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Copyright 2014 CBSE-SPOT
MATHEMATICS CBSE
2014
Class XI
Time : 3 hours
F.M : 100
General Instructions :
(i)
All questions are compulsory.
(ii)
Q. 1 to Q. 10 of Section A are of 1 mark each.
(iii)
Q. 11 to Q. 22 of Section B are of 4 marks each.
(iv)
Q. 23
Insurance Industry Analytics
Dimensions: The various aspects on which data can be evaluated
Metrics / KPIs the key output from the data (Key performance index) like mean median gross margin ,PE
ratio etc
Data Mart
Company who specialize providing data or
Dataset: houseprices.csv
Description of the dataset:
A real estate agent is trying to understand the nature of housing stock and
home prices in and around a medium sized town in upstate New York. She
has collected data from a random sample of 1047 homes s
Question 1(a) Show that the vectors [1,1]. 1,cfw_l, 31,2 and cfw_Lil lm a basis of the
vector space Hamil].
Solution. Since EllIncfw_R3 = 3. it is enough to prove that these are linearly independent.
If possible: let
ecfw_1,.1+ em, 3. 2 + c[1, 2.1] = G
Th
Question 1(a) I. Show that the function
musingi, I 3321]
rtecfw_0, 1:0
is continuous at .
'1'
2. tana is not continuous at :r = 3
Solution.
1. Given 6 2% (1', let :5 = E1 then Im| E J =:- |mgsingi| E |33| 6: E, because IsinziI ii 1.
Thus |:.I:IJI s: :5: J
Using (1), (2), (3) we get
0 =
=
=
0 =
0 =
0 =
2
2
H
H
H
2 H
y
y
x
x
x2
y 2
x
y
2
2
2
2 2 H
4 H
2 H
3 H
+ 4x y
+ 4y
2y
+ 8xy
u
u2
u v
v 2
2
2
H
H
2H
H
H
+2x2
4x2 y 2 2 + 8x3 y
4x4 2 2(x2 + y 2 )
u
u
u v
v
u
2
2
H
H
+ 4(y 4 x4 ) 2
8xy(x2 + y 2 )
u v
3 4
Question 2(b) Find the eigenvalues and eigenvectors of the matrix A =
.
4 3
3
3 4
4
= 0 (9
Solution. The characteristic equation of A =
is
4 3
4
3
2 ) 16 = 0 2 25 = 0 = 5, 5.
2 4
x1
If (x1 , x2 ) is an eigenvector for = 5, then
= 0 2x1 4x2 =
1 0 0
Question 2(a) If A = 1 0 1 show that for all integers n 3, An = An2 + A2 I.
0 1 0
50
Hence determine A .
Solution. Characteristic equation of A is
1 0 0
1
1 = 0
0
1
or (1)(2 1) = 3 2 +1 = 0. From the Cayley-Hamilton theorem, A3 A2 A+I =
0 A3
Question 5(b) Define a positive definite form. State and prove a necessary and sufficient
condition for a quadratic form to be positive definite.
Solution. See 1992 question 2(c).
Question 5(c) Show that the mapping T : R3 R3 defined by T (x, y, z) = (x y
Range of T = T(V), kernel of T = cfw_v | T(v) = 0. If w1 , w2 T(V), then w1 =
T(v1 ), w2 = T(v2 ) for some v1 , v2 V, w1 + w2 = T(v1 ) + T(v2 ) = T(v1 + v2 ).
But v1 + v2 V w1 + w2 T(V), thus T(V) is a subspace of W. Note that
T(V) 6= 0 T(V) so T(V) is a
Question 1(c) Find the least perimeter of an isoceles triangle in which a circle of radius r
can be inscribed.
Solution.
A
Let be the semi-vertical angle. D is the
midpoint of BC. E is the point of contact
of AC and the circle or radius r inscribed
OE
= t
Question 2(d) Use Cayley-Hamilton theorem
0
A= 1
3
to find the inverse of the following matrix
1 2
2 3
1 1
Solution. Characteristic polynomial is given by |xI A| = 0, where I is the 3 3 unit
matrix.
x
1
2
1 x 2 3 = 0
3 1 x 1
x[x2 3x + 2 3] + 1[x + 1 9] 2[