- (c) : S
Q
ab
aba
bSQ Q QS
S=
−(^222) ⇒ = +± +
2
()
- (a) : Note that sides are ()sinab− α
2
and
()cosab− α
2- A → q ; B → p, q, r ; C → r
(A) ∵ max | |,| |{}xy= 12 /
|| /, || /
|| /, || /xy
yx=<
=<⎧
⎨
⎩12 12
12 12if
ifRequired area = 1 × 1
= –1/= –1/= 1/= 1/= 1 sq. unit
(B) The line y = x cuts
the lines |x + y| = 6
i.e, x + y = ± 6
At x = ± 3, ⇒ (x, y) ≡ (–3, –3) and (3, 3)
then –3 < a < 3
∴ 0 ≤ |a| < 3
⇒ [|a|] = 0, 1, 2
(C) Since (0, 0) and (1, 1) lie on the same side.
So, a^2 + ab + 1 > 0
∵ Coefficient of a^2 > 0 ∴ D < 0
b^2 – 4 < 0 or – 2 < b < 2
⇒ b = –1, 0, 1
∴ Number of non-zero integral values of b are 2.
(b = –1 and b = 1)
- (2) : If x ∈ (0, 1)
Then –1 ≤ x + y < 0
And if x ∈ [1, 2)
0 ≤ x + y < 1
2
1(^123)
+=–1+= 0+= 1Required area
= 4 ⎛⎝⎜^1 ⋅⋅ ⎞⎠⎟
212
4sinπ= 2 sq. units- (3) : Let, D = (2, –1) be
the reflection of A in x-axis,
and let E = (1, 2) be the
reflection in the line y = x.
Then AB = BD and AC = CE,
so the perimeter of ABC is
DB + BC + CE ≥ DE =^10 - (4) : 2(a + b) = x (a constant)
(^)
Area of PQRS = (bsinθ + acosθ)(asinθ + bcosθ)
=+ab ab+ ≤ ab x x+ = ∴ = ⇒x=
22 2 2 2
2
2
288
sin θ () 32 16
- (1)
!
"#
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