Solve the following system of inequalities: 2 x + 1 7 x − 1 > 5 , x + 7 x − 8 > 2

This is a long answer type question as classified in NCERT Exemplar L e t u s t a k e t h e i n e q u a t i o n 2 x + 1 7 x − 1 > 5 ⇒ 2 x + 1 7 x − 1 − 5 > 0 ⇒ 2 x + 1 − 5 ( 7 x − 1 ) 7 x − 1 > 0 ⇒ 2 x + 1 − 3 5 x + 5 7 x − 1 > 0 ⇒ − 3 3 x + 6 7 x − 1 > 0 ⇒ 3 3 x − 6 7 x − 1 2 ⇒ x + 7 x − 8 − 2 > 0 ⇒ x + 7 − 2 x + 1 6 x − 8 > 0 ⇒ − x + 2 3 x − 8 > 0 ⇒ x − 2 3 x − 8 < 0 S o , x ∈ ( 8 , 2 3 ) … ( 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 .

If f(x) = ∫ [5x⁸ + 7x⁶] / [(x⁷ + x² + 1)²] dx, (x > 0), f(0) = 0 and f(1) = 1/K, then the value of K is ______.

f (x) = ∫ (5x? + 7x? ) / (x² + 1 + 2x? ) dx seems to have a typo in the denominator. Based on the solution, the denominator is (x? + 1/x? + 2)² or similar. Let's follow the solution's steps. It seems the denominator is (x? (2 + 1/x? + 1/x? )² = x¹? (2 + 1/x? + 1/x? )². f (x) = ∫ (5x? + 7x? ) / (x¹? (2 + 1/x? + 1/x? )²) dx The solution simplifies the integrand to: f (x) = ∫ (5/x? + 7/x? ) / (2 + 1/x? + 1/x? )² dx Let t = 2 + 1/x? + 1/x? dt = (-5/x? - 7/x? ) dx = - (5/x? + 7/x? ) dx. The integral becomes: f (x) = ∫ -dt / t² = 1/t + C. f (x) = 1 / (2 + 1/x? + 1/x? ) + C. Given f (0)=0, this form has a division by zero. Let's re-examine the original problem. If x≥0, as x→0, the integrand goes to 0. So f (0)=0 implies C=0. Let's assume the function is defined as the limit. However, f (x) = x? / (2x? + x² + 1) + C is a more likely result. Given f (1) = 1/k: f (1) = 1 / (2 + 1 + 1) = 1/4. So, 1/k = 1/4 => k = 4.

The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is: ______.

Initial mean of 25 observations is 40. X? = (Σx? )/25 = 40 => Σx? = 25 * 40 = 1000. A teacher of age 60 retires. The new sum of ages for the remaining 24 people is 1000 - 60 = 940. A new teacher of age x joins. The new sum for 25 people is 940 + x. The new mean is 39. (940 + x) / 25 = 39 940 + x = 39 * 25 = 975 x = 975 - 940 = 35. The new teacher's age is 35.

The missing value in the following figure is ______. (A diagram with numbers in a circular pattern is shown).

1 = (2-1)¹ (The n is likely 1). 3? = (7-4)³ (This seems to be a pattern matching (a-b)^c). 4²? = (12-8)? ! = 4²? The blank space must be (5-3)² = 2² = 4.

If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is.

MOTHER 1? E 2? H 3? M 4? O 5? R 6? T So position of word MOTHER in dictionary 2*5!+2*4!+3*3!+2!+1 =240+48+18+2+1 =309

Let a, b and c be three unit vectors such that |a-b|²+|a-c|²=8. Then |a+2b|²+|a+2c|² is equal to.

|a|=|b|=|c|=1 |a-b|²+|a-c|²=8 ⇒|a|²+|b|²-2a.b+|a|²+|c|²-2a.c=8 ⇒4-2 (a.b+a.c)=8 ⇒a.b+a.c=-2 |a+2b|²+|a+2c|² =|a|²+4|b|²+4a.b+|a|²+4|c|²+4a.c =10+4 (a.b+a.c) =10-8=2

If lim(x→1) (x+x²+x³+...+xⁿ-n)/(x-1) = 820, (n∈N) then the value of n is equal to.

lim (x→1) (x+x²+.+x? -n)/ (x-1) = 820 ⇒ lim (x→1) (x-1)/ (x-1)+ (x²-1)/ (x-1)+.+ (x? -1)/ (x-1) = 820 ⇒ 1+2+.+n = 820 ⇒ n (n+1)=2×820 ⇒ n (n+1)=40×41 Since n∈N, so n=40

(i) If −4x≥12 , then x…−3 . (ii) If 34−x≤−3, then x…4 . (iii) If 2x+2>0, then x…−2 . (iv) If x>−5 , then 4x…−20 . (v) If x>y and z 0 and q 5 , then x…−7 or x…3 . (viii)If −2x+1≥9 , then x…−4 .

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar ( i ) − 4 x ≥ 1 2 ⇒ x ≤ − 3 H e n c e , t h e f i l l e r i s ( ≤ ) ( i i ) I f − 3 4 x ≤ − 3 ⇒ x ≥ 3 x × 4 3 ⇒ x ≥ 4 H e n c e , t h e f i l l e r i s ( ≥ ) ( i i i ) I f 2 x + 2 > 0 ⇒ x > − 2 H e n c e , t h e f i l l e r i s ( > ) ( i v ) I f x > − 5 ⇒ 4 x > − 2 0 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 − 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 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 ⇒ x 5 − 2 ⇒ x 3 S o , x ∈ ( − ∞ , − 7 ) ∪ ( 3 , ∞ ) H e n c e , t h e f i l l e r i s ( ) ( v i i i ) I f − 2 x + 1 ≥ 9 t h e n ⇒ − 2 x ≥ 9 − 1 ⇒ − 2 x ≥ 8 ⇒ 2 x ≤ − 8 ⇒ x ≤ − 4 H e n c e , t h e f i l l e r i s ( ≤ )

If x − 5 (d) − x ≥ − 5

This is an Objective Type Questions as classified in NCERT Exemplar I f 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 ) .

Given that x , y and b are real numbers and x y b (d) x b ≥ y b

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t x 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 ) .

If x is a real number and ? x ? , then (a) x ≥ 3 (b) − 3 (c) x ≤ − 3 (d) − 3 ≤ x ≤ 3

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t | x | < 3 ⇒ − 3 < x < 3 [ ? | x | < a ⇒ − 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 ) .

(i) If x yb. (ii) If xy>0 , then x>0 and y 0 , then x 2 , then x∈(−5,2) . (vii) If x>−2 and x 5 , then x∈(−∞,−5)∪(5,∞). (ix) If ?x?≤4 , then x∈[−4,4]. (x) Graph of x<3 corresponds to Fig 6.12. (xi) Graph of x≥0 corresponds to Fig 6.13. (xii) Graph of y≤0 corresponds to Fig 6.14. (xiii) Solution set of x≥0 and y≤0 corresponds to Fig 6.15. (xiv) Solution set of x≥0 and y≤1 corresponds to Fig 6.16. (xv) Solution set of x+y≥0 corresponds to Fig 6.17.

This is a True or False Type Questions as classified in NCERT Exemplar ( i ) I f x 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 t h e n x 0 o r x > 0 , y 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 5 t h e n x 5 ⇒ x ∈ ( − ∞ , − 5 ) ∪ ( 5 , ∞ ) 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 | ≤ 4 t h e n − 4 ≤ x ≤ 4 ⇒ x ∈ [ − 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 ≤ 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 ≥ 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 ≥ 0 H e n c e , s t a t e m e n t ( x i i ) i s &thins

Let S be the set of all λ∈R for which the system of linear equations 2x-y+2z=2 x-2y+λz=-4 x+λy+z=4 has no solution. Then the set S

Contains more than two elements

Is an empty set

The contrapositive of the statement "If I reach the station in time, then I will catch the train" is:

If I will not catch the train, then I do not reach the station in time.

If I do not reach the station in time, then I will catch the train.

If I do not reach the station in time, then I will not catch the train.

If I will catch the train, then I reach the station in time.

The value of 3 + 1/(4 + 1/(3 + 1/(4 + 1/(3 + ...∞)))) is equal to:

3 + 2√3

2 + √3

1/(3²-1) + 1/(5²-1) + 1/(7²-1) + ... + 1/((201)²-1) is equal to:

25/101

99/400

101/408

If p(x) be a polynomial of degree three that has a local maximum value 8 at x=1 and a local minimum value 4 at x=2; then p(0) is equal to

-24

If the functions are defined as f(x) = √x and g(x) = √(1-x), then what is the common domain of the following functions: f + g, f - g, f/g, g/f, g - f where (f+g)(x) = f(x) + g(x), (f/g)(x) = f(x)/g(x)

0 ≤ x < 1

0 < x < 1

0 < x ≤ 1

If f(x) = { 1/|x| ; |x| ≥ 1, ax² + b ; |x| < 1 } is differentiable at every point of the domain, then the values of a and b are respectively:

-1/2, 3/2

1/2, 3/2

1/2, -1/2

Let X={x∈N: 1≤x≤17} and Y={ax+b: x∈X and a,b∈R, a>0}. If mean and variance of elements of Y are 17 and 216 respectively then a+b is equal to

-27

The real function f(x) = arccosec(x) / √(x - [x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to:

All non-integers except the interval [-1, 1]

All integers except 0, -1, 1

All reals except the interval [-1, 1]

If |x| < 1, |y| < 1 and x ≠ y, then the sum to infinity of the following series (x+y) + (x²+xy+y²) + (x³+x²y+xy²+y³) + ... is

(x+y-xy)/((1-x)(1-y))

(x+y+xy)/((1-x)(1-y))

If R={(x,y): x,y∈Z, x²+3y²≤8} is a relation on the set of integers Z, then the domain R?¹ is:

{-2,-1,0,1,2}

{-2,-1,1,2}

If α, β are natural numbers such that 100α - 199β = (100)(100) + (99)(101) + (98)(102) + ... + (1)(199), then the slope of the line passing through (α, β) and origin is:

550

510

530

540

Choose the correct statement about two circles whose equations are given below: x² + y² - 10x - 10y + 41 = 0 x² + y² - 22x - 10y + 137 = 0

Circles have only one meeting point

Circles have no meeting point

Circles have two meeting points

If the equation a|z|² + α'z + αz' + d = 0 represents a circle where a, d are real constants, then which of the following condition is correct?

α|² - ad > 0 and a ∈ R - {0}

|α|² - ad ≠ 0

α = 0, a, d ∈ R⁺

|α|² - ad > 0 and a ∈ R

The value of ((1+sin(2π/9)+icos(2π/9))/(1+sin(2π/9)-icos(2π/9)))³ is

-1/2(√3-i)

-1/2(1-i√3)

1/2(√3-i)

If the tangent to the curve y = x + siny at a point (a,b) is parallel to the line joining (0, 3/2) and (1/2, 2), then:

b = π/2 + a

|a+b|=1

The integral ∫ [(2x-1)cos√( (2x-1)² + 5 )] / √(4x² - 4x + 6) dx is equal to: (where c is a constant of integration)

½ sin√((2x-1)² + 5) + c

½ sin√((2x+1)² + 5) + c

½ cos√((2x+1)² + 5) + c

The equation of one of the straight lines which passes through the point (1,3) and makes an angle tan?¹(√2) with the straight line, y + 1 = 3√2x is:

4√2x + 5y - 4√2 = 0

4√2x - 5y - (5 + 4√2) = 0

5√2x + 4y - (15 + 4√2) = 0

4√2x + 5y - (15 + 4√2) = 0

The differential equation satisfied by the system of parabolas y² = 4a(x + a) is:

y(dy/dx)² + 2x(dy/dx) - y = 0

y(d²y/dx²) - 2x(dy/dx) + y = 0

y(dy/dx)² - 2x(dy/dx) - y = 0

y(dy/dx)² + 2x(dy/dx) - y = 0

Let P(h,k) be a point on the curve y = x² + 7x + 2, nearest to the line, y = 3x - 3. Then the equation of the normal to the curve at P is

x + 3y + 26 = 0

x + 3y - 62 = 0

x - 3y - 11 = 0

Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box I is:

8/17

2/3

2/5

The sum of the first three terms of a G.P is S and their product is 27. Then all such S lie in

(-∞, -3] U [9, ∞)

(-∞, 9]

(-∞, -9] U [3, ∞)

Let y = y(x) be the solution of the differential equation, ((2+sinx)/(y+1))dy/dx = -cosx, y>0, y(0)=1. If y(π)=a and dy/dx at x=π is b, then the ordered pair (a,b) is equal to

(2,1)

Maths NCERT Exemplar Solutions Class 11th Chapter Six: Overview, Questions, Preparation

Q: A company manufactures cassettes. Its cost and revenue functions are C(x)=26,000+30x and R(x)=43x, where x is the number of cassettes produced and sold in a week? How many cassettes must be sold by the company to realize some profit?

This is a short answer type question as classified in NCERT Exemplar Given that:Cost function, C(x)=26,000+30x a n d r e v e n u e f u n c t i o n R ( x ) = 4 3 x N o w f o r p r o f i t P ( x ) , R ( x ) > C ( x ) ⇒ 4 3 x > 2 6 0 0 0 + 3 0 x ⇒ 2 6 0 0 0 + 3 0 x 2 6 0 0 0 ⇒ x > 2 0 0 0 H e n c e , n u m b e r o f c a s s e t t e s t o b e m a n u f a c t u r e d f o r s o m e p r o f i t m u s t b e m o r e t h a n 2 0 0 0 .

Q: The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH readings are 8 . 4 8 and 8 . 3 5 , find the range of pH values for the third reading that will result in the acidity level being normal.

This is a short answer type question as classified in NCERT Exemplar L e t t h e t h i r d p H v a l u e b e x . G i v e n t h a t f i r s t p H v a l u e = 8 . 4 8 and second pH value=8.35 ∴ A v e r a g e v a l u e o f p H = 8 . 4 8 + 8 . 3 5 + x 3 B u t a v e r a g e v a l u e o f p H l i e s b e t w e e n 8 . 2 a n d 8 . 5 ∴ 8 . 2 < 8 . 4 8 + 8 . 3 5 + x 3 < 8 . 5 ⇒ 2 4 . 6 < 1 6 . 8 3 + x < 2 5 . 5 ⇒ 2 4 . 6 − 1 6 . 8 3 < x < 2 5 . 5 − 1 6 . 8 3 ⇒ 7 . 7 7 < x < 8 . 6 7 H e n c e , t h e t h i r d p H v a l u e l i e s b e t w e e n 7 . 7 7 a n d 8 . 6 7 .

Updated on Sep 2, 2025 11:21 IST

By Payal Gupta, Retainer

Table of content

Linear Inequalities Short Answer Type Questions
Linear Inequalities Long Answer Type Questions
Linear Inequalities Objective Type Questions
Linear Inequalities True or False Type Questions
Linear Inequalities Fill in the blanks Type Questions
JEE Mains 2020
JEE Mains 2021

Linear Inequalities Short Answer Type Questions

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

Sol.

\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}

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

Sol.

\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}

Commonly asked questions

A company manufactures cassettes. Its cost and revenue functions are $C (x) = 26, 000 + 30 x and R (x) = 43 x,$ where $x$ is the number of cassettes produced and sold in a week? How many cassettes must be sold by the company to realize some profit?

The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two pH readings are $8.48$ and $8.35$ , find the range of pH values for the third reading that will result in the acidity level being normal.

A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there are 460 liters of the 9% solution, how many liters of the 3% solution will have to be added?

Linear Inequalities Long Answer Type Questions

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

Sol.

\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}

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

Sol.

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

Commonly asked questions

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

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

Linear Inequalities Objective Type Questions

1. If $x < 5$ , then

(a) $- x < - 5$

(b) $- x \leq - 5$

(c) $- x > - 5$

(d) $- x \geq - 5$

Sol.

\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}

2. 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) $\frac{x}{b} \geq \frac{y}{b}$

Sol.

\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}

Commonly asked questions

If $x < 5$ , then

(a) $- x < - 5$

(b) $- x \leq - 5$

(c) $- x > - 5$

(d) $- x \geq - 5$

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

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

(a) $x \geq 3$

(b) $- 3 < x < 3$

(c) $x \leq - 3$

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

Linear Inequalities True or False Type Questions

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Commonly asked questions

(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.

Linear Inequalities Fill in the blanks Type Questions

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Commonly asked questions

(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 $x \dots - 4$ .

Maths NCERT Exemplar Solutions Class 11th Chapter Six Exam

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