## Mathematics MCQ on Binomial Theorem for JEE and Engineering Exam 2022

MCQ on Binomial Theorem: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.

The Binomial Theorem is the method of expanding an expression that has been raised to any finite power. A binomial Theorem is a powerful tool of expansion, which has application in Algebra, probability, etc. In this post we are providing you MCQ on Binomial Theorem, which will be beneficial for you in upcoming JEE and Engineering Exams.

## MCQ on Binomial Theorem

Q1. If the coefficients of three consecutive terms in the expansion of (1+x)n are in the ratio 1:7:42 then find the value of n.
a) 35
b) 45
c) 55
d) 65

Let (r – 1)th, (r)th and (r + 1)th be the three consecutive terms.

Then, the given ratio is 1:7:42

Now (nCr-2 / nCr – 1) = (1/7)

(nCr-2 / nCr – 1) = (1/7) ⇒ [(r – 1)/(n − r+2)] = (1/7) ⇒ n−8r+9=0 → (1)

And,

(nCr-1 / nCr) = (7/42) ⇒ [(r)/(n – r +1)] =(1/6) ⇒ n−7r +1=0 → (2)

From (1) & (2), n = 55

Q2. Find the remainder when 7103 is divided by 25.
a) 53
b) 18
c) 35
d) 58

(7103 / 25) = [7(49)51 / 25)] = [7(50 − 1)51 / 25]

= [7(25K − 1) / 25] = [(175K – 25 + 25−7) / 25]

= [(25(7K − 1) + 18) / 25]

∴ The remainder = 18

Q3. Find the last two digits of the number (13)10
a) 89
b) 63
c) 49
d) 35

(13)10 = (169)= (170 − 1)5

= 5C0 (170)− 5C(170)+ 5C(170)− 5C(170)+ 5C(170) − 5C5

= 5C(170)− 5C(170)+ 5C(170)− 5C(170)+ 5(170) − 1

A multiple of 100 + 5(170) – 1 = 100K + 849

∴ The last two digits are 49

Q4. Find the last three digits of 2726.
a) 399
b) 499
c) 699
d) 799

By reducing 2726 into the form (730 – 1)n and using simple binomial expansion we will get required digits.

We have 27= 729

Now 2726 = (729)13 = (730 – 1)13

13C(730)13 – 13C(730)12 + 13C(730)11 – . . . . . – 13C10 (730)3 + 13C11(730)2 – 13C12 (730) + 1

= 1000m + [(13 × 12)]/2] × (14)2 – (13) × (730) + 1

Where ‘m’ is a positive integer

= 1000m + 15288 – 9490 = 1000m + 5799

Thus, the last three digits of 17256 are 799.

Q5. Find the positive value of λ for which the coefficient of xin the expression x2[x + (λ/x2)]10 is 720.
a) ±2
b) ±4
c) ±6
d) ±8

⇒ x[10Cr . (√x)10-r . (λ/x2)r] = x2 [10C. λr . x(10-r)/2 . x-2r]

= x2 [10C. λr . x(10-5r)/2]

Therefore, r = 2

Hence, 10C2 . λ2 = 720

⇒ λ2 = 16

⇒ λ = ±4

Q6. Find the coefficient of x9 in the expansion of (1 + x) (1 + x2 ) (1 + x3) . . . . . . (1 + x100).
a) 16
b) 8
c) 4
d) 9

x9 can be formed in 8 ways.

i.e., xx1+8 x2+7 x3+6 x4+5, x1+3+5, x2+3+4

∴ The coefficient of x= 1 + 1 + 1 + . . . . + 8 times = 8

Q7. If the fractional part of the number (2403 / 15) is (K/15), then find K.
a) 8
b) 10
c) 15
d) 17

(2403 / 15) = [2(24)100 / 15]

= 8/15 (15 + 1)100 = 8/ 15 (15λ + 1) = 8λ + 8/15

∵ 8λ is an integer, fractional part = 8/15

So, K = 8

Q8. Let (x + 10)50 + (x – 10)50 = a0 +a1x + a2 x2 + . . . . . + a50 x50 for all x ∈R, then a2/a0 is equal to?
a) 18.50
b) 19.75
c) 13.50
d) 12.25

⇒ (x + 10)50 + (x – 10)50:

a2 = 2 × 50C2 × 1048

a0 = 2 × 1050

⇒ a2/a0 = 50C2/102 = 12.25

Q9. The sum of the real values of x for which the middle term in the binomial expansion of (x3/3  + 3/x)8 equals 5670 is?
a) ± √3
b) ± √8
c) ± √5
d) ± √7

T5 = 8C4 × (x12/81) × (81/x4) = 5670

⇒ 70 x8 = 5670

⇒ x = ± √3

Q10. The coefficients of three consecutive terms of (1 + x)n+5 are in the ratio 5:10:14, find n.
a)  4
b) 7
c) 6
d) 9

Let Tr-1, Tr, Tr+1 are three consecutive terms of (1 + x)n+5

⇒  Tr-1 = (n+5) Cr-2 . xr-2

⇒  Tr = (n+5) Cr-1 . xr-1

⇒  Tr+1 = (n+5) Cr . xr

Given

(n+5) Cr-2 : (n+5) Cr-1 : (n+5) Cr = 5 : 10 : 14

Therefore, [(n+5) Cr-2]/5= [(n+5) Cr-1]/10 = (n+5) Cr/14

Comparing first two results we have n – 3r = -9 . . . . . . (1)

Comparing last two results we have 5n – 12r = -30 . . . . . .  (2)

From equation (1) and (2) n = 6

Q11. Find the number of terms free from the radical sign in the expansion of (√5 + 4√n)100.
a) 18
b) 26
c) 35
d) 74

Tr+1 = 100Cr . 5(100 – r)/2 nr/4

Where r = 0, 1, 2, . . . . . . , 100

r must be 0, 4, 8, … 100

Number of rational terms = 26

Q12. Find the total number of terms in the expansion of (x + a)100 + (x – a)100.
a) 17
b) 27
c) 37
d) 51

⇒ (x + a)100 + (x – a)100 = 2[100C0 x100100C x98 . a2 + . . . . . . +  100C100 a100]

∴ Total Terms = 51

Q13. Find the degree of the polynomial [x + {√(3(3-1))}1/2]5 + [x + {√(3(3-1))}1/2]5.
a) 3
b) 5
c) 7
d) 9

[x + { √(3(3-1)) }1/2 ]5:

= 2 [5C0 x5 + 5C2 x5 (x3 – 1) + 5C4 . x . (x3 – 1)2]

Therefore, the highest power = 7

Q14. Find the coefficient of in the expansion of (1 + x + x+x3)11.
a) 660
b) 770
c) 880
d) 990

By expanding given equation using expansion formula we can get the coefficient x4

i.e. 1 + x + x+ x= (1 + x) + x(1 + x) = (1 + x) (1 + x2)

⇒ (1 + x + x+ x3) x11 = (1+x)11 (1+x2)11

= 1+ 11Cx11Cx11Cx11Cx. . . . . . .

= 1 + 11Cx11Cx+ . . . . . .

To find term in from the product of two brackets on the right-hand-side, consider the following products terms as

= 1 × 11Cx11Cx2 × 11Cx11Cx4

11C11C× 11C11C] x4

⇒ [55 + 605 + 330] x4 = 990x4

∴ The coefficient of x4 is 990

Q15. Find the number of irrational terms in (8√5 + 6√2)100.
a) 63
b) 93
c) 97
d) 57

Tr + 1 = 100Cr (8√5)100 – r . (6√2)= 100C. 5[(100 – r)/8] .2r/6.

∴ r = 12,36,60,84

The number of rational terms = 4

Number of irrational terms = 101 – 4 = 97

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