<p><strong>35. y < – 2</strong></p>

Ask & Answer Question

Ncert Solutions Maths class 11th Maths Class 11th Linear Inequalities

35. y < – 2

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

35. For the given equation inequality y< 2, the equation of line is y= 2

This line devides the xy-plane into two planer I and II. We take a point (0,0) to check the correctness of the inequality.

So, 0< 2

0< 2 which is false.

So, the solution of the region is II which does not include the origin.

The dotted line indicates that any point on the line does not satisfy the inequality.

0 Upvotes 0 Downvotes

Similar Questions for you

If one of the diameters of the circle $x^{2} + y^{2} - 2 \sqrt[]{2} x - 6 \sqrt[]{2} y + 14 = 0$ is a chord of circle. ${(x - 2 \sqrt[]{2})}^{2} + {(y - 2 \sqrt[]{2})}^{2} = r^{2},$ then the value of r² is equal to…………….

alok kumar singh

$x^{2} + y^{2} - 2 \sqrt[]{2} x - 6 \sqrt[]{2} y + 14 = 0$

centre $(\sqrt[]{2}, 3 \sqrt[]{2})$

radius ${({(\sqrt[]{2})}^{2} + {(3 \sqrt[]{2})}^{2} - 14)}^{1 / 2}$

$= {(2 + 18 - 14)}^{1 / 2} = (\sqrt[]{6})$

${(x - 2 \sqrt[]{2})}^{2} + {(y - 2 \sqrt[]{2})}^{2} = r^{2}$

centre $(2 \sqrt[]{2}, 2 \sqrt[]{2})$

OA = $\sqrt[]{{(2 \sqrt[]{2} - \sqrt[]{2})}^{2} + {(2 \sqrt[]{2} - 3 \sqrt[]{2})}^{2}} = \sqrt[]{2 + 2} = 2$

r² = ${(\sqrt[]{6})}^{2} + {(2)}^{2} = 6 + 4 = 10$

LE the common tangents to the curves $4 (x^{2} + y^{2}) = 9$ and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac{l}{e^{2}}$ is equal to

Payal Gupta

Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

and length of latus rectum

$= \frac{2 b^{2}}{a} = 3$

Hence

$\frac{l}{e^{2}} = \frac{3}{3 / 4} = 4$

if $z^{2} + z + 1 = 0, z \in C, t h e n | \sum_{n = 1}^{15} {(z^{n} + {(- 1)}^{n} \frac{1}{z^{n}})}^{2} |$ $^{} {^{}^{} \frac{}{^{}}}^{}$ is equal to……….

Vishal Baghel

$z^{2} + z + 1 = 0 \Rightarrow z = w, w^{2}$

$| \sum_{n = 1}^{15} {(z^{n} + {(- 1)}^{n} \frac{1}{z^{n}})}^{2} | = | \sum_{n = 1}^{15} (z^{2 n} + \frac{1}{z^{2 n}} + 2 {(- 1)}^{n}) | = | \sum_{n = 1}^{15} w^{2 n} + \frac{1}{w^{2 n}} + 2 {(- 1)}^{n} |$

$= | \frac{w^{2} (1 - 1)}{1 - w^{2}} + \frac{\frac{1}{w^{2}} (1 - 1)}{1 - \frac{1}{w^{2}}} - 2 | = 2$

(i) If $- 4 x \geq 12$ , then $x \dots - 3$ .

(ii) If $\frac{3}{4} - x \leq - 3,$ then $x \dots 4$ .

(iii) If $\frac{2}{x + 2} > 0,$ then $x \dots - 2$ .

(iv) If $x > - 5$ , then $4 x \dots - 20$ .

(v) If $x > y$ and $z < 0$ , then $- x z \dots - y z$ .

(vi) If $p > 0$ and $q < 0$ , then $p - q \dots p$ .

(vii) If $x + 2 > 5$ , then $x \dots - 7$ or $x \dots 3$ .

(viii)If $- 2 x + 1 \geq 9$ , then $x \dots - 4$ .

Payal Gupta

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

$\begin{array}{l} (i) - 4 x \geq 12 \Rightarrow x \leq - 3 \\ H e n c e, t h e f i l l e r i s (\leq) \\ (i i) I f - \frac{3}{4} x \leq - 3 \Rightarrow x \geq 3 x \times \frac{4}{3} \Rightarrow x \geq 4 \\ H e n c e, t h e f i l l e r i s (\geq) \\ (i i i) I f \frac{2}{x + 2} > 0 \Rightarrow x > - 2 \\ H e n c e, t h e f i l l e r i s (>) \\ (i v) I f x > - 5 \Rightarrow 4 x > - 20 \\ H e n c e, t h e f i l l e r i s (>) \\ (v) I f x > y a n d z < 0, t h e n \\ x z < y z \Rightarrow - x z > - y z \\ H e n c e, t h e f i l l e r i s (>) \\ (v i) I f p > 0 a n d q < 0, t h e n \\ p - q > p \\ H e n c e, t h e f i l l e r i s (>) \\ (v i i) I f | x + 2 | > 5 t h e n \\ x + 2 < - 5 o r x + 2 > 5 \\ \Rightarrow x < - 5 - 2 o r x > 5 - 2 \\ \Rightarrow x < - 7 o r x > 3 \\ S o, x \in (- \infty, - 7) \cup (3, \infty) \\ H e n c e, t h e f i l l e r i s (<) o r (>) \\ (v i i i) I f - 2 x + 1 \geq 9 t h e n \\ \Rightarrow - 2 x \geq 9 - 1 \Rightarrow - 2 x \geq 8 \Rightarrow 2 x \leq - 8 \Rightarrow x \leq - 4 \\ H e n c e, t h e f i l l e r i s (\leq) \end{array}$

(i) If $x < y$ and $b < 0$ , then $\frac{x}{b} > \frac{y}{b} .$

(ii) If $x y > 0$ , then $x > 0$ and $y < 0$ .

(iii) If $x y > 0$ , then $x < 0$ and $y < 0$ .

(iv) If $x y < 0$ , then $x < 0$ and $y < 0$ .

(v) If $x < - 5$ and $x < - 2$ , then $x \in (- \infty, - 5)$ .

(vi) If $x < - 5$ and $x > 2$ , then $x \in (- 5, 2)$ .

(vii) If $x > - 2$ and $x < 9$ , then $x \in (- 2, 9)$ .

(viii) If $? x ? > 5$ , then $x \in (- \infty, - 5) \cup (5, \infty) .$

(ix) If $? x ? \leq 4$ , then $x \in [- 4, 4] .$

(x) Graph of $x < 3$ corresponds to Fig 6.12.

(xi) Graph of $x \geq 0$ corresponds to Fig 6.13.

(xii) Graph of $y \leq 0$ corresponds to Fig 6.14.

(xiii) Solution set of $x \geq 0$ and $y \leq 0$ corresponds to Fig 6.15.

(xiv) Solution set of $x \geq 0$ and $y \leq 1$ corresponds to Fig 6.16.

(xv) Solution set of $x + y \geq 0$ corresponds to Fig 6.17.

Payal Gupta

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

$\begin{array}{l} (i) I f x < y a n d b < 0 \\ \Rightarrow \frac{x}{b} > \frac{y}{b} \\ H e n c e, s t a t e m e n t (i) i s F a l s e . \\ (i i) I f x, y > 0 t h e n x > 0, y > 0 o r x < 0, y < 0 \\ H e n c e, s t a t e m e n t (i i) i s F a l s e . \\ (i i i) I f x y > 0 t h e n x < 0, a n d y < 0 \\ H e n c e, s t a t e m e n t (i i i) i s T r u e . \\ (i v) I f x y < 0 t h e n x < 0, y > 0 o r x > 0, y < 0 \\ H e n c e, s t a t e m e n t (i v) i s F a l s e . \\ (v) I f x < - 5 a n d x < - 2 \Rightarrow x \in (- \infty, - 5) \\ H e n c e, s t a t e m e n t (v) i s T r u e . \\ (v i) I f x < - 5 a n d x > 2 t h e n x h a s n o v a l u e . \\ H e n c e, s t a t e m e n t (v i) i s F a l s e . \\ (v i i) I f x > - 2 a n d x < 9 \Rightarrow x \in (- 2, 9) \\ H e n c e, s t a t e m e n t (v i i) i s T r u e . \\ (v i i i) I f | x | > 5 t h e n x < - 5 o r x > 5 \\ \Rightarrow x \in (- \infty, - 5) \cup (5, \infty) \\ H e n c e, s t a t e m e n t (v i i i) i s F a l s e . \\ (i x) I f | x | \leq 4 t h e n - 4 \leq x \leq 4 \\ \Rightarrow x \in [- 4, 4] \\ H e n c e, s t a t e m e n t (i x) i s T r u e . \\ (x) T h e g i v e n g r a p h r e p r e s e n t s x \leq 3 \\ H e n c e, s t a t e m e n t (x) i s F a l s e . \\ (x i) T h e g i v e n g r a p h r e p r e s e n t s x \geq 0 \\ H e n c e, s t a t e m e n t (x i) i s T r u e . \\ (x i i) T h e g i v e n g r a p h r e p r e s e n t s y \geq 0 \\ H e n c e, s t a t e m e n t (x i i) i s &thins \end{array}$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

35. y < – 2

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question