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Class 11th

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New answer posted

a year ago

Class 11th Exemplar Solution

If $x - 1 > 5$ , then

(a) $x \in (- 4, 6)$

(b) $x \in [- 4, 6]$

(c) $x \in (- \infty, - 4) \cup (6, \infty)$

(d) $x \in (- \infty, - 4) \cup [6, \infty)$

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t | x - 1 | > 5 \\ \Rightarrow (x - 1) < - 5 o r (x - 1) > 5 \\ \Rightarrow x < - 5 + 1 o r x > 5 + 1 \\ \Rightarrow x < - 4 o r x > 6 \\ \Rightarrow x \in (- \infty, - 4) \cup (6, \infty) \\ H e r e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $b > 0$ and $? x ? > b$ , then

(a) $x \in (- b, \infty)$

(b) $x \in (- \infty, b)$

(c) $x \in (- b, b)$

(d) $x \in (- \infty, - b) \cup (b, \infty)$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t | x | > b, b > 0 \\ \Rightarrow x < - b o r x > b \\ \Rightarrow x \in (- \infty, - b) \cup (b, \infty) \\ H e r e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $x$ is a real number and $? x ? < 3$ , then

(a) $x \geq 3$

(b) $- 3 < x < 3$

(c) $x \leq - 3$

(d) $- 3 \leq x \leq 3$

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t | x | < 3 \\ \Rightarrow - 3 < x < 3 [? | x | < a \Rightarrow - a < x < a] \\ H e r e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $- 3 x + 17 < - 13$ , then

(a) $x \in (10, \infty)$

(b) $x \in [10, \infty)$

(c) $x \in (- \infty, 10]$

(d) $x \in [- 10, 10)$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t - 3 x + 17 < - 13 \\ \Rightarrow - 3 x < - 17 - 13 \Rightarrow - 3 x < - 30 \\ \Rightarrow 3 x > 30 \Rightarrow x > 10 \\ \Rightarrow x \in (10, \infty) \\ H e r e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Given that $x, y$ and $b$ are real numbers and $x < y, b < 0$ , then

(a) $\frac{x}{b} < \frac{y}{b}$

(b) $\frac{x}{b} \leq \frac{y}{b}$

(c) $\frac{x}{b} > \frac{y}{b}$

(d) $\frac{x}{b} \geq \frac{y}{b}$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t x < y, b < 0 \\ \Rightarrow \frac{x}{b} > \frac{y}{b}, b < 0 \\ H e r e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $x < 5$ , then

(a) $- x < - 5$

(b) $- x \leq - 5$

(c) $- x > - 5$

(d) $- x \geq - 5$

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f x < 5 t h e n - x > - 5 \\ H e r e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Show that the following system of linear inequalities has no solution: $x + 2 y \leq 3, 3 x + 4 y \geq 12, x \geq 0, y \geq 1 .$

0 Follower 1 View

Payal Gupta

Contributor-Level 10

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

0 0

New answer posted

a year ago

Class 11th Exemplar Solution

Find the linear inequalities for which the shaded region in the given figure is the solution set.

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t u s f i r s t c o n s i d e r t h e l i n e x + y = 4 \\ W e o b s e r v e t h a t t h e s h a d e d r e g i o n a n d t h e o r i g i n (0, 0) b o t h o n t h e o p p o s i t e s i d e a n d (0, 0) \\ s a t i s f y t h e constraint x + y \leq 4 . \\ S o, x + y \geq 4 i s t h e l i n e a r i n e q u a l i t y . \\ N o w, t a k e t h e l i n e a r e q u a t i o n x + y = 8 . \\ I n t h i s c a s e, t h e s h a d e d r e g i o n a n d t h e o r i g i n (0, 0) b o t h l i e o n t h e s a m e s i d e o f t h e g r a p h \\ a n d (0, 0) s a t i s f y t h e constraint x + y \leq 8 . \\ S o, x + y \leq 4 i s a n o t h e r l i n e a r i n e q u a l i t y . \\ C o n s i d e r, x = 5 . I t i s c l e a r t h a t t h e s h a d e d a r e a l i e s t o w a r d s t h e o r i g i n . S o, x \leq 5 i s t h e \\ constraints a n d s i m i l a r l y y \leq 5 i s a l s o t h e constraints. W e a l s o o b s e r v e t h a t t h e s h a d e d r e g i o n \\ l i e s i n f i r s t q u a d r a n t, S o x \geq 0 a n d y \geq 0 . \\ H e n c e, t h e r e q u i r e d l i n e a r i n e q u a l i t i e s a r e \\ x + y \geq 4, x + y \leq 8, x \leq 5, y \leq 5, x \geq 0 a n d y \geq 0 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the linear inequalities for which the shaded region in the given figure is the solution set.

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

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

\begin{array}{l} L e t u s c o n s i d e r t h e l i n e a r e q u a t i o n 3 x + 2 y = 48 . W e o b s e r v e t h a t t h e s h a d e d r e g i o n a n d t h e \\ o r i g i n b o t h o n t h e s a m e s i d e o f t h e g r a p h o f t h e l i n e a n d (0, 0) s a t i s f y t h e constraint 3 x + 3 y \leq 48 . \\ S i m i l a r l y, w e o b s e r v e t h a t t h e s h a d e d r e g i o n a n d t h e o r i g i n b o t h o n t h e s a m e s i d e o f t h e g r a p h \\ o f t h e l i n e a n d (0, 0) s a t i s f y t h e constraint x + y \leq 20 . \\ W e s e e t h a t t h e s h a d e d r e g i o n i s i n t h e f i r s t q u a d r a n t w h e r e x \geq 0 a n d y \geq 0 \\ H e n c e, t h e r e q u i r e d l i n e a r i n e q u a l i t i e s a r e \\ 3 x + 3 y \leq 48, x + y \leq 20, x \geq 0 a n d y \geq 0 . \end{array}

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New answer posted

a year ago

Class 11th Exemplar Solution

Solve the following system of inequalities: $\frac{2 x + 1}{7 x - 1} > 5, \frac{x + 7}{x - 8} > 2$

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t u s t a k e t h e i n e q u a t i o n \\ \frac{2 x + 1}{7 x - 1} > 5 \Rightarrow \frac{2 x + 1}{7 x - 1} - 5 > 0 \\ \Rightarrow \frac{2 x + 1 - 5 (7 x - 1)}{7 x - 1} > 0 \Rightarrow \frac{2 x + 1 - 35 x + 5}{7 x - 1} > 0 \\ \Rightarrow \frac{- 33 x + 6}{7 x - 1} > 0 \Rightarrow \frac{33 x - 6}{7 x - 1} < 0 \\ S o, x \in (\frac{1}{7}, \frac{2}{11}) \dots (i) \\ N o w f r o m second i n e q u a l i t y, w e h a v e \\ \Rightarrow \frac{x + 7}{x - 8} > 2 \Rightarrow \frac{x + 7}{x - 8} - 2 > 0 \\ \Rightarrow \frac{x + 7 - 2 x + 16}{x - 8} > 0 \Rightarrow \frac{- x + 23}{x - 8} > 0 \\ \Rightarrow \frac{x - 23}{x - 8} < 0 \\ S o, x \in (8, 23) \dots (i i) \\ N o w n o t h i n g i s c o m m o n t o t h e t w o v a l u e s o f x i . e ., n u l l s e t . \\ H e n c e, t h e g i v e n i n e q u a l i t i e s h a s n o s o l u t i o n . \end{array}$

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