# JEE Main

Explore popular questions from Linear Inequalities for JEE Main. This collection covers Linear Inequalities previous year JEE Main questions hand picked by experienced teachers.

## Mathematics

Linear Inequalities

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Q 1. If {tex} x {/tex} is real number and {tex} | x | < 3 , {/tex} then

A

{tex} x \geq 3 {/tex}

{tex} - 3 < x < 3 {/tex}

C

{tex} x \leq - 3 {/tex}

D

{tex} - 3 \leq x \leq 3 {/tex}

##### Explanation

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Q 2. Given that {tex} x ,\ y {/tex} and {tex} b {/tex} are real numbers and {tex} x < y ,\ b < 0 , {/tex} then

A

{tex} \frac { x } { b } < \frac { y } { b } {/tex}

B

{tex} \frac { x } { b } \leq \frac { y } { b } {/tex}

{tex} \frac { x } { b } > \frac { y } { b } {/tex}

D

{tex} \frac { x } { b } \geq \frac { y } { b } {/tex}

##### Explanation

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Q 3. Solution of a linear inequality in variable {tex} x {/tex} is represented on number line is

A

{tex} x \in ( - \infty , 5 ) {/tex}

B

{tex} x \in ( - \infty , 5 ] {/tex}

C

{tex} x \in [ 5 , \infty ) {/tex}

{tex} x \in ( 5 , \infty ) {/tex}

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Q 4. If {tex} | x + 3 | \geq 10 {/tex}, then

A

{tex} x \in ( - 13,7 ] {/tex}

B

{tex} x \in ( - 13,7 ) {/tex}

C

{tex} x \in ( - \infty , 13 ] \cup [ - 7 , \infty ) {/tex}

{tex} x \in ( - \infty , - 13 ] \cup [ 7 , \infty ) {/tex}

##### Explanation

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Q 5. Let {tex}\mathrm{ \frac { C } { 5 } = \frac { F - 32 } { 9 } } {/tex}. If {tex}\mathrm C {/tex} lies between {tex}10{/tex} and {tex}20{/tex}, then :

A

{tex} 50 < \mathrm{ F } < 78 {/tex}

{tex} 50 < \mathrm { F } < 68 {/tex}

C

{tex} 49 < \mathrm{ F } < 68 {/tex}

D

{tex} 49 < \mathrm { F } < 78 {/tex}

##### Explanation

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Q 6. The solution set of the inequality {tex} 4 x + 3 < 6 x + 7 {/tex} is

A

{tex} [ - 2 , \infty ) {/tex}

B

{tex} ( - \infty , - 2 ) {/tex}

{tex} ( - 2 , \infty ) {/tex}

D

None of these

##### Explanation

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Q 7. Which of the following is the solution set of {tex} 3 x - 7 > 5 x - 1\ \forall \ x \in R {/tex} ?

{tex} ( - \infty , - 3 ) {/tex}

B

{tex} ( - \infty , - 3 ] {/tex}

C

{tex} ( - 3 , \infty ) {/tex}

D

{tex} ( - 3,3 ) {/tex}

##### Explanation

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Q 8. The solution set of the inequality {tex} 37 - ( 3 x + 5 ) \geq 9 x - 8 ( x - 3 ) {/tex} is

A

{tex} ( - \infty , 2 ) {/tex}

B

{tex} ( - \infty , - 2 ) {/tex}

{tex} ( - \infty , 2 ] {/tex}

D

{tex} ( - \infty , - 2 ] {/tex}

##### Explanation

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Q 9. The graphical solution of {tex} 3 x - 6 \geq 0 {/tex} is

B

C

D

##### Explanation

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Q 10. The solutions of the system of inequalities {tex} 3 x - 7 < 5 + x {/tex} and {tex} 11 - 5 x \leq 1 {/tex} on the number line is

A

C

D

None of the above

##### Explanation

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Q 11. The solution set of the inequalities {tex} 3 x - 7 > 2 ( x - 6 ) {/tex} and {tex} 6 - x > 11 - 2 x , {/tex} is

A

{tex} ( - 5 , \infty ) {/tex}

B

{tex} [ 5 , \infty ) {/tex}

{tex} ( 5 , \infty ) {/tex}

D

{tex} [ - 5 , \infty ) {/tex}

##### Explanation

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Q 12. If {tex} \frac { 3 x - 4 } { 2 } \geq \frac { x + 1 } { 4 } - 1 , {/tex} then {tex} x \in {/tex}

{tex} [ 1 , \infty ) {/tex}

B

{tex} ( 1 , \infty ) {/tex}

C

{tex} ( - 5,5 ) {/tex}

D

{tex} [ - 5,5 ] {/tex}

##### Explanation

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Q 13. If {tex} - 5 \leq \frac { 5 - 3 x } { 2 } \leq 8 , {/tex} then {tex} x \in {/tex}

{tex} \left[ - \frac { 11 } { 3 } , 5 \right] {/tex}

B

{tex} [ - 5,5 ] {/tex}

C

{tex} \left[ - \frac { 11 } { 3 } , \infty \right) {/tex}

D

{tex} ( - \infty , \infty ) {/tex}

##### Explanation

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Q 14. Solutions of the inequalities comprising a system in variable {tex} x {/tex} are represented on number lines as given below, then

{tex} x \in ( - \infty , - 4 ] \cup [ 3 , \infty ) {/tex}

B

{tex} x \in [ - 3,1 ] {/tex}

C

{tex} x \in ( - \infty , - 4 ) \cup [ 3 , \infty ) {/tex}

D

{tex} x \in [ - 4,3 ] {/tex}

##### Explanation

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Q 15. The inequality {tex} \frac { 2 } { x } < 3 {/tex} is true, when {tex} x {/tex} belongs to

A

{tex} \left[ \frac { 2 } { 3 } , \infty \right) {/tex}

B

{tex} \left( - \infty , \frac { 2 } { 3 } \right] {/tex}

{tex} ( - \infty , 0 ) \cup \left( \frac { 2 } { 3 } , \infty \right) {/tex}

D

None of these

##### Explanation

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Q 16. Solution of {tex} | 3 x + 2 | < 1 {/tex} is

A

{tex} \left[ - 1 , - \frac { 1 } { 3 } \right] {/tex}

B

{tex} \left\{ - \frac { 1 } { 3 } , - 1 \right\} {/tex}

{tex} \left( - 1 , - \frac { 1 } { 3 } \right) {/tex}

D

None of these

##### Explanation

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Q 17. Solution of {tex} | x - 1 | \geq | x - 3 | {/tex} is

A

{tex} x \leq 2 {/tex}

{tex} x \geq 2 {/tex}

C

{tex} [ 1,3 ] {/tex}

D

None of these

##### Explanation

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Q 18. If {tex} - 3 x + 17 < - 13 , {/tex} then

{tex} x \in ( 10 , \infty ) {/tex}

B

{tex} x \in [ 10 , \infty ) {/tex}

C

{tex} x \in ( - \infty , 10 ] {/tex}

D

{tex} x \in [ - 10,10 ) {/tex}

##### Explanation

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Q 19. If {tex} | x + 2 | \leq 9 , {/tex} then

A

{tex} x \in ( - 7,11 ) {/tex}

{tex} x \in [ - 11,7 ] {/tex}

C

{tex} x \in ( - \infty , - 7 ) \cup ( 11 , \infty ) {/tex}

D

{tex} x \in ( - \infty , - 7 ) \cup [ 11 , \infty ) {/tex}

##### Explanation

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Q 20. If 3x + 22x ≥ 5x, then the solution set for x is

( − ∞, 2]

B

[2, ∞)

C

[0, 2]

D

{2}

##### Explanation

We have,

3x + 22x ≥ 5x

$\Rightarrow \left( \frac{3}{5} \right)^{x} + \left( \frac{4}{5} \right)^{x} \geq 1$

$\Rightarrow \left( \frac{3}{5} \right)^{x} + \left( \frac{4}{5} \right)^{x} \geq \left( \frac{3}{5} \right)^{2} + \left( \frac{4}{5} \right)^{2}$

$\Rightarrow x \leq 2 \Rightarrow x \in \left( - \infty,2 \right\rbrack\text{\ \ \ \ \ }\begin{bmatrix} \text{If\ }a^{x} + b^{x} \geq 1\ \text{and\ }a^{2} + b^{2} = 1, \\ \text{then\ }x \in ( - \infty,2) \\ \end{bmatrix}$

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Q 21. x2 − 3|x| + 2 < 0, then x belongs to

A

(1, 2)

B

( − 2, − 1)

(−2,−1) ∪ (1, 2)

D

( − 3, 5)

##### Explanation

x2 − 3|x| + 2 < 0

⇒ |x|2 − 3|x| + 2 < 0

⇒ (|x|−1)(|x|−2) < 0

⇒ 1 < |x| < 2

⇒ − 2 < x < − 1 or 1 < x < 2

∴ x ϵ(−2,−1) ∪ (1, 2)

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Q 22. Solution of $2^{x} + 2^{|x|} \geq 2\sqrt{2}$ is

A

$( - \infty,\log_{2}{(\sqrt{2} + 1))}$

B

(0, 8)

C

$\left( \frac{1}{2},\log_{2}{(\sqrt{2} - 1)} \right)$

$( - \infty,\log_{2}{(\sqrt{2} - 1)) \cup \left\lbrack \left. \ \frac{1}{2},\infty \right) \right.\ }$

##### Explanation

We have, $2^{x} + 2^{|x|} \geq 2\sqrt{2}$

If x ≥ 0,then$\ 2^{x} + 2^{x} \geq 2\sqrt{2}$

$\Rightarrow 2^{x} \geq \sqrt{2}\ \Rightarrow x \geq \frac{1}{2}$

and if x < 0, then$\ 2^{x} + 2^{- x} \geq 2\sqrt{2}$

$\Rightarrow t + \frac{1}{t} \geq 2\sqrt{2}\text{\ \ \ \ }$(where t = 2x)

$\Rightarrow t^{2} - 2\sqrt{2}t + 1 \geq 0$

$\Rightarrow \left( t - \left( \sqrt{2} - 1 \right) \right)\left( t - \left( \sqrt{2} + 1 \right) \right) \geq 0$

$\Rightarrow t \leq \sqrt{2} - 1\$or $t \geq \sqrt{2} + 1$

But t > 0

$\Rightarrow 0 < 2^{x} \leq \sqrt{2} - 1$

Or $2^{x} \geq \sqrt{2} + 1$

$\Rightarrow - \infty < x \leq \log_{2}{(\sqrt{2} - 1)}\$

Or $x \geq \log_{2}\left( \sqrt{2} + 1 \right)$

Which is not possible, because x > 0

$\therefore x \in ( - \infty,\log_{2}{(\sqrt{2} - 1)) \cup \left\lbrack \left. \ \frac{1}{2},\infty \right) \right.\ }$

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Q 23. If x1, x2, …, xn are real numbers, then the largest value of the expression sin x1cos x2 + sin x2cos x3 + … + sin xncos x1 is

A

n

$\frac{n}{2}$

C

$\frac{n}{4}$

D

$\sqrt{n^{2} - 1}$

##### Explanation

Using G.M. ≤ A.M., we have

$\sin x_{i}\cos{x_{i + 1} \leq \frac{\sin^{2}{x_{i} + \cos^{2}x_{i + 1}}}{2}\ \text{for\ }i = 1,2,3,\ldots,n,}$

where xn + 1 = x1

∴ sin x1cos x2+sin x2cos x3 + … + sin xncos xn + 1

$\leq \frac{\sin^{2}{x_{1} + \cos^{2}x_{2}}}{2} + \frac{\sin^{2}{x_{2} + \cos^{2}x_{3}}}{2} + \ldots + \frac{\sin^{2}{x_{n} + \cos^{2}x_{1}}}{2}$

$\Rightarrow \sin{x_{1}\cos{x_{2} + \sin{x_{2}\cos{x_{3} + \ldots + \sin{x_{n}\cos{x_{1} \leq \frac{n}{2}}}}}}}$

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Q 24. If a < b, then the solution x2 + (a+b)x + ab < 0 is given by

A

(a, b)

B

(−∞,a) ∪ (b, ∞)

( − b, − a)

D

(−∞,−b) ∪ ( − a, ∞)

##### Explanation

We have, x2 + (a+b)x + ab < 0

⇒ (x+a)(x+b) < 0

⇒ − b < x < − a

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Q 25. If log (x3+y3) − log10(x2 + y2 − xy) ≤ 2, then the maximum value of xy, for all x ≥ 0, y ≥ 0 is

2500

B

3000

C

1200

D

3500

##### Explanation

Given, log10(x3+y3) − log10(x2+y2−xy) ≤ 2

$\Rightarrow \ \log_{10}{\frac{(x^{3} + y^{3})}{x^{2} + y^{2} - \text{xy}} \leq 2}$

⇒ log10(x+y) ≤ 2 ⇒ x + y ≤ 100

Using AM ≥ GM

$\therefore\ \ \frac{x + y}{2} \geq \sqrt{\text{xy}}$

$\Rightarrow \ \ \ \sqrt{\text{xy}} \leq \frac{x + y}{2} \leq \frac{100}{2}$

⇒ xy ≤ 2500