JEE Advanced (One or More Correct Option): Mathematical induction & Binomial Theorem | Chapter-wise Tests for JEE Main & Advanced PDF Download

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 t₄ is numerically the greatest term,
| t₃ |<| t₄| and  | t₅ |<| t₄ |
⇒
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
=⁶C₂ 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 + x²)n = a₀ + a₁x + a₂x² + …… + a_2nx²n, then
(a) a₀ – a₂ + a₄ – a₆ + …… = 0, if n is odd
(b) a₁ – a₃ + a₅ – a₇ + …… = 0, if n is even
(c) a₀ – a₂ + a4 – a₆ + …… = 0, if n = 4p, p ∈ I+
(d) a₁ – a₃ + a₅ – a₇ + …… = 0, if n = 4p + 1, p ∈ I+

Correct Answer is options (a, b)

Putting . Then, we get

⇒
If n is odd, then Re(iⁿ) = 0
⇒ a₀ – a₂ + a₄ – a₆ + …… = 0
If n is even, then Im(iⁿ) = 0
⇒ a₁ – a₃ + a₅ – a₇ + …… = 0

Q.6. The value of the expression  is   (Here C_k = ⁿC_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)^n-1C_n-1, 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 – x²)^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 ( ⁿC_r )³ 
(d) Sum of coefficient in E is 2³ⁿ

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  x²
(b) there does not appear a term with the power  x²
(c) there appears a term with the power  x^-3
(d) the ratio of the co-efficient of x³ 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) x²
(c) 2x3
(d) All of these

Correct Answer is option (b)
(1 + x)ⁿ = 1 + nx + ([n (n − 1)] / [2!]) * x² + ([n (n − 1) (n − 2)] / [3!]) * x³ + . . . . .
(1 + x)ⁿ − nx − 1 = x² [([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 x².
Trick: (1 + x)ⁿ − nx − 1, put n = 2 and x = 3;
Then 4² − 2 * 3 − 1 = 9 is not divisible by 6, 54 but divisible by 9, which is given by option (b) i.e., x² = 9.