<p><strong>41. 2x – y >1, x – 2y < – 1</strong></p>

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

41. 2x – y >1, x – 2y < – 1

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Payal Gupta | Contributor-Level 10

4 months ago

41. The given system of inequality is

2x – y> 1 - (1)

x – 2y< –1- (1)

So the corresponding equations are

2x – y=1

x	0	0.5
y	–1	0

and x – 2y= –1

x	–1	0
y	0	0.5

Putting (x, y)= (0,0) in (1) and (2) to cheek the inequality

2 × 0 – 0 > 1

0 > 1 which is not true.

and 0 – 2 × 0< –1

0< –1 which is not true.

So, the solution of plane of inequality (1)and (2) does not include the plane with point (0,0) or origin.

? The reqd. solution of the given system of inequality is the shaded region.

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Let y = mx + c is the common tangent

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(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 $x \dots - 4$ .

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

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(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 $y < 0$ .

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

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(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.

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