Maths Solved 2008

8/6/2019 Maths Solved 2008

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Mon, August 9, 2010 3:10:37 PMtest paperFrom: Dinesh KUMAR Yadav

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Math 2008 Set 1 Close

Subjective Test

(i) All questions are compulsory.(ii) The question paperconsists of 29 questions divided into three sections A, B andC. Section A comprises of 10 questions of one mark each, Section B comprises of 12questions offour marks each, and Section C comprises of 7 questions of six marks each.(iii) All questions in section A are to be answered in one word, one sentence or as perthe exact requirements of the question.(iv) There is no overall choice. However, internal choice has been provided in 4questions of four marks each and 2 questions of six marks each. You have to attemptonly one of the alternatives in all such questions.

(v) Use of calculators is not permitted.

Question 1 ( 1.0 marks)

Iff(x) =x + 7 andg(x) =x 7, find (fog) (7)

Solution:


Evaluate: sin

Solution:

We know that the domain and range of the principal value branch of function sin1 isdefined as:
http://cbse.meritnation.com/sample-papers-model-tests/predefinefulltest/Math/nD7PNd8xqbN2AScdDt4G5A!!http://cbse.meritnation.com/sample-papers-model-tests/predefinefulltest/Math/nD7PNd8xqbN2AScdDt4G5A!!


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Find the value ofx andy if:

Solution:

On comparing the respective elements of the matrices on both sides, we get:

Thus, the value of bothx andy is 3.


Evaluate:

Solution:


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Find the co-factor ofa12 in the following:

Solution:


Evaluate:

Solution:

Let

Differentiating both sides with respect tox:


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Evaluate:

Solution:

Whenx = 0,

Whenx = 1,


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For what value of are the vectors perpendicular to

each other?Solution:

If and are perpendicular to each other, then must be 0.

Thus, the value of is


(i) Is thebinary operation* defined on set N, given by for all ,commutative?

(ii) Is the above binary operation* associative?

Solution:


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(i) For all

Now,

Thus, the binary operation* is commutative.

(ii) Let

Thus, the binary operation * is not associative.


Prove the following:

Solution:

LHS =


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Thus, the given result is proved.


Let .Express A as the sum of two matrices such that one issymmetric and the other is skew symmetric.

OR


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If , verify that A2 4A 5I = 0

Solution:

Now, A can be written as,


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Thus, is a symmetric matrix.

Thus, is a skew symmetric matrix.

OR


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For what value ofkis the following function continuous atx = 2?

Solution:


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The given functionf(x) will be continuous atx = 2, if:

Left hand limit off(x) atx = 2, is

Right hand limit off(x) atx = 2,

Thus, fork= 5, the given function is continuous atx = 2.


Differentiate the following with respect tox:

Solution:

Letx = cos 2 (1)


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Find the equation of tangent to the curvex = sin 3t,y = cos 2t, at t=

Solution:

Therefore, the tangent at the point is:


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Solution:

Whenx = , t= cos = 1

When x = 0, t= cos 0 = 1


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Solve the following differential equation:

(x2 y2) dx + 2xy dy = 0

Given thaty = 1 whenx = 1

OR


, ify = 1 whenx = 1Solution:

It is a homogeneous differential equation.

Substituting (2) and (3) in (1), we get:


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Integrating both sides, we get:

It is given that whenx = 1,y = 1

(1)2+ (1)2 = C (1)

C = 2

Thus, the required equation isy2 +x2 = 2x.

OR

It is a homogeneous differential equation.


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Substituting (2) and (3) in (1), we get:


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Ify = 1 thenx = 1

Therefore, from (4) and (5) we get:


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Solution:

It is a linear differential equation of the first order.

Comparing it with , we get:

P= sec2x and

Integration factor=

The solution of this linear differential equation can be given as:


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Adding (5) and (6), we get:

Thus, the vector is .

OR


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Thus, the given result is proved.


Find the shortest distance between the following lines:

and

OR

Find the point on the line at a distance from the point (1, 2, 3).

Solution:

The vector form of this equation is:

The vector form of this equation is:


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Therefore,

Now, the shortest distance between these two lines is given by:

OR

Let

Therefore, the general point on the line is given as:

The distance of this point from point (1, 2, 3) =


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When :

Thus, when the point is and when the point is

(2, 1, 3).


A pair of dice is thrown 4 times. If getting a doublet is considered a success, find theprobability distribution of the number of successes.

Solution:

Letx denote the number of doublets.

The possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).

Now,Xcan take the value 0, 1, 2, 3 or 4.

Probability of getting a doublet

Probability of not getting a doublet

P(X= 0) =P(no doublet)

P(X= 1)


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Thus, the probability distribution is:

X 0 1 2 3 4

P(X)


Using properties of determinants, prove the following

Solution:

Consider LHS:


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Show that the rectangle of maximum area that can be inscribed in a circle is a square.

OR

Show that the height of the cylinder of maximum volume that can be inscribed in a cone

of height h is .

Solution:

Let a rectangle ABCD be inscribed in a circle with radius r.


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LetDBC =

In right BCD:

LetA be the area of rectangle ABCD.

Therefore, by the second derivative test, is the point of local maxima ofA.


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So, the area of rectangle ABCD is the maximum at

Hence, the rectangle of the maximum area that can be inscribed in a circle is a square.

OR

Let a cylinder be inscribed in a cone of radiusR and height h.

Let the cylinders radius be rand its height be h1.

It can be easily seen that AGI and ABD are similar.

Volume (V) of the cylinder = r2h1


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Therefore, by the second derivative test, is the point of local maxima ofV.

So, the volume of the cylinder is the maximum at .

Hence, the height of the cylinder of the maximum volume that can be inscribed in a cone

of height h is .



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Using integration find the area of the region bounded by the parabolay2 = 4x and thecircle 4x2 + 4y2 = 9.

Solution:

The respective equations for the parabola and the circle are:

Equation (1) is a parabola with vertex (0, 0) which opens to the right and equation (2) is

a circle with centre (0, 0) and radius .

Therefore, the given curves intersect at .


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Required area of the region bound by the two curves


Evaluate:

Solution:

Now,


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Find the equation of the plane passing through the point (1, 1, 2) and perpendicular toeach of the following planes:

OR

Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and

parallel to the line

Solution:

The equation of the plane passing through the point (1, 1, 2) is:

Where a, b, c are the direction ratios of the normal to the plane

It is given that the plane (1) is perpendicular to the planes

Equations (2) and (3) can be solved as:

So the direction ratios of the normal to the required plane are multiples of 9, 17, and 23.

Thus, the equation of the required plane is:


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OR

Equation of the plane passing through the point (3, 4, 1) is:

Where a, b, c are the direction ratios of the normal to the plane

It is given that the plane (1) passes through the point (0, 1, 0). So this point satisfies theequation (1).

It is also given that the plane (1) is parallel to the line .So, this line is perpendicular to the normal of the plane (1).

Equations (2) and (3) can be solved as:

So, the direction ratios of the normal to the required plane are multiples of 8, 13, 15.Therefore, equation (1) becomes:


A factory owner purchases two types of machines, A and B for his factory. Therequirements and the limitations for the machines are as follows:

MachineArea

occupied

Labour

force

Daily output (in units)

A 1000 m2 12 men 60

B 1200 m2 8 men 40

He has maximum area of 9000 m2 available, and 72 skilled labourers who can operate


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both the machines. How many machines of each type should he buy to maximise thedaily output?

Solution:

Letx andy be the numbers of machines A and B respectively.

Now, according to the given information, the mathematical formulation of the givenproblem is:

MaximiseZ= 60x + 40y

Subject to the constraints

The inequalities (1) to (3) can be graphed as:

It is seen that the shaded portion OABC is the feasible region and the values ofZat thecorner points are given by the following table.

We can find the value of Z at vertices 0, A, B and C as follows:

Corner point Z= 60x+ 40y

O (0, 0) 0


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300

360Maximum

360 Maximum

The maximum value ofZis 360 units, which is attained at and .

It is clear that the number of machines cannot be in fraction.

Thus, to maximise the daily output, 6 machines of the A type need to be bought.Question 29 ( 6.0 marks)

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truckdrivers. The probability of an accident involving a scooter, a car and a truck are 0.01,0.03 and 0.15 respectively. One of the insured persons meets with an accident. What isthe probability that he is a scooter driver.

Solution:

Let E1, E2 and E3 be the events of a driver being a scooter driver, car driver and truckdriver respectively. Let A be the event of a vehicle meeting an accident.

Since there are 2000 insured scooter drivers, 4000 insured car drivers and 6000 insuredtruck drivers, total number of insured vehicle drivers = 2000 + 4000 + 6000 = 12000

It is given that:

Now, the probability that the insured person who meets with an accident is a scooter

driver is .

Therefore, by using Bayes theorem, we obtain:


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Maths Solved 2008

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