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Maths NCERT Exemplar Solutions Class 11th Chapter Four

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Maths NCERT Exemplar Solutions Class 11th Chapter Four Class 11th

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Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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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Vishal Baghel

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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Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

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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3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

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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Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

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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Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

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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3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

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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3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : \frac{n^{5}}{5} + \frac{n^{3}}{3} + \frac{7 n}{15}, \forall n \in N . \\ S t e p 1 : P (1) : \frac{1^{5}}{5} + \frac{1^{3}}{3} + \frac{7.1}{15} = \frac{3 + 5 + 7}{15} = \frac{15}{15} = 1 w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7 k}{15} . L e t i t b e t r u e f o r P (k) . A n d l e t \frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7 k}{15} = λ \\ S t e p 3 : P (k + 1) : \frac{{(k + 1)}^{5}}{5} + \frac{{(k + 1)}^{3}}{3} + \frac{7 (k + 1)}{15} \\ = \frac{1}{5} [k^{5} + 5 k^{4} + 10 k^{3} + 10 k^{2} + 5 k + 1] + \frac{1}{3} [k^{3} + 3 k^{2} + 3 k + 1] + \frac{7 k}{15} + \frac{7}{15} \\ = (\frac{k^{5}}{5} + \frac{k^{3}}{3} + \frac{7 k}{15}) + (k^{4} + 2 k^{3} + 3 k^{2} + 5 k) + \frac{1}{5} + \frac{1}{3} + \frac{7}{15} \\ = λ + k^{4} + 2 k^{3} + 3 k^{2} + 2 k + 1 [F r o m S t e p 2] \\ = p o s i t i v e integers = n a t u r a l n u m b e r \\ W h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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