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Class 11 Chapter 6 Linear Inequalities Exercise 6.3 Solutions

38. 3x + 2y ≤ 12, x ≥ 1, y ≥ 2

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Payal Gupta | Contributor-Level 10

4 months ago

38.

2. The given system of inequalities are

3x+2y≤ 12- (1)

x≥ 1- (2)

y≥ 2- (3)

We draws the graphs of the lines 3x+2y=12 using points and as 3 × 0 + 2 × 0 ≤ 12

The solution is plane which includes the origin (0, 0).

0 ≤ 12

$\begin{array}{l} x \\ y \end{array} | \begin{array}{l} 0 \\ 6 \end{array} | \begin{array}{l} 4 \\ 0 \end{array} |$

and x = 1 and y = 2.

The inequality (1), (2) and (3) represents the region between these three lines including the points on the respective lines. So, every point on the shaded region in first quadrant represents a solution of the given system of inequalities.

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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

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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

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

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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

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

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

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(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

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

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Class 11 Chapter 6 Linear Inequalities Exercise 6.3 Solutions

38. 3x + 2y ≤ 12, x ≥ 1, y ≥ 2