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New answer posted•a year ago

Exemplar Solution Class 11th

Solve: $4 x + 3 \geq 2 x + 17, 3 x - 5 < - 2$

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t 4 x + 3 \geq 2 x + 17 \dots (i) \\ 3 x - 5 < - 2 \dots (i i) \\ F r o m e q n . (i) w e g e t \\ 4 x + 3 \geq 2 x + 17 \Rightarrow 4 x - 2 x \geq 17 - 3 \\ \Rightarrow 2 x \geq 14 \Rightarrow x \geq 7 \\ F r o m e q n . (i i) w e g e t \\ \Rightarrow 3 x - 5 < - 2 \Rightarrow 3 x < 3 \\ \Rightarrow x < 1 \\ W e s e t t h a t t h e s o l u t i o n x \geq 7 a n d x < 1 i s n o t p o s s i b l e . H e n c e t h e r e w i l l b e n o s o l u t i o n o f x . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Solve: $- 5 \leq \frac{2 x - 3}{4} \leq 9$

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t - 5 \leq \frac{2 - 3 x}{4} \leq 9 \\ \Rightarrow - 20 \leq 2 - 3 x \leq 36 \Rightarrow - 22 \leq - 3 x \leq 34 \\ \Rightarrow 22 \geq 3 x \geq - 34 \\ \Rightarrow \frac{22}{3} \geq x \geq \frac{- 34}{3} \\ H e n c e x \in [\frac{- 34}{3}, \frac{22}{3}] \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Solve: $| x - 1 | \leq 5, x \geq 2$

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t | x - 1 | \leq 5 a n d | x | \geq 2 \\ \Rightarrow - 5 \leq x - 1 \leq 5 a n d x \leq - 2 o r x \geq 2 \\ \Rightarrow - 5 + 1 \leq x \leq 5 + 1 a n d x \leq - 2 o r x \geq 2 \\ \Rightarrow - 4 \leq x \leq 6 a n d x \leq - 2 o r x \geq 2 \\ H e n c e x < [- 4, - 2] \cup [2, 6] \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Solve: $\frac{1}{x - 3} \leq \frac{1}{2}$

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t \frac{1}{| x | - 3} \leq \frac{1}{2} \\ \Rightarrow | x | - 3 \geq 2 [? \frac{1}{x} < \frac{1}{y} \Rightarrow x > y] \\ \Rightarrow | x | \geq 5 \\ \Rightarrow | x | \leq - 5 o r | x | \geq 5 \\ S o, x \in (- \infty, - 5] \cup [5, \infty) \dots (i) \\ H e r e, | x | - 3 \neq 0 \\ \Rightarrow | x | - 3 < 0 o r | x | - 3 > 0 \\ \Rightarrow | x | < 3 o r | x | > 3 \\ \Rightarrow - 3 < x < 3 o r x < - 3 o r x > 3 \dots (i i) \\ F r o m i n e q n . (i) a n d (i i) w e g e t \\ x \in (- \infty, - 5] \cup (- 3, 3) \cup [5, \infty) \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Solve: $\frac{x - 2}{x - 1} \leq 0$

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t \frac{| x - 2 | - 1}{| x - 2 | - 2} \leq 0 \\ P u t | x - 2 | = y \\ ∴ \frac{y - 1}{y + 2} \leq 0 \\ \Rightarrow y - 1 > 0, y - 2 < 0 \Rightarrow y > 1, y < 2 \\ \Rightarrow 1 < y < 2 \Rightarrow 1 < | x - 2 | < 2 \\ \Rightarrow 1 < | x - 2 |, | x - 2 | < 2 \\ \Rightarrow x - 2 < - 1 o r x - 2 > 1 a n d - 2 < x - 2 < 2 \\ \Rightarrow x < 1 o r x > 3 a n d - 2 + 2 < x < 2 + 2 \\ \Rightarrow x < 1 o r x > 3 a n d 0 < x < 4 \\ H e n c e, t h e s o l u t i o n i s (0, 1) \cup (3, 4) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Solve: $\frac{4}{x} \leq 6 and \frac{3}{x + 1} \geq 1, (x > 0)$

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t \frac{4}{x + 1} \leq 3 \leq \frac{6}{x + 1}, x > 0 \\ \Rightarrow 4 \leq 3 (x + 1) \leq 6 \Rightarrow 4 \leq 3 x + 3 \leq 6 \\ \Rightarrow 4 - 3 \leq 3 x \leq 6 - 3 \Rightarrow 1 \leq 3 x \leq 3 \\ \Rightarrow \frac{1}{3} \leq x \leq 1 \\ H e n c e, t h e s o l u t i o n i s \frac{1}{3} \leq x \leq 1 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Match the following items under column  to their correct answers under column :

i. $s i n (x + y) s i n (x - y)$ (i) $c o s^{2} x - s i n^{2} y$ 

ii. $c o s (x + y) c o s (x - y)$ (ii) $\frac{1 - t a n θ}{1 + t a n θ}$ 

iii. $c o t (\frac{π}{4} + θ)$ (iii) $\frac{1 + t a n θ}{1 - t a n θ}$ 

iv. $t a n (\frac{π}{4} + θ)$ (iv) $s i n^{2} x - s i n^{2} y$

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 True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} (a) s i n (x + y) s i n (x - y) = s i n^{2} x - s i n^{2} y \\ ∴ (a) \leftrightarrow (i v) \\ (b) c o s (x + y) c o s (x - y) = c o s^{2} x - c o s^{2} y \\ ∴ (b) \leftrightarrow (i) \\ (c) c o t (\frac{π}{4} + θ) = \frac{c o t \frac{π}{4} c o t θ - 1}{c o t θ + c o t \frac{π}{4}} = \frac{c o t θ - 1}{c o t θ + 1} = \frac{1 - t a n θ}{1 + t a n θ} \\ ∴ (c) \leftrightarrow (i i) \\ (d) t a n (\frac{π}{4} + θ) = \frac{t a n \frac{π}{4} + t a n θ}{1 - t a n \frac{π}{4} t a n θ} = \frac{1 + t a n θ}{1 - t a n θ} \\ ∴ (d) \leftrightarrow (i i i) \\ H e n c e, (a) \leftrightarrow (i v), (b) \leftrightarrow (i), (c) \leftrightarrow (i i) a n d (d) \leftrightarrow (i i i) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $t a n (π c o s θ) = c o t (π s i n θ)$ , then $c o s (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt[]{2}}$ .

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 True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n (π c o s θ) = c o t (π s i n θ) \\ \Rightarrow t a n (π c o s θ) = t a n (\frac{π}{2} - π s i n θ) \\ \Rightarrow π c o s θ = \frac{π}{2} - π s i n θ \\ \Rightarrow π c o s θ + π s i n θ = \frac{π}{2} \\ \Rightarrow c o s θ + s i n θ = \frac{1}{2} \\ \Rightarrow \frac{1}{\sqrt[]{2}} c o s θ + \frac{1}{\sqrt[]{2}} s i n θ = \frac{1}{2 \sqrt[]{2}} \\ \Rightarrow c o s \frac{π}{4} c o s θ + s i n \frac{π}{4} s i n θ = \frac{1}{2 \sqrt[]{2}} \\ \Rightarrow c o s (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt[]{2}} [? c o s (θ - \frac{π}{4}) o r c o s (\frac{π}{4} - θ)] \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If  $z$ and  $w$ are two complex numbers such that  $? z w ? = 1$ and $arg (z) - arg (w) = \frac{π}{2}$ , then show that $z^{} w = - i$ .

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

$\begin{array}{l} L e t z = r_{1} (c o s θ_{1} + i s i n θ_{1}) a n d w = r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ z w = r_{1} r_{2} [(c o s θ_{1} - i s i n θ_{1})] [(c o s θ_{2} + i s i n θ_{2})] \\ | z w | = r_{1} r_{2} = 1 [G i v e n] \\ N o w a r g (z) - a r g (w) = \frac{π}{2} \\ \Rightarrow θ_{1} - θ_{2} = \frac{π}{2} \Rightarrow a r g (\frac{z}{w}) = \frac{π}{2} \\ \bar{z} w = r_{1} (c o s θ_{1} - i s i n θ_{1}) r_{2} (c o s θ_{2} + i s i n θ_{2}) \\ = r_{1} r_{2} [c o s θ_{1} c o s θ_{2} + i c o s θ_{1} s i n θ_{2} - i s i n θ_{1} c o s θ_{2} - i^{2} s i n θ_{1} s i n θ_{2}] \\ = r_{1} r_{2} [(c o s θ_{1} c o s θ_{2} + s i n θ_{1} s i n θ_{2}) + i (c o s θ_{1} s i n θ_{2} - s i n θ_{1} c o s θ_{2})] \\ = r_{1} r_{2} [c o s (θ_{2} - θ_{1}) + i s i n (θ_{2} - θ_{1})] \\ = r_{1} r_{2} [c o s (- \frac{π}{2}) + i s i n (- \frac{π}{2})] \\ = r_{1} r_{2} [c o s \frac{π}{2} - i s i n \frac{π}{2}] = 1 . [0 - i] = - i \\ H e r e \bar{z} w = - i . H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $t a n θ + t a n 2 θ + 3 t a n θ t a n 2 θ = 3$ , then $θ = \frac{n π}{3} + \frac{π}{9}$ .

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 True or False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t : t a n θ + t a n 2 θ + \sqrt[]{3} t a n θ t a n 2 θ = \sqrt[]{3} \\ \Rightarrow t a n θ + t a n 2 θ = \sqrt[]{3} - \sqrt[]{3} t a n θ t a n 2 θ \\ \Rightarrow t a n θ + t a n 2 θ = \sqrt[]{3} (1 - t a n θ t a n 2 θ) \\ \Rightarrow \frac{t a n θ + t a n 2 θ}{1 - t a n θ t a n 2 θ} = \sqrt[]{3} \\ \Rightarrow t a n (θ + 2 θ) = \sqrt[]{3} \\ \Rightarrow t a n 3 θ = t a n \frac{π}{3} ∴ 3 θ = n π + \frac{π}{3} \\ S o, θ = \frac{n π}{3} + \frac{π}{9} \\ H e n c e, t h e s t a t e m e n t i s' T r u e' . \end{array}$

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