MCQ

MCQ 11 Mark

If p(n): 2n < (1 × 2 × 3 × ... × n). Then the smallest positive integer for which p(n) is true is:

A
1
B
2
C
3
✓
4

Answer

Correct option: D.

Solution:
The smallest positive integer for which $P(n)$ is 4 .
$P(4)=2^4<(1 \times 2 \times 3 \times \ldots \times 4)$
$P(4)=16<24$.

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MCQ 21 Mark

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

✓
5
B
3
C
7
D
1

Answer

Correct option: A.

Given,
$10\text{n }+3\times^{\text{n}+2}+\lambda$ is divisible by 9,
$\text{P}(1)=10^1+3\times4^{1+2}+\lambda$ is exactly divisible by 9 then the value of $\lambda$ is 5.

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MCQ 31 Mark

If $\text{i}^2=-1,$ then the sum $\text{i}+\text{i}^2+\text{i}^3+...$ upto 1000 terms is equal to:

A
1
B
-1
C
i
✓
0

Answer

Correct option: D.

$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+...\text{i}^{1000}$
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4 \ [\because\text{i}^2=-1,\text{i}^3=-\text{i and} \ \text{i}^4=1]$
$=\text{i}-1-\text{i}+1$
$=0$
Similarly, the sum of the next four terms of the series will be equal to 0. This is because the powers of i follow a cyclicity of 4. Hence, the sum of all terms, till 1000, will be zero.
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+...\text{i}^{1000}=0$

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MCQ 41 Mark

A student was asked to prove a statement p(n) by induction. He proved p(K + 1) is true whenever p(k) is true for all $\text{k}>5\in\text{N}$ and also p(5) is true. On the basis of this he could conclude that p(n) is true.

A
For all $\text{n}\in\text{N}$
B
For all n > 5
✓
For all $\text{n}\geq5$
D
For all n > 5

Answer

Correct option: C.

For all $\text{n}\geq5$

P(n) is true for all positive integer n, i.e. $\text{n}\geq5,$
Where P(n) is a Propositional function, complete two steps:
Basic Step: Verify that the proposition P(1) is true.
Inductive Step: Show the conditional statement,
$\big[\text{P}(1) \wedge \text{P}(2) \wedge-\wedge\text{P}(\text{k})\big]\rightarrow\text{P(k+1)}$ holds for all positive integer.

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MCQ 51 Mark

If $x^n-1$ is divisible by $x-\lambda$, then the least positive integral value of $\boldsymbol{\lambda}$ is:

✓
1
B
2
C
3
D
4

Answer

Correct option: A.

Solution:
Given
$x^n-1$
We know that
$x=k$ is the root of the equation $(x-1)$
$\Rightarrow x^n-1=0$
$\Rightarrow x^n=1$
Hence, the least positive integral value of $\lambda$ is 1 .

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MCQ 61 Mark

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

A
19
✓
17
C
23
D
25

Answer

Correct option: B.

Solution:
$3.5^{2 n+1}+2^{3 n+1}$ is divisible by $17, n \in N$
Step 1: $3.5^{2(1)+1}+2^{3(1)+1}$
$3.5^3+2^4=391$
Step 2: Assuming True for $\mathrm{n}=\mathrm{k}$
Hence, it is proved that $3.5^{2 n+1}+2^{3 n+1}$ is divisible by 17 .

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MCQ 71 Mark

If p(n): $49^\text{n}+16^{\text{n}}\lambda$ is divisible by 64 for $\text{n}\in\text{N}$ is true, then the least negative integral value of $\lambda$ is:

A
-3
B
-2
✓
-1
D
-4

Answer

Correct option: C.

-1

-1

Solution:
$(49)^n+16 n-1$
$\Rightarrow(1+48)^n+16 n-1$
$\Rightarrow 1+48 n+\ldots 48^n+16 n-1$
$\Rightarrow 64 n+n C_2(48)^2+n C_3(48)^3+\ldots+(48)^n$
$\Rightarrow 64\left(n+n C_2(6)^2+n C_3(6)^3 48+\ldots+(6)^n 8^{n-2}\right)$
$\therefore 49^n+16 n-1 \text { is divisible by } 64$

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