(A) We have x + y = |a| and ax – y = 1. Therefore,
− = 1 (a > 0)
Now, x + y = a and
- y = 1
For intersection, at the first quadrant, we have
a ∊
(mAB,mAC) mAB =
=
And a ∊ (1 , ∫)
= a ⇒
a2 = 1 ⇒ a = 1 , - 1
Therefore, a = 1.
mAB = 1 and
mAC = ∞. Hence, the value of
() =
Therefore, (A)→(Q).
(B) If the point (α, β, γ) lies on the plane x + y + z = 2, then α + β + γ = 2. Now,
=
×(×) =
(.) -
(.) =
-
= -
= 0
Therefore,
αi + β j = 0 ⇒ α = 0 = β ⇒ γ = 2
Therefore, (B)→(P).
(C) We have
I =
|(1−y2)dy| +
|(y2−1)dy| =
|(1−y2)dy| +
|−(y2−1)dy| |(1−y2)dy| +
|(1−y2)dy| =
2|(1−y2)dy| =
2‌(1−y2)dy = 2
(y−)|01 =
2(1−) =
I2 =
|√1−xdx|+|√1+xdx| =
=
2|√1−xdx| 2‌√1−xdx = - 4
t2 dt = 4
t2 dt =
⇒
I1 =
I2 Therefore,
1 – x =
t2 dx = −2t dt
Therefore, (C)→(R).
(D) We have
sinAsinBsinC + cosAcosB = 1
Therefore,
sin C =
0 < sin C ≤ 1
0 <
≤ 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
Therefore, (D)→(S).