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Karnataka 2nd PUC Maths Model Question Paper 3 with Answers
Time. 3 Hrs 15 Min
Max. Marks. 100
Instructions
- The question paper has five parts namely A, B, C, D and E. Answer all the parts
- 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. 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