12th Grade > Mathematics
BINOMIAL THEOREM MCQs
Binomial Theorem
Total Questions : 74
| Page 1 of 8 pages
Answer: Option B. -> 12
:
B
mc2=−18⇒m(m−1)2=−18⇒4m2−4m+1=0⇒(2m−1)2=0⇒m=12
:
B
mc2=−18⇒m(m−1)2=−18⇒4m2−4m+1=0⇒(2m−1)2=0⇒m=12
Answer: Option C. -> 2a2a2+a3
:
C
Let a1,a2,a3,a4 be respectively the coefficients of (r+1)th,(r+2)th,(r+3)thand(r+4)th terms in the expansion of (1+x)n.
Then a1=nCr,a2=nCr+1,a3=nCr+2,a4=nCr+3
Now,
a1a1+a2+a3a3+a4=nCrnCr+nCr+1+nCr+2nCr+2+nCr+3=nCrn+1Cr+1+nCr+2n+1Cr+3=nCrn+1r+1nCr+nCrn+1r+3nCr+2=r+1n+1+r+3n+3=2(r+2)n+1
Also, solving the R.H.S, we get
2a2a2+a3=2nCr+1nCr+1+nCr+2=2nCr+1n+1Cr+2=2(r+2)n+1
:
C
Let a1,a2,a3,a4 be respectively the coefficients of (r+1)th,(r+2)th,(r+3)thand(r+4)th terms in the expansion of (1+x)n.
Then a1=nCr,a2=nCr+1,a3=nCr+2,a4=nCr+3
Now,
a1a1+a2+a3a3+a4=nCrnCr+nCr+1+nCr+2nCr+2+nCr+3=nCrn+1Cr+1+nCr+2n+1Cr+3=nCrn+1r+1nCr+nCrn+1r+3nCr+2=r+1n+1+r+3n+3=2(r+2)n+1
Also, solving the R.H.S, we get
2a2a2+a3=2nCr+1nCr+1+nCr+2=2nCr+1n+1Cr+2=2(r+2)n+1
Answer: Option B. -> 198
:
B
(x+a)n + (x−a)n = 2[xn+nC2xn−2a2+nC4xn−4a4+nC6xn−6a6+.......]
Here, n = 6, x =√2, a = 1;
6C2 = 15,6C4 = 15,6C6 = 1
∴ (√2+1)6 + (√2−1)6 = 2[(√2)6+15.(√2)4.1+15(√2)2.1+1.1]
=2[8+15×4+15×2+1]=198
:
B
(x+a)n + (x−a)n = 2[xn+nC2xn−2a2+nC4xn−4a4+nC6xn−6a6+.......]
Here, n = 6, x =√2, a = 1;
6C2 = 15,6C4 = 15,6C6 = 1
∴ (√2+1)6 + (√2−1)6 = 2[(√2)6+15.(√2)4.1+15(√2)2.1+1.1]
=2[8+15×4+15×2+1]=198
Answer: Option C. -> 9
:
C
Tr =16Cr−1 (x4)16−r(1x3)r−1 =16Cr−1x67−7r
→67−7r=4→r=9
:
C
Tr =16Cr−1 (x4)16−r(1x3)r−1 =16Cr−1x67−7r
→67−7r=4→r=9
Answer: Option C. -> 23 and 9
:
C
nc1.ax=6x;nc2(ax)2=16x2⇒an=6;n(n−1)2a2=16⇒a=23⇒n=9
:
C
nc1.ax=6x;nc2(ax)2=16x2⇒an=6;n(n−1)2a2=16⇒a=23⇒n=9
Answer: Option C. -> 9
:
C
16=nC6(213)n−6(3−13)6nCn−6(213)6(3−13)n−6
6−1=613(n−12)
n−12=−3
n=9
:
C
16=nC6(213)n−6(3−13)6nCn−6(213)6(3−13)n−6
6−1=613(n−12)
n−12=−3
n=9
Answer: Option A. -> 1
:
A
The numerator is of the form
a3 + b3+3ab(a+b)=(a+b)3
∴ N' = (18+7)3 = 253
∴ D' = 36+6C1⋅35⋅21+6C2⋅34⋅22+6C3⋅33⋅23+6C4⋅32⋅24+6C5⋅31⋅25+6C6⋅26
This is clearly the expansion of (3+2)6=56=(25)3
N′D′=(25)3(25)3=1
:
A
The numerator is of the form
a3 + b3+3ab(a+b)=(a+b)3
∴ N' = (18+7)3 = 253
∴ D' = 36+6C1⋅35⋅21+6C2⋅34⋅22+6C3⋅33⋅23+6C4⋅32⋅24+6C5⋅31⋅25+6C6⋅26
This is clearly the expansion of (3+2)6=56=(25)3
N′D′=(25)3(25)3=1
Answer: Option B. -> 6
:
B
In the expansion of(y15+x110)55, the general
term is
Tr+1 =55Cr(y15)55−r(x110)r =55Cry11−r5xr10.
This Tr+1 will be independent of radicals if the exponents r5 and r10 are integers, for
0≤ r≤ 55 which is possible only when
r = 0, 10, 20, 30, 40, 50.
∴ There are six terms viz. T1,T11,T21,T31,T41,T51
which are independent of radicals.
:
B
In the expansion of(y15+x110)55, the general
term is
Tr+1 =55Cr(y15)55−r(x110)r =55Cry11−r5xr10.
This Tr+1 will be independent of radicals if the exponents r5 and r10 are integers, for
0≤ r≤ 55 which is possible only when
r = 0, 10, 20, 30, 40, 50.
∴ There are six terms viz. T1,T11,T21,T31,T41,T51
which are independent of radicals.
Answer: Option B. -> 198
:
B
(x+a)n + (x−a)n = 2[xn+nC2xn−2a2+nC4xn−4a4+nC6xn−6a6+.......]
Here, n = 6, x =√2, a = 1;
6C2 = 15,6C4 = 15,6C6 = 1
∴ (√2+1)6 + (√2−1)6 = 2[(√2)6+15.(√2)4.1+15(√2)2.1+1.1]
=2[8+15×4+15×2+1]=198
:
B
(x+a)n + (x−a)n = 2[xn+nC2xn−2a2+nC4xn−4a4+nC6xn−6a6+.......]
Here, n = 6, x =√2, a = 1;
6C2 = 15,6C4 = 15,6C6 = 1
∴ (√2+1)6 + (√2−1)6 = 2[(√2)6+15.(√2)4.1+15(√2)2.1+1.1]
=2[8+15×4+15×2+1]=198
Answer: Option D. -> 6
:
D
18C2r+3 =18Cr−3 ⇒ 2r + 3 + r - 3 = 18 ⇒ r = 6
:
D
18C2r+3 =18Cr−3 ⇒ 2r + 3 + r - 3 = 18 ⇒ r = 6