2nd PUC Maths Model Question Paper 3 with Answers

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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:
2nd PUC Maths Model Question Paper 3 with Answers 1

Question 3.
Define a diagonal matrix.
Answer:
A square matrix in which except the principle diagonal elements all the elements are zero.
2nd PUC Maths Model Question Paper 3 with Answers 3

KSEEB Solutions

Question 4.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} \right]\) then find the value of |2A|.
Answer:
2nd PUC Maths Model Question Paper 3 with Answers 2

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

KSEEB Solutions

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.

KSEEB Solutions

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:
2nd PUC Maths Model Question Paper 3 with Answers 4

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
2nd PUC Maths Model Question Paper 3 with Answers 5
∴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:
2nd PUC Maths Model Question Paper 3 with Answers 6
2nd PUC Maths Model Question Paper 3 with Answers 7

KSEEB Solutions

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
2nd PUC Maths Model Question Paper 3 with Answers 8

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.
2nd PUC Maths Model Question Paper 3 with Answers 9
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

KSEEB Solutions

Question 19.
Evaluate \(\int_{0}^{1} \frac{1}{1+x^{2}} \cdot d x\)
Answer:
2nd PUC Maths Model Question Paper 3 with Answers 10

Question 20.
Find the order and degree of differential equation.
2nd PUC Maths Model Question Paper 3 with Answers 11
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.
2nd PUC Maths Model Question Paper 3 with Answers 12

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:
2nd PUC Maths Model Question Paper 3 with Answers 13

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

KSEEB Solutions

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.

KSEEB Solutions

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
2nd PUC Maths Model Question Paper 3 with Answers 14
2nd PUC Maths Model Question Paper 3 with Answers 15

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
2nd PUC Maths Model Question Paper 3 with Answers 16

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) ∋
2nd PUC Maths Model Question Paper 3 with Answers 17
⇒ 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.
2nd PUC Maths Model Question Paper 3 with Answers 18
= 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

KSEEB Solutions

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
2nd PUC Maths Model Question Paper 3 with Answers 19

Question 32.
Integrate x2.ex w.r.t x.
answer;
2nd PUC Maths Model Question Paper 3 with Answers 20

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

2nd PUC Maths Model Question Paper 3 with Answers 21
∴ The Required area of the
2nd PUC Maths Model Question Paper 3 with Answers 23

KSEEB Solutions

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
2nd PUC Maths Model Question Paper 3 with Answers 24
Substitute the value of a from solution. (2) in eqn. (1), we get.
2nd PUC Maths Model Question Paper 3 with Answers 25

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:
2nd PUC Maths Model Question Paper 3 with Answers 26

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:
2nd PUC Maths Model Question Paper 3 with Answers 27

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:
2nd PUC Maths Model Question Paper 3 with Answers 28
Hence the shortest distance b/w the given lines is given by.
2nd PUC Maths Model Question Paper 3 with Answers 29

KSEEB Solutions

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
2nd PUC Maths Model Question Paper 3 with Answers 30
= \(\frac{35}{68}\)

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)]
2nd PUC Maths Model Question Paper 3 with Answers 31
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:
2nd PUC Maths Model Question Paper 3 with Answers 32
2nd PUC Maths Model Question Paper 3 with Answers 33

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
2nd PUC Maths Model Question Paper 3 with Answers 34
∴ 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.
2nd PUC Maths Model Question Paper 3 with Answers 35
on equating, we get x = 2, y = 1, z = 3

KSEEB Solutions

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
2nd PUC Maths Model Question Paper 3 with Answers 36
2nd PUC Maths Model Question Paper 3 with Answers 37

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)
2nd PUC Maths Model Question Paper 3 with Answers 38
b) A Area of.rectangle = xy
⇒ A = x.y
2nd PUC Maths Model Question Paper 3 with Answers 39

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:
2nd PUC Maths Model Question Paper 3 with Answers 40
2nd PUC Maths Model Question Paper 3 with Answers 41
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]
2nd PUC Maths Model Question Paper 3 with Answers 42
2nd PUC Maths Model Question Paper 3 with Answers 43
2nd PUC Maths Model Question Paper 3 with Answers 44

KSEEB Solutions

Question 46.
Solve \((x \log x) \frac{d y}{d x}+y=\frac{2}{x}(\log x)\)
Answer:
2nd PUC Maths Model Question Paper 3 with Answers 45

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
2nd PUC Maths Model Question Paper 3 with Answers 46
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
2nd PUC Maths Model Question Paper 3 with Answers 47
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 = \(\frac{1}{2}\)
2nd PUC Maths Model Question Paper 3 with Answers 48

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:
2nd PUC Maths Model Question Paper 3 with Answers 49
Let t = a + b – x then dt – dx
When x = a and x = b
t = b t = a
2nd PUC Maths Model Question Paper 3 with Answers 50
2nd PUC Maths Model Question Paper 3 with Answers 51
(b) Determine the value of k, if
2nd PUC Maths Model Question Paper 3 with Answers 52
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
2nd PUC Maths Model Question Paper 3 with Answers 53

KSEEB Solutions

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).
2nd PUC Maths Model Question Paper 3 with Answers 54
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:
2nd PUC Maths Model Question Paper 3 with Answers 55

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