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## Karnataka 2nd PUC Maths Model Question Paper 2 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.

A Relation R on A = {1,2,3} defined by R = {(1,1) (1,2) (3,3)} is not symmetric, Why?

Answer:

The Relation R is defined by R = {.(1,1) (1,2) (3,3)} for Symmetric Relation if (1,2)∈R then (2,1) ∈R but (2,1) ∉ R?

∴ R is not Symmetric Relation.

Question 2.

Write the domain of f(x) = cos^{-1} x.

Answer:

The domain of cos^{-1}x is [-1,1]

Question 3.

Define a scalar matrix.

Answer:

A diagonal matrix in which all the Principle diagonal Elements are equal is called scalar matrix.

Question 4.

If A = \(\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right] \) find |2A|.

Answer:

Question 5.

If y = log (sin x) find \(\frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}\)

Answer:

y = log(sin x)

Differentiae w.r.t.x, we get

\(\frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}\) = \(\frac{1}{\sin x}\) . cosx = cotx

\(\frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}\) = cot x

Question 6.

Evaluate ∫( sin x + cos x) .dx.

Answer:

∫(sinx + cosx).dx

= ∫ sin x.dx + ∫ cos x.dx.

= -cosx + sin x + c

= sinx – cosx + c

Question 7.

Find the direction cosines of the vector \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)

Answer:

\(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)

Question 8.

Find the equation of the plane with intercept 4 on z-axis and parallel to xoy plane.

Answer:

∴ The Required Equation of the plane is z =4.

Question9.

Define feasible region.

Answer:

The Common Region determined by all the cons traints including the noh-negative constraints (x ≥ 0, y ≥ 0) of a linear programming problem is called the feasible region.

Question 10.

If P(A) = \(\frac{4}{5}\) and P(B/A) = \(\frac{2}{5}\) find

P(A∩B)

Answer:

P(A /B) = \(\frac{P(A \cap B)}{P(A)}\)

∴ P(A∩B) = P(B/A).P(A)

= \(\frac{2}{5} \cdot \frac{4}{5}=\frac{8}{25}\)

∴ P(A∩B) = \(\frac{8}{25}\)

Part – B

II. Answer any TEN questions : ( 10 × 2 = 20 )

Question 11.

Verify whether the operation * defined on Q by a*b = \(\frac{\mathrm{ab}}{2}\) is associative or not.

Answer:

* is defined on Q by a*b = \(\frac{\mathrm{ab}}{2}\) for associative we have to prove that

a* (b* c) = (a* b)* c

∴ a*(b*c) = a * \(\frac{\mathrm{ab}}{2}\) b*c = \(\frac{\mathrm{bc}}{2}\)

= \(\frac{\mathrm{abc}}{4}\) ……………….. (1)

(a*b)*c = \(\frac{\mathrm{ab}}{2}\) *c

= \(\frac{\mathrm{abc}}{4}\) ……………….. (2)

∴ from (1) and (2)

∴ * is Associative

∴ * Satisfies the associative property.

Question 12.

Write the simplest form of

Answer:

Question 13.

Evaluate

Answer:

Question 14.

Find the equation of line joining (1,2) and (3,6) using determinants.

Answer:

Let P (x,y) be any point on AB. Then Area of triangle ABP is is “O”

x(2 – 6) – y(1 – 3) + 1(6 – 6) = 0

-4x + 2y = 0

2x – y = 0

Question 15.

If y + siny = cosx find \(\frac{d y}{d x}\)

Answer:

y + siny = cos x

Differentiate w.r.t. x

Question 16.

If y = x^{x} find \(\frac{d y}{d x}\).

Answer:

y = x^{x}

Take logarithm on Both side, we get

logy = logx^{x} ⇒ logy=xlogx

Differentiate wr.t. x

\(\frac{1}{y} \cdot \frac{d y}{d x}\) = x x \(\frac{1}{x}\) + log x

\(\frac{d y}{d x}\) = y[1 + logx]

\(\frac{d y}{d x}\) = x^{x}[1 + logx]

Question 17.

Find the approximate charge in the volume v of a cube of side x metres caused by increasing the sind by 2%.

Answer:

w.r.t volume of a cube = v = x^{3}.

⇒ \(\frac{d v}{d x}\) = 3x^{2} dx

Δ_{x} = 0.02 x

∴ dv = ( \(\frac{d v}{d x}\) ) Δ_{x} = (3x^{2}) Δ_{x} = 3x^{2} x 0.02x

= 0.06 x^{3}m^{3}

Question 18.

Evaluate \(\int \frac{\sin ^{2} x}{1+\cos x} d x\)

Answer:

Question 19.

Evaluate \(\int_{1}^{e} \frac{1}{x} \cdot d x\)

Answer:

Question 20.

Find the order and degree of the differential equation.

Answer:

Order -2, Degree – 1

Question 21.

If \(\overrightarrow{\mathrm{a}}\) is a Unit Vector such that

( \(\vec{x}-\vec{a})(\vec{x}+\vec{a}\)) = 8 find | \(\overrightarrow{\mathbf{x}}\) )

Answer:

Question 22.

Find the area of a parallelogram whose adjacent sides are given by the vectors

\(\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \cdot \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)

Answer:

The area of a parallelogram with \(\vec{a}\) and

\(\vec{b}\) as its adjacent sides is given by | \(\vec{a} \times \vec{b}\) |

Question 23.

Find the angle between the pair of lines.

\(\vec{r}\) = 3î + 2 ĵ – 4k̂ + λ (î + 2ĵ + 2k̂) and

\(\vec{r}\) = 5î + 2 ĵ + λ (3î + 2 ĵ + 6k̂)

Answer:

Here

\(\vec{b}_{1}\) = î + 2 ĵ + 2k̂ and \(\vec{b}_{1}\) = 3î + 2 ĵ + 6k̂

The angle θ between the two lines is given

Question 24.

Find the probability ditribution of number of heads in two tosses of a coin.

Answer:

When a coin is tossed twice the sample space is S = {HH,HT, TH,TT}

Let x denote the number of heads in any out come in s.

∴x(HH) = 2

x(HT) = 1

x(TH) = 1

x(TT) = 0

P(x = 0) = P(TT) = \(\frac{1}{4}\)

P(x = 1) P(HT,TH) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

P(x = 2) P(HT,TH) = \(\frac{1}{4}\)

Probability distribution of x is

Part – C

III. Answer any TEN questions : ( 10 × 3 = 30 )

Question 25.

Find gof and fog if 1:R → R and g: R → R are given by f(x) = cosx and g(x) = 3x^{2} show that gof ≠ fog.

Answer:

f(x) = cosx

g(x)=3x^{2}

gof(x) = g[f(x)]

= g[cosx]

= 3cos^{2}x

fog(x) = f[g(x)]

= f[3x^{2}]

= cos3x^{2}

fog ≠ gof

Question 26.

P.T. 3cos^{-1}x=cos^{-1}(4x^{3} – 3x) x ∈ \(\frac{1}{2}\) , 1

Answer:

cos^{-1} x = A Now \(\frac{1}{2}\) ≤ x ≤ 1

cosA = x

cos3A = 4cos^{3} A – 3cosA ⇒\(\frac{1}{2}\) ≤ A < 1

cos3A=4x^{3} – 3x

3A = cos ^{-1}(4x^{3} – 3x) ⇒ 0 ≤ A ≤ \(\frac{\pi}{3}\)

3cos^{-1}x = cos^{-1} [4x^{3} – 3x] ⇒ 0 ≤ 3A ≤ π

Question 27.

By using Elementary transformations

find the iii,e of the matrix

A = missing

In order to use elementary row operation

A = IA

Question 28.

Find \(\frac{d y}{d x}\) if x = a(cosθ + θsinθ) and dx

y = a(sinθ – θcosθ)

Answer:

x = a(cosθ + θsinθ)

Diff w.r.t θ we have

\(\frac{d x}{d \theta}\) = a(-sinθ + θcosθ + sinθ )

\(\frac{d x}{d \theta}\) = a[θcosθ]

y = a(sinθ – θcosθ)

w.r.t θ

\(\frac{d y}{d \theta}\)

= a[cosθ – θ x sin θ – cosθ]

\(\frac{d y}{d \theta}\) = a[sinθ]

∴ \(\frac{d y}{d x}\) = tan θ

Question 29.

Verify Rolle’s theorem for the function y = x^{2} + 2, x ∈[-2,2]

Answer:

The function y = x^{2} +2 is continuous in [-2,2], and differentiable in (-2,2)

Also f (-2) ⇒ f (x) = x^{2} + 2

f(- 2) = (-2)^{2} + 2

= 4 + 2 = 6

f(2) = 2^{2} + 2 = 6

∴ f(-2) = f(2) = 6

∴ The value of f(x) at -2 and 2 coincide.

∴ Rolle’s Theorem states that there is a point

c ∈ (-2,2) where f'(c) = 0

Since f'(x) = 2x, we have

f’ (c) = 2c

2c = 0

c = 0

∴ we have c = 0 ∈ (-2,-2)

Question 30.

Find the interval in which the function f given by f (x) = x^{2} – 4x + 6 is

(i) Strictly increasing.

(ii) Strictly decreasing.

Answer:

f(x) = x^{2} – 4x + 6

Differentiate w.r.t x

f'(x) = 2x – 4

∴ f'(x) = 0 gives

0 = 2x – 4

2x = 4

x = 2

∴ The point x = 2 divides the real line into two disjoint intervas namely, (-∞,2) and (2,∞).

The interval (-∞, 2) f'(x) = 2x – 4 < 0.

∴ f is strictly decreasing in (-∞, 2). Also in the interval (2,∞) f'(x)>0 and so the function f is strictly increasing in (2,∞).

Question 31.

Find \(\int \frac{\left(x^{2}+1\right) \cdot e^{x}}{(x+1)^{2}} \cdot d x\)

Answer:

Question 32.

Evaluate ∫tan ^{-1}x.dx.

Answer:

∫tan ^{-1}x.dx.

= ∫xtan ^{-1}x. ∫tan ^{-1}dx + c

Apply Rule of integration by parts by taking tan^{-1 }x as the first function.

Question 33.

Find the area of the region bounded by the curve y = x^{2} and the line y = 4.

Answer:

y = x^{2} represents the parabola symmetrical about y – axis

∴ The Area of the region AOBA is given by 2 \(\int_{0}^{4} x \cdot d y=2\) [Area of the region BONB bounded by curve, y-axis and the lines y=0 and y=4]

Question 34.

Form the differential equation representing the family of curves y=mx where m is arbitrary constant.

Answer:

The given family of curves represents the equation of straight lines.

∴ y = mx ……………………. (1)

Differentiate w.r.t x, we get dy

\(\frac{d y}{d x}\) = m

∴ substitute m = \(\frac{d y}{d x}\), we have

y = mx

⇒ y = \(\frac{d y}{d x}\) x

\(\frac{d y}{d x}\) = \(\frac{y}{x}\)

\(\frac{d y}{d x} – \frac{y}{x}\) = 0

which is the required differential equation.

Question 35.

Prove that

Answer:

Question 36.

If \(\vec{a}=2 \vec{i}+2 \hat{j}+3 \hat{k} \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} and [latex]\overrightarrow{\mathbf{c}}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}\) are such that \(\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{b}}\) is perpendicular to \(\overrightarrow{\mathrm{c}}\). Then find the value of λ.

Answer:

Question 37.

Find the distance of a point (2, 5, -3) from the plane \(\vec{r} \cdot(6 \hat{i}-3 \hat{j}+2 \hat{k})=4\)

Answer:

\(\vec{a}=2 \hat{i}+5 \hat{j}-3 \hat{k}\)

\(\vec{N}=6 \hat{i}-3 \hat{j}+2 \hat{k}\)

d = 4

distance of a point from a plane is

∴ The distance of the point (2, 5 – 3) from the

given plane is

Question 38.

A die is tossed thrice. Find the probability of getting an odd number at least once.

Answer:

When a dice is tossed thrice, the sample space, is S = {(x,y,z):x,y z e (1,2,3,4,5,6}}

which contains 6 x 6 x 6 = 216 Equally likely sample points.

Let E : “an odd number at least once”

E^{1} : not an odd number any time

E^{1} : “An even number on all throws”

E^{1} = {(x,y,z):x,y,ze{2,4,6}j

e^{1} contains 3x3x3=27 equally likely simple events.

Required probability = P(E) = 1 – P(E).

= \(1-\frac{27}{216}=1-\frac{1}{8}=\frac{7}{8}\)

Part – D

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

Question 39.

Prove the function f : N → Y defined by f(x) = 4x + 3, where Y = {y : Y = 4x + 3, x∈N} is invertible. Also write the inverse of f(x).

Answer:

f:N→y defined by

f(x) = 4x + 3

and Assume a an inverse f^{-1}:y → N. is defined by f^{-12}y) = x.

∴ wk.t

f(x) = 4x + 3

y = 4x + 3

y – 3 = 4x

x = \(\frac{y-3}{4}\)

and wk 7

f^{-1}(y) = x

f^{-1}(y) = \(\frac{y-3}{4}\) f^{-1} = g

or

g(y) = \(\frac{y-3}{4}\)

∴ To S.T. it exist inverse.

To prove it is inversible.

We have to prove that

fog(x) = x

gof(y) = y

To find the inverse of a function :

f^{-1}:y → N. by f^{-1}(y) = x

f^{-1}(y) = \(\frac{y-3}{4}\)

Question 40.

If A = \(\left[ \begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix} \right]\) B = \(\left[ \begin{matrix} 2 & 0 \\ 1 & 3 \end{matrix} \right]\) and C = \(\left[ \begin{matrix} 1 & 1 \\ 2 & 3 \end{matrix} \right]\) Calculate AC BC and (A+B).C. Also verify that (A+B). C = AC+BC.

Answer:

From (1) and (2)

(A+B). C = AC+BC.

Question 41.

Solve the following system of equations by matrix method. x-y + z = 4, 2x + y – 3y = 0 andx + y + z =2.

Answer:

The given system can be written as AX = B

The given system is consistent and has a unique Solution.

Question 42.

If y = 3e^{2x}+2e^{3x} P.T. \(\frac{d^{2} y}{d x^{2}}\) – \(\frac{d y}{d x}\) + 6y = 0

y = 3e^{2x} + 2e^{3x}

Differentiate w.r.t x.

\(\frac{d y}{d x}\) = 6e^{2x} + 6e^{3x}

Again Diff w.r.t x

\(\frac{d^{2} y}{d x^{2}}\) = 12e^{2x} + 18e^{3x}

= 12e^{2x} +18^{3x} – 5 ((5e^{2x} + 6e^{3x}) + 6(3e^{2x }+ 2e^{3x})

= 12e^{2x} + 18^{3x} – 30e^{2x} – 30e^{3x} +18e^{2x }+ 12e^{3x}

= 30e^{2x} + 30e^{3x} – 30e^{2x} – 30e^{3x}

Question 43.

A ladder 24 ft long leans against a vertical wall. The lower end is moving away at the rate of 3ft/sec. Find the rate at which the top of the ladder is moving down wars. If its foot is 8ft from the wall.

Answer:

The foot of the ladder is at a distance x from the wall and the top is at a height y at instant of time t. then

(24)^{2} = x^{2} + y^{2}

Diff w.r.t time, we get

given \(\frac{d x}{d t}\) = 3ft / sec. dt

and x = 8 ft.

y^{2} = (24)^{2} – x^{2}

= (24)^{2} – (8)^{2}

= 576 – 64

y^{2} = 512

y = √512

0 = 2 × 8 × 3 + 2 × √5I2 \(\frac{d y}{d t}\)

= 48 + 2√512

∴ The ladder is moving downwards at rate of \(\frac{24}{\sqrt{512}}\) ft/ sec. or 1.06100 ft / sec.

Question 44.

Find \(\int \frac{d x}{x^{2}-a^{2}}\) and hence evaluate \(\int \frac{d x}{3 x^{2}+13 x-10}\)

Answer:

Question 45.

Find the area of the region bounded by the two parabolas y = x^{2} and y^{2} = x.

Answer:

The given two curves are parabola y = x^{2} and y^{2} = x

The point of intersection of these two parabolas are 0 (0,0) and A (1,1) as shown in the fig.

We can set

y^{2} =x or y = √x – f (x)

y – x^{2} = g(x) where

f(x) ≥ g(x) in [0,1]

∴ The required area of the Shaded Region

∴The Required area of the Region is = \(\frac{1}{3}\) units.

Question 46.

Find the general solution of the differential equation. e/tany. dx + (1 -e^{x}). sec^{2}y.dy = 0

Answer:

e^{x}. tany dx + (1-e^{x}). sec^{2} y dy = 0 (1)

Separating the variables and integrating, we get

This can be done by divide throughout (l-e^{x})t any Eqn. (1) becomes and integrate

Which is the required solution.

Question 47.

Derive the equation of the plane in normal form both in the cartesian and vector form.

Answer:

Consider a plane whose perpendicular distance from the origin is d. where d ≠ 0.

If \(\overrightarrow{\mathrm{ON}}=d \hat{\mathrm{n}}\)

Let p be any point on the plane Therefore NP.OP=0(l)

lx + my + nz = d

This is the Required Equation of a plane in normal form.

To find P and Q :

x^{2 }= 4y and x = 4y – 2

(4y – 2)^{2} = 4y

⇒ 16y^{2 }+ 4 – 16y = 4y

⇒ 16y^{2} – 20y + 4 =0

⇒ (y – 1) (16y – 4) =0

⇒ y =1 (or) y = 1/4

Ify=1; x = 2 ⇒ 2 (2,1)

and y = 1/4 x = -1 (-1, 1/4)

Thus , the points of Intersection are

(2,1) 1 (-1.,1/4)

∴ Area of the Region O B Q C = Area of BCQ+ Area of QCO.

Question 48.

A person buys a lottery ticket in 50 lotteries in each of which his chance of winning a prize is \(\frac{1}{100}\) what is the probability that he will win a prize exactly once?

Answer:

Given it is a lax of Bemoullian trials with n = 50 and P = \(\frac{1}{100}[latex]

Hence q = 1 – P = 1 – [latex]\frac{1}{100}\) = \(\frac{99}{100}\)

The Probability that he will win a prize exactly once is

Part – E

V. Answer any ONE questions : ( 1 × 10 = 10 )

Question 49.

a) Minimize and maximize Z = x + 2y subject to the constraints,

x + 2y ≥ 100

2x – y ≤ 0

2x + y ≤ 200

x, y ≥ 0 by graphical method.

Our problem is to minimize and maximize

Z = x + 2y …………..(1)

Subject to constraints are

x + 2y ≥ 100 …………….(2)

2 x – y ≤ 0 ……………..(3)

2x + y ≤ 200…………….(4)

x> 0, x>0…………….(5)

∴ x + 2y- 100 which passes through the points A( 100,0) and B(0,50)

2x – y = 0 which passes through the points (0,0) and (20,4 0)

2x + y = 200 which passes through the points A(100,0)&C (0,200)

∴ The comer points of feasible region are 5(0,50) P(20,40) 0(50,100) and C(0,200).

At B (0,50) ⇒ Z = 0 + 2 × 50 = 100

At P(20,40) ⇒ Z = 20 + 2 × 40 – 100

At Q(5O,1OO) ⇒ Z = 50 + 2 × 100 = 250

At C(0,200) Z = 0 + 2 × 200 = 400

Hence max Z = 400 at (0, 200) and

Min Z = 100 at B and also at P.

(b)

R1 R1 – (R2 + R3)

= 0 [(c + a) (a + b) – bc) + 2c (b (a + b) – bc) – 2b (bc – (c + a)]

= 0 + 2c [ab + b^{2} -bc] – 2b [bc – c^{2} -ac]

= 2abc + 2cb^{2} – 2bc^{2} – 2b^{2}c + 2bc^{2} + 2abc

= 4 abc = RHS

Question 50.

a) Prove that \(\int_{0}^{a} f(x) \cdot d x=\int_{0}^{a} f(a-x) \cdot d x\) and hence evaluate \(\int_{0}^{\pi / 4} \log (1+\tan x) \cdot d x\)

Answer:

put t = a- x

x = a – t

than dt = -dx

when x = 0 x = a

t = a t = 0

then

(b) Find the value of k, if

Answer: