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51.

What is the last digit of `3^(3^(4n)) +1`, where n is a natural number?A. 2B. 7C. 8D. None of these

Answer» Correct Answer - D
In `3^(n)`, last digit is 3, if n=1,9 if n=2,7 if n=3 and 1 if
n=4 and it is repeated after than
Given expression is `3^(3^(4n))+1`
Let `x=3^(3^(4n))+1=3^(8"In")+1`
`rArr" "x=3^(80n).3^(n)+1`
Last digit of x will be decided by `3^(n)" since "3^(80n)` has power multiple of 4.
If n=1 last digit is 3+1=4
n=2 last digit is `3^(2)+1=9+1=10`
So, last digit is zero.
n=3 last digit is `3^(3)+1=27+1=28`
last digit is 8.
If n=4 last digit is `3^(4)+1=81+1=82`
last digit is 2.
So, there is no definite value of last digit.
Discussion
52.

What is the coefficient of `x^(4)` in the expansion of `((1-x)/(1+x))^(2)` ?A. `-16`B. 16C. 8D. `-18`

Answer» Correct Answer - B
Consider `((1-x)/(1+x))^(2)=(1-x)^(2)(1+x)^(-2)`
`=(1-2x+x^(2))(1+x)^(-2)`
`=(1-2x+x^(2))(1-2x+3x^(2)-4x^(3)+5x^(4)-.....)`
`therefore` Coefficient of `x^(4)` in `((1-x)/(1+x))^(2)=5 + 8 + 3 = 16`
Discussion
53.

What is the middle term in the expansion of `((xsqrt(y))/(3)-(3)/(ysqrt(x)))^(12)` ?A. `C(12,7)x^(3)y^(-3)`B. `C(12,6)x^(-3)y^(3)`C. `C(12,7)x^(-3)y^(3)`D. `C(12,6)x^(3)y^(-3)`

Answer» Correct Answer - D
In the expansion of `((xsqrt(y))/(3)-(3)/(ysqrt(x)))^(12)`, then middle term
is `(12)/(2)+1=7^(th)" term ". (r+1)_(th)" term, "`
`T_(r+1)=""^(12)C_(r)[(xsqrt(y))/(3)]^(12-r)cdot(-(3)/(ysqrt(x)))^(r)`
`therefore" "T_(7)=T_(6+1)=""^(12)C_(6)((xsqrt(y))/(3))^(6)(-(3)/(ysqrt(x)))^(6)`
`=""^(12)C_(6)(x^(6)y^(3))/(y^(6)x^(2))=""^(12)C_(6)x^(3)y^(-3)=C(12,6)x^(3)y^(-3)`
Discussion
54.

The coefficient of `x^3` in the expansion of `(3-2x)/((1+3x)^3)` isA. `-272`B. `-540`C. `-870`D. `-918`

Answer» Correct Answer - D
`((3-2x))/((1+3x)^(3))=(3-2x)(1+3x)^(-3)`
`=(3-2x)(1-9x+((-3)(-4))/(2!).9x^(2)+((-3)(-4)(-5))/(3!).27 x^(3)+....)`
`["Expanding"(1+3x)^(-3)]`
`=(3-2x)(1-9x+54x^(2)-270x^(3)+…..)`
`therefore" Coefficient of "x^(3)=-270xx3-2xx54`
`=-810-108=-918`
Discussion
55.

What is the coefficient of `x^(5)` in the expansion `(1-2x+3x^(2)-4x^(3)+……oo)^(-5)` ?A. `(10!)//(5!)^(2)`B. `5^(-5)`C. `5^(5)`D. `10!//{6!)(4!)}`

Answer» Correct Answer - A
`1-2x+3x^(2)-4x^(3)+…..`
`=(1+x)^(2), so (1-2x+3x^(2)-4x^(3)+….oo)^(-5)`
`=((1+x^(-2))=(1+x)^(10) rArr T_(r+1)=""^(10)C_(r)x^®`
Putting r=5, coefficient of `x^(5)=""^(10)C_(5)=(10!)/(5!5!)=(10!)/((5!)^(2))`
Discussion
56.

What are the last two digits of the number `9^(200)`A. 19B. 21C. 41D. 1

Answer» Correct Answer - D
Using binomial theorem `9^(200)=(1+8)^(200)`
`=1+8.200+(200xx199)/(2!)xx8^(2)+…..`
`=1+1600+1273600+….`
From above, it is clear that the last two digits of the number `9^(200)` are 0.1
Discussion
57.

For any positive integer n, if `4^(n)-3n` is divided by 9, then what is the remainder ?A. 8B. 6C. 4D. 1

Answer» Correct Answer - D
Using binomial theorem. `4^(n)-3n=(1+3)^(n)-3n`
`=1+n.3+(n(n-1))/(2!)3^(2)+…-3`
`=1+(n(n+1))/(2!).3^(2)+(n(n-1)(n-2))/(3!).3^(3)+....`
`rArr" "4^(n)-3n=9{(n(n-1))/(2!)+....}+1`
Thus, when `4^(n)-3n` is divided by 9, the remainder is 1.
Discussion