March 2011 - Career Point

14. If
Sol. y =
Equation of the chord having (α, – α) as mid-points is
T = S 1
⎛ ⎞
⇒ xα + y(–α) – ⎜
1+
2a ⎛
⎟ (x + α) – ⎝ 4
⎟ ⎞
⎜
1 – 2a (y – α)
⎠ ⎝ 4 ⎠
⎛ ⎞
= α 2 + (– α) 2 – ⎜
1+
2a ⎛
⎟ α – ⎝ 2
⎟ ⎞
⎜
1–
2a
(– α)
⎠ ⎝ 2 ⎠
⇒ 4xα – 4yα – (1 + 2 a) x – (1 + 2 a)α
– (1 – 2 a) y + (1 – 2 a) α
= 4α 2 + 4α 2 – (1 + 2 a). 2α + (1 + 2 a).2α
⇒ 4αx – 4αy – (1 + 2 a) x – (1 – 2 a)y
= 8α 2 – (1 + 2 a)α + (1 – 2 a)α
But this chord will pass through the point
⎛ ⎞
⎜
1+
2a
1– 2a
, ⎟
⎝ 2 2 ⎠
⎛ ⎞
4α ⎜
1+
2a
⎛ ⎞
⎟ – 4α ⎜
1–
2a
⎟
⎝ 2
⎠ ⎝ 2 ⎠
( 1+
2a )(1 + 2a)
( 1– 2a)(1–
2a)
–
–
2
2
= 8α 2 – 2 2 aα
⇒ 2α [(1 + 2 a – 1 + 2 a)] = 8α 2 – 2 2 aα
⇒ 4 2 aα – 2
1 [2 + 2 ( 2 a) 2 ] = 8α 2 – 2 2 aα
[Q (a + b) 2 + (a – b) 2 = 2a 2 + 2b 2 ]
⇒ 8α 2 – 6 2 aα + 1 + 2a 2 = 0
But this quadratic equation will have two distinct
roots if
(6 2 a) 2 – 4(8) (1 + 2a 2 ) > 0
⇒ 72a 2 – 32 (1 + 2a 2 ) > 0
⇒ 72a 2 – 32 – 64a 2 > 0
⇒ 8a 2 – 32 > 0
⇒ a 2 – 4 > 0
⇒ a 2 > 4 ⇒ a < – 2 ∪ a > 2
Therefore, a ∈ (– ∞ , – 2) ∪ (2, ∞).
2
ax
bx c
+
+ + 1,
( x – a)(
x – b)(
x – c)
( x – b)(
x – c)
( x – c)
Prove that y
y'
= x
1
2
⎛ a b c ⎞
⎜ + + ⎟
⎝ a – x b – x c – x ⎠
[IIT-1998]
ax
bx
+
+
( x – a)(
x – b)(
x – c)
( x – b)(
x – c)
ax
=
+
( x – a)(
x – b)(
x – c)
ax
=
+
( x – a)(
x – b)(
x – c)
2
2
c
x
bx
+
( x – b)(
x – c)
x
( x – c)
⎛
⎜
⎝
b
x – b
+ 1
– c
x
x – c
⎞
+1⎟
⎠
ax
=
+
( x – a)(
x – b)(
x – c)
ax
=
+
( x – a)(
x – b)(
x – c)
=
=
2
2
2
x
( x – c)(
x – b)
2
⎛
⎜
⎝
a
x – a
x ⎛ b + x – b ⎞
⎜ ⎟
( x – c)
⎝ x – b ⎠
x
.
( x – c)
⎞
+1⎟
⎠
x ⎛ a + x – a ⎞
⎜ ⎟
( x – c)(
x – b)
⎝ x – a ⎠
⇒ y =
⇒
x
( x – a)(
x – b)(
x – c)
log y = log
3
x
( x – a)(
x – b)(
x – c)
3
x
( x – b)
⇒ log y = log x 3 – log (x – a) (x – b) (x – c)
⇒ log y = 3 log x – log (x – a)
– log (x – b) – log (x – c)
⇒
⇒
⇒
⇒
⇒
⇒
y'
y
3 1
= – x x – a
1 1
– –
x – b x – c
y' ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
= ⎜ – ⎟⎠ + ⎜ – ⎟ + ⎜ – ⎟
y ⎝ x x – a ⎝ x x – b ⎠ ⎝ x x – c ⎠
y' x – a – x x – b – x x – c – x
= + +
y x(
x – a)
x(
x – b)
x(
x – c)
y'
y
y'
y
y'
y
– a
= –
x(
x – a)
a
= +
x( a – x)
b
–
x( x – b)
b
+
x( b – x)
1 ⎛ a b c ⎞
= ⎜ + + ⎟
x ⎝ a – x b – x c – x ⎠
c
x( c – x)
c
x( c – x)
15. Two planes P 1 and P 2 pass through origin. Two lines L 1
and L 2 also passing through origin are such that L 1 lies
on P 1 but not on P 2 , L 2 lies on P 2 but not on P 1 , A, B, C
are three points other than origin, then prove that the
permutation [A'B'C'] of [ABC] exists. Such that :
(i) A lies on L 1 , B lies on P 1 not on L 1 , C does not lie
on P 1 .
(ii) A' lies on L 2 , B lies on P 2 not on L 2 , C' does not lies
on P 2 .
[IIT-2004]
Sol. Here 6 of permutations are possible as, A
corresponds to one of A', B', C' and B corresponds
to one of remaining A', B', C' and corresponds to
third of A', B', C'
∴ A lies on L 1 , B lies on the line of intersection of
P 1 and P 2 and C lies on the line L 2 on the plane P 2 .
∴ A' lies on L 2 = C
B' lies on the line of intersection of P 1 and P 2 = B
C' lies of L 1 on the plane P 1 = A
Thus, [A' B' C'] = [CBA]
i.e., permutation of [A'B'C'] is [ABC].
XtraEdge for IIT-JEE 16 MARCH 2011