# Math_2010_AI_QP_Sol by nehalwan

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Math 2010 Set 2                                                                                   Close

Subjective Test

(i)     All questions are compulsory.

(ii)    The question paper consists of 29 questions divided into three sections A, B and C. Section
A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four 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 per the exact
requirements of the question.

(iv)   There is no overall choice. However, internal choice has been provided in 4 questions of four
marks each and 2 questions of six marks each. You have to attempt only one of the alternatives in all
such questions.

(v)     Use of calculators is not permitted.

Question 1 ( 1.0 marks)

Evaluate:

Solution:

Let log x = t ...(1)

Differentiating both sides:

Question 2 ( 1.0 marks)

If                       , then for what value of α is A an identity matrix?

Solution:

If A is an identity matrix, then:

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Thus, for α = 0°, A is an identity matrix.

Question 3 ( 1.0 marks)

What is the principal value of               ?

Solution:

Let

Then,

Thus, the principal value of                        .

Question 4 ( 1.0 marks)
What is the cosine of the angle w hich the vector               makes with y-axis?

Solution:
The vector form of y-axis is     or

The cosine of angle (θ) between two vectors        and   is given by

Here,                   and

Thus, the cosine of angle between y-axis and the vector                is

Question 5 ( 1.0 marks)
Write a vector of magnitude 15 units in the direction of vector

Solution:
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Solution:
The given vector is

The unit vector       along the direction of vector    is given by

Thus, the unit vector along the direction of vector                  is        or

The magnitude of the unit vector is 1.

Thus, the vector whose magnitude is 15 and direction along the vector               is

Question 6 ( 1.0 marks)

What is the range of the function                  ?

Solution:

This function f(x) is not defined at x = 1.

For x ≥ 1,

For

Thus, the range of the function is −1 or 1 at all points excluding x = 1.

Question 7 ( 1.0 marks)
Find the minor of the element of second row and third column (a23 ) in the following determinant:

Solution:

The given determinant is                       .

Minor of the element of second row and third column a23 is M23 .

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Thus, the minor of the element of second row and third column is 13.

Question 8 ( 1.0 marks)
Write the vector equation of the following line:

Solution:

The given equation of line is                         .

This equation can be written as

Comparing this equation with the standard form                                  :

x1 = 5, y1 = −4, z1 = 6, a = 3, b = 7, c = −2

Thus, the required line passes through the points (5, −4, 6) and is parallel to the vector          .

It is known that the equation of a line passing through a point whose position vector is     and
parallel to a vector   is given by

Thus, the vector form of the given line is

Question 9 ( 1.0 marks)
What is the degree of the following differential equation?

Solution:
The degree of a differential equation is the highest power of the highest order derivative in it.

The given differential equation is

The highest order derivative in the equation is           and its power is 1.

Thus, the degree of the given differential equation is 1.

Question 10 ( 1.0 marks)

If                                           , then write the value of k.

Solution:

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When two matrices are equal, their corresponding elements are the same.

Thus, the value of k is 17.

Section B

Question 11 ( 4.0 marks)
Find all points of discontinuity of f, where f is defined as follows:

OR

Find      , if

Solution:

The given function f is

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < −3

Case II:

Therefore, f is continuous at x = −3

Case III:

Therefore, f is continuous in (−3, 3).

Case IV:

If c = 3, then the left hand limit of f at x = 3 is,

The right hand limit of f at x = 3 is,

It is observed that the left and right hand limit of f at x = 3 do not coincide.

Therefore, f is not continuous at x = 3

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Case V:

Therefore, f is continuous at all points x, such that x > 3

Hence, x = 3 is the only point of discontinuity of f.

OR

Let

Differentiating both sides with respect to x, it is obtained:

Differentiating both sides with respect to x, it is obtained:

From equations (1), (2) and (3), it is obtained:

Question 12 ( 4.0 marks)
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Question 12 ( 4.0 marks)
Prove the following:

OR

Prove the following:

Solution:

Thus, the given result is proved.

OR

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Now, consider LHS:

Thus, the given result is proved.

Question 13 ( 4.0 marks)
On a multiple choice examination with three possible answers (out of which only one is correct) for
each of the five questions, what is the probability that a candidate would get four or more correct

Solution:
The repeated guessing of correct answers from multiple choice questions are Bernoulli trials.

Let X represent the number of correct answers by guessing in the set of multiple choice questions.

Clearly, X has a binomial distribution with n = 5 and     .

P (guessing more than 4 correct answers)

= P(X ≥ 4)

= P(X = 4) + P(X = 5)

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Thus, the required probability is       .

Question 14 ( 4.0 marks)
Let * be a binary operation on Q defined by

Show that * is commutative as well as associative. Also find its identify element, if it exists.

Solution:

Given that * is a binary operation on Q defined by               , where a, b ∈ Q

Now,

As ab = ba,

Thus, the binary operation * is commutative.

Let a, b, c ∈ Q

From equations (1) and (2):

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

Thus, the binary operation * is associative.

For a binary operation *: A × A → A, if an element e ∈ A such that a * e = e * a = a &mnForE; a∈A,
then e is called the identity element.

Now,

Also,

Now,                    and         .

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Thus,    is the identity element for the binary operation *.

Question 15 ( 4.0 marks)
Using elementary row operations, find the inverse of the following matrix:

Solution:

Let

The inverse of A can be found by using elementary row operations as:

It is known that A =I.A

Question 16 ( 4.0 marks)
Find the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, − 1, 2) and

parallel to the line                    .

Solution:
Let the equation of plane be ax + by + cz + d = 0

It is given that the plane passes through the point A (0, 0, 0).

∴ a× 0 +b× 0 +c× 0 +d=0

⇒ d = 0 … (1)

∴ Equation of the plane = ax + by + cz =0 … (2)

It is also given that the plane passes through the point B (3, −1, 2)

∴ a × 3 + b × (−1) + c × 2 = 0

⇒ 3a − b + 2c = 0 … (3)

Equation of the line is                      or

Thus, the vector form of the line is,

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This equation is of the form            , where                  and                    is the normal
vector.

The required plane is parallel to the given line.

Therefore, the normal to this plane is perpendicular to the given line.

∴ a × 1 + b ×(−4)+ c × 7 = 0

⇒ a − 4b + 7c = 0 … (4)

Cross multiplying equations (3) and (4), we obtain:

Substituting the values of a, b and c in equation (3), we obtain the equation of plane as:

kx − 19ky − 11kz = 0

⇒ k (x − 19y − 11z) = 0

⇒ x − 19y − 11z = 0

Thus, the equation of the plane is x − 19y − 11z = 0

Question 17 ( 4.0 marks)
Find the position vector of a point R which divides the line joining two points P and Q whose position
vectors are         and           respectively, externally in the ratio 1:2. Also, show that P is the
midpoint of the line segment RQ.

Solution:
The position vectors of the points P and Q are                          respectively.

It is given that the point R divides the line joining P and Q externally in the ratio 1:2.

Thus the position vector of R is given by

This information can be represented using a figure as:

It can be observed that:

Thus, P is the mid-point of line segment RQ.

Question 18 ( 4.0 marks)

Evaluate:

Solution:
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Solution:                          Subjective Test Paper - Math - Meritnati…

Let I =                ... (i)

Question 19 ( 4.0 marks)

Evaluate:

OR

Evaluate:

Solution:

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OR

It can be seen that the given integrand is not a proper fraction.

On dividing (1 − x2 ) by x(1 − 2x), it is obtained:

Substituting x = 0 and    , it is obtained:

A = 2 and B = 3

Substituting in equation (1), it is obtained:

Question 20 ( 4.0 marks)

Find the equations of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y
+4 =0

Solution:
The equation of the given curve is y = x3 + 2x + 6.

The slope of the tangent to the given curve at any point (x, y) is given by

∴ Slope of the normal to a curve at any point (x, y) =

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∴ Slope of normal of given curve

It is given that normals to the given curve are parallel to the line x + 14y + 4 = 0.

x + 14y + 4 = 0

⇒

∴ Slope of the given line =

When x = 2, y = 8 + 4 + 6 = 18.

When x = −2, y = −8 − 4 + 6 = −6.

Therefore, there are tw o normals to the given curve with slope      and passing through the points

(2, 18) and (−2, −6).

Thus, the equation of the normal through (2, 18) is given by:

The equation of the normal through (−2, −6) is given by:

Thus, the equations of the normals to the given curve are

Question 21 ( 4.0 marks)
Find the particular solution of the differential equation satisfying the given conditions:

x2 dy + (xy + y2 ) dx = 0; y = 1 when x = 1.

Solution:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx
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y = vx

Substituting the values of y and       in equation (1), we get:

Integrating both sides, we get:

Now, y = 1 at x = 1.

Substituting           in equation (2), we get:

This is the required solution of the given differential equation.

Question 22 ( 4.0 marks)
Find the general solution of the differential equation

OR
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Find the particular solution of the differential equation satisfying the given conditions:

, given that y = 1 when x = 0

Solution:
The given differential equation is:

This equation is in the form of a linear differential equation as:

The general solution of the given differential equation is given by the relation:

Substituting the value of                  in equation (1), we get:

This is the required general solution of the given differential equation.

OR

Integrating both sides, we get:

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Substituting C = 1 in equation (1), we get:

y = sec x

Thus, the required particular solution of given equation is y = sec x.

Section C

Question 23 ( 6.0 marks)

Evaluate                as limit of sums.

OR

Using integration, find the area of the following region:

Solution:

It is known that:

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From equations (2) and (3), we obtain:

OR

The given equation of the ellipse can be represented as:

The equation of the curve is

It can be observed that ellipse is symmetrical about x-axis and y-axis.

∴ Area bounded by ellipse = 4 × area of OAB

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∴ Area of OAB

Therefore, the area bounded by ellipse is         = 6π square units.

Question 24 ( 6.0 marks)
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured
per day is atmost 24. It takes 1 hour to make a ring the 30 minutes to make a chain. The maximum
number of hours available per day is 16. If the profit on a ring is Rs. 300 and that on a chain is Rs.
90, find the number of rings and chains that should be manufactured per day, so as to earn the
maximum profit. Make it as an L.P.P. and solve it graphically.

Solution:
Let the firm manufacture x gold rings and y chains per day.

Therefore, x ≥ 0 and y ≥ 0.

It is given that the total number of gold rings and chains manufactured per day is at most 24.

∴ x + y ≤ 24

The gold ring takes 1 hour to make and chain takes 30 min, that is, 0.5 hour to make.

The maximum number of hours available per day is 16 hours.

∴ x + 0.5y ≤ 16

⇒ 2x + y ≤ 32

The profit on a ring is Rs 300 and on a chain is Rs 190.

∴Total profit, z = 300x + 190y

Thus, the mathematical formulation of the given problem is:

Maximise z = 300x + 190y ... (i)

Subject to the constraints:

x + y ≤ 24 ... (ii)

2x + y ≤ 32 ... (iii)

x, y ≥ 0 ... (iv)

The feasible region determined by the system constraints is as follows:

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The corner points are A(16, 0), B(8, 16) and C(0, 24).

The value of z at these corner points are as follows:

Corner point     z = 300x + 190y

A(16, 0)           4800

B(8, 16)           5440           → Maximum

C(0, 24)           4560

Thus, the maximum value of z is 5440 at (8, 16)

Thus, 8 gold rings and 16 chains should be manufactured per day to maximise the profits.

Question 25 ( 6.0 marks)
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at
random and are found to both clubs. Find the probability of the lost card being of clubs.

OR

From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the
probability distribution of the number of defective bulbs.

Solution:
Let E 1 and E 2 be the respective events of choosing a club card and a card which is not club.

Let A denote the event of drawing a lost card.

Out of 52 cards, 13 cards are club and 39 cards are not club.

When one club card is lost, there are 12 club cards out of the 51 cards.

Two cards can be drawn out of 12 club cards in          ways.

Similarly, 2 club cards can be drawn out of 51 cards in         ways.

The probability of getting two cards, when one club card is lost, is given by P (A|E 1 ).

When the lost card is not a club, there are 13 club cards out of 51 cards.

Two cards can be drawn out of 13 club cards in          ways whereas 2 cards can be drawn out of 51

cards in        ways.

The probability of getting two cards, when the lost card is not club, is given by P (A|E 2 ).

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The probability that the lost card is club is given by P (E1 |A).

By using Bayes’ theorem:

OR

It is given that out of 10 bulbs, 3 are defective.

⇒ Number of non-defective bulbs = 10 − 3 = 7

2 bulbs are drawn from the lot with replacement.

∴ Probability of selecting a defective bulb =

⇒ Probability of selecting a non-defective bulb =

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.
Then, X can take the values 0, 1, or 2.

P (X = 0) = P (non-defective and non-defective)

= P (non-defective) × P (non-defective)

P (X = 1) = P (non-defective and defective or defective and non-defective)

= P (non-defective and defective) + P (defective and non-defective)

P (X = 2) = P (defective and defective)

= P (defective) × P (defective)

Therefore, the required probability distribution is as follows:

X      0      1      2

P (X)

Question 26 ( 6.0 marks)
Using properties of determinants, show the following:

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Solution:

Applying R1 → aR1 , R2 → bR2 , R3 → cR3 to ∆ and dividing by abc:

Taking out common factors a, b, c from C 1 , C 2 and C 3 respectively:

Applying C 2 → C 2 − C 1 , C 3 → C 3 − C 1 :

Taking out common factor (a + b + c) from C 2 and C 3 :

Applying R1 → R1 − (R2 + R3 ):

Applying C 2 →                and C 3 →            :

Finally expanding along R1 :

∆ = (a + b + c)2 (2bc) [(a + c) (a + b) − bc]

⇒ ∆ = (a + b + c)2 (2bc) (a2 + ab + ac)

⇒ ∆ = (a + b + c)3 (2abc)

⇒ ∆ = 2abc (a + b + c)3

Hence proved

Question 27 ( 6.0 marks)
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Find the values of x for w hich f(x) = [x(x − 2)]2 is an increasing function. Also, find the points on the
curve, where the tangent is parallel to x-axis.

Solution:

The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals:

In intervals                      ,           .

∴ f(x) is strictly decreasing in intervals                      .

However, in intervals (0, 1) and (2, ∞),

∴ f(x) is strictly increasing in intervals (0, 1) and (2, ∞).

Thus, f(x) is strictly increasing for 0 < x < 1 and x > 2.

It is also known that a tangent to a curve y = f(x) is parallel to the x-axis at point x = a if          .

From equation (1), it is obtained that               at x = 0, 1, and 2.

Thus, at the points x = 0, 1, and 2; the tangent to the given curve is parallel to the x-axis.

Question 28 ( 6.0 marks)
Show that the right circular cylinder, open at the top, and of given surface area and maximum
volume is such that its height is equal to the radius of the base.

Solution:
Let r and h be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder that opens at the top is given by:

Let V be the volume of the cylinder.

Therefore, by second derivative test, the volume is the maximum when                    .

meritnation.com/…/J6x0a81vDz1rkTsQ7…                                                                            23/24
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Thus, the volume of a cylinder is the maximum when its height is equal to the radius of its base.

Question 29 ( 6.0 marks)
Write the vector equations of the following lines and hence determine the distance between them:

Solution:
The given equations of lines are:

Comparing (1) and (2) with the standard form                                  :

For equation (1), x1 = 1, y1 = 2, z1 = −4, a = 2, b = 3, c = 6

For equation (2), x1 = 3, y1 = 3, z1 = −5, a = 4, b = 6, c = 12

Therefore, the vector equation of the line (1) is:

The vector of equation of the line (2) is:

Here, µ = 2α

From equations (3) and (4) it is observed that lines (1) and (2) are parallel.

Comparing (3) and (4) with               and              respectively:

The distance betw een two parallel lines is given by                      .

Thus, the distance between the given lines is            units.

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