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New answer posted•a year ago

Exemplar Solution Class 11th

$n (n^{2} + 5)$ is divisible by 6, for each natural number $n$ .

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : n (n^{2} + 5) \\ S t e p 1 : P (1) : 1 (1 + 5) = 6 w h i c h i s d i v i s i b l e b y 6 . \\ S o, i t i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : k (k^{2} + 5) = 6 λ L e t i t b e t r u e . \\ \Rightarrow k^{3} + 5 k = 6 λ \\ \Rightarrow k^{3} = 6 λ - 5 k \dots (i) \\ S t e p 3 : P (k + 1) : (k + 1) [{(k + 1)}^{2} + 5] \\ = (k + 1) [k^{2} + 1 + 2 k + 5] \\ = (k + 1) [k^{2} + 2 k + 6] \\ = k^{3} + 2 k^{2} + 6 k + k^{2} + 2 k + 6 \\ = k^{3} + 3 k^{2} + 8 k + 6 \\ = k^{3} + 5 k + 3 k^{2} + 3 k + 6 \\ = 6 λ - 5 k + 5 k + 3 (k^{2} + k + 2) [F r o m (i)] \\ = 6 λ + 3 (k^{2} + k + 2) \\ W e k n o w t h a t k^{2} + k + 2 i s d i v i s i b l e b y 2 f o r e a c h v a l u e o f k \in N, \\ S o, l e t k^{2} + k + 2 = 2 m \\ S o, P (k + 1) = 6 λ + 3.2 m = 6 (λ + m) w h i c h i s d i v i s i b l e b y 6 . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

$n^{3} - n$ is divisible by 6, for each natural number $n \geq 2$ .

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : n^{3} - n \\ S t e p 1 : P (2) : 2^{3} - 2 = 8 - 2 = 6 w h i c h i s d i v i s i b l e b y 6 . \\ S o, i t i s t r u e f o r P (2) . \\ S t e p 2 : P (k) : k^{3} - k = 6 λ L e t i t b e t r u e f o r k \geq 2 . \\ \Rightarrow k^{3} = 6 λ + k \dots (i) \\ S t e p 3 : P (k + 1) : {(k + 1)}^{3} - (k + 1) \\ = k^{3} + 1 + 3 k^{2} + 3 k - k - 1 \\ = k^{3} - k + 3 (k^{2} + k) \\ = 6 λ + 3 (k^{2} + k) [f r o m (i)] \\ W e k n o w t h a t 3 (k^{2} + k) i s d i v i s i b l e b y 6 f o r e v e r y v a l u e o f k \in N . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

For any natural numbers $n, x^{n} - y$ is divisible by 5.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : x^{n} - y^{n} \\ S t e p 1 : P (1) : x^{1} - y^{1} = x - y w h i c h i s d i v i s i b l e b y x - y . \\ S o, i t i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : x^{k} - y^{k} = (x - y) λ L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) : x^{k + 1} - y^{k + 1} \\ = x^{k + 1} + x^{k} . y - x^{k} . y - y^{k + 1} = (x^{k + 1} - x^{k} . y) + (x^{k} . y - y^{k + 1}) \\ = x^{k} (x - y) + y . (x^{k} - y^{k}) \\ = x^{k} (x - y) + y (x - y) λ (f r o m S t e p 2) \\ = (x - y) (x^{k} + y λ) w h i c h i s d i v i s i b l e b y (x - y) . \\ S o, i t i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Find the general solution of the equation  $5 c o s^{2} θ + 7 s i n^{2} θ - 6 = 0 .$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} 5 c o s^{2} θ + 7 s i n^{2} θ - 6 = 0 \\ \Rightarrow 5 c o s^{2} θ + 7 (1 - c o s^{2} θ) - 6 = 0 \\ \Rightarrow 5 c o s^{2} θ + 7 - 7 c o s^{2} θ - 6 = 0 \Rightarrow - 2 c o s^{2} θ + 1 = 0 \\ \Rightarrow 2 c o s^{2} θ + 1 = 0 \Rightarrow c o s^{2} θ = \frac{1}{2} \\ \Rightarrow c o s^{2} θ = c o s^{2} \frac{π}{4} \\ ∴ θ = n π \pm \frac{π}{4} [\begin{array}{l} ? I f c o s^{2} θ = c o s^{2} α \\ ∴ θ = n π \pm α \end{array}] \\ H e n c e, t h e g e n e r a l s o l u t i o n o f θ = n π \pm \frac{π}{4}, n \in Z . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

For any natural numbers $n, 7^{n} - 2^{n}$ is divisible by x-y where x and y are any integers with x≠y.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : 7^{n} - 2^{n} \\ S t e p 1 : P (1) : 7^{1} - 2^{1} = 7 - 2 = 5 w h i c h i s d i v i s i b l e b y 5 . \\ S o, i t i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : 7^{k} - 2^{k} = 5 λ L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) : 7^{k + 1} - 2^{k + 1} \\ = 7^{k + 1} + 7^{k} . 2 - 7^{k} . 2 - 2^{k + 1} = (7^{k + 1} - 7^{k} . 2) + (7^{k} . 2 - 2^{k + 1}) \\ = 7^{k} (7 - 2) + 2 . (7^{k} - 2^{k}) \\ = 5 . 7^{k} + 2.5 λ (f r o m S t e p 2) \\ = 5 (7^{k} + 2 λ) w h i c h i s d i v i s i b l e b y 5 . \\ S o, i t i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

$3^{2 n} - 1$ is divisible by 8, for all natural numbers $n$ .

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : 3^{2 n} - 1 \\ S t e p 1 : P (1) : 3^{2} - 1 = 9 - 1 = 8 w h i c h i s d i v i s i b l e b y 8 \\ S o, i t i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : 3^{2 k} - 1 = 8 λ L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) : 3^{2 (k + 1)} - 1 \\ = 3^{2 k + 2} - 1 = 3^{2} . 3^{2 k} - 9 + 8 = 9 (3^{2 k} - 1) + 8 \\ = 9.8 λ + 8 (f r o m S t e p 2) \\ = 8 [9 λ + 1] w h i c h i s d i v i s i b l e b y 8 . \\ S o, i t i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Find the value of the expression $c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) + c o s^{4} (\frac{5 π}{8}) + c o s^{4} (\frac{7 π}{8}) .$

Hint : Simplify the expression to 2 (  $c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) = 2 [{(c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8})}^{2} - c o s^{2} \frac{π}{8} c o s^{2} \frac{3 π}{8} .$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) + c o s^{4} (\frac{5 π}{8}) + c o s^{4} (\frac{7 π}{8}) . \\ = c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) + c o s^{4} (π - \frac{3 π}{8}) + c o s^{4} (π - \frac{π}{8}) \\ = c o s^{4} \frac{π}{8} + c o s^{4} \frac{3 π}{8} + c o s^{4} \frac{3 π}{8} + c o s^{4} \frac{π}{8} \\ = 2 c o s^{4} \frac{π}{8} + 2 c o s^{4} \frac{3 π}{8} = 2 [c o s^{4} \frac{π}{8} + c o s^{4} \frac{3 π}{8}] \\ = 2 [c o s^{4} \frac{π}{8} + c o s^{4} (\frac{π}{2} - \frac{π}{8})] = 2 [c o s^{4} \frac{π}{8} + s i n^{4} \frac{π}{8}] \\ = 2 [c o s^{4} \frac{π}{8} + s i n^{4} \frac{π}{8} + 2 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8} - 2 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8}] \\ = 2 [{(c o s^{2} \frac{π}{8} + s i n^{2} \frac{π}{8})}^{2} - 2 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8}] \\ = 2 [1 - 2 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8}] = 2 - 4 s i n^{2} \frac{π}{8} . c o s^{2} \frac{π}{8} \\ = 2 - {(2 s i n \frac{π}{8} . c o s \frac{π}{8})}^{2} = 2 - {(s i n \frac{π}{4})}^{2} \\ = 2 - {(\frac{1}{\sqrt[]{2}})}^{2} = 2 - \frac{1}{2} = \frac{3}{2} \\ H e n c e, t h e r e q u i r e d v a l u e o f t h e expression = \frac{3}{2} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

$n^{3} - 7 n + 3$ is divisible by 3, for all natural numbers $n$ .

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : n^{3} - 7 n + 3 \\ S t e p 1 : P (1) : {(1)}^{3} - 7 (1) + 3 = 1 - 7 + 3 = - 3 w h i c h i s d i v i s i b l e b y 3 \\ S o, i t i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : {(k)}^{3} - 7 k + 3 = 3 λ L e t i t b e t r u e . \\ \Rightarrow k^{3} = 3 λ + 7 k - 3 \\ S t e p 3 : P (k + 1) : {(k + 1)}^{3} - 7 (k + 1) + 3 \\ = k^{3} + 1 + 3 k^{2} + 3 k - 7 k - 7 + 3 \\ = k^{3} + 3 k^{2} - 4 k - 3 \\ = (3 λ + 7 k - 3) + 3 k^{2} - 4 k - 3 (f r o m S t e p 2) \\ = 3 k^{2} + 3 k + 3 λ - 6 \\ = 3 (k^{2} + k + λ - 2) w h i c h i s d i v i s i b l e b y 3 . \\ S o, i t i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

$2 3^{n} - 1$ is divisible by 7, for all natural numbers $n$ .

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : 2^{3 n} - 1 \\ S t e p 1 : P (1) : 2^{3.1} - 1 = 8 - 1 = 7 w h i c h i s d i v i s i b l e b y 7, s o i t i s t r u e . \\ S t e p 2 : P (k) : 2^{3 k} - 1 = 7 λ L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) = 2^{3 (k + 1)} - 1 \\ = 2^{3 k + 3} - 1 = 2^{3} . 2^{3 k} - 8 + 7 = 8 . 2^{3 k} - 8 + 7 \\ = 8 (2^{3 k} - 1) + 7 (f r o m S t e p 2) \\ = 8.7 λ + 7 \\ = 7 [8 λ + 1] w h i c h i s t r u e a s i t i s d i v i s i b l e b y 7 \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

$4^{n} - 1$ is divisible by 3, for each natural number $n$ .

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : 4^{n} - 1 \\ S t e p 1 : P (1) : 4 - 1 = 3 w h i c h i s d i v i s i b l e b y 3, s o i t i s t r u e . \\ S t e p 2 : P (k) : 4^{k} - 1 = 3 λ L e t i t b e t r u e . \\ S t e p 3 : P (k + 1) = 4^{k + 1} - 1 \\ = 4^{k} . 4 - 1 = 4 . 4^{k} - 4 + 3 = 4 . (4^{k} - 1) + 3 \\ = 4 (3 λ) + 3 (f r o m S t e p 2) \\ = 3 [4 λ + 1] w h i c h i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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