Class 11th

Let $P (n)$ be a statement and let P(k + 1) for some natural number $k$ . Then $P (n)$ is true for all $n \in ℕ$ .

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Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n t h a t : P (k) \Rightarrow P (k + 1) \\ P (1) \Rightarrow P (2) w h i c h i s n o t t r u e . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted

4 months ago

If $P (n) : 2^{n} < n!$ , $n \in ℕ$ , then $P (n)$ is true for all $n \geq$ ____.

0 Follower 4 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n t h a t P (n) : 2 n < n! \forall n \in N \\ F o r n = 1 2 < 1 (N o t T r u e) \\ F o r n = 2 2 * 2 < 2! \Rightarrow 4 < 2 (N o t T r u e) \\ F o r n = 3 2 * 3 < 3! \Rightarrow 6 < 3.2.1 \Rightarrow 6 < 6 (N o t T r u e) \\ F o r n = 4 2 * 4 < 4! \Rightarrow 8 < 4.3.2.1 \Rightarrow 8 < 24 (T r u e) \\ F o r n = 5 2 * 5 < 5! \Rightarrow 10 < 5.4.3.2.1 \Rightarrow 10 < 120 (T r u e) \\ S o, P (n) i s t r u e f o r n \geq 4 . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 4 . \end{array}$

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New answer posted

4 months ago

The value of $c o s 1^{\circ} \cdot c o s 2^{\circ} \cdot c o s 3^{\circ} \dots c o s 17 9^{\circ}$ is:

(a) $\frac{1}{\sqrt[]{2}}$

(b) 0

(c) 1

(d) -1

0 Follower 3 Views

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o s 1^{0} . c o s 2^{0} . c o s 3^{0} \dots c o s 17 9^{0} \\ = c o s 1^{0} . c o s 2^{0} . c o s 3^{0} \dots c o s 9 0^{0} . c o s 9 1^{0} \dots c o s 17 9^{0} \\ = 0 [? c o s 9 0^{0} = 0] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

4 months ago

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

(a) 1

(b) 2

(c) 3

(d) 4

0 Follower 3 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t P (n) = x^{n} - 1 i s d i v i s i b l e b y x - k . \\ P (1) = x - 1 i s d i v i s i b l e b y x - k . \\ Since k = 1 i s t h e p o s s i b l e l e a s t integral v a l u e o f k . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

4 months ago

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

(a) 19

(b) 17

(c) 23

(d) 25

0 Follower 4 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t P (n) = 3 . 5^{2 n + 1} + 2^{3 n + 1} \\ P (1) = 3 . 5^{2.1 + 1} + 2^{3.1 + 1} = 3 . 5^{3} + 2^{4} = 3 (125) + 16 = 375 + 16 \\ = 391 = 23 * 17 \\ S o, i t i s d i v i s i b l e b y 17 a n d 23 b o t h . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) a n d (c) . \end{array}$

0 0

New answer posted

4 months ago

The value of $\frac{1 - t a n^{2} 1 5^{\circ}}{1 + t a n^{2} 15 \circ}$ is:

(a) 1

(b) $\sqrt[]{3}$

(c) $\frac{\sqrt[]{3}}{2}$

(d) 2

0 Follower 3 Views

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : \frac{1 - t a n^{2} 1 5^{0}}{1 + t a n^{2} 1 5^{0}} \\ L e t θ = 1 5^{0} ∴ 2 θ = 3 0^{0} \\ c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ} \\ \Rightarrow c o s 3 0^{0} = \frac{1 - t a n^{2} 1 5^{0}}{1 + t a n^{2} 1 5^{0}} \\ \Rightarrow \frac{\sqrt[]{3}}{2} = \frac{1 - t a n^{2} 1 5^{0}}{1 + t a n^{2} 1 5^{0}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

4 months ago

If $1 0^{n} + 3 \cdot 4^{n + 2} + k$ is divisible by 9 for all $n \in ℕ$ , then the least positive integral value of $k$ is:

(a) 5

(b) 3

(c) 7

(d) 1

0 Follower 6 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t P (n) = 1 0^{n} + 3 . 4^{n + 2} + k i s d i v i s i b l e b y 9, \forall n \in N \\ P (1) = 1 0^{1} + 3 . 4^{1 + 2} + k = 10 + 3.64 + k \\ = 10 + 192 + k = 202 + k m u s t b e d i v i s i b l e b y 9 . \\ I f (202 + k) i s d i v i s i b l e b y 9 t h e n k m u s t b e e q u a l t o 5 \\ 202 + 5 = 207 w h i c h i s d i v i s i b l e b y 9 \\ = \frac{207}{9} = 23 \\ S o, t h e l e a s t p o s i t i v e integral v a l u e o f k = 5 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New question posted

4 months ago

The value of $\frac{1 - t a n^{2} 1 5^{?}}{1 + t a n^{2} 15 ?}$ is:

(a) 1

(b) $\sqrt[]{3}$

(c) $\frac{\sqrt[]{3}}{2}$

(d) 2

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New answer posted

4 months ago

The value of $t a n 1^{\circ} \cdot t a n 2^{\circ} \cdot t a n 3^{\circ} \dots t a n 8 9^{\circ}$ is:

(a) 0

(b) 1

(c) 2

(d) Not defined

0 Follower 3 Views

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n 1^{0} t a n 2^{0} t a n 3^{0} \dots t a n 8 9^{0} \\ = t a n 1^{0} t a n 2^{0} t a n 3^{0} \dots t a n 4 5^{0} . t a n {(90 - 44)}^{0} . t a n {(90 - 43)}^{0} \dots t a n {(90 - 1)}^{0} \\ = t a n 1^{0} c o t 1^{0} . t a n 2^{0} c o t 2^{0} . t a n 3^{0} c o t 3^{0} \dots t a n 8 9^{0} . c o t 8 9^{0} \\ = 1.1.1.1 \dots 1.1 = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

0 0

New answer posted

4 months ago

Which of the following is not correct?

(a) $s i n θ = - \frac{1}{5}$

(b) $c o s θ = 1$

(c) $s e c θ = 2$

(d) $t a n θ = 20$

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