JEE Advanced Model Paper 19 with solutions for online practice

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 <

31x

1+x

< 16
x ∊ (

15

16

,

16

15

)
C. Let the sides be a, ar, ar2 and angles opposite to a and ar2 be θ and 2θ, the using sine rule

sin2θ

ar2

=

sinθ

a

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
→

r4

4

< 3/4 r < 31∕4
But r > 1 ∴ r ∊ (1,31∕4)
D . f (x) =

x4−x2−2x−8

x−x2−2x+4

= 1 +

4

(x2−1)2+(x−1)2+2

f(x) takes maximum value 3 when x = 1
and f(x) → 1 as x → ± ∞
∴ f(x) ∈ (1, 3].