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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

$2^{n} < (n + 2)!$ , for all natural numbers $n$ .

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : 2 n < (n + 2)! f o r a l l k \in N . \\ S t e p 1 : P (1) : 2.1 < (1 + 2)! \Rightarrow 2 < 3! \Rightarrow 2 < 6 w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : 2 k < (k + 2)! L e t i t b e t r u e f o r P (k) \\ S t e p 3 : P (k + 1) : 2 (k + 1) < (k + 1 + 2)! \\ F r o m S t e p 2, w e g e t \\ 2 k < (k + 2)! \\ \Rightarrow 2 k + 2 < (k + 2)! + 2 \\ \Rightarrow 2 (k + 1) < (k + 2)! + 2 \\ Also, (k + 2)! + 2 < (k + 3)! \\ ∴ 2 (k + 1) < (k + 3)! \\ \Rightarrow 2 (k + 1) < (k + 2 + 1)! w h i c h i s t r u e f o r P (k + 1) \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

KAU - Kerala Agricultural University B.Sc.

What is the fee structure for BSc courses at KAU?

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Abhishek Hazarika

Contributor-Level 10

The total fee for BSc courses at KAU includes several components, such as tuition fees, admission fees, etc. The total tuition fee of the course ranges from INR 71,700 to INR 71,900. The fee amount mentioned is sourced from the official website/sanctioning body and is subject to change. Therefore, it is indicative.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

Find the general solution of the equation $s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x .$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : s i n x - 3 s i n 2 x + s i n 3 x = c o s x - 3 c o s 2 x + c o s 3 x \\ \Rightarrow (s i n 3 x + s i n x) - 3 s i n 2 x = (c o s 3 x + c o s x) - 3 c o s 2 x \\ \Rightarrow 2 s i n (\frac{3 x + x}{2}) . c o s (\frac{3 x - x}{2}) - 3 s i n 2 x = 2 c o s (\frac{3 x + x}{2}) . c o s (\frac{3 x - x}{2}) - 3 c o s 2 x \\ \Rightarrow 2 s i n 2 x . c o s x - 3 s i n 2 x = 2 c o s 2 x . c o s x - 3 c o s 2 x \\ \Rightarrow 2 s i n 2 x . c o s x - 2 c o s 2 x . c o s x = 3 s i n 2 x - 3 c o s 2 x \\ \Rightarrow 2 c o s x (s i n 2 x - c o s 2 x) - 3 (s i n 2 x - c o s 2 x) = 0 \\ \Rightarrow (s i n 2 x - c o s 2 x) (2 c o s x - 3) = 0 \\ \Rightarrow s i n 2 x - c o s 2 x = 0 a n d 2 c o s x - 3 \neq 0 [? - 1 \leq c o s x \leq 1] \\ \Rightarrow \frac{s i n 2 x}{c o s 2 x} - 1 = 0 \Rightarrow t a n 2 x = 1 \\ \Rightarrow t a n 2 x = t a n \frac{π}{4} \Rightarrow 2 x = n π + \frac{π}{4} ∴ x = \frac{n π}{2} + \frac{π}{8} \\ H e n c e, t h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n i s \\ x = \frac{n π}{2} + \frac{π}{8}, n \in Z . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

$n^{2} < 2^{n}$ , for all natural numbers $n \geq 5$ .

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t P (n) : n^{2} < 2^{n} f o r a l l n a t u r a l n u m b e r s, n \geq 5 \\ S t e p 1 : P (5) : 1^{5} < 2^{5} \Rightarrow 1 < 32 w h i c h i s t r u e f o r P (5) . \\ S t e p 2 : P (k) : k^{2} < 2^{k} . L e t i t b e t r u e f o r k \in N . \\ S t e p 3 : P (k + 1) : {(k + 1)}^{2} < 2^{k + 1} \\ F r o m S t e p 2, w e g e t \\ k^{2} < 2^{k} \\ k^{2} + 2 k + 1 < 2^{k} + 2 k + 1 \\ {(k + 1)}^{2} < 2^{k} + 2 k + 1 \dots (i) \\ Since (2 k + 1) < 2^{k} \\ S o k^{2} + 2 k + 1 < 2^{k} + 2^{k} \\ k^{2} + 2 k + 1 < 2 . 2^{k} \\ k^{2} + 2 k + 1 < 2^{k + 1} (i i) \\ F r o m e q n . (i) a n d (i i), w e g e t {(k + 1)}^{2} < 2^{k + 1} . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e f o r k \in N, n \geq 5 . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Two Exemplar Solution

If $f (x) = a x + b$ , where $a$ and $b$ are integers, $f (- 1) = - 5$ and $f (3) = 3$ , then $a$ and $b$ are equal to

(a) $a = - 3, b = - 1$

(b) $a = 2, b = - 3$

(c) $a = 0, b = 2$

(d) $a = 2, b = 3$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = a x + b \\ \Rightarrow f (- 1) = a (- 1) + b \\ \Rightarrow - 5 = - a + b \\ \Rightarrow a - b = 5 \dots (i) \\ \Rightarrow f (3) = 3 a + b \\ \Rightarrow 3 = 3 a + b \\ \Rightarrow 3 a + b = 3 \dots (i i) \\ O n s o l v i n g e q n . (i) a n d (i i), w e g e t a = 2, b = - 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

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

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alok kumar singh

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

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Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

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 ( $c o s^{4} (\frac{π}{8}) + c o s^{4} (\frac{3 π}{8}) = 2 [{(c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8})}^{2} - c o s^{2} \frac{π}{8} c o s^{2} \frac{3 π}{8} .$

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alok kumar singh

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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9 months ago

KAU - Kerala Agricultural University B.Sc.

What is the eligibility criteria for admissions to BSc courses at KAU?

0 Follower 4 Views

Rachit Kumar

Contributor-Level 10

The minimum eligibility criteria for admission to BSc courses at KAU are that candidates must complete their Class 12 from the PCB or PCM/ PCB stream with an aggregate of 50%. Additionally, for the selection process, the university and its affiliated college accept the scores of entrance exams, such as KEAM Medical, ICAR CUET, etc. Moreover, the eligibility requirements and accepted entrance exam can change according to the selected specialisation.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Three Exemplar Solution

If $θ$ lies in the first quadrant and $c o s θ = \frac{8}{17}$ , then find the value of $c o s (3 0^{\circ} + θ) + c o s (4 5^{\circ} - θ) + c o s (12 0^{\circ} - θ) .$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t : c o s θ = \frac{8}{17} \\ ∴ s i n θ = \sqrt[]{1 - {(\frac{8}{17})}^{2}} = \sqrt[]{1 - \frac{64}{289}} = \sqrt[]{\frac{289 - 64}{289}} = \sqrt[]{\frac{225}{289}} = \frac{15}{17} \\ B u t θ l i e s i n I q u a d r a n t . \\ ∴ s i n θ = \frac{15}{17} \\ N o w c o s (3 0^{0} + θ) + c o s (4 5^{0} - θ) + c o s (12 0^{0} - θ) \\ = c o s 3 0^{0} c o s θ - s i n 3 0^{0} s i n θ + c o s 4 5^{0} c o s θ + s i n 4 5^{0} s i n θ + c o s 12 0^{0} c o s θ + s i n 12 0^{0} s i n θ \\ = \frac{\sqrt[]{3}}{2} c o s θ - \frac{1}{2} s i n θ + \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ - \frac{1}{2} c o s θ + \frac{\sqrt[]{3}}{2} s i n θ \\ = (\frac{\sqrt[]{3}}{2} c o s θ + \frac{\sqrt[]{3}}{2} s i n θ) - \frac{1}{2} (s i n θ + c o s θ) + \frac{1}{\sqrt[]{2}} (c o s θ + s i n θ) \\ = \frac{\sqrt[]{3}}{2} (c o s θ + s i n θ) - \frac{1}{2} (s i n θ + c o s θ) + \frac{1}{\sqrt[]{2}} (c o s θ + s i n θ) \\ = (\frac{\sqrt[]{3}}{2} - \frac{1}{2} + \frac{1}{\sqrt[]{2}}) (c o s θ + s i n θ) = (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) (\frac{8}{17} + \frac{15}{17}) \\ = (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) (\frac{23}{17}) \\ = \frac{23}{17} (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) . \\ H e n c e, t h e r e q u i r e d s o l u t i o n = \frac{23}{17} (\frac{\sqrt[]{3} - 1}{2} + \frac{1}{\sqrt[]{2}}) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

Give an example of a statement $P (n)$ which is true for all $n$ . Justify your answer.

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} T h e r e q u i r e d s t a t e m e n t i s P (n) : 1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2} \\ J u s t i f i c a t i o n : P (1) : 1 = \frac{1 (1 + 1)}{2} \\ P (k) : 1 + 2 + 3 + \dots + k = \frac{k (k + 1)}{2} . L e t i t b e t r u e . \\ P (k + 1) : 1 + 2 + 3 + \dots + k + (k + 1) = \frac{k (k + 1)}{2} + (k + 1) \\ = \frac{(k + 1) (k + 2)}{2} \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Four Exemplar Solution

Give an example of a statement $P (n)$ which is true for all $n \geq 4$ but $P (1)$ , $P (2)$ , and $P (3)$ are not true. Justify your answer.

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} T h e r e q u i r e d s t a t e m e n t i s P (n) = 2 n < n! \\ J u s t i f i c a t i o n : P (n) : 2 n < n! \\ P (1) : 2.1 < 1! \Rightarrow 2 < 1 n o t t r u e \\ P (2) : 2.2 < 2! \Rightarrow 4 < 2.1 \Rightarrow 4 < 2 n o t t r u e \\ P (3) : 2.3 < 3! \Rightarrow 6 < 3.2.1 \Rightarrow 6 < 6 n o t t r u e \\ P (4) : 2.4 < 4! \Rightarrow 8 < 4.3.2.1 \Rightarrow 8 < 24 T r u e \\ P (5) : 2.5 < 5! \Rightarrow 10 < 5.4.3.2.1 \Rightarrow 10 < 120 T r u e \\ H e n c e, P (n) : 2 n < n! i s n o t t r u e f o r P (1), P (2) a n d P (3) b u t i t i s t r u e f o r a l l v a l u e s o f n \geq 4 . \end{array}$

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9 months ago

SOU Devibai Narayandas Chhabada Rural Education Society Colleges in India

Are admissions to SOU Devibai Narayandas Chhabada Rural Education Society entrance based or merit based?

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saurya snehal

Contributor-Level 10

Admissions to the courses at SOU Devibai Narayandas Chhabada Rural Education Society can either be merit-based or entrance-based. Candidates are required to check the selection criteria for the course they wish to apply to. Courses such as BTech and MBA makes it mandatory for candidates to appear for entrance exams such as JEE Mains or CAT.

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