Class 11th

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 6 Views

Contributor-Level 10

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

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

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New answer posted

a year 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:

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

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New answer posted

a year 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:

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

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New question posted

a year 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

0 Follower 2 Views

New answer posted

a year 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:

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

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New answer posted

a year 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) $t a n θ = 20$

0 Follower 2 Views

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exempar

Sol:

$\begin{array}{l} s i n θ = - \frac{1}{5} i s c o r r e c t . ? - 1 \leq s i n θ \leq 1 \\ S o (a) i s c o r r e c t . \\ c o s θ = 1 i s c o r r e c t . ? c o s 0^{0} = 1 \\ S o (b) i s c o r r e c t . \\ s e c θ = \frac{1}{2} \Rightarrow c o s θ = 2 i s n o t c o r r e c t . ? - 1 \leq c o s θ \leq 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

If $t a n θ = \frac{1}{2} =$ and $t a n ϕ = 1 / 3$ , then the value of $θ + ϕ$ is:

(a) $\frac{π}{6}$

(b) $π$

(c) 0

(d) $\frac{π}{4}$

0 Follower 3 Views

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t \\ t a n (θ + ?) = \frac{t a n θ + t a n ?}{1 - t a n θ t a n ?} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} * \frac{1}{3}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \\ \Rightarrow t a n (θ + ?) = t a n \frac{π}{4} \\ ∴ θ + ? = \frac{π}{4} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Prove that the number of subsets of a set containing $n$ distinct elements is $2^{n}$ , for all $n \in ℕ$ .

0 Follower 4 Views

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : N u m b e r o f s u b s e t s o f a s e t c o n t a i n i n g n d i s t i n c t e l e m e n t s i s 2^{n}, f o r a l l n \in N . \\ S t e p 1 : I t i s c l e a r t h a t P (1) i s t r u e f o r n = 1 . N u m b e r o f s u b s e t s = 2^{1} = 2 . W h i c h i s t r u e . \\ S t e p 2 : P (k) i s a s s u m e d t o b e t r u e f o r n = k . Since t h e n u m b e r o f s u b s e t s = 2^{k} . \\ S t e p 3 : P (k + 1) = 2^{k + 1} \\ W e k n o w i f o n e n u m b e r (i . e ., e l e m e n t) i s a d d e d t o t h e e l e m e n t s o f a g i v e n s e t, t h e n u m b e r \\ o f s u b s e t s b e c o m e d o u b l e . \\ ∴ N u m b e r o f s u b s e t s o f a s e t h a v i n g (k + 1) d i s t i n c t e l e m e n t s = 2 * 2^{k} = 2^{k + 1} w h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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New answer posted

a year ago

Prove that $1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n}} > \frac{1}{1 + \frac{1}{n}}$ , for all $n > 1$ .

0 Follower 5 Views

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : \frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{2 n} > \frac{13}{24}, \forall n \in N . \\ S t e p 1 : P (2) : \frac{1}{2 + 1} + \frac{1}{2 + 2} > \frac{13}{24} \Rightarrow \frac{1}{3} + \frac{1}{4} > \frac{13}{24} \\ \Rightarrow \frac{7}{12} > \frac{13}{24} \Rightarrow \frac{14}{24} > \frac{13}{24} w h i c h i s t r u e f o r P (2) . \\ S t e p 2 : P (k) : \frac{1}{k + 1} + \frac{1}{k + 2} + \dots + \frac{1}{2 k} > \frac{13}{24} . L e t i t b e t r u e f o r P (k) \\ S t e p 3 : P (k + 1) : \frac{1}{k + 1} + \frac{1}{k + 2} + \dots + \frac{1}{2 k} + \frac{1}{2 (k + 1)} > \frac{13}{24} \\ Since \frac{1}{k + 1} + \frac{1}{k + 2} + \dots + \frac{1}{2 k} > \frac{13}{24} \\ S o, \frac{1}{k + 1} + \frac{1}{k + 2} + \dots + \frac{1}{2 k} + \frac{1}{2 (k + 1)} > \frac{13}{24} \\ W h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

0 0

New answer posted

a year ago

Show that $\frac{5^{n}}{3^{n}} + \frac{7^{n}}{1 5^{n}} + \dots$
is a natural number, for all $n \in ℕ$ .

0 Follower 4 Views