Answer: Option A. -> 11
:
A
Option (a)
There are 11 digits in this number
When sum of digits is 2, there are 2 options
There are 2 ones, with one 1 fixed in the first position
The question in now based on the arrangement of 10 zeroes and 1 one
Number of possibilities = $\frac{10!}{9!1!}$ = 10 ways
When 2 is the first digit = one possibility.
Total number of possibilities = 11

Answer: Option A. -> 11
:
D
Option (d)nth term, This equation can be rewritten as S = =

Answer: Option A. -> 80
:
A
Option (a)
Calculating “n” which does not satisfy the requisite condition (i.e. the value of “n” which will not yield subsets with an integral multiple of 6).
The terms will fall in an AP with n =1 as the first term. The general form of the AP is 6k+1.
The last term will be 91. Thus, total number of sets which do not satisfy the condition= $\frac{90}{6}$ = 15+1 = 16.
Total number of terms = 96.
Thus, number of sub sets possible = 96 - 16 = 80. Option (a).

Answer: Option A. -> 80
:
C
Option (c) From the given condition in questionArea and perimeter of S₁ = a², 4aArea and perimeter of S₂ = Area and perimeter of S₃ =Area and perimeter of S₄ These are 2 infinite GPs , which can be solved as followsRequired ratio=== → →

Answer: Option D. -> b ≠ 1
:
D
(d) Let f(x) = x³ – ax² + bx – a =0
In the given equation, there are 3 sign changes, therefore there are at most 3 positive roots (Descarte’s Rule). If f(-x), there is no sign change. Thus there is no negative real root, i.e., if α, β and γ are the roots then they are all positive and we have
f(x) = (x - α)(x – β)(x - γ) = 0
→ b = αβ + βγ +γα → a = α + β + γ = αβγ
→
→ αβ, βγ, γα > 1 → b > 3.
Thus b ≠ 1.

Answer: Option D. -> 652
:
D
Ans. (d)Since sum of roots is given to you as 12 and product as 64. the six roots are 2,2,2,2,2,2.Sum of roots taken two at a time (a) = ⁶C₂x4=60Sum of roots taken three at a time(b)= ⁶C₃x8=160Sum of roots taken 4 at a time (c)= ⁶C₄x16=240Sum of roots taken 5 at a time(d)=⁶C₅x32=192a+b+c+d=652

Answer: Option D. -> 65
:
D
OPTION (d)g(2,3)= 85= 2² +3⁴g(5,4)= 281= 5² +4⁴therefore g(7,2)= 7²+2⁴= 65

Answer: Option B. -> 455
:
B
Similar to different grouping( permutation and combination)
a,b,c,d will all be some power of 5 from 0 to 12.
This question is same as a + b + c + d = 12
Solutions = ¹⁵C₃
12 identical things to 4 distinct groups =¹⁵ C₃ = 455

Answer: Option C. -> 2
:
C
Answer = cProduct of roots = 2e^2ln(k) -1 = 7; 2e^2ln(k) = 8; e^2ln(k) = 4; k² = 4; k = ±2For the roots to be real, k>0 k=2. to recheck CHECK for D>0; D= 9k² -4 (2e^2ln(k) -1) = 8>0

Answer: Option B. -> have –ve real parts
:
B
option (b)x=(-b±√D)/2aD=n²-4mp<n² (as m>0 and p>0), the roots of this equation will be –ve. If D<0, then the roots will have negative real parts