1. Xtra Edge February 2012 - Career Point

MATHEMATICAL CHALLENGES
1. as φ (a) = φ (b) = φ (c)
so by Rolle’s theorem there must exist at least a point
x = α & x = β each of intervals (a, c) & (c, b) such
that φ′(α) = 0 & φ′(β) = 0. Again by Rolle’s theorem,
there must exist at least a point x = µ such that
α < µ < β where φ′(µ) = 0
2 f ( a)
2 f ( b)
so
+
( a − b)
( a − c)
( b − c)
( b − a)
2 f ( c)
+
– f ′′ (µ) = 0
( c − a)
( c − b)
f ( a)
f ( b)
so
+
( a − b)
( a − c)
( b − c)
( b − a)
f ( c)
1
+
= f ′′ (µ)
( c − a)
( c − b)
2
where a < µ < b.
2. Required probability
r
5 5 5 5 1 5
1 . . . ........ . =
6 6 6 6 6 6
−
⎛ ⎞
⎜ ⎟
⎝ ⎠
2
1
. (r – 2) times
6
Note : any number in 1st loss
same no. does not in 2nd (any other comes).
Now 3rd is also diff. (and in same r − 2 times)
Now (r − 1) th & r th must be same.
3. 2s = a + b + c
ON = − BN + BO
Let BN = x
2BN + 2CN + 2AR = 2s
x + (a − x) + (b − a + x) = s
x = s − b
A
B
M
I (h,k)
N
r
a
so h = ON = − (s − b)
2
−2s
+ a + 2b
b − c
=
= & r = k.
2 2
SOLUTION FOR JANUARY ISSUE (SET # 9)
R
O
C
∆
so r = k = =
s
s ( s − a)(
s − b)(
s − c)
s
r = k =
s ( s − a)(
s − b)(
s − c)
s
2sk = s( s − a)(
a − b + c)(
a + b − c)
= s ( s − a)(
a − 2x)(
a + 2x)
2sk = s( s − a)(
a − 4h
)
required locus is
4s 2 y 2 = A(a 2 – 4x 2 )
⇒ s2y2 + Ax2 Aa
=
4
where A is = s (s – a)
here h 2 < as so it is an ellipse
4. f (0) = c
f (1) = a + b + c & f (−1) = a − b + c
solving these,
1
a = [f (1) + f (−1) − 2 f (0)] ,
2
1
b = [f (1) − f (−1)] & c = f (0)
2
so f (x) =
x(
x + 1)
XtraEdge for IIT-JEE 37 FEBRUARY 2012
2
2
2
2
f (1) + (1− x 2 ) f (0) +
x(
x −1)
2
f(−1)
2 | f (x) | < | x | | x + 1 | + 2| 1 − x 2 | + | x | | x − 1| ;
as | f (1) | , | f (0) |, | f (−1) | ≤ 1.
2 | f (x) | ≤ | x | (x + 1) + 2 (1 − x 2 ) + | x | (1 − x) as
x ∈ [−1, 1]
so 2 | f (x) | ≤ 2 (|x| + 1 − x 2 5 5
) ≤ 2 . so | f (x) | ≤
4
4
Now as g (x) = x 2 1
f (1/x) = (1 + x) f (1)
2
+ (x 2 1
− 1) f (0) + (1 − x) f (−1)
2
so 2 | g (x) | ≤ | x + 1 | + 2 | 1 − x 2 | + | 1 − x|
⇒ 2 | g (x) | ≤ x + 1 + 2 (1 − x 2 ) | + 1 − x ;
as x ∈ [−1, 1]
⇒ 2 | g (x) | < − 2x 2 + 4 ≤ 4.
⇒ |g (x) | ≤ 2.
5. Oil bed is being shown by the plane A′ PQ. θ be the
angle between the planes A′ PQ & A′ B′ C′. Let A′ B′
C′ be the x − y plane with x-axis along A′ C′ and
origin at A′. The P.V.s of the various points are
defined as follows