MCQ

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 $4.$
$P(4) = 2^4 < (1 \times 2 \times 3 \times ... \times 4)$
$P(4) = 16 < 24.$

View full question & answer→

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.

$5$

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 $5$.

View full question & answer→

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.

$0$

$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+...\text{i}^{1000}$
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4 \ [\because\text{i}^2=-1,\text{i}^3=- 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.
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4+...\text{i}^{1000}=0$

View full question & answer→

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.

View full question & answer→

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 $1.$

View full question & answer→

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 $17$.

View full question & answer→

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 $64$

View full question & answer→

Maths MCQ

Take a timed test