Maths question paper with solution

MATHS | 29th Jan 2023 _ Shift-1

,
, Section A
61. Let 𝛼 and 𝛽 be real numbers. Consider a 3 × 3 matrix A such that 𝐴2 = 3𝐴 + 𝛼𝐼. If 𝐴4 = 21𝐴 + 𝛽𝐼,
then
(1) 𝛽 = −8 (2) 𝛽 = 8 (3) 𝛼 = 4 (4) 𝛼 = 1
Sol. 1
A2 = 3A +  I …… (1)
and A = 21A +  I
4
……..(2)
4 2 2
Now A = A . A
A4 = (3A + I) . (3A + I) {from (1)}
A4 = 9A2 + 6A + 2I ……..(3)
From (2) and (3)
9A2 + 6A + I = 21 A +  I
putting value of A2 from (1)
9(3A + I) + 6A + 2I = 21 A +  I
(27 + 6)A + (9 + 2) I = 21A +  I
by comparison
27 + 6 = 21 and 9 + 2 = 
 6 = – 6 putting  = –1
  = –1  = – 8

62. Let 𝑥 = 2 be a root of the equation 𝑥 2 + 𝑝𝑥 + 𝑞 = 0 and
1 − cos⁡(𝑥 2 − 4𝑝𝑥 + 𝑞 2 + 8𝑞 + 16)
, 𝑥 ≠ 2𝑝
𝑓(𝑥) = { (𝑥 − 2𝑝)4
0, , 𝑥 = 2𝑝
lim𝑥→2𝑝+ [𝑓(𝑥)]
where [·] denotes greatest integer function, is
(1) 0 (2) −1 (3) 2 (4) 1
Sol. 1
 1  cos(x2  4px  q2  8q  16)
 ,x  2p
f(x) =  (x  2p)4
0 , x  2p

 x = 2 is a root of equation x2 + px + q =0
 4 + 2p + q = 0
 2p = –q – 4
 4p2 = (q + 4)2 = q2 + 8q + 16 …….(1)
1  cos(x – 4px  4p )
2 2
Now lim f(x)  lim (from (1))
x 2p x 2p (x  2p)4
 1  cos(x  2p)2 
= lim  
 {(x  2p) } 
2 2
x 2p

1  1  cos  1 
=  lim  
2  x 0 2 2
1 
 lim[f(x)]     0
x 2p 2