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 (nC_{r-2} / nC_{r – 1}) = (1/7)
(nC_{r-2} / nC_{r – 1}) = (1/7) ⇒ [(r – 1)/(n − r+2)] = (1/7) ⇒ n−8r+9=0 → (1)
And,
(nC_{r-1} / nC_{r}) = (7/42) ⇒ [(r)/(n – r +1)] =(1/6) ⇒ n−7r +1=0 → (2)
From (1) & (2), n = 55
Q2. Find the remainder when 7^{103} is divided by 25.
a) 53
b) 18
c) 35
d) 58
(7^{103 }/ 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)^{5 }= (170 − 1)^{5}
= 5C_{0} (170)^{5 }− 5C_{1 }(170)^{4 }+ 5C_{2 }(170)^{3 }− 5C_{3 }(170)^{2 }+ 5C_{4 }(170) − 5C_{5}
= 5C_{0 }(170)^{5 }− 5C_{1 }(170)^{4 }+ 5C_{2 }(170)^{3 }− 5C_{3 }(170)^{2 }+ 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 27^{26}.
a) 399
b) 499
c) 699
d) 799
By reducing 27^{26} into the form (730 – 1)^{n} and using simple binomial expansion we will get required digits.
We have 27^{2 }= 729
Now 27^{26} = (729)^{13} = (730 – 1)^{13}
= ^{13}C_{0 }(730)^{13} – ^{13}C_{1 }(730)^{12} + ^{13}C_{2 }(730)^{11} – . . . . . – ^{13}C_{10 }(730)^{3} + ^{13}C_{11}(730)^{2} – ^{13}C_{12 }(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 17^{256} are 799.
Q5. Find the positive value of λ for which the coefficient of x^{2 }in the expression x^{2}[√x + (λ/x^{2})]^{10} is 720.
a) ±2
b) ±4
c) ±6
d) ±8
⇒ x^{2 }[^{10}C_{r} . (√x)^{10-r} . (λ/x^{2})^{r}] = x^{2} [^{10}C_{r }. λ^{r} . x^{(10-r)/2} . x^{-2r}]
= x^{2} [^{10}C_{r }. λ^{r} . x^{(10-5r)/2}]
Therefore, r = 2
Hence, ^{10}C_{2} . λ^{2} = 720
⇒ λ^{2} = 16
⇒ λ = ±4
Q6. Find the coefficient of x^{9} in the expansion of (1 + x) (1 + x^{2} ) (1 + x^{3}) . . . . . . (1 + x^{100}).
a) 16
b) 8
c) 4
d) 9
x^{9} can be formed in 8 ways.
i.e., x^{9 }x^{1+8 }x^{2+7 }x^{3+6 }x^{4+5}, x^{1+3+5}, x^{2+3+4}
∴ The coefficient of x^{9 }= 1 + 1 + 1 + . . . . + 8 times = 8
Q7. If the fractional part of the number (2^{403} / 15) is (K/15), then find K.
a) 8
b) 10
c) 15
d) 17
(2^{403} / 15) = [2^{3 }(2^{4})^{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} = a_{0} +a_{1}x + a_{2} x_{2} + . . . . . + a_{50} x^{50} for all x ∈R, then a_{2}/a_{0} is equal to?
a) 18.50
b) 19.75
c) 13.50
d) 12.25
⇒ (x + 10)^{50 }+ (x – 10)^{50}:
a_{2} = 2 × ^{50}C_{2} × 10^{48}
a_{0} = 2 × 10^{50}
⇒ a_{2}/a_{0} = ^{50}C_{2}/10^{2} = 12.25
Q9. The sum of the real values of x for which the middle term in the binomial expansion of (x^{3}/3 + 3/x)^{8} equals 5670 is?
a) ± √3
b) ± √8
c) ± √5
d) ± √7
T_{5} = ^{8}C_{4} × (x^{12}/81) × (81/x^{4}) = 5670
⇒ 70 x^{8} = 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 T_{r-1}, T_{r}, T_{r+1 }are three consecutive terms of (1 + x)^{n+5}
⇒ T_{r-1} = (n+5) C_{r-2} . x^{r-2}
⇒ T_{r} = (n+5) C_{r-1} . x^{r-1}
⇒ T_{r+1} = (n+5) C_{r} . x^{r}
Given
(n+5) C_{r-2} : (n+5) C_{r-1 }: (n+5) C_{r} = 5 : 10 : 14
Therefore, [(n+5) C_{r-2}]/5= [(n+5) C_{r-1}]/10 = (n+5) C_{r}/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
T_{r+1} = ^{100}C_{r} . 5^{(100 – r)/2} n^{r/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[^{100}C_{0} x^{100}. ^{100}C_{2 } x^{98} . a^{2} + . . . . . . + ^{100}C_{100 }a^{100}]
∴ 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 [^{5}C_{0} x^{5} + ^{5}C_{2} x^{5} (x^{3} – 1) + ^{5}C_{4} . x . (x^{3} – 1)^{2}]
Therefore, the highest power = 7
Q14. Find the coefficient of in the expansion of (1 + x + x^{2 }+x^{3})^{11}.
a) 660
b) 770
c) 880
d) 990
By expanding given equation using expansion formula we can get the coefficient x^{4}
i.e. 1 + x + x^{2 }+ x^{3 }= (1 + x) + x^{2 }(1 + x) = (1 + x) (1 + x^{2})
⇒ (1 + x + x^{2 }+ x^{3}) x^{11 }= (1+x)^{11 }(1+x^{2})^{11}
= 1+ ^{11}C_{1 }x^{2 }+ ^{11}C_{2 }x^{2 }+ ^{11}C_{3 }x^{3 }+ ^{11}C_{4 }x^{4 }. . . . . . .
= 1 + ^{11}C_{1 }x^{2 }+ ^{11}C_{2 }x^{4 }+ . . . . . .
To find term in from the product of two brackets on the right-hand-side, consider the following products terms as
= 1 × ^{11}C_{2 }x^{4 }+ ^{11}C_{2 }x^{2} × ^{11}C_{1 }x^{2 }+ ^{11}C_{4 }x^{4}
= ^{11}C_{2 }+ ^{11}C_{2 }× ^{11}C_{1 }+ ^{11}C_{4 }] x^{4}
⇒ [55 + 605 + 330] x^{4} = 990x^{4}
∴ The coefficient of x^{4} is 990
Q15. Find the number of irrational terms in (^{8}√5 + ^{6}√2)^{100}.
a) 63
b) 93
c) 97
d) 57
T_{r + 1} = 100C_{r} (^{8}√5)^{100 – r }. (^{6}√2)^{r }= 100C_{r }. 5[(100 – r)/8] .2^{r/6}.
∴ r = 12,36,60,84
The number of rational terms = 4
Number of irrational terms = 101 – 4 = 97
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