M.C.Q (1 Marks)

MCQ 11 Mark

If $x^n - 1$ is divisible by $x - k,$ then the least positive integral value of $k$ is:

✓
$1$
B
$2$
C
$3$
D
$4$

Answer

Correct option: A.

$1$

Let $\text{P}(n) = x^n - 1$ is divisible by $x - k$
$\text{P}(1) = x - 1$ is divisible by $x - k.$
Since $x - 1$ is divisible by $x - 1,$ the least integral value of $k$ is $1.$

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MCQ 21 Mark

For all $n \in N, 3 \times 5^{2n+1} + 2^{3n+1}$ is divisible by:

A
$19$
✓
$17$
C
$23$
D
$25$

Answer

Correct option: B.

$17$

Let $\text{P(n): }3\times5^{2\text{n}+1}+2^{3\text{n}+1}$
For $\text{P}(1):\ 3\times5^{2\times1+1}+2^{3\times1+1}=3.5^3+2^4$
$=3\times(125)+16=375+16$
$=391=23\times17$
So $,\text{P}(1)$ is divisible by $17.$
Assume $\text{P}((k)$ is divisible by $17\Rightarrow\text{P(k): }3\times5^{2\text{k}+1}+2^{3\text{k}+1}=17\lambda_1,\lambda_1\in\text{N}$
So$, 3\times5^{2\text{k}+1}=17\lambda_1-2^{3\text{k}+1}\ ....(\text{i})$
Now we have to prove $P(k + 1)$ is divisible by $17\Rightarrow\text{P(k+1): }3\times5^{2\text{k}+3}+2^{3\text{k}+4}=17\lambda_2,\lambda_2\in\text{N}$
$=\big[17\lambda_1-2^{3\text{k}+1}\big]\times5^2+2^{3\text{k}+1}\cdot2^{3}$
$=17\lambda_1\cdot25-25\cdot2^{3\text{k}+1}+8\cdot2^{3\text{k}+1}$
$=17\lambda_1\cdot25-17\cdot2^{3\text{k}+1}$
$=17\big[25\lambda_1-2^{3\text{k}+1}\big]$
$=17\lambda_2,\lambda_2=25\lambda_1-2^{3\text{k}+1}$
So$, \text{P}((k + 1)$ is divisible by $17$ whenever $\text{P}((k)$ is divisible by $17.$
Hence by the principle of mathematical induction we have $\text{P}((n): 3 \times 5^{2n+1} + 2^{3n+1}$ is divisible by $17,\forall\text{n}\in\text{N}$

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MCQ 31 Mark

If $10^n + 3 \times 4^{n+2} + k$ is divisible by $9$ for all $n \in N,$ then the least positive integral value of $k$ is:

✓
$5$
B
$3$
C
$7$
D
$1$

Answer

Correct option: A.

$5$

Let $P(n) = 10n + 3 × 4n+2 + k$ is divisible by $9, \forall\text{ n}\in\text{N}$
$P(1) = 101 + 3 × 41+2 + k = 10 + 3 × 64 + k$
$= 10 + 192 + k = 202 + k$ must be divisible by $9.$
If $(202 + k) $is divisible by 9 then k must be equal to $5.$
$202 + 5 = 207$ which is divisible by $9.$
$=\frac{207}{9}=23$
So$,$ the least positive integral value of $k = 5$

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