IIT JEE Advanced 2006 Paper 1

(A) We have x + y = |a| and ax – y = 1. Therefore,

x

1 a

−

y

1

= 1 (a > 0)
Now, x + y = a and

x

( 1 a )

- y = 1

For intersection, at the first quadrant, we have
a ∊ (mAB,mAC)
mAB =

0+1

a−0

=

1

a

And a ∊ (1 , ∫)

1

a

= a ⇒ a2 = 1 ⇒ a = 1 , - 1
Therefore, a = 1. mAB = 1 and mAC = ∞. Hence, the value of
(

2ac

3

) =

2

3

Therefore, (A)→(Q).
(B) If the point (α, β, γ) lies on the plane x + y + z = 2, then α + β + γ = 2. Now,

^

a

=

α ^ i +β ^ j +γ ^ k

√α2+β2γ2

^

k

×(

^

k

×

^

a

) = (

^

k

.

^

a

)

^

k

- (

^

k

.

^

k

)

^

a

=

γ ^ k

√α2+β2+γ2

-

(α ^ i +β ^ j +γ ^ k )

√α2+β2+γ2

= -

(α ^ i +β ^ j )

√α2+β+γ2

= 0
Therefore,
αi + β j = 0 ⇒ α = 0 = β ⇒ γ = 2
Therefore, (B)→(P).
(C) We have
I = |

1

∫

0

(1−y2)dy| + |

0

∫

1

(y2−1)dy|
= |

1

∫

0

(1−y2)dy| + |−

1

∫

0

(y2−1)dy|
|

1

∫

0

(1−y2)dy| + |

1

∫

0

(1−y2)dy|
= 2|

1

∫

0

(1−y2)dy|
= 2‌

1

∫

0

(1−y2)dy = 2 (y−

y3

3

)|01 = 2(1−

1

3

) =

4

3

I2 = |

1

∫

0

√1−xdx|+|

0

∫

−1

√1+xdx|
= = 2|

1

∫

0

√1−xdx|
2‌

1

∫

0

√1−xdx
= - 4

0

∫

1

t2 dt = 4

1

∫

0

t2 dt =

4

3

⇒ I1 = I2
Therefore,
1 – x = t2
dx = −2t dt
Therefore, (C)→(R).
(D) We have
sinAsinBsinC + cosAcosB = 1
Therefore,
sin C =

1−cosAcosB

sinBsinA

0 < sin C ≤ 1
0 <

1−cosAcosB

sinBsinA

≤ 1
1 – cosAcosB ≤ sinAsinB
1 ≤ cosAcosB + sinAsinB
cos(A – B) ≥ 1
Therefore, it is possible only if
cos(A – B) = 1
A – B = 0
A = B
Therefore,
sin C =

1−cos2A

sin2a

= 1
Therefore, (D)→(S).