Binomial Theorem MCQ AP Inter 1st Year Maths Chapter 7

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Practice AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Binomial Theorem MCQ

Question 1.
The 3rd term of (3x – \(\frac{y^3}{6}\))4
1) \({ }^4 C_2(3 x)^2\left(\frac{-y^3}{6}\right)^2\)
2) \({ }^4 C_2(3 x)^2\left(\frac{-y^2}{6}\right)^2\)
3) \({ }^4 C_2(3 x)^3\left(\frac{-y^3}{6}\right)^2\)
4) \({ }^4 C_2(3 x)^3\left(\frac{-y^2}{6}\right)^2\)
Answer:
1) \({ }^4 C_2(3 x)^2\left(\frac{-y^3}{6}\right)^2\)

Explanation:
Given (3x – \(\frac{y^3}{6}\))4, Here a = 3x, b = \(\frac{-y^3}{6}\), n = 4, Tr+1 = nCr an-r.br

T3 = T2+1 = 4C2 (3x)4-2\(\left(\frac{-y^3}{6}\right)^2\)
= 4C2 (3x)2\(

Question 2.
C0 + 3.C1 + 32.C2 + …………. + 3n.Cn =
1) 2n
2) 3n
3) 4n
4) 5n
Answer:
3) 4n

Explanation:
We know that (1 + x)n = C0 + C1x + C1x2 + C3x3 + … + Cnxn … (1)
Put x = 3 in eq (1) we get C0 + C13 + C232 + … + Cn3n = (1 + 3)n
C0 + 3.C1 + 32.C2 + …… + 3n.Cn = 4n

Question 3.
The middle term in the expansion of (4 + 2x)6 is
1) 11240 x2
2) 10240 x3
3) 12240 x4
4) 10340 x4
Answer:
2) 10240 x3

Explanation:
Given (4 + 2x)6 contains 7 terms
The middle term T4 = T3+1 = 6C3 (4)6-3. (2x)3 = 20 × 64 × 8 × x3 = 10240 x3

Question 4.
The coefficient of x2y2 in the expansion of (x + 1)2 (y + 1)3 is
1) 3
2) 5
3) 2
4) 10
Answer:
1) 3

Explanation:
(x + 1)2 (y + 1)3 = (x2 + 2x + 1) (y3 + 3y2 + 3y + 1)
Coefficient of x2y2 is “3”

Binomial Theorem MCQ AP Inter 1st Year Maths Chapter 7

Question 5.
In the expansion of (1 + x)20, if the coefficients of rth and (r + 4)th terms are equal, then r is
1) 7
2) 8
3) 9
4) 10
Answer:
3) 9

Explanation:
Given expansion is (1 + x)20
Given condition is coefficient of r* term = coeff of (r + 4)th term
⇒ 20Cr-1 = 20Cr+3
⇒ r – 1 + r + 3 = 20 [∵ nCr = nCs ⇒ r + s = n]
⇒ 2r = 18
⇒ r = 9

Question 6.
The remainder when 848 is divided by 63 is
1) 4
2) 2
3) 1
4) 7
Answer:
3) 1

Explanation:
848 = (64)24 = (63 + 1)24 = 63q + 1 [∵ q is some Integer]
∴ Remainder = 1.

Question 7.
The expansion of (x + y)1000 is
1) [latex]\sum_{\mathrm{r}=0}^{1000} 1000 \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}-1000} \mathrm{y}^{\mathrm{r}}\)
2) \(\sum_{\mathrm{r}=0}^{1000} 1000 \mathrm{C}_{\mathrm{r}} \mathrm{x}^{1000-\mathrm{r}^2} \mathrm{y}^{\mathrm{r}}\)
3) \(\sum_{r=0}^{999}{ }^{1000} C_r x^{r-1000} y^r\)
4) \(\sum_{r=0}^{999}{ }^{1000} C_r x^{1000-r} y^r\)
Answer:
2) \(\sum_{\mathrm{r}=0}^{1000} 1000 \mathrm{C}_{\mathrm{r}} \mathrm{x}^{1000-\mathrm{r}^2} \mathrm{y}^{\mathrm{r}}\)

Explanation:
Given expansion = (x + y)1000 = \(\sum_{\mathrm{r}=0}^{1000} 1000 \mathrm{C}_{\mathrm{r}} \mathrm{x}^{1000-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}\)

Question 8.
The coefficients of the first and the last terms of the expansion (x + y)1000 are
1) 2
2) 3
3) 1
4) 7
Answer:
2) 3

Explanation:
Given expansion is (x + y)n
(x + y)n = nC0xn + nC1xn-1y + … + nCn. yn
Coefficient of first and last terms are nC0 and nCn
nC0 = nCn = 1.

Question 9.
The expansion of (xy + 2)2 is
1) x2 + y2 + 4
2) x2 y2 + 4xy + 4
3) xy2 + 4 + 2xy
4) x2y2 + 2xy + 4
Answer:
2) x2 y2 + 4xy + 4

Explanation:
(xy + 2)2 = (xy)2 + 2(xy) . 2 + 22 = x2y2 + 4xy + 4

Question 10.
The term 91C2x89 belongs to
1) x89
2) (x – 2)90
3) (x – 1)91
4) (x + 1)90
Answer:
3) (x – 1)91

Explanation:
Term 91C2x89 is belongs to (x – 1)91
T3 of (x – 1)91 = 91C2x91-2.(-1)2
= 91C2x89

Binomial Theorem MCQ AP Inter 1st Year Maths Chapter 7

Question 11.
The cofficient of x8y10 in the expansion f (x + y)18 is
1) 18C8
2) 18P10
3) 218
4) None of these
Answer:
1) 18C8

Explanation:
Given expansion is (x + y)18
The coefficient of x18y18 = 18C10
= 18C18-10
= 18C8