1. Xtra Edge February 2012 - Career Point

10.
11
∴ n(E) = 10 + 9 + 8 + ... + 2 + 1 = = 55
2
∴ the probability of subtracting without borrowing
55
in each column = .
100
⎛ 55 ⎞ ⎛ 11 ⎞
∴ the required probability = ⎜ ⎟⎠ = ⎜ ⎟⎠ .
⎝100
⎝ 20
4. Let S be the coefficients of x 49 in given expression
f(x) and if P be product of roots of the equation
S
f(x) = 0, then find the value of , given that :
P
f(x) = (x – 1) 2 ⎛ x ⎞ ⎛ 1 ⎞ ⎛ x ⎞ ⎛ 1 ⎞
⎜ − 2⎟
⎜ x − ⎟ ⎜ − 3⎟
⎜ x − ⎟ ,
⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠
⎛ x ⎞ ⎛ 1 ⎞
......... ⎜ − 25⎟
⎜ x − ⎟
⎝ 25 ⎠ ⎝ 25 ⎠
Sol. Here we can write f(x) as :
⎧ ⎛ x ⎞⎛
x ⎞ ⎛ x ⎞⎫
f(x) = ⎨(
x −1) ⎜ − 2⎟⎜
− 3⎟...
⎜ − 25⎟⎬
⎩ ⎝ 2 ⎠⎝
3 ⎠ ⎝ 25 ⎠⎭
⎧ ⎛ 1 ⎞⎛
1 ⎞ ⎛
× ⎨(
x −1)
⎜ x − ⎟⎜
x − ⎟...
⎜ x −
⎩ ⎝ 2 ⎠⎝
3 ⎠ ⎝
Now roots of f(x) = 0 are;
1 2 , 2 2 , 3 2 , ..... , 25 2 1 1 1
and 1, , , .....,
2 3 25
Now f(x) is the polynomial of degree 50,
So coefficient of x 49 will be :
S = – (sum of roots)
= – (1 2 + 2 2 + ... + 25 2 ⎛ 1 1 1 ⎞
) – ⎜1
+ + + .... + ⎟
⎝ 2 3 25 ⎠
6
25
6
1 ⎞⎫
⎟⎬
25 ⎠⎭
⎧25×
26×
51 ⎫
1
= – ⎨ + K⎬
where, K =
⎩ 6
∑ ⎭
n
n=
1
⇒ S = –(K + 5525).
Product of roots :
1 2 . 2 2 . 3 2 .... 25 2 1 1 1
. 1 . . .... = 1 . 2 . 3 ...25
2 3 25
∴ P = 25 !
25
S −(
K + 5525)
1
Hence =
, where K =
P 25!
∑ n
n=
1
5. A traveller starts from a certain place on a certain day
and travels 1 km on the first day and on subsequent
days, he travels 2 km more than the previous day.
After 3 days, a second traveller sets out from the
same place and on his first day he travels 12 km and
on subsequent days he travels 1 km more than the
previous day. On how many days will the second
traveller be ahead of the first?
Sol. The first traveller travels on different days as follow
(in km) 1, 1 + 2, 1 + 2 + 2, ... .
After 3 days the first traveller is already ahead by
(1 + 3 + 5) km, i.e., 9 km.
1 3 5
The second traveller travels on different days as
follows : 0, 0, 0, 12, 13, 14, ...
After n days from the day the second traveller starts,
the distance covered by the first
= 1 + 3 + 5 + (7 + 9 + ... to n terms)
= 1 + 3 + 5 + ... to (n + 3) terms
= (n + 3) 2
and the distance covered by the second
= 12 + 13 + 14 + ... to n terms
n n(
n + 23)
= {24 + (n – 1).1} =
2
2
The second traveller is ahead of the first on the n th
day (after the second sets off) if
n(
n + 23)
> (n + 3)
2
2
or n 2 + 23n > 2(n 2 + 6n + 9)
or n 2 – 11n + 18 < 0
or (n – 2) (n – 9) < 0.
So n – 2 > 0 and n – 9 < 0 ...(i)
or n – 2 < 0 and n – 9 > 0 ...(ii)
(i) ⇒ n > 2 and n < 9
(ii) ⇒ n < 2 and n > 0 (absurd)
Thus, from the begining of the 3 rd day to the end of
the 9 th day the second traveller is ahead of the first.
So, the second is ahead of the first on the 3 rd , 4 th , 5 th ,
..., 9 th days (after the second sets off).
Hence, the required number of days = 7.
6. Let P(x) be a polynomial of degree n such that
i
P(i) = for i = 0, 1, 2 ..... n. If n is odd than find
i + 1
the value of P(n + 1).
Sol. Let Q(x) = (x + 1) P(x) – x
clearly Q(x) is polynomial of degree n + 1. Also
i
Q(i) = (i + 1) – i = 0 for i = 1, 2, 3 .....n
i + 1
Thus we can assume
Q(x) = kx(x – 1) (x – 2) ...... (x – n) where k is a constant.
Now Q(–1) = k(–1)(–2)(–3) ...... (–1 – n)
1 = (–1) n + 1 k(n + 1) !
⇒ k =
XtraEdge for IIT-JEE 41 FEBRUARY 2012
( n
1
+
1)
!
(Q n is odd)
Thus P(x) =
1 ⎡ x(
x −1)(
x − 2)....(
x − n)
⎤
⎢
+ x⎥
,
x + 1 ⎣ ( x + 1)
! ⎦
where n is odd , ∴ P(n + 1) = 1