IIT JEE Advanced 2006 Paper 1

(A) We have,y=(sin‌x)cos‌x
ln‌y=cos‌x×ln(sin‌x)
Now,
y′=y(cos‌x‌cot‌x−sin‌x‌ln‌sin‌x)
y′=(sin‌x)cos‌x(cos‌x‌cot‌x−sin‌x‌ln‌sin‌x)
Therefore,


Therefore, (A)→(P)

(B) For the curves, we have
y2=‌

−x

4

y2=−‌

1

5

(x−1)
That is,
‌‌−‌

x

4

=−‌

1

5

(x−1)
5x=4x−4
x=−4
y2=1⇒y=−1,1 Therefore, the area bounded is given by,

1

∫

−1

(1−5y2+4y2)dy‌‌
=2‌

1

∫

0

(1−y2)dy
=2(y−‌

y3

3

)|01
‌‌=2(1−‌

1

3

)=‌

4

3

Therefore (B) → (S)

(C) The point of intersection of the curve is
y3=3x−1(ln‌x)
and y=xx−1 is (1,0). Now, for the first curve, we have
y′=3x−1(‌

1

x

)+(3x−1‌ln‌x‌ln‌3)
y(1,0)′=1+0=1
and for the second curve, we have
y′=(y+1)(1+ln‌x)
y(1,0)′=1
Therefore, the angle between the curves is 0 .
Therefore, (C)→(Q).

(D) We have
‌

dy

dx

=‌

2

x+y

⇒x+y=t
1+‌

dy

dx

=‌

dt

dx

That is,
‌

dt

dx

−1=‌

2

t

⇒‌

dt

dx

=‌

2

t

+1=‌

t+2

t

⇒‌

t

t+2

dt=dx
⇒∫(‌

t+2−2

t+2

)dt=∫dx+c
⇒∫(1−‌

2

t+2

)dt=∫dx+c
⇒t−2‌ln(t+2)=x+c
⇒(x+y)−2‌ln(x+y+2)=x+c
⇒1−2‌ln‌3=1+c[‌

dy

dx

=‌

2

x+y

‌ passes through ‌(1,0)]
⇒c=−2‌ln‌3
⇒y−2‌ln(x+y+2)=−2‌ln‌3
⇒y=2‌ln(‌

x+y+2

3

)
Therefore,
x+y+z=3ey∕2
Therefore, (D)→(R).