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Exemplar Solution Class 11th

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) $\frac{π}{4}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Exemplar Solution Class 11th

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

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

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

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

If $A \times B = {(a, x), (a, y), (b, x), (b, y)}$ , then $A = {a, b}$ , $B = {x, y}$ .

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = {a, b}, B = {x, y} \\ ∴ A \times B = {(a, x), (a, y), (b, x), (b, y)} \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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Exemplar Solution Class 11th

If  $(x - 2, y + 5) = (- 2, \frac{1}{3})$ are two equal ordered pairs, then $x = 4$ , $y = \frac{14}{3}$ .

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : (x - 2, y + 5) = (- 2, \frac{1}{3}) \\ \Rightarrow x - 2 = - 2 \Rightarrow x = 0 \\ a n d y + 5 = \frac{1}{3} \Rightarrow y = \frac{1}{3} - 5 = - \frac{14}{3} \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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Exemplar Solution Class 11th

Prove that $s i n θ + s i n 2 θ + s i n 3 θ + \dots + s i n n θ = \frac{s i n \frac{n θ}{2} \cdot s i n \frac{(n + 1) θ}{2}}{s i n \frac{θ}{2}}$ , for all $n \in ℕ$ .

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t P (n) : s i n θ + s i n 2 θ + s i n 3 θ + \dots + s i n n θ = \frac{s i n \frac{n θ}{2} . s i n (\frac{n + 1}{2}) θ}{s i n \frac{θ}{2}}, f o r a l l n \in N . \\ S t e p 1 : P (1) : s i n θ = \frac{s i n \frac{θ}{2} . s i n (\frac{1 + 1}{2}) θ}{s i n \frac{θ}{2}} = \frac{s i n \frac{θ}{2} . s i n θ}{s i n \frac{θ}{2}} = s i n θ \\ \Rightarrow s i n θ = s i n θ w h i c h i s t r u e f o r P (1) . \\ S t e p 2 : P (k) : s i n θ + s i n 2 θ + s i n 3 θ + \dots + s i n k θ = \frac{s i n \frac{k θ}{2} . s i n (\frac{k + 1}{2}) θ}{s i n \frac{θ}{2}} L e t i t b e t r u e f o r P (k) . \end{array}$

$\begin{array}{l} S t e p 3 : P (k + 1) : s i n θ + s i n 2 θ + s i n 3 θ + \dots + s i n k θ + s i n (k + 1) θ \\ = \frac{s i n \frac{k θ}{2} . s i n (\frac{k + 1}{2}) θ}{s i n \frac{θ}{2}} + s i n (k + 1) θ \\ = \frac{s i n \frac{k θ}{2} . s i n (\frac{k + 1}{2}) θ + s i n (k + 1) θ . s i n \frac{θ}{2}}{s i n \frac{θ}{2}} \\ = \frac{2 s i n \frac{k θ}{2} . s i n (\frac{k + 1}{2}) θ + s i n (k + 1) θ . s i n \frac{θ}{2}}{2 s i n \frac{θ}{2}} \\ = \frac{c o s (\frac{k θ}{2} - \frac{k + 1}{2} θ) - c o s (\frac{k θ}{2} + \frac{k + 1}{2} θ) + c o s [(k + 1) θ - \frac{θ}{2}] - c o s [(k + 1) θ + \frac{θ}{2}]}{2 s i n \frac{θ}{2}} \\ = \frac{c o s (- \frac{θ}{2}) - c o s (k θ + \frac{θ}{2}) + c o s (k θ + \frac{θ}{2}) - c o s (k θ + \frac{3 θ}{2})}{2 s i n \frac{θ}{2}} \\ = \frac{c o s (\frac{θ}{2}) - c o s (k θ + \frac{3 θ}{2})}{2 s i n \frac{θ}{2}} \\ = \frac{- 2 s i n (\frac{\frac{θ}{2} + k θ + \frac{3 θ}{2}}{2}) . s i n (\frac{\frac{θ}{2} - k θ - \frac{3 θ}{2}}{2})}{2 s i n \frac{θ}{2}} [? c o s A - c o s B = - 2 s i n \frac{A + B}{2} . s i n \frac{A - B}{2}] \\ = \frac{- 2 s i n (\frac{k θ + 2 θ}{2}) . s i n (\frac{- k θ - θ}{2})}{2 s i n \frac{θ}{2}} \\ = \frac{s i n (\frac{k θ + 2 θ}{2}) . s i n (\frac{k θ + θ}{2})}{s i n \frac{θ}{2}} \\ = \frac{s i n [\frac{(k + 1) + 1}{2}] θ . s i n [\frac{k + 1}{2}] θ}{s i n \frac{θ}{2}} w h i c h i s t r u e f o r P (k + 1) . \\ H e n c e, P (k + 1) i s t r u e w h e n e v e r P (k) i s t r u e . \end{array}$

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Exemplar Solution Class 11th

If $f (x) = c o s^{2} x + s e c^{2} x$ , then:

(a) $f (x) < 1$

(b) $f (x) = 1$

(c) $2 < f (x) < 1$

(d) f(x) $\geq 2$

[Hint: A.M ≥ G.M.]

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : f (x) = c o s^{2} x + se c^{2} x \\ W e k n o w t h a t A M \geq G M \\ \Rightarrow \frac{c o s^{2} x + se c^{2} x}{2} \geq \sqrt[]{c o s^{2} x . se c^{2} x} \\ \Rightarrow \frac{c o s^{2} x + se c^{2} x}{2} \geq 1 \Rightarrow c o s^{2} x + se c^{2} x \geq 2 \\ \Rightarrow f (x) \geq 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Class 11th

If $A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}$ , then $(A \times B) \cup (A \times C)$ $= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}$

10 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = {1, 2, 3}, B = {3, 4} a n d C = {4, 5, 6} \\ ∴ A \times B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} \\ a n d A \times C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} \\ (A \times B) \cup (A \times C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)} \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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Exemplar Solution Class 11th

Prove that $c o s θ \cdot c o s 2 θ \cdot c o s 4 θ \dots c o s 2^{n - 1} θ = \frac{s i n (2^{n} θ)}{2^{n} s i n θ}$ , for all $n \in ℕ$ .

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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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) $\frac{π}{4}$

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

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

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

If $A \times B = {(a, x), (a, y), (b, x), (b, y)}$ , then $A = {a, b}$ , $B = {x, y}$ .

If  $(x - 2, y + 5) = (- 2, \frac{1}{3})$ are two equal ordered pairs, then $x = 4$ , $y = \frac{14}{3}$ .

Prove that $s i n θ + s i n 2 θ + s i n 3 θ + \dots + s i n n θ = \frac{s i n \frac{n θ}{2} \cdot s i n \frac{(n + 1) θ}{2}}{s i n \frac{θ}{2}}$ , for all $n \in ℕ$ .

If $f (x) = c o s^{2} x + s e c^{2} x$ , then:

(a) $f (x) < 1$

(b) $f (x) = 1$

(c) $2 < f (x) < 1$

(d) f(x) $\geq 2$

[Hint: A.M ≥ G.M.]

If $A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}$ , then $(A \times B) \cup (A \times C)$ $= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}$

Prove that $c o s θ \cdot c o s 2 θ \cdot c o s 4 θ \dots c o s 2^{n - 1} θ = \frac{s i n (2^{n} θ)}{2^{n} s i n θ}$ , for all $n \in ℕ$ .