Mathematics Times – July 2019

apply the Lagrange MVT to the interval

  x,1,1 to get some c x 1  1 ,1 such that

 

 1 

1

,

0

'

1

y x

y c

x





 (4)

The R.H.S. is negative. So there is atleast one

point on  where the negative sign holds for

the R.H.S. of (3). But then because of

continuity, the negative sign holds at all points

of  and so we get

xy'   1 x^2 (5)

at all points of . Hence (B) is true and (D) is

false.

We can now rewrite the differential equation

as

1 2

'

x

y

x



  (6)

and solve it by integrating both the sides.

1 x^2

y dx

x



 

cos^2

sin

d







 

1 sin^2

sin

d









 

 (cosecsin ) d

  ( log (cote cosec ) cos )   c

1 cos

log cos

e sin

c







    

     

   

12.Sol:

2

log^1112

e

x

x c

x

   

   

 

(7)

We are given that the point (1,0) lies on .
Hence c 0. That makes (A) true and (C)
false.
12.Sol: The major importance of the adjoint
adj M of a square matrix M comes from the
following equation.

adjM M M adjM I     (1)

Where is the determinant of M and I is the

identity matrix.

We are given the adjoint of M, but not the

matrix M itself. Two of the nine entries of M

are the unknowns a and b. If we write down

its adjoint, many entires will have these

unknowns. So, the system of nine equations

that would result by equating the corresponding

entries of the unknown and the given

expression for adj M will be a complicated

one to solve. Instead, we use (1). Of course,

multiplying two3 3 matrices fully is not a very

enviable job. But in the present case, note

that the R.H.S. is a diagonal matrix. The

six non-diagonal entires of it are all 0. As there

are only two unknowns In M, we need only

two equations to find them. Fortunately, if we

start multiplying adj M and M by writing

1 1 1 0 1

8 6 2 1 2 3

5 3 1 3 1

a

b

   

    

   

    

     2 1 b a 2  (2)

we need not go any further. the two non-

diagonal entries is the first row of the matrix

on the R.H.S. must vanish. This given

a 2 and b 1. hence (A) is true.

Now that we know the matrix M completely,

viz.

0 1 2

1 2 3

3 1 1

M

 

 

 

 

(3)

we can answer the remaining options by brute

force if necessary. But rarely brute force is

the best method. For example, take (D). Since

M is invertible, (as  0 ), we can find  , ,

 by writing

1

1

2

3

M









   

   

   

     

(4)