M.C.Q (1 Marks)

MCQ 11 Mark

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

A
$1$
B
$2$
C
$3$
✓
$4$

Answer

Correct option: D.

$4$

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

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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 $\text{x}-\lambda,$ then the least positive integral value of $\lambda$ is:

✓
$1$
B
$2$
C
$3$
D
$4$

Answer

Correct option: A.

$1$

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

A
$19$
✓
$17$
C
$23$
D
$25$

Answer

Correct option: B.

$17$

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

$-1$

$(49)^n + 16n - 1$
$\Rightarrow (1 + 48)^n + 16n - 1$
$\Rightarrow 1 + 48n + ... 48^n + 16n - 1$
$\Rightarrow 64n + nC_2(48)^2 + nC_3(48)^3 + ... + (48)^n$
$\Rightarrow 64(n + nC_2(6)^2 + nC_3(6)^348 + ... + (6)^n 8^{n-2})$
$\therefore 49^n + 16n - 1$ is divisible by $64$

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Maths M.C.Q (1 Marks)

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