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Exemplar Solution Class 11th

If $f (x) = \frac{x - 1}{x + 1}$ , then show that

(i) $f (\frac{1}{x}) = - f (x)$  

(ii) $f (\frac{1}{- x}) = \frac{- 1}{f (x)}$

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : f (x) = \frac{x - 1}{x + 1} \\ (i) f (\frac{1}{x}) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1} = \frac{1 - x}{1 + x} = \frac{- (x - 1)}{x + 1} = - f (x) \\ H e n c e, f (\frac{1}{x}) = - f (x) \\ (i i) f (\frac{- 1}{x}) = \frac{- \frac{1}{x} - 1}{- \frac{1}{x} + 1} = \frac{- (\frac{1}{x} + 1)}{- (\frac{1}{x} - 1)} = \frac{1 + x}{1 - x} = \frac{1}{\frac{1 - x}{1 + x}} \\ = \frac{1}{- (\frac{x - 1}{x + 1})} = \frac{- 1}{f (x)} \\ H e n c e, f (- \frac{1}{x}) = \frac{- 1}{f (x)} . \end{array}$

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Exemplar Solution Class 11th

Redefine the function $f (x) = | x - 2 | + | 2 + x |, - 3 \leq x \leq 3$

13 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : f (x) = | x - 2 | + | 2 + x |, - 3 \leq x \leq 3 \\ Since | x - 2 | = - (x - 2), x < 2 \\ a n d | x - 2 | = (x - 2), x \geq 2 \\ | 2 + x | = - (2 + x), x < - 2 \\ a n d | 2 + x | = (2 + x), x \geq - 2 \\ N o w f (x) = | x - 2 | + | 2 + x |, - 3 \leq x \leq 3 \\ = {\begin{cases} - (x - 2) - (2 + x), - 3 \leq x < - 2 \\ - (x - 2) + (2 + x), - 2 \leq x < 2 \\ (x - 2) + (2 + x), 2 \leq x \leq 3 \end{cases} \\ ∴ f (x) = {\begin{cases} - 2 x, - 3 \leq x < - 2 \\ 4, - 2 \leq x < 2 \\ 2 x, 2 \leq x \leq 3 \end{cases} \end{array}$

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Exemplar Solution Class 11th

Find the range of the following functions given by

(i) $f (x) = \frac{3}{2 - x^{2}}$  

(ii) $f (x) = 1 - | x - 2 |$  

(iii) $f (x) = | x - 3$  

(iv) $f (x) = 1 + 3 c o s 2 x$

(Hint:- $1 \leq c o s 2 x \leq 1 \Rightarrow - 3 \leq 3 c o s 2 x \leq 3 \Rightarrow - 2 \leq 1 + 3 c o s 2 x \leq 4)$

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} (i) G i v e n t h a t : f (x) = \frac{3}{2 - x^{2}} \\ L e t y = f (x) ∴ y = \frac{3}{2 - x^{2}} \\ \Rightarrow y (2 - x^{2}) = 3 \Rightarrow 2 y - y x^{2} = 3 \\ \Rightarrow y x^{2} = 2 y - 3 \\ \Rightarrow x^{2} = \frac{2 y - 3}{y} \Rightarrow x = \sqrt[]{\frac{2 y - 3}{y}} \\ H e r e, x i s r e a l i f 2 y - 3 \geq 0 a n d y \geq 0 \\ \Rightarrow y \geq \frac{3}{2} \\ H e n c e, t h e r a n g e o f f = [\frac{3}{2}, \infty) . \\ (i i) G i v e n t h a t : f (x) = 1 - | x - 2 | \\ W e k n o w t h a t | x - 2 | = - (x - 2) i f x < 2 \\ a n d | x - 2 | = (x - 2) i f x \geq 2 \\ ∴ - | x - 2 | \leq 0 \Rightarrow 1 - | x - 2 | \leq 1 \\ H e n c e, t h e r a n g e o f f = (- \infty, 1] . \\ (i i i) G i v e n t h a t : f (x) = | x - 3 | \\ W e k n o w t h a t | x - 3 | \geq 0 \Rightarrow f (x) \geq 0 \\ H e n c e, t h e r a n g e o f f = [0, \infty) \\ (i v) G i v e n t h a t : f (x) = 1 + 3 c o s 2 x \\ W e k n o w t h a t - 1 \leq c o s 2 x \leq 1 \\ \Rightarrow - 3 \leq 3 c o s 2 x \leq 3 \Rightarrow - 3 + 1 \leq 1 + 3 c o s 2 x \leq 3 + 1 \\ \Rightarrow - 2 \leq 1 + 3 c o s 2 x \leq 4 \Rightarrow - 2 \leq f (x) \leq 4 \\ H e n c e, t h e r a n g e o f f = {- 2, 4} . \end{array}$

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Exemplar Solution Class 11th

If $s e c x c o s 5 x + 1 = 0,$ where $0 < x \leq \frac{π}{2}$ , then find the value of $x$ .

2 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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : s e c x c o s 5 x + 1 = 0 \\ \Rightarrow \frac{1}{c o s x} . c o s 5 x + 1 = 0 \\ \Rightarrow c o s 5 x + c o s x = 0 \\ \Rightarrow 2 c o s (\frac{5 x + x}{2}) . c o s (\frac{5 x - x}{2}) = 0 \\ \Rightarrow c o s 3 x . c o s 2 x = 0 \\ \Rightarrow c o s 3 x = 0 o r c o s 2 x = 0 \\ \Rightarrow 3 x = \frac{π}{2} o r 2 x = \frac{π}{2} \\ \Rightarrow x = \frac{π}{6} o r x = \frac{π}{4} \\ H e n c e, t h e v a l u e o f x a r e \frac{π}{6}, \frac{π}{4} . \end{array}$

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Exemplar Solution Class 11th

If $2 s i n^{2} θ = 3 c o s θ$ , where $0 \leq θ \leq 2 π$ , then find the value of $θ$ .

2 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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : 2 s i n^{2} θ = 3 c o s θ \\ \Rightarrow 2 (1 - c o s^{2} θ) = 3 c o s θ \\ \Rightarrow 2 - 2 c o s^{2} θ - 3 c o s θ = 0 \\ \Rightarrow 2 c o s^{2} θ + 3 c o s θ - 2 = 0 \\ \Rightarrow 2 c o s^{2} θ + 4 c o s θ - c o s θ - 2 = 0 \\ \Rightarrow 2 c o s θ (c o s θ + 2) - 1 (c o s θ + 2) = 0 \\ \Rightarrow (c o s θ + 2) (2 c o s θ - 1) = 0 \\ \Rightarrow c o s θ + 2 = 0 o r 2 c o s θ - 1 = 0 \\ \Rightarrow c o s θ \neq - 2 [- 1 \leq c o s θ \leq 1] \\ ∴ 2 c o s θ - 1 = 0 \\ \Rightarrow c o s θ = \frac{1}{2} \\ \Rightarrow c o s θ = c o s \frac{π}{3}, c o s (2 π - \frac{π}{3}) \\ \Rightarrow c o s θ = c o s \frac{π}{3}, c o s \frac{5 π}{3} \\ ∴ θ = 2 n π \pm \frac{π}{3} a n d θ = 2 n π \pm \frac{5 π}{3} (n \in Z) \\ H e n c e, t h e v a l u e o f θ a r e \frac{π}{3} a n d \frac{5 π}{3} . \end{array}$

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Exemplar Solution Class 11th

Find the domain of each of the following functions given by

(i) $f (x) = \frac{1}{\sqrt[]{1 - c o s x}}$

(ii) $f (x) = \frac{1}{\sqrt[]{x + | x |}}$

(iii) $f (x) = | x |$

(iv) $f (x) = \frac{x^{3} - x + 3}{x^{2} - 1}$

(v) $f (x) = \frac{3 x}{2 x - 8}$

7 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} (i) G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{1 - c o s x}} \\ W e k n o w t h a t - 1 \leq c o s x \leq 1 \\ \Rightarrow 1 \geq - c o s x \geq - 1 \\ \Rightarrow 1 + 1 \geq 1 - c o s x \geq - 1 + 1 \\ \Rightarrow 2 \geq 1 - c o s x \geq 0 \\ \Rightarrow 0 \leq 1 - c o s x \leq 2 \\ F o r r e a l v a l u e o f d o m a i n \\ 1 - c o s x \neq 0 \Rightarrow c o s x \neq 1 \\ \Rightarrow x \neq 2 n π \forall n \in Z \\ H e n c e, t h e d o m a i n o f f = R - {2 n π, n \in Z} \\ (i i) G i v e n t h a t : f (x) = \frac{1}{\sqrt[]{x + | x |}} \\ ? x + | x | = x + x = 2 x i f x \geq 0 \\ a n d x + | x | = x - x = 0 i f x < 0 \\ S o f a r x < 0, f i s n o t d e f i n e d . \\ H e n c e, t h e d o m a i n f = R^{+} . \\ (i i i) G i v e n t h a t : f (x) = x | x | \\ I t i s c l e a r t h a t f (x) i s d e f i n e d f o r a l l x \in R . \\ H e n c e, t h e d o m a i n f = R . \\ (i v) G i v e n t h a t : f (x) = \frac{x^{3} - x + 3}{x^{2} - 1} \\ H e r e, f (x) i s o n l y d e f i n e d i f x^{2} &min \end{array}$

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Exemplar Solution Class 11th

Is $g = {(1, 1), (2, 3), (3, 5), (4, 7)}$ a function? Justify. If this is described by the relation, $g (x) = α x + β$ , then what values should be assigned to $α$ and $β$ ?

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : g = {(1, 1), (2, 3), (3, 5), (4, 7)} \\ Since e v e r y e l e m e n t o f t h e d o m a i n i n t h i s r e l a t i o n s h a s u n i q u e i m a g e, s o g i s a f u n c t i o n . \\ N o w g (x) = α x + β \\ F o r (1, 1) g (1) = α (1) + β = 1 \Rightarrow α + β = 1 \dots (i) \\ F o r (2, 3) g (2) = α (2) + β = 3 \Rightarrow 2 α + β = 3 \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i) w e h a v e \\ α = 2 a n d β = - 1 \\ H e n c e, t h e v a l u e o f α = 2 a n d β = - 1 . \end{array}$

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Exemplar Solution Class 11th

If $c o t θ + t a n θ = 2 c s c θ$ , then find the general value of $θ$ .

5 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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : c o t θ + t a n θ = 2 c o s e c θ \\ \Rightarrow \frac{c o s θ}{s i n θ} + \frac{s i n θ}{c o s θ} = \frac{2}{s i n θ} \\ \Rightarrow \frac{c o s^{2} θ + s i n^{2} θ}{s i n θ c o s θ} = \frac{2}{s i n θ} \\ \Rightarrow \frac{1}{s i n θ c o s θ} = \frac{2}{s i n θ} \\ \Rightarrow 2 s i n θ c o s θ = s i n θ \\ \Rightarrow 2 s i n θ c o s θ - s i n θ = 0 \\ \Rightarrow s i n θ (2 c o s θ - 1) = 0 \\ \Rightarrow s i n θ \neq 0 o r 2 c o s θ - 1 = 0 o r c o s θ = \frac{1}{2} \\ \Rightarrow c o s θ = c o s \frac{π}{3} \\ ∴ θ = 2 n π \pm \frac{π}{3} \\ H e n c e, t h e g e n e r a l v a l u e o f θ a r e 2 n π \pm \frac{π}{3} . \end{array}$

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Exemplar Solution Class 11th

Find the most general value of $θ$ satisfying the equation $t a n θ = - 1 and c o s θ = \frac{1}{\sqrt[]{2}} .$

2 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 Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ = - 1 a n d c o s θ = \frac{1}{\sqrt[]{2}} \\ t a n θ = - 1 \\ \Rightarrow t a n θ = t a n (\frac{- π}{4}) \\ \Rightarrow t a n θ = t a n (2 π - \frac{π}{4}) \Rightarrow t a n θ = t a n \frac{7 π}{4} \\ ∴ θ = \frac{7 π}{4} \\ N o w, c o s θ = \frac{1}{\sqrt[]{2}} \Rightarrow c o s θ = c o s \frac{π}{4} \\ \Rightarrow c o s θ = c o s (2 π - \frac{π}{4}) \\ \Rightarrow c o s θ = c o s \frac{7 π}{4} \\ ∴ θ = \frac{7 π}{4} [t a n θ a n d c o s θ a r e p o s i t i v e i n 4 t h q u a d r a n t] \\ H e n c e, t h e m o s t g e n e r a l v a l u e o f θ a r e 2 n π + \frac{7 π}{2} . \end{array}$

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Exemplar Solution Class 11th

Given A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2}. Then A = B

4 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} H e r e, A = {0, 1, 2}, B i s a s e t h a v i n g a l l r e a l n u m b e r s f r o m 0 t o 2 . S o A \neq B . \\ H e n c e, t h e s t a t e m e n t i s' F a l s e' . \end{array}$

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Redefine the function $f (x) = | x - 2 | + | 2 + x |, - 3 \leq x \leq 3$

If $s e c x c o s 5 x + 1 = 0,$ where $0 < x \leq \frac{π}{2}$ , then find the value of $x$ .

If $2 s i n^{2} θ = 3 c o s θ$ , where $0 \leq θ \leq 2 π$ , then find the value of $θ$ .

Find the domain of each of the following functions given by

(i) $f (x) = \frac{1}{\sqrt[]{1 - c o s x}}$

(ii) $f (x) = \frac{1}{\sqrt[]{x + | x |}}$

(iii) $f (x) = | x |$

(iv) $f (x) = \frac{x^{3} - x + 3}{x^{2} - 1}$

(v) $f (x) = \frac{3 x}{2 x - 8}$

Is $g = {(1, 1), (2, 3), (3, 5), (4, 7)}$ a function? Justify. If this is described by the relation, $g (x) = α x + β$ , then what values should be assigned to $α$ and $β$ ?

If $c o t θ + t a n θ = 2 c s c θ$ , then find the general value of $θ$ .

Find the most general value of $θ$ satisfying the equation $t a n θ = - 1 and c o s θ = \frac{1}{\sqrt[]{2}} .$

Given A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2}. Then A = B