A. Let
tan−1 x = 0 then
≤ θ
as
≤ x ≤
cos−1() + tan
=
cos−1 () +
tan−1 () cos−1 (sin2θ) +
tan−1 (tan2θ) = π/2 - 2θ + 2θ = π/2
∴ Given integral =
using formula
=
{ +
} =
B. PA + PB < 2 → P lies inside the ellipse
+
= 1
and PB + PC < 2 → P lies inside the ellipse
+
= 1
Both should hold according to given condition.
∴ Required to find should area DQ ERD
D ≡ (1, 0) B ≡ (3/2, 0).
From the second ellipse y =
√1−(x−2)2
Required area = 4
y dx =
2√3 √1−(x−2)2 dx
=
2√3 [√1−(x−2)2 +
sin−1(x−2)]13∕2 =
√3(−) C. Given ∫ f (x)
sec2 x dx = f(x) sec x + c → f(x)
sec2 x = f(x) sec x + tan x + sin f′(x).
→ f(x) sec x = f(x) tan x + f′(x) or
= sec x - tan x
=
=
=
D. Hyperbola is
-
= 1 , latus rectum = 1/2
⇒
= 1/2
⇒
× sin θ =
2 sin θ cos θ =
→ sin2θ =
⇒ 2θ =
or
∴ θ =
or