A. Let f(x) =
x2 – 2kx +
k2 – 1 = 0
use f(–3) > 0, f(4) > 0 and
discriminant > 0 to get –1 < k < 1.
B. In the expansion of
(1+x)30, 16th term is middle term and it will have greatestcoefficient. It should be the greatest term → 15 <
< 16
x ∊
(,) C. Let the sides be a, ar,
ar2 and angles opposite to a and
ar2 be θ and 2θ, the using sine rule
=
or
r2 = 2 cos θ . Obviously 2θ > 60° as 2θ is the largestangle.
∴ cos 2θ < 1/2
∴
2cos2θ - 1 < 1/2 or
cos2θ < 3/4
But
r2 = 32cosθ
→
< 3/4 r <
31∕4 But r > 1 ∴ r ∊
(1,31∕4) D . f (x) =
= 1 +
f(x) takes maximum value 3 when x = 1
and f(x) → 1 as x → ± ∞
∴ f(x) ∈ (1, 3].