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New answer posted•a year ago

Exemplar Solution Class 11th

The ordered pair $(5, 2)$ belongs to the relation $R = {(x, y) : y = x - 5, x, y \in ?}$ .

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : R = {(x, y) : y = x - 5, x, y \in Z} \\ F o r (5, 2), y = x - 5 \\ P u t x = 5, y = 5 - 5 = 0 \neq 2 \\ S o, (5, 2) i s n o t t h e o r d e r e d p a i r o f R . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If ${[x]}^{2} - 5 [x] + 6 = 0$ , where [.] denotes the greatest integer function, then

(a) $x \in [3, 4]$

(b) $x \in (2, 3]$

(c) $x \in [2, 3]$

(d) $x \in [2, 4]$

9 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e {[x]}^{2} - 5 [x] + 6 = 0 \\ \Rightarrow {[x]}^{2} - 3 [x] - 2 [x] + 6 = 0 \\ \Rightarrow [x] ([x] - 3) - 2 ([x] - 3) = 0 \\ \Rightarrow ([x] - 3) ([x] - 2) = 0 \Rightarrow [x] = 2, 3 \\ S o, x \in [2, 3] \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Let $n (A) = m$ , and $n (B) = n$ . Then the total number of non-empty relations that can be defined from $A$ to $B$ is:

(a) $m^{n}$

(b) $n^{m} - 1$

(c) $m^{n} - 1$

(d) $2^{m n} - 1$

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : n (A) = m a n d n (B) = n \\ ∴ n (A \times B) = n (A) . n (B) = m n \\ S o, t h e t o t a l n u m b e r o f r e l a t i o n s f r o m A t o B = 2^{m n} - 1 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Let $f = {(2, 4), (5, 6), (8, - 1), (10, - 3)}$ $g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)$ be two real functions. Then Match the following:

(a) $f - g$ (i) ${(2, \frac{4}{5})} (8, \frac{- 1}{4}) (10, \frac{- 3}{13})$

(b) $f + g$ (ii) ${(2, 20), (8, - 4), (10, - 39)}$

(c) $f \cdot g$ (iii) ${(2, - 1), (8, - 5), (10, - 16)}$

(d) $\frac{f}{g}$ (iv) ${(2, 9), (8, 3), (10, 10)}$

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f = {(2, 4), (5, 6), (8, - 1), (10, - 3)} \\ a n d g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)} \\ f - g, f + g, f . g, \frac{f}{g} a r e d e f i n e d i n t h e d o m a i n \\ (d o m a i n o f f \cap d o m a i n o f g) \\ \Rightarrow {2, 5, 8, 10} \cap {2, 7, 8, 10, 11} \Rightarrow {2, 8, 10} \\ (i) (f - g) 2 = f (2) - g (2) = 4 - 5 = - 1 \\ (f - g) 8 = f (8) - g (8) = - 1 - 4 = - 5 \\ (f - g) 10 = f (10) - g (10) = - 3 - 13 = - 16 \\ ∴ (f - g) = {(2, - 1), (8, - 5), (10, - 16)} \\ (i i) (f + g) 2 = f (2) + g (2) = 4 + 5 = 9 \\ (f + g) 8 = f (8) + g (8) = - 1 + 4 = 3 \\ (f + g) 10 = f (10) + g (10) = - 3 + 13 = 10 \\ ∴ (f + g) = {(2, 9), (8, 3), (10, 10)} \\ (i i i) (f . g) 2 = f (2) . g (2) = 4.5 = 20 \\ (f . g) 8 = f (8) . g (8) = (- 1) . 4 = - 4 \\ (f . g) 10 = f (10) . g (10) = - 3.13 = - 39 \\ ∴ (f . g) = {(2, 20), (8, - 4), (10, - 39)} \\ (i v) (\frac{f}{g}) (2) = \frac{f (2)}{g (2)} = \frac{4}{5} \\ (\frac{f}{g}) (8) = \frac{f (8)}{g (8)} = \frac{- 1}{4} \\ (\frac{f}{g}) (10) = \frac{f (10)}{g (10)} = \frac{- 3}{13} \\ (\frac{f}{g}) = {(2, \frac{4}{5}), (8, \frac{- 1}{4}), (10, \frac{- 3}{13})} \\ H e n c e, t h e c o r r e c t o p t i o n i s \\ (a) \leftrightarrow (i i i), (b) \leftrightarrow (i v), (c) \leftrightarrow (i i), (d) \leftrightarrow (i) \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Let f and g be two real functions given by $f = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)}$ $g = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)}$ then the domain of $f ? g$ is given by _________.

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)} \\ a n d g (x) = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)} \\ ∴ d o m a i n o f f = {0, 2, 3, 4, 5} \\ a n d d o m a i n o f g = {1, 2, 3, 4, 5} \\ S o, d o m a i n o f f . g = D o m a i n o f f \cap D o m a i n o f g \\ = {2, 3, 4, 5} \\ H e n c e, t h e f i l l e r i s {2, 3, 4, 5} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $f (x) = y = \frac{a x - b}{c x - a}$ , then prove that $f (y) = x$ .

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = y = \frac{a x - b}{c x - a} \\ f (y) = \frac{a y - b}{c y - a} = \frac{a [\frac{a x - b}{c x - a}] - b}{c [\frac{a x - b}{c x - a}] - a} \\ \Rightarrow f (y) = \frac{a^{2} x - a b - b c x + a b}{c a x - b c - c a x + a^{2}} = \frac{a^{2} x - b c x}{a^{2} - b c} \\ \Rightarrow f (y) = \frac{(a^{2} - b c)}{a^{2} - b c} = x \\ H e n c e, f (y) = x . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Find the domain and range of the function $f (x) = \frac{1}{\sqrt[]{x - 5}}$ .

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{x - 5}} \\ H e r e, i t i s c l e a r t h a t f (x) i s r e a l w h e n x - 5 > 0 \Rightarrow x > 5 \\ H e n c e, t h e d o m a i n = (5, \infty) \\ N o w t o f i n d t h e r a n g e p u t \\ f (x) = y = \frac{1}{\sqrt[]{x - 5}} \\ \Rightarrow \sqrt[]{x - 5} = \frac{1}{y} \Rightarrow x - 5 = \frac{1}{y^{2}} \\ \Rightarrow x = \frac{1}{y^{2}} + 5 \\ F o r x \in (5, \infty), y \in R^{+} . \\ H e n c e, t h e r a n g e o f f = R^{+} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $c o s (θ + ϕ) = m c o s (θ - ϕ)$ , then prove that tan $θ = \frac{1 - m}{1 + m} . c o t ϕ$

[Hint: Express $\frac{c o s (θ + α)}{c o s (θ - α)} = \frac{m}{1}$ .]and apply componendo and dividendo.]

3 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:

$\begin{array}{l} G i v e n t h a t : c o s (θ + ?) = m c o s (θ - ?) \\ \Rightarrow \frac{c o s (θ + ?)}{c o s (θ - ?)} = \frac{m}{1} \\ Using c o m p o n e n d o a n d d i v i d e n d o t h e o r e m, w e g e t \\ \frac{c o s (θ + ?) + c o s (θ - ?)}{c o s (θ + ?) - c o s (θ - ?)} = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{2 c o s (\frac{θ + ? + θ - ?}{2}) . c o s (\frac{θ + ? - θ + ?}{2})}{- 2 s i n (\frac{θ + ? + θ - ?}{2}) . s i n (\frac{θ + ? - θ + ?}{2})} = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{c o s θ . c o s ?}{- s i n θ . s i n ?} = \frac{m + 1}{m - 1} \Rightarrow - c o t θ . c o t ? = \frac{m + 1}{m - 1} \\ \Rightarrow \frac{- c o t ?}{t a n θ} = \frac{m + 1}{m - 1} = - \frac{1 + m}{1 - m} \\ \Rightarrow t a n θ = \frac{1 + m}{1 - m} c o t ? . \\ H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $s i n (θ + α) = a$ and $s i n (θ + β) = b$ , then prove that $c o s 2 (α - β) - 4 a b c o s (α - β) = 1 - 2 a^{2} - 2 b^{2} .$

Hint: Express cos $(α - β) =$ cos (( $θ +$ $α$ ) $-$ $(θ + β)$ )]

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:

$\begin{array}{l} G i v e n t h a t : \\ s i n (θ + α) = a a n d s i n (θ + β) = b \\ c o s (α - β) = c o s [θ + α - θ - β] = c o s [(θ + α) - (θ + β)] \\ = c o s (θ + α) c o s (θ + β) + s i n (θ + α) s i n (θ + β) \\ = \sqrt[]{1 - s i n^{2} (θ + α)} \sqrt[]{1 - s i n^{2} (θ + β)} + a . b \\ = \sqrt[]{(1 - a^{2}) (1 - b^{2})} + a b = a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ N o w c o s 2 (α - β) - 4 a b c o s (α - β) \\ = 2 c o s^{2} (α - β) - 1 - 4 a b c o s (α - β) [? c o s 2 θ = 2 c o s^{2} θ - 1] \\ = 2 {[a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}]}^{2} - 1 - 4 a b [a b + \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}] \\ = 2 [a^{2} b^{2} + 1 - a^{2} - b^{2} + a^{2} b^{2} + 2 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}}] - 1 - 4 a^{2} b^{2} - 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ = 2 a^{2} b^{2} + 2 - 2 a^{2} - 2 b^{2} + 2 a^{2} b^{2} + 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} - 1 - 4 a^{2} b^{2} - 4 a b \sqrt[]{1 - a^{2} - b^{2} + a^{2} b^{2}} \\ = 1 - 2 a^{2} - 2 b^{2} \\ H e n c e, c o s 2 (α - β) - 4 a b c o s (α - β) = 1 - 2 a^{2} - 2 b^{2} \\ H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Let  $f (x) = \sqrt[]{x}$ and $g (x) = x$ be two functions defined in the domain $?^{+} \cup {0}$ .

Find

(i) $(f + g) (x)$  

(ii) $(f - g) (x)$  

(iii) $(f g) (x)$  

(iv) $(\frac{f}{g}) (x)$

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : f (x) = \sqrt[]{x} a n d g (x) = x b e t w o f u n c t i o n s d e f i n e d i n t h e d o m a i n R^{+} \cup {0} \\ (i) (f + g) (x) = f (x) + g (x) = \sqrt[]{x} + x \\ (i i) (f - g) (x) = f (x) - g (x) = \sqrt[]{x} - x \\ (i i i) (f g) (x) = f (x) . g (x) = \sqrt[]{x} . x = x^{3 / 2} \\ (i v) (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\sqrt[]{x}}{x} = \frac{1}{\sqrt[]{x}} . \end{array}$

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