MATHEMATICS
(Maximum Marks: 100)
(Time Allowed: Three Hours)
(Candidates are allowed additional 15 minutes for only reading the paper.
They must NOT start writing during this time)
The Question Paper consists of three sections A, B and C.
Candidates are required to attempt all questions from Section A and all question EITHER from Section B OR Section C
Section A: Internal choice has been provided in three questions of four marks each and two questions of six marks each.
Section B: Internal choice has been provided in two question of four marks each.
Section C: Internal choice has been provided in two question of four marks each.
All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.
The intended marks for questions or parts of questions are given in brackets [ ].
Mathematical tables and graphs papers are provided.
SECTION – A (80 Marks)
Question 1: [10 × 2]
(i) The binary operation is defined as
Find
.
(ii) If and
is symmetric matrix, show that
(iii) Solve :
(iv) Without expanding at any stage, find the value of:
(v) Find the value of constant so that the function
defined as:
is continuous at .
(vi) Find the approximate change in the volume of a cube of side
meters caused by decreasing the side by
.
(vii) Evaluate :
(viii) Find the differential equation of the family of concentric circles
(ix) If and
are events such that
and
, then find:
(a)
(b)
(x) In a race, the probabilities of A and B winning the race are and
respectively. Find the probability of neither of them winning the race.
Answer:
(i) Given
(ii) Since is symmetric, therefore
We can compare corresponding terms. We get
(iii)
(iv)
Applying and
we get
since
(v) Given is continuous at
Therefore
Therefore
Since
Therefore
(vi) Volume of a cube
Therefore
Hence change in volume
Hence change in volume decrease by
(vii)
(viii) Family of concentric circles is
Therefore Differential w.r.t.
Therefore
(ix)
(x) Let win the race be
win the race be
Question 2: If the function is invertible then find its inverse. Hence prove that
. [4]
Answer:
Let
Squaring both sides
Therefore
Now,
Therefore
Question 3: If , prove that
. [4]
Answer:
Question 4: Use properties of determinants to solve for :
and
[4]
Answer:
Given and
we get
we get
or
But , therefore
Question 5: [4]
(a) Show that the function
is continuous at
but not differentiable.
OR
(b) Verify Rolle’s theorem for the following function:
Answer:
(a) Continuity at
Therefore
Therefore is continuous at
Now differentiate at
Therefore
Hence is not differentiable at
(b)
(i) is continuous on
because
are continuous function on its domain.
(ii) and
is differentiable on
(iii)
(iv) Let be number such that
Therefore
Therefore
Therefore
Therefore
Therefore
Therefore Rolle’s theorem verified
Question 6: If
, prove that
[4]
Answer:
Therefore
differentiating both sides w.r.t.
Again differentiating both sides w.r.t.
Question 7: Evaluate:
[4]
Answer:
Put
Question 8: [4]
(a) Find the points on the curve at which the equation of the tangent is parallel to the x-axis.
OR
(b) Water is dripping out from a conical funnel of semi-vertical angle at the uniform rate of
in the surface, through a tiny hole at the vertex of the bottom. When the slant height of the water level is 4 cm, find the rate of decrease of the slant height of the water.
Answer:
(a) … … … … … (i)
Given that lines is parallel to
Therefore
Put
in equation (i)
When
then
Therefore Point
When
, then
Therefore Point
Therefore Points
and
(b) Let be the radius,
be the height and
be the volume of the funnel at any time
.
… … … … … (i)
Let be the slant height of the funnel
Given : Semi-vertical angle in the
:
… … … … … (ii)
therefore the equation (i) can be rewritten as:
Differentiate w.r.t. :
Since it is given that rate of change (decrease) of volume of water w.r.t. is
Thereofore
Question 9: [4]
(a) Solve:
OR
(b) The population of a town grows at the rate of per year. Using differential equation, find how long will it take for the population to grow
times.
Answer:
(a)
Compare with
Therefore solution of the linear differential equation
(b)
(Since increase in population speeds up with increase in population) and let
be the population at anytime
.
Therefore
(where r is proportionality constant)
Therefore
integrating both sides
, (where c is the integration constant)
Therefore
where
Here is the rate of increase and
is the initial population let
then
Given to find the time taken to attain
times population, so
Therefore
Taking log on both sides
Question 10: [6]
(a) Using matrices, solve the following system of equations :
OR
(b) Using elementary transformation, find the inverse of the matrix :
Answer:
(a) Given, the three equations:
We can write this in the form of , i.e. as follows:
We know,
Hence it is a non – singular matrix. Therefore A^{-1} exists. Let us find the (adj A) by finding the minors and co-factors
We know , then
Matrix multiplication can be done by multiplying the rows of matrix with the column of matrix
.
Hence ,
and
(b) Let
Therefore exists.
Interchanging and
and
Therefore
Question 11: speaks truth in
of the cases, while
is
of the cases. In what percent of cases are they likely to contradict each other in stating the same fact ? [4]
Answer:
speaks truth
speaks truth
They contradict each other
of cases they likely to contradict each other
Question 12: A cone is inscribed in a sphere of radius . If the volume of the cone is maximum, find its height. [6]
Answer:
Let be a cone of greatest volume inscribed in a sphere of radius
. It is obvious that for maximum volume the axis of the cone must be along a diameter of the sphere. Let
be the axis of the cone and
be the center of the sphere such that
.
Then, height of cone. Applying Pythagoras theorem,
Let be the volume of the cone, then
… … … … … (i)
Now,
Thus, is maximum when
. Putting
in (i), we obtain
Height of the cone
Question 13: [6]
(a) Evaluate:
OR
(b) Evaluate:
Answer:
(a) Let
Therefore
and
(b)
… … … … … (i)
Using
Therefore
… … … … … (ii)
Adding equation (i) and (ii)
Put and
When
When
Question 14: From a lot of items containing
defective items, a sample of
items are drawn at random. Let the random variable
denote the number of defective items in the sample. If the sample is drawn without replacement, find :
(a) The probability distribution of
(b) Mean of
(c) Variance of [6]
Answer:
In items
defective and
non-defective. Let
is the probability of defective items
Let number of defective items. Therefore
Therefore
(i) Mean
(ii) Variance
SECTION B (20 Marks)
Question 15: [3 × 2]
(a) Find if the scalar projection of
and
is
units.
(b) The Cartesian equation of line is : . Find the vector equation of a line passing through
and parallel to the given line.
(c) Find the equation of the plane through the intersection of the planes and
and passing through the origin.
Answer:
(a) Projection on on
is
Given and
Therefore
Therefore
(b) Cartesian equation of a line is
i.e.
Dividing by throughout we get
Therefore D.r.s of the above line is . Now, equation of a line passing through point
and parallel to the above line whose d.r.s. is
is
(c) Equation of 1st plane is:
i.e.
… … … … … (i)
Equation of 2nd plane is:
i.e.
… … … … … (ii)
Now, equation of a plane passing through intersection of given planes is:
Since plane is passing through the origin
Question 16: [4]
(a) If are three non- collinear points with position vectors
respectively, then show that the length of the perpendicular from
is
OR
(b) Show that the four points A,B, C and D with position vectors and
Answer:
(a) Let be a triangle and let
be the position vectors of its vertices
respectively. Let
be the perpendicular from
on
. Then,
Area of
Also, Area of
Area of
Therefore
(b) Given:
are coplanar if
i.e.
Therefore
Therefore are coplanar.
Therefore Points and
are coplanar
Question 17: [4]
(a) Draw a rough sketch of the curve and find the area of the region bounded by curve and the line
.
OR
(b) Sketch the graph of . Using integration, find the area of the region bounded by the curve
and
and
.
Answer:
(a) Given equation is
Comparing with
We get
Given
Also, meets
Therefore and
are their point of intersection.
Required are
sq. units.
(b) , if
, if
For when when Points are |
For when when Points are |
Therefore the required area:
sq. units
Question 18: Find the image of a point having position vector : in the plane
[6]
Answer:
Let be root of point
in the plane
can of
is
substitute and
in plane
Therefore
Therefore by mid point formula
SECTION C (20 Marks)
Question 19: [3 × 2]
(a) Given the total cost function of units of a commodity as
Find: (i) Marginal cost function (ii) Average cost function
(b) Find the coefficient of correlation from the regression lines: and
.
(c) The average cost function associated with producing and marketing units of an item is given by
. Find the range of values of the output
, from which
is increasing.
Answer:
(a)
Marginal cost function
Average cost function
(b) Let be of
on
.
Therefore
Similarly, let be of
on
Therefore
Hence correlation
(c) Given
Now
Therefore for will keep increasing.
Question 20: [4]
(a) Find the line of regression of on
from the following table.
1 | 2 | 3 | 4 | 5 | |
7 | 6 | 5 | 4 | 3 |
Hence estimate the value of when
.
OR
(b) From the given data:
Variable | ||
Mean | 6 | 8 |
Standard Deviation | 4 | 6 |
And the correlation coefficient: . Find
(i) Regression coefficients and
(ii) Regression line on
(iii) Most likely value of when
Answer:
(a)
1 | 7 | 0 | 2 | 0 | 0 | 4 |
2 | 6 | 1 | 1 | 1 | 1 | 1 |
3 | 5 | 2 | 0 | 0 | 4 | 0 |
4 | 4 | 3 | -1 | -3 | 9 | 1 |
5 | 3 | 4 | -2 | -8 | 16 | 4 |
|
|
Hence there is high negative correlation between and
.
Therefore equation of line:
When we get
(b) To Be Solved
Question 21: [4]
(a) A product can be manufactured at a total cost
, where
is the number of units produced. The price at which each unit can be sold is given by
. Determine the production level
at which the profit is maximum.What is the price per unit and profit at the level of production.
OR
(b) A manufacturer’s marginal cost function is . Find the cost involved to increase production from
units to
units.
Answer:
(a) Total Cost
Price
If units are produced, then
For maximizing profits
units.
units
(b)
The cost involved to increase production from units to
units:
units
Question 22: A manufacturing company produces two type of teaching aids and
of Mathematics for Class X. Each type of
requires
hours for fabricating and
hour for finishing. Each type of
requires
hours for fabricating and
hours for finishing. For fabricating and finishing, the maximum labor hours available for a week are
and
respectively. The company makes a profit of
on each piece of type
and
on each piece of type
. How many pieces of type
and type
should be manufactured per week to get a maximum profit? Formulate this as a linear programming problem and solve it. Identify the feasible region from the rough sketch. [6]
Answer:
Let quantity of teaching aid and Quantity of teaching aid
Each type of requires
hours for fabricating and
hour for finishing.
Therefore … … … … … (i)
Each type of requires
hours for fabricating and
hours for finishing.
Therefore … … … … … (ii)
We know the company makes a profit of on each piece of type
and
on each piece of type
.
We also know
Therefore Profit … … … … … (iii)
We need to maximize . We will solve it graphically.
We see that the coordinates of the vertices of the feasible region are . Now calculate Profit for each of the three coordinates:
For
For
For
Therefore, the quantity of teaching aid and Quantity of teaching aid
Solution of Q.no. 19(b) is wrong, because r² <= 1 (always) and yours is greater.
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