1. If A, B, C be the centres of three co-axial circles
   and t 1 , t 2 , t 3 be the lengths of the tangents to them
   from any point, then prove that
   BC t⋅ 12 +CA t⋅^22 +=AB t 32 0
   Kishor, U.P.
   Ans. Let the equations of three circles are
   x^2 + y^2 + 2gix + c = 0, i = 1, 2, 3.
   According to the question
   A ≡ (–g 1 , 0), B ≡ (–g 2 , 0), C ≡ (–g 3 , 0)
   Let any point be P(h, k)
   thkghc 1 =++ +^2221

thkghc 2 =++ +^2222

thkghc 3 =++ +^2223

and AB=(g 12 −g )

BC=()g^23 −g

and CA=(g 31 −g)

Now BC t⋅ 12 +CA t⋅ 22 +AB t⋅ 32

= Σ(g 2 – g 3 ) (h^2 + k^2 + 2g 1 h + c)

= (h^2 + k^2 + c) Σ (g 2 – g 3 ) + 2hΣg 1 (g 2 – g 3 )

= (h^2 + k^2 + c) (g 2 – g 3 + g 3 – g 1 + g 1 – g 2 )

+2h {g 1 (g 2 – g 3 ) + g 2 (g 3 – g 1 ) + g 3 (g 1 – g 2 )}.

= (h^2 + k^2 + c) (0) + 2h(0)

= 0

which proves the result.

2. If f(x) = |x + 1| {|x| + |x – 1|}, then draw the graph
   of f(x) in the interval [–2, 2] and discuss the
   continuity and differentiability in [–2, 2].
   Karan Sharma, New Delhi

Do you have a question that you just can’t get

answered?

Use the vast expertise of our MTG team to get to the

bottom of the question. From the serious to the silly,

the controversial to the trivial, the team will tackle the

questions, easy and tough.

The best questions and their solutions will be printed in

this column each month.

Ans. Here, f(x) = |x + 1| {|x| + |x – 1|}

fx

xx x

xx x

xx

x

()

()( );

()( );

();

(

=

+ −−≤<−

− + −−≤<

+ ≤ <

+

12 1 2 1

12 1 1 0

101

1))(xx−≤≤); 21 1 2

⎧

⎨

⎪⎪

⎩

⎪

⎪

Thus the graph of f(x) is;

which is clearly, continuous for x ∈[–2, 2] and

differentiable for x ∈[–2, 2] – {–1, 0, 1}

3. If n is a positive integer, show that
   n
   xx xn

C

x

C

x

!

( ) ( )....( )

(!)

++ +

=

+

−

+

+

12 1

2

2

1 2

(!) (^3) .... ( ) (!)
3
C 3 1 1
x
nC
xn
n n

• − +−


• 
  


• 
  

  Mehul, Assam
  Ans. If the L.H.S of the given expression is decomposed
  into partial fractions, we have
  n
  xx xn
  A
  x
  A
  x
  A
  xn
  ! n
  ()( )....()
  ....
  ++ +

  
  


• 
  


• 
  


• 
  

  ++
  12 1 2 +
  12
  i.e., n! = A 1 (x + 2) (x + 3)····(x + n) + A 2 (x + 1)
  (x + 3) ···· (x + n) + ···· + An(x + 1) (x + 2)
  ····(x + n – 1) ...(i)
  Putting x = –1 in (i), we have
  n! = A 1 · 1 · 2 · 3 ········(n – 1)
  ie A n
  n
  .., .......! nCn
  (^11123) () 1

  
  

  ⋅⋅ −

  
  

  Similarly, putting x = –2, –3, ········, –n in (i)
  respectively, we have
  A n
  n
  22 nn nC
  12 2
  = 12
  −⋅ −
  .......! =−−=−
  ()
  () (!)
  A n
  n
  33 nn n nC
  12 3
  = 12 3
  ⋅−
  .......! = −−=
  ()
  ()( ) (!)
  and An = (–1)n + 1 nCn (n!)
  Hence, we have
  n
  xx xn
  C
  x
  C
  x
  !
  ()( )( )
  (!)
  ++⋅⋅⋅⋅ +

  
  


• 
  

  −

  
  


• 
  


• 
  

  12 1
  2
  2
  12
  (!)
  .... ( )
  3 (!)
  3
  3 1 1
  C
  x
  nC
  xn
  n n

  
  


• 
  

  − +−

  
  


• 
  


• 
  

  which is the desired result. ””

  
  

YUASK

WE ANSWER