Q.1. Given that the 4th term in the expression of has the maximum numerical value, then x can lie in the interval(s)
(a)
(b) ( 2, 2)
(c)
(d)
Correct Answer is options (a, c)
Since t_{4} is numerically the greatest term,
 t_{3} < t_{4} and  t_{5} < t_{4} 
⇒
But
⇒
and
⇒
Q.2. For a positive integer n, if the expansion of has a term independent of x, then n can be
(a) 18
(b) 21
(c) 27
(d) 99
Correct Answer is options (a, b, c, d)
Let (r + 1)^{th} term of be independent of x, we have
For this term to be independent of x, 6r  2n = 0 or n = 3r
∴ Each of 18, 21, 27, 99 is divisible by 3.
Q.3. Let , where x_{n}, y_{n} are integers, then
(a)
(b)
(c)
(d)
Correct Answer is options (a, b, d)
We have (1)
(2)
From (1) and (2), we get
Next,
=
Thus, and
Q.4. Which of the following is/are true
(a)
(b)
(c)
(d)
Correct Answer is options (a, b, c)
(A) No. of onto functions from a set containing 6 elements to a set containing 5 elements
=^{6}C_{2} L5
(B) No .of onto functions from a set containing 5 elements to a set containing 6 elements = 0
(C) No. of onto function from a set containing 6 elements to a set containing 6 elements
= L6 = 720.
Q.5. If (1 + x + x^{2})n = a_{0} + a_{1}x + a_{2}x^{2} + …… + a_{2}_{n}x^{2}n, then
(a) a_{0} – a_{2} + a_{4} – a_{6} + …… = 0, if n is odd
(b) a_{1} – a_{3} + a_{5} – a_{7} + …… = 0, if n is even
(c) a_{0} – a_{2} + a4 – a_{6} + …… = 0, if n = 4p, p ∈ I+
(d) a_{1} – a_{3} + a_{5} – a_{7} + …… = 0, if n = 4p + 1, p ∈ I+
Correct Answer is options (a, b)
Putting . Then, we get
⇒
If n is odd, then Re(i^{n}) = 0
⇒ a_{0} – a_{2} + a_{4} – a_{6} + …… = 0
If n is even, then Im(i^{n}) = 0
⇒ a_{1} – a_{3} + a_{5} – a_{7} + …… = 0
Q.6. The value of the expression is (Here C_{k} = ^{n}C_{k})
(a) 0, if n is odd
(b) (1)^{n}, if n is odd
(c) (1)^{n/2} C_{n/2}, if n is even
(d) (1)^{n1}C_{n1}, if n is even
Correct Answer is options (a, c)
When n is odd, taken n = 2m + 1, so that
=
But
etc.
Therefore S = 0
When n is even, we take n = 2m. In this case
= Coefficient of constant term in
= Coefficient of constant term in
= Coefficient of x^{2m} in (1 – x)^{2m}(1 + x)^{2m} = Coefficient of x^{2m} in (1 – x^{2})^{2m}
=
Q.7. Let an expression E be given by E = (1 + x)^{n} (1 + y)^{n} (1 + z)^{n} then
(a) number of dissimilar terms in E will be (n + 1)^{3}
(b) number of dissimilar terms in E will be n3
(c) coefficient of n r in E is ( ^{n}C_{r} )^{3}
(d) Sum of coefficient in E is 2^{3n}
Correct Answer is options (a, d)
Each individual expansion will have n +1 terms.
Put x = y = z = 1
Q.8. The cube of any whole number when divided by 9 may yield the reminder
(a) 0
(b) 2
(c) 1
(d) 8
Correct Answer is options (a, c, d)
Any whole number is either 9K, 9K+1, ……, 9K+8 When we cube them reminder will be 0, 1 or 8 only
Q.9. In the expansion of
(a) there appears a term with the power x^{2}
(b) there does not appear a term with the power x^{2}
(c) there appears a term with the power x^{3}
(d) the ratio of the coefficient of x^{3 }to that of x^{3} is
Correct Answer is options (b, c, d)
Now 33 – 6r = 2 ⇒ 6r = 31 (not possible)
33 – 6r = –3 ⇒ r = 6
33 – 6r = 3 ⇒ r = 5
∴
Q.10. 1 + x)n − nx − 1 is divisible by (where n∈N)
(a) 2x
(b) x^{2}
(c) 2x3
(d) All of these
Correct Answer is option (b)
(1 + x)^{n }= 1 + nx + ([n (n − 1)] / [2!]) * x^{2 }+ ([n (n − 1) (n − 2)] / [3!]) * x^{3 }+ . . . . .
(1 + x)^{n }− nx − 1 = x^{2} [([n (n − 1)] / [2!]) + ([n (n − 1) (n − 3)] / [3!]) * x + . . . . .]
From above it is clear that (1 + x)^{n }− nx − 1 is divisible by x^{2}.
Trick: (1 + x)^{n }− nx − 1, put n = 2 and x = 3;
Then 4^{2 }− 2 * 3 − 1 = 9 is not divisible by 6, 54 but divisible by 9, which is given by option (b) i.e., x^{2 }= 9.
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