

Question 32
Integrate x2. ex w. r. t x
answer;

Question 33
Determine the area of the region bounded by y2 = x and the lines x = 1 and x = 4 and the x-axis in the first quadrant
Answer
→ Given that y2 = x
Its a Equation of the Parabola and the Lines x = 1 and x = 4

∴ The Required area of the

Question 34
Form the differential Equation of the family of circles touching the x-axis at origin
Answer
→ Let C denote the family of circles touching x – axis at origin. Let (0,a) be the co-ordinates of the centre of any member of the family
∴ Equation of family C is. x2 + (y – a)2 = a2
x2 + y2 = 2 ay
Where a is an arbitary constant
Differentiating both side w. r. t x, we get

Substitute the value of a from solution. (2) in eqn. (1), we get

Question 35
If the two vector \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) such that \(. \overrightarrow{\mathrm{a}}. \) = 2
\(. \overrightarrow{\mathrm{b}}. \) = 3 and \(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=4\) final \(. \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}. \)
Answer

Question 36
Find a Unit Vector perpendicular to each of the vector \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\)
Where \(\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\)\(\overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
Answer

Question 37
Find the shortest distance between the lines l1and l2 whose vector equations are \(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) and \vec{r}=2 \hat{i}+\hat{j}-\hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\)
Answer

Hence the shortest distance b/w the given lines is given by

Question 38
Bag-I contains 3 red and 4 black balls while another Bag-II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag-II
Answer
→ Let E1 be the event of choosing the bag I. E2 be the event of choosing the bag II and A be the event of drawing a red ball
Then P(E1) = P(E1) = \(\frac{1}{2}\)
P(A / E1) = P (drawing a red ball from Bag I) = \(\frac{3}{7}\)
P(A / E2) = P (drawing a red ball from Bag II) = \(\frac{5}{11}\)
Now, the probability of drawing a ball from Bag – III. being given that it is red, is P(E2/A)
By using Baye’s theorem, we have

= \(\frac{35}{68}\)

Part – D

IV. Answer the following questions

Question 39
Prove that the function f. N → N defined by ‘ f(x) = x2 where y = {y. y = x2, x ∈ N} is invertible. Also find the inverse of f
Answer
f . N → N defined by f(x) = x2 then ∃ another function g . N →N by g(y) = x
f(x) = x2
⇒ y = x2
x = √y
∴ g(y) =√y
To show that f is invertible
fog(x) = x
gof(y) = y
fog(x) = f[g(x)]

gof(y) = g[f/(y)]
= g[y2] = y ∈N
∴ gof{y) = y ∈ N
∴ f is invertible
∴  inverse of f exist
∴ f-1 . N → N f-1(y) = x
f-1(y) = √y

Question 40
If A = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{matrix} \right]\) then show that A3 – 23 A – 40 I = 0
Answer

Question 41
Solve the system of linear equation using matrix method,
x – y + 2z = 7
3x + 4y – 5z = -5
2x – y + 3z = 12
Answer
The given system of equation can be expressed as AX = B, where

∴ A is non-singular matrix so A-1 exists
A11 = 7, A12 = -19, A13 =-11, A21 =1, A22= -1, A23 = -1, A31 = -3, A32 = 11, A33 = 7

on equating, we get x = 2, y = 1, z = 3

Question 42
Given y = 3 cos (logx) + 4 sin (logx)
x2y2 + xy1 + y = 0
Answer
y = 3 cos (logx) + 4 sin (logx)
Diff w. r. t x

Question 43
The length x of a rectangle is decreasing at the rate 5 cm/min and width y is increasing at the rate of 4cm/ min when x = 8cm and y = 6cm. Find the rate of change of (i) The perimeter (ii) The rate of the rectangle
Answer
Given \(\frac{d x}{d t}\) = -5 cm /min and \(\frac{d y}{d t}\) = 4 cm /min
a) perimeter = 2(x+ 5) ⇒ P = 2 (x + 4)

b) A Area of. rectangle = xy
⇒ A = x. y

Question 44
Find the integral of \(\frac{1}{\sqrt{x^{2}+a^{2}}^{2}}\) with respect to x and hence. Evaluate \(\int \frac{1}{\sqrt{x^{2}+7}} \cdot d x\)
Answer

Question 45
Find the area of the region enclosed b/w the two Θles. x2 + y2 = 4 and (x – 2)2 + y2 = 4
Answer
Equations of the given circles are
x2 + y2 = 4 ……………. (1)
(x – 2)2 + y2 = 4 ……………………. (2)
Equation (1) is a circle with centre O at the origin and radius 2. Eqn. (2) is a circle with centre C (2,0) and radius 2. Solving Eqn. (1) and (2), we have
(x – 2 )2 + y2 = x2 + y2
x2 – 4x + 4 + y2 = x2 + y2
+ 4x = -4 ⇒ x = 1
x = 1 which given y = ± √3
Thus the points of intersection of the given circles are A(1,√3) and A1( 1,-√3} as shown in the fig
Required area of the enclosed region OACA1O between circle s =2 [area of the region ODCAO]
=2 [area of the region ODAO + area of the region on DCAD]

Question 46
Solve \((x \log x) \frac{d y}{d x}+y=\frac{2}{x}(\log x)\)
Answer

Question 47
Derive equation of a line in a space through a given point and parallel to a vector both in the vector and cartesian form
Answer
Let \(\vec{a}\) be the position vector of the given point A with respect to the origion O of the rectangular co-ordinate system. Let l be the line which passes through the point A and is parallel to a given vector \(\vec{b}\) Let \(\vec{r}\) be the position vector of an arbitary point P on the line

Then \(\vec{AP}\) is parallel to the vector \(\vec{b}\) i. e. \(\overrightarrow {A P}=\lambda \vec{b}\) where λ is some real number
But \(\overrightarrow{A P}=\overrightarrow{O P}-\overrightarrow{O A}\)
\(\lambda \vec{b}=\vec{r}-\vec{a}\)
Conversely for each value of the parameter λ, this equation gives the position vector of a point P on the line
Hence the vector equantion of the line is given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\)
Derivation of cartesian form from vector form
Let the co-ordinates of the given point A be (x, y, z) and the direction ratios of the line a b c consider the co-ordinates of any point P be (x,y,z) then

Substituting these values in (1) and equating the co-efficients i, \(\hat{i}\) and \(\hat{r}\) we get
x = x1 + λa ; y = y1 + λb ; z = z1 + λc ,
These are parametric equation of the line. Eliminating the Parameter λ from (2) we get

This is the cartesian equating of the line

Question 48
If a fair coins is tossed 10 times. Find the probability of
(i) Exactly six heads and
(ii) at least six heads
Answer
The repeated toss of a coin are Bernoullis trial s. Let X denote the number of heads in the experiment of 10 trials
Clearly, X has the binomial distribution with
n = 10 and P = \(\frac{1}{2}\)
∴ P(X = x) = nCxqn-x – px. x = 1,2,3 ………………. n
n = 10 and P = \(\frac{1}{2}\) q = 1 – p = \(\frac{1}{2}\)

Part – E

V. Answer any One question

Question 49
a) Prove that \(\int_{a}^{b} f(x) \cdot d x=\int_{a}^{b} f(a+b-x) d x\) and hence. Evaluate \(\int_{\pi / 3}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x\)
Answer

Let t = a + b – x then dt – dx
When x = a and x = b
t = b t = a


(b) Determine the value of k, if

is continuous x = \(\pi / 2\)
Answer
If f is continuous at x = \(\pi / 2\)
⇒ \(\lim _{x \rightarrow \pi / 2} f(x)=f(\pi / 2)\)
Given at x = \(\pi / 2\)
f( \(\pi / 2\) ) = 3

Question 50
a) Minimize and maximize
Z = 5x + 10y
subject to the constraints
x + 2y ≤ 120
x + y ≥ 60,
x – 2y ≥ and x ≥ 0, y ≥ 0 by graphical method
Answer
Given . Z = 5x + 10y ……………. (1)
subject to constraints x + 2y ≤ 120 ……………. (2)
x + y > 60 ……………. (3)
x – 2y ≥ 0 ……………. (4)
x ≥ 0, y ≥ 0 ……………. (5)
First we locate the region represented by (2), (3), (4) & (5) for this consides the lines ;
x + 2y = 120 which passes through A(120, 0) & B(0, 60)
x + y = 60 which passes through C (60, 0) & B (0, 60) and x – 2y = 0 which passes through 0 (0, 0) & P (40, 20)
Now that P(40, 20) lies on BC also. Also x – 2y = 0&x + 2y= 120 meet in Q (60, 30)
The feasible region is shown shaded in the figure (indicated as feasible region) note that (0,0) does nopt lie in this region as it doesnot satisfy (3) The point C(60,0) lie this region as it saties fies all the constraints (2) (3) (4) and (5)

The corner points of the feasible region which are to be examined for optimum solution are C(60,0) A( 120,0) Q(60, 30) and P(40,20)
At C(60, 0), Z = 5 × 60 + 10 × 0 = 300
At A(120, 0), Z = 5 × 120 + 10 × 0 = 600
At Q(60, 30), Z = 5 × 60 + 10 × 30 = 600
At P(40, 20), Z = 5 × 40 + 10 × 20 = 400
Hence Z is minimum at C(60,0) and min Z = 300 and max Z = 600 at A and also at Q ⇒ max Z will be attained at every point of segment AQ
b) Find the value of k, if
f(x) = \(\frac{1-\cos 2 x}{1-\cos x}\) , x≠ 0
f(x) = k ,x= 0
is continuous at x = 0
Answer