March 2011 - Career Point

As 'm' of VP =
2
1
t
t
1
– 0
– 0
2
= t 1
t2
– 0
and 'm' of VQ = = t 2 ,
t2
– 0
VP ⊥ VQ ⇒ t 1 .t 2 = – 1 ...(i)
The equation of the normal to a curve at (x 1 , y 1 ) is
–1
y – y 1 =
. (x – x 1 ).
⎛ dy ⎞
⎜ ⎟
⎝ dx ⎠
Here, y = x 2
( x1,y1
)
dy
∴ = 2x. dx
∴ the equation of the normal at P(t 1 ,
y –
2
t 1 =
–1
(x – t 1 )
2t
1
or 2t 1 y + x = 2 t 1
3 + t 1
2
t 1 ) is
...(ii)
Similarly, the equation of the normal at Q(t 2 ,
2
t 2 ) is
2t 2 y + x = 2 t 3 2 + t 2
...(iii)
Eliminating t 1 , t 2 from (i), (ii) and (iii) we get the
locus of M.
(ii) – (iii) ⇒ 2y(t 1 – t 2 ) = 2( t 3 1 – 3
t 2 ) + (t 1 – t 2 )
or 2y = 2( t 1
2 + t 1 t 2 +
2
t 2 ) + 1
Also, t 2 × (ii) – t 1 × (iii)
⇒ (t 2 – t 1 )x = (2 t 1
3 + t 1 )t 2 – (2 t 2
3 + t 2 )t 1
= 2t 1 t 2 ( t 1
2 –
2
t 2 )
or x = – 2t 1 t 2 (t 1 + t 2 )
From (i) and (v), x = 2(t 1 + t 2 )
or t 1 + t 2 = 2
1 x
...(iv)
...(v)
...(vi)
From (iv), 2y = 2{t 1 + t 2 ) 2 – t 1 t 2 } + 1
⎪⎧
2
⎛ 1 ⎞ ⎪⎫
= 2⎨⎜
x ⎟ + 1⎬
+ 1, using (i) and (vi)
⎪⎩ ⎝ 2 ⎠ ⎪⎭
∴ the equation of the required locus is
2
x
2y = + 3
2
or x 2 = 2(2y – 3), which is a parabola.
nπ+θ
4. Show that
∫| sin x | dx = 2n + 1 – cos θ, where
n ∈ N and 0 ≤ θ < π.
nπ+θ
Sol. I =
∫|
sin x | dx
0
0
0
nπ
nπ+θ
=
∫| sin x | dx +
∫| sin x | dx = I 1 + I 2 .
nπ
nπ
π
Now, I 1 =
∫| sin x | dx = n
∫| sin x | dx ,
0
0
(Q |sin x| is a periodic function of the period π)
π
= n
∫
sin x dx
0
(Q sin x is positive in the interval)
x
x 0
= n [– cos ] = n (1 + 1) = 2n.
nπ+θ
I 2 =
∫| sin x | dx . Putting x = nπ + z,
nπ
θ
I 2 =
∫| sin( nπ + z)
| dz = | sin z | dz
0
θ
∫
0
= | sin x | dx
θ
∫
0
= sin x dx (Q in 0 ≤ θ < π, |sin x| = sin x)
θ
θ
∫
0
= [– cos x ] 0 = – cos θ + 1
∴ I 1 + I 2 = 2n + 1 – cos θ.
5. The decimal parts of the logarithms of two numbers
taken at random are found to six places of decimal.
What is the chance that the second can be subtracted
from the first without "borrowing"?
Sol. For each column of the two numbers,
n(S) = number of ways to fill the two places by the
digits 0, 1, 2, ... , 9
= 10 × 10 = 100.
x
× × × × × ×
y
× × × × × ×
Let E be the event of subtracting in a column without
borrowing. If the pair of digits be (x, y) in the column
where x is in the first number and y is in the second
number then
E = {(0, 0), (1, 0), (2, 0), .. ,(9, 0),
(1, 1), (2, 1), ..., (9, 1),
(2, 2), (3, 2), ..., (9, 2),
(3, 3), (4, 3), ..., (9, 3),
......
(8, 8), (9, 8),
(9, 9)}
10.11
∴ n(E) = 10 + 9 + 8 + ... + 2 + 1 = = 55
2
∴ the probability of subtracting without borrowing
55
in each column = . 100
XtraEdge for IIT-JEE 44 MARCH 2011